MMorikawa’s Unsolved Problem
Jan E. Holly and David Krumm
Abstract.
By combining theoretical and computational techniques from geometry, calculus,group theory, and Galois theory, we prove the nonexistence of a closed-form algebraic solutionto a Japanese geometry problem first stated in the early nineteenth century. This resolves anoutstanding problem from the sangaku tablets which were at one time displayed in templesand shrines throughout Japan.
1. INTRODUCTION.
During the Edo Period of Japanese history (1603–1867) theredeveloped a curious practice of hanging wooden tablets with mathematical contentfrom the eaves of Buddhist temples and Shinto shrines. Many of these tablets, knownas sangaku , have been lost to history, but close to 900 of them have been preserved[ ]. The problems inscribed on the surviving sangaku are mostly of a geometric na-ture, and they range in difficulty from trivial to unsolved. Solutions to many of theseproblems can be found in the books by Fukagawa and Pedoe [ ] and Fukagawa andRothman [ ]; the latter reference also discusses various historical aspects surroundingthe mathematics of the Edo Period.Unsolved sangaku problems seem to be rare; in fact, we are aware of only two suchproblems listed in the literature, both in Fukagawa and Rothman [ , Chapter 7]. One ofthe problems mentioned in that text was originally proposed in 1821 and has recentlybeen solved [ ]. The present article concerns the other unsolved problem, which wasproposed by Jihei Morikawa during the same time period.The main objects involved in Morikawa’s problem are illustrated in Figure 1. Givena line L and circles C and C r of radii 1 and r ≥ , respectively, such that C and C r are tangent to each other and to L , the problem asks us to express, in terms of r ,the minimum side length µ ( r ) of a square that can be inscribed in the region between C , C r , and L . Here, “inscribed” means touching all three of C , C r , and L . rµ ( r ) L r C C Figure 1.
The circles C and C r , the line L , and the minimal inscribed square with side length µ ( r ) . Theoriginal statement of the problem involves circles of radii a and b with b ≥ a , but a scaling of the planereduces this general case to the case a = 1 . a r X i v : . [ m a t h . HO ] J u l surviving travel diary of mathematician Kanzan Yamaguchi, a contemporaryof Morikawa, includes an entry with some additional information about Morikawa’sproblem and the sangaku containing it. Based on that entry, Fukagawa and Rothman[ , p. 265] report the following.The tablet contained no solution, but Morikawa had written, “I will be veryhappy if someone can solve this problem.” And so, says Yamaguchi, “I wentto Morikawa’s home with my friend Takeda and asked him what the answer is.He said that he could not solve the problem yet.” Neither does Yamaguchi’s di-ary contain a solution and, like Morikawa, we would be very happy if someonesolves this problem. [ , p. 265]It seems surprising that Morikawa’s problem would have frustrated all attempts ata solution, considering that the mathematicians of the Edo Period had a strong under-standing of geometry, were adept in the use of algebra, and even had some knowledgeof basic calculus. One begins to wonder whether the problem can in fact be solved.The purpose of this article is to address the question of the existence of a closed-form expression for µ ( r ) . From our analysis in Section 4 it follows that µ ( r ) is a rootof a polynomial whose coefficients are polynomials in r ; in light of this fact, a naturalquestion is whether µ ( r ) is expressible by radicals in terms of r . Precise terminologyis defined below, but the question can be stated intuitively as follows: Is there a radicalexpression, such as √ r − π · √ r − r + 2 i + (cid:112) i + √ r − − √ r + 9 r − , that for every real number r ≥ can be evaluated to yield µ ( r ) ?We provide here a negative answer to this question, thus showing that a closed-formalgebraic solution does not exist in the classical sense. In order to state our resultswe introduce the following terminology. Recall that if J ⊆ C is a nonempty set and f : J → C is a function, we say that f is an algebraic function if there exists a nonzeropolynomial q ∈ C [ k, x ] such that q ( c, f ( c )) = 0 for every c ∈ J . If, moreover, thiscondition is satisfied by a polynomial q whose Galois group over the field C ( k ) issolvable — or equivalently, whose splitting field is contained in a radical extensionof C ( k ) — then we say that f is a radical function .We can now state our main result. Theorem (see Theorem 5.4).
The function µ : [1 , ∞ ) → R is not radical. In fact,there is no infinite subset J ⊆ [1 , ∞ ) such that µ : J → R is radical. The proof of this theorem makes critical use of computational tools in Galois theorythat have only recently become available due to work of N. Sutherland [ ]. In partic-ular, our argument relies on Sutherland’s implementation in the system M AGMA [ ]of an algorithm for computing geometric Galois groups. Besides this algorithm, theproof uses elementary geometry as well as calculus and Galois theory.This article is organized as follows. In Section 2 we provide notation and basic re-sults about inscribed squares. In Section 3 we show that any minimal inscribed squaremust be positioned as in Figure 1, with a corner on each of C , C r , and L , and with noside of the square tangent to these objects. In Section 4 we derive an explicit formulafor a function whose minimum value is µ ( r ) ; as a byproduct we obtain a numericalmethod for approximating µ ( r ) given the radius r . In addition, we show that µ ( r ) canbe expressed in terms of a root of a certain polynomial of degree 10. Finally, in Section5 we use Galois theory to study this polynomial and thus prove the main theorem.2 . CONFIGURATIONS FOR INSCRIBED SQUARES. Figure 2 shows the nota-tion that will be used throughout the article, regardless of which inscribed square isunder discussion. Denoted are the circles ( C , C r ), centers of the circles ( O , O r ),line ( L ), vertices of the square ( V , V r , V up , V dn ), and angle between L and the lowerright side of the square ( θ ∈ [0 , π/ ). If θ = 0 , then V dn is the lower left vertex. Tofacilitate phrasing, the line is considered horizontal as shown, with C on the left.Most of the discussion in Sections 2 and 3 assumes an arbitrary but fixed valueof r ≥ . Basic geometric facts [ ] are used throughout. O r O r V V line L r C circle: on line L dn V up V : highest corner C : center of r C : center of C (radius of = ) C circle r C (radius of = r )> θ Figure 2.
Notation for the line, two circles, and square.
Lemma 2.1.
For inscribed squares (for fixed r ≥ ), let θ ∈ [0 , π/ be as in Fig-ure 2.(i) For every θ ∈ [0 , π/ , there is a unique inscribed square.(ii) There exists a minimum side length over the set of inscribed squares. (Thus,Morikawa’s problem is well-defined.)Note: Throughout the rest of the article, the side length of the inscribed square at angle θ ∈ [0 , π/ will be denoted s ( θ ) .Proof. To prove (i), fix θ ∈ [0 , π/ . We find the inscribed square at angle θ as fol-lows, stated somewhat informally to avoid excessive technicalities.Consider all squares, inscribed or not, that make angle θ with L as in Figure 2. Let s be a side length under consideration for being that of an inscribed square. Imaginesliding such a square — with angle θ and side length s — along L until the square is tothe right of C but is just touching C . If s is too small, then the square will not reach C r . If s is too large, then the square will overlap C r . By a continuity and monotonicityargument, there is a unique s such that the C -touching square will exactly reach C r .The existence and uniqueness of an inscribed square follows from that of s above,along with the fact that the inscribed square clearly cannot be moved left or right andstill be inscribed.Subsequently, (ii) follows from the facts that s ( θ ) is continuous on [0 , π/ and lim θ → π/ − s ( θ ) = s (0) . 3enerally in this article, “the square” will mean the inscribed square as given by thecontext, unless stated otherwise. Figure 3 shows the types of intersections between thesquare and circles as θ increases from to π/ , i.e., as the square rotates (and changessize as necessary). Each combination of such intersections, as shown in Figure 3, willhenceforth be referred to as a configuration . The configurations as illustrated in Fig-ure 3 will be denoted Con1, Con2, Con3, etc. LEGENDtangent,not at cornerintersect at corner,not tangenttangent at corner ===
Con1 Con2 Con3Con10 Con9 Con14 Con4Con5Con6Con11 Con12Con7Con8 Con13 Con15Con16Con17Con18Con19 to Con1 to Con2 to Con3
Figure 3.
Possible configurations, in terms of the square’s types of intersections with the line and circles. InSection 3 we prove that a minimal square can occur only in Con6, or in Con19 with r = 1 , by eliminating all ofthe others: Con1,2,3 (Lemma 3.2), Con4,5 (Lemma 3.3), Con7,10,11 (Lemma 3.4), Con12,13,14 (Lemma 3.5),Con15 (Lemma 3.6), Con16,18 (Lemma 3.7), Con19 (Lemma 3.8), Con8,9,17 (Lemma 3.16). emma 2.2. Figure 3 shows all possible steps through the configurations as θ in-creases from to π/ .Proof. The steps through the configurations obviously depend upon r . For example,from Con6, the value of r determines which circle first becomes tangent to a side ofthe square as θ increases. Small r leads to Con7, large r leads to Con12, and a certainintermediate value of r leads to Con11. Also note that only r = 1 gives Con9.The fact that these are the steps through the configurations is generally clear, withtwo exceptions: from Con8 to Con9 and Con10, and from Con15 to Con16.For Con8 to Con9 and Con10, the question is whether the upper right side of thesquare could instead rotate past tangency with C r before the upper left side of thesquare becomes tangent to C . This can happen only if the lower left side of the squarehas steeper slope than the line through V dn and O — i.e., the lower left side “pointsabove” O — and the lower right side points above O r . However, this is impossiblebecause the circle of radius ( r + 1) / through O and O r , as shown in Figure 4, istangent to L . Every angle inscribed in a semicircle is a right angle, so since L is belowthe new circle except at the point of tangency, Q , the lower sides of the square cannotboth point above their respective circles’ centers. At best, the lower sides can pointexactly at the centers, but only if r = 1 and V dn = Q . circle ofradius( r + ) / Q O r OL r C C r Figure 4.
Circle with diameter O -to- O r is tangent to L . For Con15 to Con16, the question is whether V up can instead touch C before thelower right side of the square touches C r . This can happen only if the square withupper right side tangent at V r to C r has V up touching C . However, this is impossible.Any square with upper right side tangent at V r to C r must have V dn at or to the left of Q in Figure 4 in order for the square to reach C , but then the square is angled suchthat V up cannot touch C .
3. CONFIGURATIONS FOR MINIMAL SQUARES.
In this section we prove thata minimal square — i.e., an inscribed square with side length that is minimal over all orientations — exists only in Con6, and has V lower than O . An exception occurs if r = 1 , where Con19 is the reflection of Con6 and thus also has a minimal square. The5roof consists of a sequence of lemmas showing that a minimal square cannot be inany other configuration. The final result is given by Proposition 3.17.Additional notation is used throughout this section, for the lines perpendicular toeach of C , C r , and L at the points of contact with the square under consideration. Asillustrated in Figure 5, these (dashed) lines are denoted T , T r , and T L , respectively.As before, unless stated otherwise we assume an arbitrary but fixed value of r ≥ . Lemma 3.1.
For a given inscribed square, if T intersects T L above T r (respectively,below T r ), then s is a strictly decreasing (respectively, increasing) function of θ at thatsquare’s angle.Proof. If T intersects T L above T r , then the lines form a triangle to the right of T L ,such as in Figure 5. r C C T L T r T O r OL θ Figure 5.
Notation: Lines T , T r , and T L through points of intersection. Consider fixing the size of the square, and rotating it counterclockwise about a pointinside the triangle. As the square begins to rotate, it starts to overlap with each of C , C r , and L . Formally, the rate of change of the following are positive: (1) the radius of C minus the distance from O to the square, (2) the radius of C r minus the distancefrom O r to the square, and (3) the distance below L to the lowest point on the square.This means that the difference between the side lengths of an inscribed square and afixed-size square is a strictly decreasing function of θ ; thus s is a strictly decreasingfunction of θ . Similarly, if T intersects T L below T r , then s is a strictly increasingfunction of θ . Lemma 3.2.
A minimal square cannot be in Con1, Con2, or Con3.Proof.
Each of these configurations corresponds to θ = 0 . For small enough (cid:15) > ,it is easy to see that for all θ ∈ (0 , (cid:15) ) , T intersects T L above T r . Therefore byLemma 3.1, s ( θ ) is strictly decreasing for θ ∈ (0 , (cid:15) ) and thus by continuity s (0) isnot minimal. 6 emma 3.3. A minimal square cannot be in Con4 or Con5, and a minimal squarecannot be in Con6 with V as high or higher than O .Proof. Figure 6 illustrates Con4, and the same reasoning applies to the other two con-figurations. First, note that V is higher than O in Con4 and Con5 because some partof the left side of the square is tangent to C , and that left side has negative slope.Consider the line L (cid:48) tangent to C r at V up , and the line L (cid:48)(cid:48) bisecting the angle between L and L (cid:48) . Line L (cid:48) must intersect C (in order to “escape” the region between C , C r ,and L ), so L (cid:48)(cid:48) is below O . O line L line L ’’ line L ’ dn V up V V T L T r T r C C Figure 6.
Illustration for Lemma 3.3. The line L (cid:48)(cid:48) bisecting the angle between L and L (cid:48) is below O , so V is higher than L (cid:48)(cid:48) , leading to an application of Lemma 3.1. Because V is higher than O and therefore higher than L (cid:48)(cid:48) , the square is tilted insuch a way that V up is closer than V dn to the intersection of L (cid:48) and L . Therefore, T r in-tersects L (cid:48)(cid:48) to the right of where T L intersects L (cid:48)(cid:48) ; thus the intersection of T L and T r isbelow the intersection of T L and L (cid:48)(cid:48) . Meanwhile, T has nonnegative slope and there-fore intersects T L above the intersection of T L and L (cid:48)(cid:48) , thus above the intersection of T L and T r . Hence by Lemma 3.1, s ( θ ) is a strictly decreasing function and is thereforenot minimal for these values of θ . Lemma 3.4.
A minimal square cannot be in Con7, Con10, or Con11.Proof.
In each of these configurations, T intersects T L below the intersection between T L and T r . Therefore by Lemma 3.1, s ( θ ) is strictly increasing and therefore notminimal at the corresponding θ . up V dn V T L T r T r C C θ Figure 7.
Illustration for Lemma 3.5. In Con12, Con13, and Con14, θ > π/ , so T L is to the right of V up ,leading to an application of Lemma 3.1. The proof that a minimal square cannot be in Con8, Con9, or Con17 is substantiallymore complicated than the other proofs, and is saved for the end of the section.7 emma 3.5.
A minimal square cannot be in Con12, Con13, or Con14.Proof.
In each of these configurations, some part of the upper left side of the square istangent to C . Therefore, in order for V up to reach C r , θ cannot be particularly small,and certainly θ > π/ : As seen in Con9 (Figure 3) which has θ = π/ and r = 1 , asquare with θ = π/ and upper left side tangent to C can only just barely reach —and not even with V up — the smallest possible C r , that with r = 1 . Then decreasing θ with the square’s upper left side still tangent to C would shrink and move the squareaway from C r , so θ ≤ π/ is not possible here when r = 1 . It is also not possiblewhen r > because any larger C r would further prevent the square from reaching it.Therefore, T L is to the right of V up , as illustrated in Figure 7 for the case of Con12.Thus by Lemma 3.1, s ( θ ) is a strictly increasing function of θ and is therefore notminimal for these values of θ . Lemma 3.6.
A minimal square cannot be in Con15.Proof.
We prove this by showing that as the square rotates counterclockwise throughthe range of Con15, the intersection between T and T L moves upward strictly mono-tonically, and the intersection between T r and T L moves downward strictly monotoni-cally. Because it is clear in Con15 that T L ’s intersection with T is below that with T r for the smaller values of θ , and above that with T r for the larger values of θ , Lemma 3.1implies that s ( θ ) strictly increases then strictly decreases as a function of θ , so the re-sult follows. (This seems to imply that there is a maximal square in Con15. However,this article does not address maximal squares.)We give the proof that the intersection between T and T L moves upward strictlymonotonically; an analogous proof shows that the intersection between T r and T L moves downward strictly monotonically. T L T r T O qp f ( p, q ) g ( p, q ) Figure 8.
For Lemma 3.6, the rectangle determined by the point ( p, q ) . Consider not just inscribed squares, but inscribed rectangles in general, determinedby the pair ( p, q ) as shown in Figure 8, where p is the horizontal distance between O and the bottom vertex, and q is the height above L of the intersection between T and T L . Let f ( p, q ) = length of lower left side of rectangle, g ( p, q ) = length of lower right side of rectangle.For each p there is a unique q , say q = h ( p ) , such that the rectangle is a square, i.e., f ( p, q ) = g ( p, q ) . Our goal is to show that h ( p ) is a strictly increasing function of p .We will do this by showing that 81) f ( p, q ) is a strictly increasing function of p ,(2) g ( p, q ) is a strictly decreasing function of p ,(3) f ( p, q ) is a strictly decreasing function of q ,(4) g ( p, q ) is a strictly increasing function of q .This will complete the proof because if point ( p, q ) gives a square, then (1) and (2)imply that increasing p will cause the lower left side to become larger than the lowerright side, so by (3) and (4), q must be increased in order to restore equality of the sidelengths. ℓ O mp q Figure 9.
For Lemma 3.6, notation for the proof that f ( p, q ) is a strictly increasing function of p . For (1), fix q and note that f ( p, q ) is (cid:96) + m − in Figure 9. By similar triangles, m/q = (1 − q ) /(cid:96) . Define a function j ( (cid:96) ) = (cid:96) + m − (cid:96) + q (1 − q ) (cid:96) − , which equals f ( p, q ) . An easy calculation shows that j (cid:48) ( (cid:96) ) > because (cid:96) > / , andsince p and (cid:96) increase together, this gives ∂f∂p > , thus proving (1). lower leftside of rectangle O Q T Figure 10.
For Lemma 3.6, geometry for the proof that g ( p, q ) is a strictly decreasing function of p . For (2), note that g ( p, q ) is exactly r less than the distance from O r to the lower leftside of the rectangle. That side is on a line through a point Q below O . If p increaseswhile q stays fixed, then as shown in Figure 10, the lower left side of the rectangle is9 O lower leftside of rectangle lower rightside of rectangle T Figure 11.
For Lemma 3.6, geometry for proofs that f ( p, q ) is strictly decreasing, and g ( p, q ) is strictlyincreasing, as functions of q . still aligned with Q because that side is always parallel to T . Therefore, as p increases,the distance from O r to the lower left side of the rectangle decreases, thus proving (2).For (3) and (4), note that f ( p, q ) is exactly less than the distance from O to thelower right side of the rectangle, and g ( p, q ) is exactly r less than the distance from O r to the lower left side of the rectangle. If p is fixed and q increases, then as shownin Figure 11, the lower left side of the square (which is parallel to T ) rotates awayfrom O r , and the lower right side (which is perpendicular to T ) rotates toward O .This proves (3) and (4), thus completing the proof. Lemma 3.7.
A minimal square cannot be in Con16 or Con18.Proof.
In each of these configurations, T intersects T L above the intersection between T L and T r . Therefore by Lemma 3.1, s ( θ ) is strictly decreasing and therefore notminimal at the corresponding θ .If r = 1 , then Con19 is simply the reflection of Con6, so for Con19 we focus onthe case r (cid:54) = 1 . Lemma 3.8. If r (cid:54) = 1 , then a minimal square cannot be in Con19.Proof. Given a square at angle θ in Con19, we claim that the inscribed square inthe reflected orientation, i.e., at angle π/ − θ , is smaller. This fact can be seen bymapping the original square to a horizontal reflection that is positioned so that theimage P (cid:48) of vertex P ( = V ) is on C r , as shown in Figure 12. In the process, vertex Q ( = V r ) is mapped to a point Q (cid:48) inside the disk bounded by C for the followingreason. The acute angles φ made with the horizontal are the same for the line through P and Q (cid:48) and the line through P (cid:48) and Q , while circle C is steeper than C r at everyheight above L between and .Therefore, the inscribed square at angle π/ − θ must be smaller than the reflectedsquare and thus the original square.To prove that a minimal square cannot be in Con8, Con9, or Con17, preliminaryresults and definitions are helpful. Definition 3.9.
Con10 + will refer to Con10 but will also allow r ≤ . Con8 + willrefer to the union of Con7, Con8, Con9, and Con10, and will also allow r ≤ . Remark 3.10.
A proof that a minimal square cannot be in Con8 + constitutes a prooffor Con8, Con9, and Con17 (as well as Con7, Con10, Con16, and Con18, for whichproofs have already been given) because Con17 with r = r ≥ is equivalent to aversion of Con8 with r = 1 /r ≤ . 10 C C P ’ Q ’ Q (= ) r VP (= ) up V φφ θ Figure 12.
Illustration for Lemma 3.8 showing the horizontal reflection of the square that sends point P on C to a point P (cid:48) on C r . (Such a reflection is not necessarily about the midline of the square.) Lemma 3.11.
The square in Con10 + has side length less than or equal to M , where M = 2 rr + √ √ r + 1 , with equality if and only if r = 1 .Proof. Let P be the point on line L between C and C r such that the distance from P to the closest point on C (along the line through P and O ) equals the distancefrom P to the closest point on C r . The line segment from P to O can be viewedas the hypotenuse of a right triangle, as can the line segment from P to O r , so theequality of distance can be written using the Pythagorean theorem as follows. Let p bethe distance from P to the intersection of L and C . Because √ r is the distance fromthe intersection of L and C to the intersection of L and C r (see Figure 15), we have (cid:112) p − (cid:113) r + (2 √ r − p ) − r. By basic geometry, the square in Con10 + must have V dn to the left of P , unless r = 1 ,in which case V dn = P . Therefore the square has side length less than or equal to √ p − , with equality exactly if and only if r = 1 .It is straightforward to confirm, by substituting the following into the equationabove, and by noting that this is the desired solution because it lies in (0 , √ r ) for r > , that p = √ r + 2 √ rr + √ √ r + 1 , and then that (cid:112) p − rr + √ √ r + 1 , which is M in the statement of the lemma.11 emma 3.12. For a square in Con17, if the distance from L to the midpoint of theupper right side of the square is less than or equal to r − r/ √ , then the square is notminimal.Proof. Given such a square in Con17 as shown in Figure 13, we claim that reflectingthe square about its midline — the vertical line through its center point — will cause itto overlap both C and C r . This will prove the lemma because an inscribed square inthe same orientation as the reflected square is thus smaller than the original square. Weprove the claim by showing that reflecting the square takes vertex P (= V up ) as shownin Figure 13 to vertex P (cid:48) in the interior of the disk with boundary C r , and takes Q (= V r ) to vertex Q (cid:48) such that either Q (cid:48) or part of the left side of the reflected square is inthe interior of the disk with boundary C . P ’ Q ’ Q (= ) r VP (= ) up V V r C C height of midpointof upper right sideof squaremidline of square αα ” α ’ centerof squareslope = α (= ) α (= ) Figure 13.
For Lemma 3.12, the horizontal reflection of the square about its midline.
First, the line through P and Q (cid:48) has slope , and the line through Q and P (cid:48) has slope − , by the following reasoning: In Figure 13, the fact that α (cid:48) = α is clear because ofthe right angle shown, and α (cid:48)(cid:48) = α because of the symmetry of the reflection. Theline segment between P and Q (cid:48) is seen to be the base of an isoceles triangle since P and Q (cid:48) are the same distance from the center of the square. Because α (cid:48) = α (cid:48)(cid:48) , theisoceles triangle is symmetric about a line of slope − through the center of the square.Therefore, the base of the triangle, and thus the line through P and Q (cid:48) , has slope .Similar reasoning shows that the line through Q and P (cid:48) has slope − .Next, we claim that the secant line segment intersecting C r at Q and the pointnearest horizontally to P (cid:48) has slope shallower than − . (This secant would lie almostexactly along C r in Figure 13, and would not be distinguishable in the figure.) To provethe claim, we note that the midpoint of that secant is at the same height as the upperright side of the square, no higher than r − r/ √ by hypothesis, because the secant’sendpoints are at the heights of vertices P and Q . A secant has as a perpendicularbisector a radial line segment of C r ; such a radial line segment for this secant mustreach below a height of r − r/ √ since it passes through the secant’s midpoint. Thus,this radial line segment must have slope steeper than , because at slope it would12each down only to height r − r/ √ with its length r and top endpoint at height r .The secant, which is perpendicular to it, must therefore have slope shallower than − .Therefore, P (cid:48) is in the interior of the disk with boundary C r .To show the overlap with C , we first note that Q (cid:48) is at height greater than − / √ . This is because V r of a square in Con17 must be higher than that in Con9 orCon10 (whichever applies, depending on whether r = 1 ) by the basic geometry ofthe counterclockwise rotation toward Con17. Among these three configurations, Con9with its r = 1 has the lowest V r , at height − / √ since θ = π/ .Now, Q (cid:48) lies on the line of slope through P , a line whose lower intersection with C is at height less than − / √ because − / √ is where the tangent to C hasslope . Therefore, Q (cid:48) is on the line of slope through P on either the segment from P to C or the segment that lies in the interior of the disk with boundary C . In thelatter case, the proof of the lemma is complete. The former case can only occur if theoriginal square is so large that Q (cid:48) is above a portion of the top half of C , because Q (cid:48) is straight above V . In that case, the left side of the reflected square must intersectthe interior of the disk with boundary C , completing the proof of the lemma, becausethe reflected square is also a reflection of the original square about a horizontal linethrough its center point. Since the left side of the original square is tangent to C , andthe horizontal line of reflection is below the center of C , the reflected left side mustoverlap C and thus intersect the interior of the disk with boundary C . Corollary 3.13. If r ≥ , then a minimal square cannot be in Con17.Proof. Suppose for contradiction that r ≥ and a minimal square is in Con17. Thenthe side length, s , must be less than or equal to the side lengths of all squares in all otherconfigurations, including Con10 which is addressed in Lemma 3.11. In particular, wemust have s ≤ M from Lemma 3.11. Let h be the distance from L to the midpoint ofthe upper right side of the square. We know that h can be no greater than ( √ / s ,because this is the highest that a midpoint of a side can possibly be, occurring if themidpoint is straight above V dn . In summary, h ≤ ( √ / s ≤ ( √ / M . Therefore, h ≤ r − r/ √ if √ (cid:32) rr + √ √ r + 1 (cid:33) ≤ r − r √ , which is true if r ≥ . However, Lemma 3.12 then implies that the square is not mini-mal. This gives a contradiction. Definition 3.14.
For a family { J ( t ) } t ∈ I of lines or line segments, where I is an in-terval in R , the pivot at a given t ∈ I is the point about which the line or line segmentis pivoting at t , if such a point exists. In other words, for a given t ∈ I , if there exists (cid:15) > such that u ∈ I ∩ ( t − (cid:15), t + (cid:15) ) implies that J ( u ) ∩ J ( t ) consists of a singlepoint, then the pivot, P , at t is defined by P = lim u → t J ( u ) ∩ J ( t ) if the limit exists. (The limit is one-sided if t is an endpoint of I .) Lemma 3.15.
Let L φ be the line through the origin having angle φ ∈ (0 , π/ withthe positive x -axis as in Figure 14, define an interval I = ( π/ − φ, π/ , and fix (cid:96) > . Let { J ( β ) } β ∈ I be the family of line segments of length (cid:96) with endpoints on the -axis and L φ , and for which J ( β ) intersects the x -axis at angle β as in Figure 14.Then the following hold. P ℓ bc β K ( β ) K J ( β ) J φ L γ xy φ Figure 14.
Notation for Lemma 3.15. (i) For { J ( β ) } β ∈ I , the position of the pivot P at β is determined by b (cot β ) = c (cot γ ) , where b and c are the distances along J ( β ) from P to the x -axis and L φ , re-spectively, and γ = π − φ − β is the angle between J ( β ) and L φ .(ii) Let { K ( β ) } β ∈ I be the family of lines such that K ( β ) is the line perpendic-ular to J ( β ) through J ( β ) ’s pivot. Then for { K ( β ) } β ∈ I , the pivot at β has y -coordinate (cid:96) (cid:18) cos γ + cos( γ − β ) cos β sin( γ + β ) (cid:19) . Proof.
Given such L φ , (cid:96) , { J ( β ) } β ∈ I , and { K ( β ) } β ∈ I , fix a value β ∈ I .To prove (i), let P be the pivot at β for { J ( β ) } β ∈ I , and let b and c be thedistances along J ( β ) from P to the x -axis and L φ , respectively. Define a family { H ( β ) } β ∈ (0 ,π/ of variable-length line segments as those for which H ( β ) is theline segment through P with endpoints on the x -axis and L φ , and that intersects the x -axis at angle β ; e.g., H ( β ) = J ( β ) . It follows that d ( length of H ( β )) /dβ = 0 at β = β because H ( β ) and J ( β ) have the same pivot at β , so the instantaneous ratesof change of location of their respective endpoints (on the x -axis and L φ ) are the same.Let B ( β ) and C ( β ) be the distances along H ( β ) from P to the x -axis and L φ , respec-tively, so B (cid:48) ( β ) + C (cid:48) ( β ) = 0 . Basic trigonometry shows that B ( β ) = h B / sin β and C ( β ) = h C / sin γ , where γ = π − φ − β , and h B and h C are the (shortest) dis-tances from P to the x -axis and L φ , respectively. Putting all of this together, (i) followsbecause b = B ( β ) and c = C ( β ) .For (ii), we note that K ( β ) is the line y = cot β (cid:18) x − (cid:18) (cid:96) sin γ sin φ − b cos β (cid:19)(cid:19) + b sin β, i.e., y = (cot β ) x − (cid:96) (cid:18) sin( φ + 2 β )sin φ sin β (cid:19) , where the second form uses γ = π − φ − β , as well as a substitution for b stemmingfrom (i) which gives b (cot β ) = ( (cid:96) − b )(cot γ ) .14or all β , the slope of K ( β ) is greater than zero, so the pivot for { K ( β ) } β ∈ I at β can be found as follows. For any given y , define the function f y ( β ) giving the valueof x such that ( x, y ) is on the line K ( β ) : f y ( β ) = (tan β ) y + (cid:96) (cid:18) sin( φ + 2 β )sin φ cos β (cid:19) . Then the pivot at β must have y -coordinate such that f (cid:48) y ( β ) = 0 . Solving this equa-tion for y gives (ii). Lemma 3.16.
A minimal square cannot be in Con8, Con9, or Con17.Proof.
Suppose for contradiction that a minimal square exists in Con8 + . Noting thatthe extended version of Con8 with r = r ≤ is equivalent to Con17 with r = 1 /r ,we may apply Corollary 3.13, concluding that r > / .Let I = [ β , β ] be the set of angles between L and the lower left sides of thesquares in Con8 + , and let β ∈ I be the maximum such angle of a minimal squarein Con8 + . Then we know that β (cid:54) = β , and that β (cid:54) = β unless r = 1 , because aminimal square cannot be in Con7, Con10, Con16, or Con18. Let (cid:96) be the side lengthof the minimal square, and let { G ( β ) } β ∈ I be the family of line segments of length (cid:96) with endpoints on L and C , for which G ( β ) intersects L at angle β . Let m ( β ) bethe distance from O r to G ( β ) . Then m (cid:48) ( β ) = 0 because either m ( β ) is a minimalvalue on an open interval, or the square for β is in Con9 so the shortest distance from O r to G ( β ) is the distance from O r to G ( β ) ’s endpoint at L , which is the pivot for { G ( β ) } β ∈ I at β .Let { H ( β ) } β ∈ I be the family of lines such that H ( β ) is perpendicular to G ( β ) through G ( β ) ’s pivot. Let h ( β ) be the vertical distance from O r to H ( β ) , but consid-ered negative if H ( β ) is below O r . The sign of m (cid:48) ( β ) is the same as the sign of h ( β ) ,because h ( β ) > means that G ( β ) is moving away from O r as β increases, and viceversa. Therefore, h ( β ) = 0 .Let L be the line tangent to C at G ( β ) ’s endpoint on C . It follows from thistangency of L and C that the pivot at β for { G ( β ) } β ∈ I is the same as the pivotat β for the family { J ( β ) } β ∈ I of line segments of length (cid:96) with endpoints on L and L with J ( β ) intersecting L at angle β . Intuitively, this match of the pivot is because J ( β ) acts like G ( β ) near β , and can be seen formally by standard (cid:15) - δ reasoning. Italso follows that the pivot P at β for { H ( β ) } β ∈ I is the same as the pivot at β for thefamily { K ( β ) } β ∈ I of lines such that K ( β ) is perpendicular to J ( β ) through J ( β ) ’spivot.We now apply Lemma 3.15(ii) to { K ( β ) } β ∈ I in order to show that the distancefrom L to P is less than r . This distance, the “ y -coordinate” of P , is thus given by y P = (cid:96) (cid:18) cos γ + cos( γ − β ) cos β sin( γ + β ) (cid:19) , where γ is the angle that J ( β ) makes with L . We claim that y P ≤ (cid:96) √ , and beginthe proof by noting that for angle φ between L and L we have φ + γ + β = π , alongwith φ, β ∈ (0 , π/ , γ ∈ (0 , π/ , and γ + β ∈ ( π/ , π ) . Let v = ( γ + β ) / and w = ( γ − β ) / , and let f ( v, w ) = (cid:96) (cid:18) cos( v + w ) + cos(2 w ) cos( v − w )sin(2 v ) (cid:19)
15o that y P ≤ (cid:96) √ can be proved by showing that f ( v, w ) ≤ √ on the domain de-fined by v + w ≤ π/ , v − w < π/ , and v ≥ π/ . Although v cannot take thevalue π/ in the geometric interpretation, this extension of the domain of f to include v = π/ facilitates phrasing in the proof of a bound.Straightforward computations show that ∂f∂v = − cos β − γ cos β sin ( γ + β ) < on the domain, so the maximum occurs on the border v = π/ . Let g ( w ) = f (cid:16) π , w (cid:17) = cos (cid:16) w + π (cid:17) + cos(2 w ) cos (cid:16) w − π (cid:17) . The maximum value of g ( w ) occurs at w = 0 , because g (cid:48) ( w ) = √ (cid:0) cos w − cos w − cos w sin w (cid:1) and w ∈ ( − π/ , π/ on the domain, so g ( w ) increases then decreases, and more-over, g (cid:48) ( w ) = 0 only at w = 0 . Therefore, we have f ( v, w ) ≤ f ( π/ ,
0) = √ forall ( v, w ) in the domain of f , proving that y P ≤ (cid:96) √ .To complete the proof that the distance from L to P is less than r , we note thatbecause (cid:96) is the minimum side length of a square, Lemma 3.11 implies that y P ≤ (cid:96) √ ≤ (cid:32) rr + √ √ r + 1 (cid:33) √ . Since we know from Corollary 3.13 that r > / , the fact that y P < r easily follows.Now, because H ( β ) has positive slope, and pivot P on H ( β ) is at a distance lessthan r from L , P is below and to the left of O r . Recalling that h ( β ) is the (signed)vertical distance from O r to H ( β ) , the fact that pivot P is to the left of O r impliesthat h (cid:48) ( β ) < . We already know that h ( β ) = 0 , so there exists (cid:15) > such that β ∈ ( β , β + (cid:15) ) implies h ( β ) < . However, since h ( β ) and m (cid:48) ( β ) have the samesign, this means m (cid:48) ( β ) < on ( β , β + (cid:15) ) , contradicting m ( β ) being minimal. Proposition 3.17.
A minimal square occurs only in Con6, and possibly in Con19, andmust have V lower than O . Con19 has a minimal square if and only if r = 1 .Proof. This result follows from the lemmas ruling out all other configurations, and thefact that Con19 is the reflection of Con6.
4. EQUATIONS FOR MINIMUM SIDE LENGTH.
In this section we derive equa-tions toward the pursuit of the minimum side length of an inscribed square, knowingfrom Proposition 3.17 that this minimum occurs in Con6 and has V lower than O .We continue using here the notation introduced at the beginnings of Sections 2 and 3. Lemma 4.1.
In Con6 with V lower than O , s ( θ ) decreases then increases, bothstrictly monotonically, as a function of θ . Thus, there is a unique minimal squarein Con6.Proof. As θ increases in Con6 with V lower than O , T L moves to the right and V moves down, so the intersection between T (which has negative slope) and T L moves16own. Simultaneously, V r moves up, so the intersection between T r and T L moves up.Therefore, as θ increases, the triangle formed by T L , T , and T r switches from the rightside of T L to the left side of T L at a unique value, when T and T r intersect T L at thesame point. By Lemma 3.1, s ( θ ) decreases then increases, both strictly monotonically.The uniqueness of the minimal square in Con6 follows from this and Proposi-tion 3.17.For the equations used in seeking the minimum side length of an inscribed square,the parameter representing orientation will be the distance, x , from the line L to theupper left vertex V of the square, as shown in Figure 15. It is not difficult to see that x is related strictly monotonically to θ . rr V up V O r O line L b = 2 √ r − x − (cid:127) x − x c = (cid:127) r − b (cid:127) ( r + 1) − ( r − = 2 √ r (cid:127) − (1 − x ) = (cid:127) x − x h = r − x − c xx zz Figure 15.
Equations used to compute the side length of a square in Con6, given distance x from the line L to the upper left vertex V of the square. Here the side length is denoted z . Proposition 4.2.
The minimum side length, µ ( r ) , of an inscribed square is the mini-mum value of the function z ( x ) = x + (cid:32) r − x − (cid:114) r − (cid:16) √ r − x − √ x − x (cid:17) (cid:33) (1) on the interval (1 − / √ , . The minimum occurs at a unique point, denoted x m .Moreover, the function z is strictly decreasing for x < x m and strictly increasing for x > x m .Proof. Let ( x , x ) be the interval of values of the distance x from L to V suchthat the associated square is in Con6 with V lower than O . We know from Proposi-tion 3.17 that the minimum side length corresponds to some x ∈ ( x , x ) .17f x ∈ ( x , x ) , then as illustrated in Figure 15, congruent triangles using x and z show that V up is at a horizontal distance x from V . Therefore, for x ∈ ( x , x ) , z = (cid:0) x + h (cid:1) , where h is the vertical distance from V to V up . Thus by the equations in Figure 15, theside length is given as a function of x ∈ ( x , x ) by z ( x ) in equation (1). In addition,it follows from Lemma 4.1 that on ( x , x ) the minimum occurs at a unique point, x m , and that z is strictly decreasing for x < x m and strictly increasing for x > x m .However, the stated domain (1 − / √ , extends beyond ( x , x ) ; certainly, x < by definition. Toward the claim that x > − / √ , we suppose for con-tradiction that x = 1 − / √ is possible. Because V is distance x from L , x =1 − / √ occurs exactly when the line through O and V has slope − . In thatcase, − / √ is also the horizontal distance from V to a vertical line tangent to C on the right side. Because V up is horizontal distance x ( = 1 − / √ ) from V , V up ison that vertical line. This contradicts the fact that V up is on C r which requires V up to beto the right of that vertical line. Noting that x < − / √ would result in a similarcontradiction, we conclude that x > − / √ .Now considering z in equation (1) as a function of x ∈ (1 − / √ , , we know on ( x , x ) that z represents the side length of an inscribed square, but outside ( x , x ) we only have equation (1). It is helpful to set up geometric interpretations for the cases x ∈ (1 − / √ , x ) and x ∈ ( x , in order to complete the proof by showing that z strictly decreases when x ∈ (1 − / √ , x ) and strictly increases when x ∈ ( x , . rrh xx z z T L T r T Figure 16.
Geometry showing that z given by (1) is not a minimum when x < x . Figures 16 and 17 illustrate geometric interpretations of z as given by equation (1)when x belongs to the intervals (1 − / √ , x ) and ( x , , respectively. Here, thethree consecutive vertices still lie on the line and the two circles, although the squaresare no longer inscribed. In both cases, the same equations as in Figure 15 still apply,and lead to (1). The fact that z strictly decreases when x ∈ (1 − / √ , x ) and strictly18 rrc h = r − x − ch = r − x − c xzxz T L T r T Figure 17.
Geometry showing that z given by (1) is not a minimum when x > x . increases when x ∈ ( x , can been seen by the same reasoning as in Lemma 4.1,this time using T , T r , and T L defined as before except that the “point of contact” isspecifically that with the relevant vertex, as shown in Figures 16 and 17. The reasoningthen uses a modified form of Lemma 3.1 that applies to these squares with consecutivevertices on the line and circles, instead of to inscribed squares, and whose proof isanalogous to that of Lemma 3.1. Remark 4.3 (approximating µ ( r ) ). From Proposition 4.2, one can deduce an algo-rithm for computing an approximation of µ ( r ) given the radius r . Indeed, it sufficesfor this purpose to minimize the function z , which can be achieved by applying root-finding methods to z (cid:48) . Proposition 4.4.
In the result of Proposition 4.2, i.e., that the minimum side length isgiven by µ ( r ) = z ( x m ) , the number x m is a root of the 10th degree polynomial f ( f f + f f − f f ) − ( f f + f − f f f − f f ) , (2) where, letting k = √ r , f ( x ) = − x + 4 kf ( x ) = (4 k − x + k − k f ( x ) = (6 k − x + k − k − k f ( x ) = − x + 2 xf ( x ) = 4 x − (2 k + 6 k + 7) x + (2 k + 3 k + 10 k ) x − k f ( x ) = 8 x + ( − k − x + (4 k − k + 6) x − k + 8 k − kf ( x ) = (4 k − k ) x + ( − k + 4 k − k + 40 k ) x + (2 k − k + 4 k − k + 2) x + 4 k . roof. By Proposition 4.2, µ ( r ) = z ( x m ) , where z ( x ) is given by equation (1), and x m is the unique point in the interval (1 − / √ , such that dz/dx = 0 , or equiva-lently dA/dx = 0 where A ( x ) = ( z ( x )) .It remains to show that x m is a root of (2). Let g ( x ) = ( − x + 4 k ) (cid:113) f ( x ) + (4 k − x + k − k , so that expansion in (1) gives A ( x ) = 2 (cid:16) ( x − k ) (cid:113) g ( x ) + ( − x + 2 k ) (cid:113) f ( x ) + x + ( − k + 2 k − x + k − k (cid:17) . Note that dgdx = (4 k − (cid:112) f ( x ) + 4 x − (4 k + 6) x + 4 k (cid:112) f ( x ) , which leads to dAdx = ( x − k ) (cid:16) (2 k − (cid:112) f ( x ) + 2 x − (2 k + 3) x + 2 k (cid:17)(cid:112) g ( x ) (cid:112) f ( x ) + (cid:113) g ( x )+ x − (2 k + 1) x + 2 k (cid:112) f ( x ) − (cid:113) f ( x ) + 2 x − k + 2 k − . In view of the fact that dA/dx = 0 at x = x m , set dA/dx = 0 , then multiply by (cid:112) g ( x ) (cid:112) f ( x ) and separate terms with (cid:112) g ( x ) from the rest to obtain (cid:18) (2 x − k + 2 k − (cid:113) f ( x ) + 2 x − (2 k + 3) x + 2 k (cid:19) (cid:113) g ( x )= − f ( x ) (cid:113) f ( x ) − f ( x ) . Note that g ( x ) = f ( x ) (cid:112) f ( x ) + f ( x ) , so that squaring both sides and omitting“ ( x ) ” for readability results in (cid:16) f (cid:112) f + f (cid:17) (cid:16) f (cid:112) f + f (cid:17) = f f + 2 f f (cid:112) f + f . Separating terms with √ f from the rest gives ( f f + f f − f f ) (cid:112) f = f f + f − f f f − f f . Squaring both sides and rearranging leads to the final polynomial equation, f ( f f + f f − f f ) − ( f f + f − f f f − f f ) = 0 . f and f f f , each of which has degree 6, but both f and f f f have x as their 6thdegree term, so f − f f f has degree 5.In summary, x m is a root of the polynomial in (2).
5. NONEXISTENCE OF A SOLUTION BY RADICALS.
In this section we provethat µ is not a radical function as defined in Section 1. For convenience we workinstead with the function λ defined by λ ( c ) = ( µ ( c )) . It is intuitively clear that if µ is a radical function then λ is also radical; a formal proof of this fact is given ina more general context in Lemma 5.1 below. The proof of our main result, Theorem5.4, shows that λ is not radical on any infinite subset of [1 , ∞ ) . As a consequencewe obtain the fact that µ cannot be radical on any such set. We refer the reader to [ ,Chapters 13, 14] for the algebraic background assumed in this section. Lemma 5.1.
Let J be a nonempty subset of [1 , ∞ ) , let n ∈ Z + , and let I = n √ J bethe set of positive n th roots of elements of J . Suppose that a function f : J → R isradical, and define g : I → R and h : J → R by g ( c ) = f ( c n ) and h ( c ) = ( f ( c )) n .Then g and h are radical.Proof. We begin by showing that g is radical. Let q ∈ C [ k, x ] be a nonzero polynomialsatisfying q ( c, f ( c )) = 0 for every c ∈ J , and such that there is a radical extension R/ C ( k ) containing a splitting field S of q . Let ϕ : C ( k ) → C ( k ) be the embeddinginduced by the map k (cid:55)→ k n , and let Ω be an algebraic closure of C ( k ) . By basic fieldtheory (see [ , Chapter V, §
2, Theorem 2.8]), we may extend the map ϕ to an em-bedding ϕ : R → Ω . Defining Q ( k, x ) = q ( k n , x ) we have Q ( c, g ( c )) = 0 for every c ∈ I . Moreover, since Q is the polynomial obtained by applying ϕ to the coefficientsof q , the above observations imply that Q splits in the field ϕ ( S ) , which is containedin the radical extension ϕ ( R ) / C ( k ) . Thus g is a radical function.Next we show that h is radical. The argument will be given assuming that q is monicand has no repeated root; the general case can be proved similarly. Let α , . . . , α d bethe roots of q in S , and let Q ( k, x ) = ( x − α n ) · · · ( x − α nd ) . Note that since q has coefficients in C [ k ] , the same holds for Q . (The elementary sym-metric functions of α n , . . . , α nd are polynomials in the elementary symmetric functionsof α , . . . , α d .) Moreover, as explained below, one can show that Q ( c, h ( c )) = 0 forevery c ∈ J . Since Q splits in the field C ( k, α n , . . . , α nd ) ⊆ S ⊆ R , this implies that h is radical.Fixing c ∈ J , the fact that Q ( c, h ( c )) = 0 can be seen heuristically first: Since q ( k, x ) = ( x − α ) · · · ( x − α d ) and q ( c, f ( c )) = 0 , we must have f ( c ) = α i ( c ) for some i , and thus Q ( c, h ( c )) = 0 .This argument can be made rigorous by extending the map k (cid:55)→ c to a ring homomor-phism C [ k, α , . . . , α d ] → C , so that α i ( c ) is well-defined. We refer the interestedreader to Section 3 in [ , Chapter VII] for the necessary tools.Next we prove two preliminary results needed to show that λ is not radical. Lemma 5.2.
Let p ( k, x ) be the polynomial defined by (2) considering k and x asindeterminates. As an element of the ring C ( k )[ x ] , the polynomial p is irreducible andhas Galois group isomorphic to the symmetric group S . roof. We rely on a computation carried out using the computer algebra systemM
AGMA ; the code for our computation is available in the supplemental online ma-terial. Constructing p ( k, x ) as an element of the ring Q ( k )[ x ] , we use Sutherland’salgorithm [ ] to compute a permutation representation of the Galois group of p over C ( k ) , and we obtain the group S . It follows that the Galois group acts transitivelyon the roots of p , so p is irreducible over C ( k ) .By Proposition 4.2 we may regard x m as a function of r defined on the interval [1 , ∞ ) , and moreover, we have µ ( r ) = z ( x m ( r )) for all r ≥ . (3)For convenience we will make the change of variable k = √ r and work insteadwith the function ξ : [1 , ∞ ) → R be defined by ξ ( k ) = x m ( k ) . From Proposition4.4 we deduce that p ( k, ξ ( k )) = 0 for all k ∈ [1 , ∞ ) . (4)Furthermore, (3) implies that λ ( k ) = ( z ( ξ ( k ))) . Hence, writing ξ for ξ ( k ) , we have λ ( k ) = ξ + (cid:32) k − ξ − (cid:114) k − (cid:16) k − ξ − (cid:112) ξ − ξ (cid:17) (cid:33) . By manipulating the equation above we obtain a polynomial h ∈ Q [ k, x, y ] withthe property that h ( k, ξ ( k ) , λ ( k )) = 0 for all k ∈ [1 , ∞ ) . (5)Explicitly, h is given by the formula h ( k, x, y ) = (cid:0) ( y − x − c − c + c + c ) + 4 c c − c c + 4 c c + 4 c c (cid:1) − c (cid:0) c c + 4 c ( y − x − c − c + c + c ) (cid:1) , where c = k − x , c = k , c = 2 k − x , and c = 2 x − x . Lemma 5.3.
Let Ω be an algebraic closure of the field C ( k ) . Suppose that α, β ∈ Ω satisfy p ( k, α ) = h ( k, α, β ) = 0 . Then C ( k, α ) ⊆ C ( k, β ) .Proof. We rely on a number of computations in M
AGMA ; the code used for all com-putations is available in the supplemental online material. Let F = Q ( k, α ) . To provethe lemma it suffices to show that F ⊆ Q ( k, β ) . Regarding p as an element of the ring Q ( k )[ x ] , note that p is irreducible by Lemma 5.2, and α is a root of p by hypothesis,so we may identify F with the field Q ( k )[ x ] / ( p ( k, x )) . Constructing F in M AGMA and factoring the polynomial h ( k, α, y ) over F , we find that this polynomial has tworoots in F and two roots that are quadratic over F . Note that β must be one of thesefour roots since h ( k, α, β ) = 0 .Suppose that β ∈ F . Computing the minimal polynomial of β over Q ( k ) we obtaina polynomial of degree 10; thus [ Q ( k, β ) : Q ( k )] = 10 . Since β ∈ F and [ F : Q ( k )] = deg( p ) = 10 , The algorithm used by M
AGMA to factor polynomials over algebraic function fields is discussed in [ ]. F = Q ( k, β ) . In particular, F ⊆ Q ( k, β ) as desired.Now suppose that β is quadratic over F . Then a minimal polynomial computationshows that [ Q ( k, β ) : Q ( k )] = 20 . Since [ F ( β ) : F ] = 2 and [ F : Q ( k )] = 10 , thisimplies that F ( β ) = Q ( k, β ) , so again F ⊆ Q ( k, β ) .We can now prove the main theorem of this article. Theorem 5.4.
There is no infinite subset J ⊆ [1 , ∞ ) such that µ : J → R is radical.Proof. As above, let Ω denote an algebraic closure of the field C ( k ) . By Lemma 5.1,in order to prove the theorem it suffices to show that λ is not radical on any infinitesubset of [1 , ∞ ) . Suppose for contradiction that I ⊆ [1 , ∞ ) is an infinite set such that λ : I → R is radical. Then there is a nonzero polynomial q ( k, y ) ∈ C [ k, y ] whoseGalois group over C ( k ) is solvable, and such that q ( k, λ ( k )) = 0 for all k ∈ I. (6)Regarding q and h as elements of the ring C [ k, x, y ] , let f ( k, x ) = Res y ( h ( k, x, y ) , q ( k, x, y )) , where Res y denotes the resultant as polynomials in y . By (5) and (6), for every k ∈ I the polynomials h ( k, ξ ( k ) , y ) and q ( k, ξ ( k ) , y ) have a common root, namely λ ( k ) ;hence f ( k, ξ ( k )) = 0 for all k ∈ I. (7)Similarly, letting g ( k ) = Res x ( f ( k, x ) , p ( k, x )) ∈ C [ k ] , equations (4) and (7) imply that g ( k ) = 0 for every k ∈ I . Since I is an infinite set,we must have g = 0 . Therefore, f and p have a common root α ∈ Ω . Given that f ( k, α ) = 0 , the definition of f implies that q ( k, y ) and h ( k, α, y ) have a commonroot β ∈ Ω . Note that the assumptions in Lemma 5.3 are satisfied.Let N ⊂ Ω be the splitting field of q ( k, y ) over C ( k ) and let F = C ( k, α ) . ByLemma 5.3 we have F ⊆ C ( k, β ) and therefore F ⊆ N . Letting L ⊂ Ω be the split-ting field of p ( k, x ) over C ( k ) , we have L ⊆ N since F ⊆ N and L is the Galoisclosure of the extension F/ C ( k ) . Since the extension L/ C ( k ) is Galois, the group Gal( L/ C ( k )) is a quotient of Gal( N/ C ( k )) . The definition of q ( k, y ) implies thatthe latter group is solvable, so the former is, too. This contradicts Lemma 5.2 (sincethe group S is not solvable), and thus completes the proof of the theorem. ACKNOWLEDGMENTS.
JEH received support from the Colby College Research Grant Program. The au-thors thank Gerardo Lafferriere of the Fariborz Maseeh Department of Mathematics and Statistics at PortlandState University for welcoming JEH as a Visiting Scholar during the final stages of this project.
REFERENCES
1. Bosma, W., Cannon, J., Playoust, C. (1997). The Magma algebra system. I. The user language.
J. SymbolicComput.
Introduction to Geometry , 2nd ed. New York: John Wiley & Sons, Inc.3. Dummit, David S., Foote, Richard M. (2004).
Abstract Algebra , 3rd ed. Hoboken, NJ: John Wiley & Sons,Inc. . Fukagawa, H., Pedoe, D. (1989). Japanese Temple Geometry Problems . Winnipeg, Canada: The CharlesBabbage Research Centre.5. Fukagawa, H., Rothman, T. (2008).
Sacred Mathematics: Japanese Temple Geometry . Princeton, NJ:Princeton Univ. Press.6. Kinoshita, H. (2018). An unsolved problem in the Yamaguchi’s travell diary.
Sangaku Journal of Mathe-matics.
Algebra , 3rd ed. New York, NY: Springer-Verlag.8. Rothman, T. (1998). Japanese temple geometry.
Scientific American.
Computeralgebra Rundbrief.
SYMSAC’76: Proceedings of the Third ACM Symposium on Symbolic and Algebraic Computation . New York:ACM, pp. 219–226.
JAN E. HOLLY
Department of Mathematics and Statistics, Colby College, Waterville, ME [email protected]
DAVID KRUMM