SSangaku Journal of Mathematics (SJM) c (cid:13)
SJMISSN 2534-9562Volume 3 (2019), pp.51-66Received 14 July 2019. Published on-line 04 August 2019web: c (cid:13) The Author(s) This article is published with open access . Relationships Between Six Incircles
Stanley Rabinowitz
545 Elm St Unit 1, Milford, New Hampshire 03055, USAe-mail: [email protected]:
Abstract. If P is a point inside (cid:52) ABC , then the cevians through P divide (cid:52) ABC into six smaller triangles. We give theorems about the relationship be-tween the radii of the circles inscribed in these triangles.
Keywords.
Euclidean geometry, triangle geometry, incircles, inradii, cevians.
Mathematics Subject Classification (2010).
Introduction
Japanese Mathematicians of the Edo period were fond of finding relationshipsbetween the radii of circles associated with triangles. For example, the 1781 book“Seiyo Sampo” [2, no. 84] gives the relationship r = √ r r + √ r r + √ r r betweenthe radii of the circles in Figure 1 below. This result was later inscribed on an1814 tablet in the Chiba prefecture [3, p. 30]. Figure 1. This article is distributed under the terms of the Creative Commons Attribution Licensewhich permits any use, distribution, and reproduction in any medium, provided the originalauthor(s) and the source are credited. a r X i v : . [ m a t h . HO ] A ug Relationships Between Six Incircles
In the spirit of Wasan, we investigate the relationships between the radii of sixcircles associated with a triangle and the three cevians through a point inside thattriangle. 2.
Notation
Let P be any point inside a triangle ABC . The cevians through P divide (cid:52) ABC into six smaller triangles. Circles are inscribed in these six triangles. The sixtriangles and their incircles are numbered from 1 to 6 counterclockwise as shownin Figure 2, where the first triangle is formed by the lines AP , BP , and BC . Thei-th triangle has inradius r i , semiperimeter s i , circumradius R i , and area K i . Figure 2. numberingIf X and Y are points, then we use the notation XY to denote either the linesegment joining X and Y or the length of that line segment, depending on thecontext. 3. The Orthocenter
We start with a known result [4] giving the relationship between the r i when P isthe orthocenter. Theorem 3.1. If P is the orthocenter of (cid:52) ABC (Figure 3), then r r r = r r r . Figure 3. r r r = r r r The following proof comes from [5]. tanley Rabinowitz Proof.
Note that triangles 1 and 4 in Figure 3 are similar. Corresponding lengthsin similar triangles are in proportion, so r /BP = r /AP . Triangles 2 and 5 arealso similar as well as triangles 3 and 6, giving similar proportions. Therefore r BP · r CP · r AP = r AP · r BP · r CP which implies r r r = r r r . (cid:3) The Centroid
Next, we will consider the case when P is the centroid. We start with a fewlemmas. Lemma 4.1. If P is the centroid of (cid:52) ABC (Figure 4), then the six trianglesformed all have the same area. That is, K = K = K = K = K = K . Figure 4. six triangles have same area
Proof. If XY Z is a triangle, then [
XY Z ] will denote the area of that triangle.If two triangles have the same altitude, then the ratio of their areas is the sameas the ratio of their bases. Thus [
P BD ] = [
P DC ]. Since the centroid divides amedian in the ratio 2 : 1, this means that [
CEP ] = [ CP B ] = [
CP D ]. Thustriangles 1, 2, and 3 have the same area. In the same manner, we see that all sixtriangles have the same area. (cid:3)
Lemma 4.2. If P is the centroid of (cid:52) ABC , then s + s + s = s + s + s .Proof. We have the following six equations for the perimeters of the six triangles.2 s = P B + P D + BD, s = P D + P C + DC, s = P C + P E + CE, s = P E + P A + EA, s = P A + P F + AF, s = P F + P B + F B.
Thus 2 s +2 s +2 s − (2 s +2 s +2 s ) = ( BD − DC )+( CE − EA )+( AF − F B ) = 0and the lemma follows. (cid:3) Relationships Between Six Incircles
We can now state the relationship between the r i when P is the centroid. Thistheorem is attributed to Reidt in [1, p. 618]. Theorem 4.1. If P is the centroid of (cid:52) ABC , then r + 1 r + 1 r = 1 r + 1 r + 1 r . Proof.
Recall that if r , s , and K are the inradius, semiperimeter, and area of atriangle, respectively, then r = K/s . From Lemma 4.1, the six triangles have thesame area. Call this area K . From Lemma 4.2, s + s + s = s + s + s . Divideboth sides of this equation by K and use the fact that 1 /r i = s i /K i to get thedesired identity. (cid:3) We note a similar result from [1, p. 618].
Theorem 4.2. If P is the centroid of (cid:52) ABC , then R R R = R R R . The Circumcenter
Now we will consider the case when P is the circumcenter. Lemma 5.1.
Let P be the circumcenter of (cid:52) ABC and let R be its circumradius.Let ∠ P AB = ∠ P BA = α and ∠ P BC = β (Figure 5). Then R/r = cot α +cot β . Figure 5.
Proof.
Let X be the center of incircle 1 (Figure 5). Let Y be the foot of theperpendicular from X to P B . Since BX is the bisector of ∠ P BD , ∠ P BX = β/
2. Since ∠ BP D = ∠ P AB + ∠ P BA , ∠ BP X = α . Then P Y = r cot α and Y B = r cot β . Since P B = R , this gives R = r cot α + r cot β and the resultfollows. (cid:3) We now give the relationship between the r i when P is the circumcenter. Theorem 5.1. If P is the circumcenter of (cid:52) ABC (Figure 6), then r + 1 r + 1 r = 1 r + 1 r + 1 r . tanley Rabinowitz Figure 6. r + r + r = r + r + r Proof.
Let R be the circumradius of (cid:52) ABC . Let ∠ P AB = ∠ P BA = α , ∠ P BC = ∠ P CB = β , and ∠ P CA = ∠ P AC = γ (Figure 5). By Lemma 5.1, we have thefollowing six relationships. R/r = cot α + cot β , R/r = cot γ + cot β ,R/r = cot β + cot γ , R/r = cot α + cot γ ,R/r = cot γ + cot α , R/r = cot β + cot α . Adding the equations on the left gives the same result as adding the equations onthe right. Dividing out the common factor R proves the theorem. (cid:3) We mention the relationship between the R i when P is the circumcenter. Theorem 5.2. If P is the circumcenter of (cid:52) ABC , then R = R .Proof. Let R be the circumradius of (cid:52) ABC , so that
P A = P B = P C = R .Let ∠ BDP = α and ∠ P DC = β (Figure 7). By the Extended Law of Sines in (cid:52) P BD , R/ sin α = 2 R . Similarly for (cid:52) P CD , R/ sin β = 2 R . But since α and β are supplementary, sin α = sin β . Therefore, R = R . (cid:3) The Incenter
Next, we will consider some cases when P is the incenter. Theorem 6.1. If P is the incenter of (cid:52) ABC and ∠ ABC = 60 ◦ (Figure 8), then r + 1 r + 1 r = 1 r + 1 r + 1 r . Relationships Between Six Incircles
Figure 7.
Yellow circles are congruent
Figure 8. r + r + r = r + r + r Proof.
Let ∠ BAD = ∠ DAC = a , ∠ ABE = ∠ EBC = b , and ∠ ACF = ∠ F CB = c , as shown in Figure 9. Note that a + b + c = π/
2. By the Extended Law ofSines in (cid:52)
ABC , AB/ sin 2 c = 2 R , where R is the radius of the circumcircle of (cid:52) ABC . Since the result we want to prove is invariant under scaling, without lossof generality, we may assume that R = 1 /
2. Thus AB = sin 2 c . In the samemanner, we can find AC and BC . We get the following. AB = sin 2 c, BC = sin 2 a, CA = sin 2 b. Applying the Law of Sines to (cid:52)
ABD gives
BD/ sin a = AB/ sin( a + 2 c ). Thisallows us to compute BD . In a similar manner, we get the following. BD = sin a sin 2 c sin( a + 2 c ) , CE = sin b sin 2 a sin( b + 2 a ) , AF = sin c sin 2 b sin( c + 2 b ) ,CD = sin a sin 2 b sin( a + 2 b ) , AE = sin b sin 2 c sin( b + 2 c ) , BF = sin c sin 2 a sin( c + 2 a ) . tanley Rabinowitz Figure 9.
Note that ∠ BP D = a + b . Applying the Law of Sines to (cid:52) BP D allows us tocompute the values of
P B and
P D . In the same way, we can compute
P C , P E , P A , and
P F . We get the following.
P A = sin c sin 2 b sin( c + a ) , P D = sin a sin b sin 2 c sin( a + b ) sin( a + 2 c ) ,P B = sin a sin 2 c sin( a + b ) , P E = sin b sin c sin 2 a sin( b + c ) sin( b + 2 a ) ,P C = sin b sin 2 a sin( b + c ) , P F = sin c sin a sin 2 b sin( c + a ) sin( c + 2 b ) . We now have expressions for the length of every line segment in the figure in termsof a , b , and c . Thus, the perimeters of all the triangles are known and we haveexpressed each of the s i in terms of a , b , and c . The areas of the triangles can alsobe found. For example, K = P B · BD sin b . Knowing all the s i and K i lets usfind the values of all the r i , since r i = K i /s i .We can plug these values for the r i into the expression1 r + 1 r + 1 r − (cid:18) r + 1 r + 1 r (cid:19) . Letting a = π/ − b − c and b = π/ c as theonly variable. Simplifying this expression (using a symbolic algebra system), wefind that the result is 0, thus proving our theorem. (cid:3) Theorem 6.2. If P is the incenter of (cid:52) ABC and ∠ ABC = 120 ◦ (Figure 10),then r r r + r r r + r r r = r r r + r r r + r r r . This can also be written as r + r + r r r = r + r + r r r . Proof.
This theorem can be proven using the same procedure that was used toprove Theorem 6.1. The details are omitted. (cid:3)
Open Question 1.
Is there a simple relationship between the r i that holds for alltriangles when P is the incenter? Relationships Between Six Incircles
Figure 10. r r r + r r r + r r r = r r r + r r r + r r r Note that there are two independent variables, a and b , and six equations repre-senting the values of the r i . Thus, variables a and b can be eliminated resultingin an equation relating the r i . Since there are so many more equations than vari-ables, multiple relationships can be found. For example, in the 30 ◦ –60 ◦ –90 ◦ righttriangle, we have a number of simple relationships between the r i , as shown bythe following theorem. Theorem 6.3. If P is the incenter of (cid:52) ABC and ∠ ABC = 30 ◦ and ∠ ACB =60 ◦ , then the r i are related to each other by each of the following equations. r + 8 r + 6 r + 5 r = 14 r + 20 r , r + 3 r + 3 r = 3 r + 2 r + 3 r , r + 2 r + 1 r + 1 r = 3 r + 2 r , r + 5 r + 55 r + 22 r = 4 r + 75 r , r r + 3 r r + 9 r r + 9 r r = 27 r r + r r . Proof.
For this triangle, the values of the r i are as follows. r = 14 (cid:16) √ − (cid:17) (cid:16) √ − (cid:17) ,r = 14 (cid:16) − − √ √ √ (cid:17) ,r = 14 (cid:16) − √ − √ √ (cid:17) ,r = 18 (cid:16) − − √ √ √ (cid:17) ,r = 124 (cid:16) − √ √ √ (cid:17) ,r = 14 (cid:16) √ − √ (cid:17) . These values can be substituted into the stated equations to verify the results(using computer simplification, as necessary). (cid:3) tanley Rabinowitz Here is an approach that might be used to find the general relationship betweenthe r i when P is the incenter. To avoid fractions, we will replace a , b , and c fromFigure 9 by 2 a , 2 b , and 2 c to get Figure 11. We can express P B as the sum of twolengths using circles 6 and 1 in two different ways. Equating these expressionsgives the following equation.(1) r r = cot b + cot( a + b )cot b + cot( b + c ) . Figure 11.
Since 4 a + 4 b + 4 c = 180 ◦ , c = 45 ◦ − a − b . Substitute this value of c intoequation (1). Then use the addition formula for cotangent,cot( x + y ) = cot x cot y − x + cot y , to write all trigonometric expressions in terms of cot a and cot b .In a similar manner we can form two other equations for r /r and r /r . Thisgives us the following three equations for u = r /r , v = r /r , and w = r /r interms of the two unknowns C a = cot a and C b = cot b . Relationships Between Six Incircles u = ( C a − C b + 2 C a C b − C a + C b )(1 + C a − C b + C a C b ) ,v = ( C b − C b C a − C a − C b )( C a − C a C b − C b − C a ) ,w = ( C a + C b )(1 − C a + C b + C a C b )( C b − C a + 2 C a C b − . Clearing fractions gives us three polynomial equations in the variables C a and C b .In theory, we should be able to eliminate C a and C b from these three equations,leaving us with a single equation relating u , v , and w . This equation would bethe desired relationship between the r i . I have not been able to perform thiselimination.The following may be a simpler question. Open Question 2.
Find the relationship between r , r , and r that holds for alltriangles when P is the incenter. Such a relationship should exist because we can express each of r , r , and r interms of a and b . This would give us three equations in two unknowns. In theory,we should be able to eliminate a and b from these three equations, giving us asingle equation relating r , r , and r .7. Other Points
Next, we will consider other points inside triangle
ABC . We start with an examplewhere there is a linear relationship between the r i . Theorem 7.1. If P is a point inside (cid:52) ABC , and ∠ ABP = 10 ◦ , ∠ P BC = 30 ◦ , ∠ BCP = 80 ◦ , and ∠ P CA = 20 ◦ (Figure 12), then r + 6 r + r = r + 3 r + 15 r . Proof.
We follow the same general procedure that was used in the proof of The-orem 6.1. Let ∠ ABE = a , ∠ EBC = b , ∠ BCF = c , and ∠ F CA = d , as shownin Figure 13. Note that ∠ BAC = π − a − b − c − d , ∠ AF C = a + b + c , ∠ AEB = b + c + d , and ∠ EP C = b + c .Without loss of generality, we may assume that R = 1 /
2, where R is the radius ofthe circumcircle of (cid:52) ABC . By the Extended Law of Sines we get the following. AB = sin( c + d ) , AC = sin( a + b ) , BC = sin( a + b + c + d ) . tanley Rabinowitz Figure 12. r + 6 r + r = r + 3 r + 15 r Figure 13.
We then use the Law of Sines to get the following. AF = sin( a + b ) sin d sin( a + b + c ) , BF = sin( a + b + c + d ) sin c sin( a + b + c ) ,AE = sin( c + d ) sin a sin( b + c + d ) , CE = sin( a + b + c + d ) sin b sin( b + c + d ) . Applying the Law of Sines again gives the following.
P B = sin( a + b + c + d ) sin c sin( b + c ) , P E = CE sin d sin( b + c ) ,P C = sin( a + b + c + d ) sin b sin( b + c ) , P F = BF sin a sin( b + c ) . By Ceva’s Theorem,
BDCD = BF · AEAF · CE . This gives the following. Relationships Between Six Incircles BD = BF · AE · BCBF · AE + AF · CE ,CD = AF · CE · BCBF · AE + AF · CE .
Length
P A is calculated using the Law of Cosines in triangle
AP B . We get thefollowing.
P A = √ AB + P B − · AB · P B · cos a. To find the length of
P D without introducing another square root, we can applyMenelaus’ Theorem to transversal
BP E in (cid:52) ADC . We get the following.
P D = P A · BD · CEBC · AE .
We now have expressions for the length of every line segment in the figure in termsof a , b , c , and d . We can therefore calculate all the s i , K i , and r i .We can plug these values for the r i into the expression5 r + 6 r + r − ( r + 3 r + 15 r ) . Letting a = 10 ◦ , b = 30 ◦ , c = 80 ◦ , and d = 20 ◦ gives us an expression with novariables. Simplifying this expression (using a symbolic algebra system), we findthat the result is 0, thus proving our theorem. (cid:3) Sometimes the relationship between the r i is more regular as in the following twotheorems. The proofs are similar to the proof of Theorem 7.1. When the formulasfor the lengths of the radii previously found are applied to the equation to beproved, the result is a trigonometric equation that can be proven to be an identityusing symbolic algebra computation. The details are omitted. Theorem 7.2. If P is a point inside (cid:52) ABC , and ∠ P BA , ∠ P BC , ∠ P CB , and ∠ P CA are as shown in Figure 14, then r + 1 r + 1 r = 1 r + 1 r + 1 r . Theorem 7.3. If P is a point inside (cid:52) ABC , and ∠ P BA , ∠ P BC , ∠ P CB , and ∠ P CA are as shown in any of the triangles depicted in Figure 15, then r + 1 r + 1 r = 1 r + 1 r + 1 r . Note that in each of these examples, t is an arbitrary parameter. The reader may be wondering how I came up with these results. Here is theprocedure I used.Let f ( r ) denote some function of r such as r , r , or 1 /r . I varied all combinationsof a , b , c , and d (see Figure 13) through all multiples of 1 ◦ and calculated f ( r )through f ( r ). Using the Mathematica R (cid:13) function FindIntegerNullVector , Ilooked for any linear relationships between these six numbers. If a linear relation-ship existed which involved all six numbers, I logged the quadruple (cid:104) a, b, c, d (cid:105) along tanley Rabinowitz Figure 14. r + r + r = r + r + r Figure 15. r + r + r = r + r + r with the coefficients of the relationship into a database. After all quadruples wereexamined, I looked at all pairs of entries in the database that had the same set of Relationships Between Six Incircles coefficients. If Q = (cid:104) a , b , c , d (cid:105) and Q = (cid:104) a , b , c , d (cid:105) were two such quadru-ples, I then formed the quadruple Q such that (cid:104) Q , Q , Q (cid:105) formed an arithmeticprogression in each component. Then I examined Q to see if it also satisfied thesame linear combination. If it did, then this suggested a one-parameter family ofsolutions (which had to be confirmed).Note that many solutions were found for various functions f , but no one-parameterfamilies of solutions were found except when f ( r ) = 1 /r . It is not clear why thisshould be the case.We note several related results that hold whenever P is an arbitrary point inside (cid:52) ABC . Theorem 7.4. If P is any point inside (cid:52) ABC (Figure 16), then K K K = K K K . Figure 16. K K K = K K K Proof.
Let ∠ BP D = α , ∠ DP C = β , and ∠ CP E = γ . Noting that vertical anglesare equal and using the formula for the area of a triangle in terms of two sidesand the sine of the included angle, we have the following six equations.2 K = P B · P D · sin α, K = P D · P C · sin β, K = P C · P E · sin γ, K = P E · P A · sin α, K = P A · P F · sin β, K = P F · P B · sin γ. Multiplying gives8 K K K = ( P B · P D · sin α )( P C · P E · sin γ )( P A · P F · sin β )and 8 K K K = ( P D · P C · sin β )( P E · P A · sin α )( P F · P B · sin γ ) . The right sides of both equations are equal, so K K K = K K K . (cid:3) Theorem 7.5. [6, p. 43] If P is any point inside (cid:52) ABC , then K + 1 K + 1 K = 1 K + 1 K + 1 K . Open Question 3.
Is there a simple relationship between the r i that holds for alltriangles and all points P inside the triangle? A simpler question may be the following. tanley Rabinowitz Open Question 4.
Is there a simple relationship between the r i that holds for allpoints P inside an equilateral triangle? Opportunities for Future Research
When the three cevians through a point inside a triangle are extended to thecirumcircle, other circles can be formed. Figure 17 shows some examples. Eachyellow circle associated with the triangle on the left is the incircle of a regionbounded by two cevians and the circumcircle. Each green circle associated withthe triangle on the right is the incircle of a region bounded by one side of thetriangle, one cevian, and the circumcircle.
Figure 17. sets of incircles tangent to circumcircle of (cid:52)
ABC
Open Question 5.
Investigate the relationship between the radii in each set ofsix incircles shown in Figure 17.
We can also form regions bounded by the incircle of (cid:52)
ABC . Figure 18 showssome examples. Each red circle associated with the triangle on the left is theincircle of a region formed by two cevians and the incircle of (cid:52)
ABC . Each bluecircle associated with the triangle on the right is the incircle of a region formedby one side of the triangle, one cevian, and the incircle of (cid:52)
ABC . Figure 18. sets of incircles tangent to incircle of (cid:52)
ABC
Open Question 6.
Investigate the relationship between the radii in each set ofsix incircles shown in Figure 18. Relationships Between Six Incircles
Open Question 7.
For a fixed triangle
ABC , characterize those points P suchthat r + r + r = r + r + r . There are many such points. Some of them look like they lie on a straight line.We have found relationships between the r i when P is the orthocenter, circum-center, centroid, and incenter. Open Question 8.
Investigate the relationship between the r i when P is someother notable point, such as the nine-point center, the Nagel Point, or the Ger-gonne Point. References [1] Antoine Dalle, , 8th edition,Wesmael-Charlier, Bruxelles, 1961.[2] Teisi Fujita,
Seiy¯o Samp¯o , 1781. Tohoku University Digital Collection. [3] Hidetosi Fukagawa and Dan Pedoe,
Japanese Temple Geometry Problems , Winnipeg,Canada, Charles Babbage Research Center, 1989.[4] Antonio Gutierrez,
Geometry Problem 79 , 2008, Geometry from the Land of the Incas, .[5] Hexram,
Solution to Geometry Problem 79 , 2012, Geometry from the Land of the Incas, https://gogeometry.blogspot.com/2008/05/elearn-geometry-problem-79.html .[6] Victor Mayer Am´ed´ee Mannheim,
Transformation des propi´et´es m´etriques des figures `a l’aidede la th´eorie des polaires r´eciproques , Mallet-Bachelier, Paris, 1857., Mallet-Bachelier, Paris, 1857.