Grid dissections of tangential quadrilaterals
GGRID DISSECTIONS OF TANGENTIAL QUADRILATERALS
ERICA CHOI, DAN ISMAILESCU, JIHO LEE, AND JOONSOO LEE
Abstract.
For any integer n ≥
2, a square can be partitioned into n smallersquares via a checkerboard-type dissection. Does there such a class preservinggrid dissection exist for some other types of quadrilaterals? For instance,is it true that a tangential quadrilateral can be partitioned into n smallertangential quadrilaterals using an n × n grid dissection? We prove that theanswer is affirmative for every integer n ≥ The problem
For any integer n ≥
2, a square can be naturally partitioned into n smallersquares via a checkerboard-type dissection. Are there any other properties P suchthat any convex quadrilateral Q having property P can be dissected in a grid-like manner into smaller quadrilaterals, all of which have property P as well?The answer is obvious if property P says “ Q is a parallelogram ” or “ Q is atrapezoid ” or “ Q is a convex quadrilateral ”.What if property P states: “ Q is a cyclic quadrilateral ” or ” Q is a tangential quadrilateral ” or “ Q is an orthodiagonal quadrilateral ”?For instance, figure 1 displays the case of a tangential quadrilateral partitionedinto nine smaller tangential quadrilaterals. Is such a dissection always possible?And if it is, how do we find it? Figure 1.
A 3 × a r X i v : . [ m a t h . M G ] A ug . Class preserving grid dissections of quadrilateralsDefinition 2.1.
Let
ABCD be a convex quadrilateral and let m , n be twopositive integers. Consider two sets of segments S = { s , s , . . . , s m − } and T = { t , t , . . . , t n − } with the following properties:a) If s ∈ S then the endpoints of s belong to the sides AB and CD . Similarly,if t ∈ T then the endpoints of t belong to the sides AD and BC .b) Every two segments in S are pairwise disjoint and the same is true for thesegments in T .We then say that segments s , s , . . . , s m − , t , t , . . . , t n − define an m × n griddissection of ABCD – see figure 2. Note that either S or T could be empty. Figure 2.
A 3 × × ABCD
The following problem raised in [1] is the main motivation of our paper.
Problem 2.2.
Is it true that every cyclic, orthodiagonal or tangential quadri-lateral can be partitioned into cyclic, orthodiagonal, or tangential quadrilaterals,respectively, via an m × n grid dissection? The authors call such dissections class preserving grid dissections .In [1] it is proved that if either m or n is even then an m × n grid dissectionof a cyclic quadrilateral Q into mn cyclic quadrilaterals exists only if Q is eitheran isosceles trapezoid or a rectangle. The situation is a little bit better if both m and n are odd. Theorem 2.3. ( [1] )(a) Every cyclic quadrilateral all of whose angles are greater than arccos(( √ − / ≈ . ◦ admits a × grid dissection into three cyclic quadrilaterals.(b) Let Q be a cyclic quadrilateral such that the measure of each of the arcsdetermined by the vertices of Q on the circumcircle is greater than ◦ . Then ABCD admits a × grid dissection into nine cyclic quadrilaterals. It is conjectured however, that when m and n both odd become large, a cyclicquadrilateral admits an m × n grid dissection into cyclic quadrilaterals only if itis “close” to being a rectangle.Things get even worse for class preserving grid dissections of orthodiagonalquadrilaterals. It is conjectured that with the possible exception of some specialcases, such dissections do not exist. fter these mostly negative results, it comes as a surprise that there is somehope when dealing with tangential quadrilaterals.Ismailescu and Vojdany [1] proved that every tangential quadrilateral has a2 × n × n grid dissections.Surprisingly enough, the answer is affirmative and this constitutes the mainresult of our paper. The following elementary result provides a characterizationof tangential quadrilaterals and will be used in the sequel. The direct statementis due to Pitot and the converse to Steiner. Theorem 2.4. (see e. g. [3] ) A quadrilateral is tangential if and only if the sumof two opposite sides equals the sum of the other two opposite sides. The main resultTheorem 3.1.
For any integer n ≥ and for any tangential quadrilateral Q ,there exists an n × n grid dissection of Q into n smaller tangential quadrilaterals. The idea of the proof is natural. Given a checkerboard dissected axes-parallelsquare
ABCD in the xy plane and a “target” tangential quadrilateral A (cid:48) B (cid:48) C (cid:48) D (cid:48) in the uv plane (see figure 3 below), we are trying to find a geometric transfor-mation T with the following properties: Figure 3.
The main idea behind transformation T • T maps arbitrarily small axes-parallel squares from the xy plane intotangential quadrilaterals in the uv plane. • T maps the horizontal grid lines from the xy plane into the grid linespassing through the point O (0 ,
0) in the uv plane. • T maps the vertical grid lines from the xy plane into the grid lines passingthrough P (1 ,
0) in the uv plane.Of course, at this point we do not know whether such a transformation exists,and even if it does, whether it achieves the overall goal of generating the desired n × n grid dissection of A (cid:48) B (cid:48) C (cid:48) D (cid:48) into smaller tangential quadrilaterals. . The local conditionLemma 4.1.
Suppose that there exists a transformation T : R → R from the xy plane to the uv plane defined as T : u = f ( x, y ) , v = g ( x, y ) such that T maps every square from the xy plane into a tangential quadrilateral in the uv plane. Then (1) f x ( a, b ) + g x ( a, b ) = f y ( a, b ) + g y ( a, b ) for every ( a, b ) ∈ R . Figure 4.
The local condition
Proof.
Let
ABCD be an axis-parallel square of side (cid:15) in the xy plane and let A (cid:48) = T ( A ), B (cid:48) = T ( B ), C (cid:48) = T ( C ) and D (cid:48) = T ( D ) as shown in figure 4. Itfollows that u A (cid:48) = f ( a, b ), v A (cid:48) = g ( a, b ), u B (cid:48) = f ( a + (cid:15), b ), v B (cid:48) = g ( a + (cid:15), B ) etc.Consider first the ratio A (cid:48) B (cid:48) (cid:15) = (cid:20) f ( a + (cid:15), b ) − f ( a, b ) (cid:15) (cid:21) + (cid:20) g ( a + (cid:15), b ) − g ( a, b ) (cid:15) (cid:21) from whichlim (cid:15) → (cid:18) A (cid:48) B (cid:48) (cid:15) (cid:19) = (cid:112) f x ( a, b ) + g x ( a, b ) . (2)Similarly, C (cid:48) D (cid:48) (cid:15) = (cid:20) f ( a + (cid:15), b + (cid:15) ) − f ( a, b + (cid:15) ) (cid:15) (cid:21) + (cid:20) g ( a + (cid:15), b + (cid:15) ) − g ( a, b + (cid:15) ) (cid:15) (cid:21) → lim (cid:15) → (cid:18) C (cid:48) D (cid:48) (cid:15) (cid:19) = (cid:112) f x ( a, b ) + g x ( a, b ) . (3)Analogously, we obatain the following equalities: B (cid:48) C (cid:48) (cid:15) = (cid:20) f ( a + (cid:15), b + (cid:15) ) − f ( a + (cid:15), b ) (cid:15) (cid:21) + (cid:20) g ( a + (cid:15), b + (cid:15) ) − g ( a + (cid:15), b ) (cid:15) (cid:21) ,D (cid:48) A (cid:48) (cid:15) = (cid:20) f ( a, b + (cid:15) ) − f ( a, b ) (cid:15) (cid:21) + (cid:20) g ( a, b + (cid:15) ) − g ( a, b ) (cid:15) (cid:21) , which givelim (cid:15) → (cid:18) B (cid:48) C (cid:48) (cid:15) (cid:19) = lim (cid:15) → (cid:18) D (cid:48) A (cid:48) (cid:15) (cid:19) = (cid:113) f y ( a, b ) + g y ( a, b ) . (4)Since we want A (cid:48) B (cid:48) C (cid:48) D (cid:48) to be tangential, by using the Pitot-Steiner conditionit is necessary and sufficient to have that A (cid:48) B (cid:48) + C (cid:48) D (cid:48) = B (cid:48) C (cid:48) + D (cid:48) A (cid:48) . Divideboth sides by (cid:15) and pass to the limit as (cid:15) → (cid:3) . The trapezoid case.Theorem 5.1.
There exists a transformation, T : R → R with the followingproperties(a) T maps vertical lines from the xy plane into vertical lines in the uv plane.(b) T maps horizontal lines from the xy plane into lines through the origin in the uv plane.(c) T maps any axes-parallel square from the xy plane into a tangential trapezoidin the uv plane.(d) For any given tangential trapezoid K in the uv plane, an appropriately scaledand/or rotated copy of K can be obtained as the image under transformation T of a conveniently chosen square in the xy plane.Proof. Let T : u = f ( x, y ) , v = g ( x, y ) Any vertical line x = c in the xy planeis mapped into a vertical line u = c (cid:48) , in the uv plane. This means f ( c, y ) = c (cid:48) ,where c (cid:48) depends on c only. It follows that f ( x, y ) depends on x only, that is(5) f ( x, y ) = F ( x ) . On the other hand, any horizontal y = c is mapped into a line through theorigin v = m c u which can be rewritten as g ( x, c ) = m c · f ( x, c ), from which g ( x, c ) /f ( x, c ) = m c , where m c depends only on c . This means that g ( x, y ) /f ( x, y ) = G ( y ), which after using (5) becomes(6) g ( x, y ) = F ( x ) · G ( y ) . Taking partial derivatives, we obtain f x ( x, y ) = F (cid:48) ( x ) , g x ( x, y ) = F (cid:48) ( x ) · G ( y ) , f y ( x, y ) = 0 , g y ( x, y ) = F ( x ) · G (cid:48) ( y ) . Using equation (1), we have that f x + g x = f y + g y → [ F (cid:48) ( x )] + [ F (cid:48) ( x )] · G ( y ) = F ( x ) · [ G (cid:48) ( y )] from which (cid:18) F (cid:48) ( x ) F ( x ) (cid:19) = G (cid:48) ( y ) G ( y ) . Since the left hand side of the above equation depends on x only while theright hand side depends on y only, it follow that we have system of differentialequations F (cid:48) ( x ) F ( x ) = ± k, G (cid:48) ( y ) (cid:112) G ( y ) = ± k. One set of solutions of this system is given by F ( x ) = a x , G ( y ) = a y − a − y . Using now relations (5) and (6), we have obtained(7) T : f ( x, y ) = a x , g ( x, y ) = a x ( a y − a − y )2 . We still have to verify property (c). Consider the axes-parallel square
ABCD in the xy plane with A ( x, y ) , B ( x + l, y ) , C ( x + l, y + l ) , D ( x, y + l ) where l >
0. Denote A (cid:48) := T ( A ) , B (cid:48) := T ( B ) , C (cid:48) := T ( C ) , D (cid:48) := T ( D ). It is easyto verify that A (cid:48) B (cid:48) C (cid:48) D (cid:48) is a positively oriented trapezoid. It remains to checkthat A (cid:48) B (cid:48) C (cid:48) D (cid:48) is tangential. Denote X = a x , Y = a y , L = a l . Obviously, X > , Y > , L >
1. To this end, we compute the side lengths of A (cid:48) B (cid:48) C (cid:48) D (cid:48) . (cid:48) B (cid:48) = X (1 + Y )( L − Y , C (cid:48) D (cid:48) = X (1 + Y L )( L − Y LB (cid:48) C (cid:48) = X (1 + Y L )( L − Y , D (cid:48) A (cid:48) = X (1 + Y L )( L − Y L .
It is easy to verify that A (cid:48) B (cid:48) + C (cid:48) D (cid:48) = B (cid:48) C (cid:48) + D (cid:48) A (cid:48) , thus A (cid:48) B (cid:48) C (cid:48) D (cid:48) is tangential.Finally, we show that given any tangential trapezoid K , an appropriate scaledand/or rotated copy of K can be obtained as the image under transformation T of a conveniently chosen square in the xy plane. Without loss of generality,assume that the parallel sides of K are vertical, and that the non-parallel sidesintersect at the origin - see figure 5. Figure 5.
The square
ABCD being mapped into the tangential trape-zoid A (cid:48) B (cid:48) C (cid:48) D (cid:48) . Vertical lines are mapped into vertical lines, horizontallines are mapped into lines through the origin. Let the slope of A (cid:48) B (cid:48) be m , and let the slope of C (cid:48) D (cid:48) be p , with p > m . Wehave that the slope of A (cid:48) B (cid:48) = Y − Y , and the slope of C (cid:48) D (cid:48) = Y L − Y L . Choosing X = 1, Y = m + √ m , and L = ( (cid:112) p + p )( √ m − m ) achieves thedesired goal. (cid:3) The general case
A result similar to Theorem 5.1 holds when the tangential quadrilateral to bedissected is not a trapezoid. We present the details below.
Theorem 6.1.
There exists a transformation, T : R → R with the followingproperties(a) T maps vertical lines from the xy plane into lines through the point P (1 , in the uv plane.(b) T maps horizontal lines from the xy plane into lines through the origin O (0 , in the uv plane.(c) T maps any axes-parallel square lying in the halfplane x + y > of the xy plane into a tangential quadrilateral in the uv plane. roof. Let T : u = f ( x, y ) , v = g ( x, y ). Any vertical line x = c in the xy plane is mapped into a line v = m c ( u − uv plane. This means g ( c, y ) = m c ( f ( c, y ) − g ( c, y ) / ( f ( c, y ) −
1) = m c . It follows that g ( x, y ) / ( f ( x, y ) −
1) depends on x only, that is(8) g ( x, y ) = F ( x ) · ( f ( x, y ) − . On the other hand, any horizontal y = c is mapped into a line through theorigin v = m c u . This can be rewritten as g ( x, c ) = m c · f ( x, c ), from which g ( x, c ) /f ( x, c ) = m c , where m c depends only on c . This means that(9) g ( x, y ) /f ( x, y ) = G ( y ) from which g ( x, y ) = G ( y ) · f ( x, y ) . Solving (8) and (9), we obtain(10) f ( x, y ) = F ( x ) F ( x ) − G ( y ) , g ( x, y ) = F ( x ) · G ( y ) F ( x ) − G ( y ) . Taking partial derivatives in (10), we obtain f x ( x, y ) = − F (cid:48) ( x ) · G ( y )[ F ( x ) − G ( y )] , g x ( x, y ) = G ( y ) · f x ( x, y ) ,f y ( x, y ) = − F ( x ) · G (cid:48) ( y )[ F ( x ) − G ( y )] , g y ( x, y ) = F ( x ) · f y ( x, y ) . Using the local condition (1), we have that f x + g x = f y + g y −→ ( f x ( x, y )) · [1 + G ( y )] = ( f y ( x, y )) · [1 + F ( x )] →→ ( F (cid:48) ( x )) · ( G ( y )) · [1 + G ( y )] = ( F ( x )) · ( G (cid:48) ( y )) · [1 + F ( x )] →−→ (cid:32) F (cid:48) ( x ) F ( x ) · (cid:112) F ( x ) (cid:33) = (cid:32) G (cid:48) ( y ) G ( y ) · (cid:112) G ( y ) (cid:33) . Since the left hand side of the above equation depends on x only while theright hand side depends on y only, it follows that we have the following systemof differential equations F (cid:48) ( x ) F ( x ) · (cid:112) F ( x ) = ± k ; G (cid:48) ( y ) G ( y ) · (cid:112) G ( y ) = ± k. One set of solutions of this system is given by F ( x ) = 2 a x a x − , G ( y ) = 2 a y a y − , where a > . Substituting these into (10) we obtain the desired transformation(11) T : f ( x, y ) = a x ( a y − a x + a y )( a x + y − , g ( x, y ) = 2 a x + y ( a x + a y )( a x + y − . We still have to verify property (c). Notice that transformation T is notdefined for points along the line x + y = 0. It is only the squares which areentirely contained in one of the two open half-planes determined by x + y = 0which have the desired property.Consider the square ABCD in the xy plane with A ( x, y ) , B ( x + l, y ) , C ( x + l, y + l ) , D ( x, y + l ) where l >
0. Further assume that x + y > ABCD is above the line x + y = 0. et A (cid:48) := T ( A ) , B (cid:48) := T ( B ) , C (cid:48) := T ( C ) , D (cid:48) := T ( D ) and denote X := a x , Y := a y , L := a l . Obviously, since a > x + y > X > , Y > , XY > , L >
1. Thus, the coordinates of these fourpoints can be written as A (cid:48) = (cid:18) X ( Y − X + Y )( XY − , XY ( X + Y )( XY − (cid:19) ,B (cid:48) = (cid:18) LX ( Y − LX + Y )( LXY − , LXY ( LX + Y )( LXY − (cid:19) ,C (cid:48) = (cid:18) LX ( L Y − LX + LY )( L XY − , L XY ( LX + LY )( L XY − (cid:19) ,D (cid:48) = (cid:18) X ( L Y − X + LY )( LXY − , LXY ( X + LY )( LXY − (cid:19) . It is easy to verify that A (cid:48) B (cid:48) C (cid:48) D (cid:48) is a positively oriented quadrilateral.It remains to check that A (cid:48) B (cid:48) C (cid:48) D (cid:48) is tangential. Calculate the side lengths of A (cid:48) B (cid:48) C (cid:48) D (cid:48) . A (cid:48) B (cid:48) = XY ( L − Y + 1)( X L + 1)( X + Y )( XY − XL + Y )( XY L − ,B (cid:48) C (cid:48) = XY ( L − Y L + 1)( X L + 1)( X + Y )( XL + Y )( XY L − XY L − ,C (cid:48) D (cid:48) = XY ( L − Y L + 1)( X L + 1)( X + Y )( X + Y L )( XY L − XY L − ,D (cid:48) A (cid:48) = XY (1 + X )( L − Y L + 1)( X + Y )( XY − X + Y L )( XY L − . It is easy to verify that A (cid:48) B (cid:48) + C (cid:48) D (cid:48) = B (cid:48) C (cid:48) + D (cid:48) A (cid:48) , thus by the Pitot-Steinertheorem it follows that A (cid:48) B (cid:48) C (cid:48) D (cid:48) is tangential. This proves the theorem. (cid:3) As in the trapezoid case, we can show that given any tangential quadrilateral A (cid:48) B (cid:48) C (cid:48) D (cid:48) , an appropriate scaled and/or rotated copy of A (cid:48) B (cid:48) C (cid:48) D (cid:48) in the uv plane can be obtained as the image under transformation T of a convenientlychosen square in the xy plane. However, in this case the computations are morecomplicated. We present the details in the following theorem. Theorem 6.2.
Let A (cid:48) B (cid:48) C (cid:48) D (cid:48) be a tangential quadrilateral in the uv plane. With-out loss of generality assume that after an eventual scaling and/or rotation wehave that A (cid:48) B (cid:48) ∩ C (cid:48) D (cid:48) = O (0 , and B (cid:48) C (cid:48) ∩ D (cid:48) A (cid:48) = P (1 , as shown in figure6 below. Then for any a > there exists constants x , y and l , with x + y > , l > such that the image of the axes-parallel square of side l with the bottomleft corner located at A ( x, y ) under the transformation T defined by (11) is thequadrilateral A (cid:48) B (cid:48) C (cid:48) D (cid:48) .Proof. Denote by t = tan( ∠ A (cid:48) / t = tan( ∠ B (cid:48) / t = tan( ∠ C (cid:48) /
2) and t = tan( ∠ D (cid:48) / t i > ≤ i ≤
4. Since the angles ofa quadrilateral sum up to 2 π it follows easily that the quantities t i satisfy thefollowing identity(12) t + t + t + t = t t t + t t t + t t t + t t t . From our choice of A (cid:48) B (cid:48) C (cid:48) D (cid:48) we have that ∠ A (cid:48) + ∠ B (cid:48) < π and therefore t t <
1. Similarly, since ∠ A (cid:48) + ∠ D (cid:48) < π , it follows that t t < igure 6. Proving that the transformation T is surjective: every ap-propriately scaled/rotated tangential quadrilateral is the image of someaxes-parallel square. We also need the following identity(13) t t + t t + t t − ∠ C (cid:48) / ∠ A (cid:48) /
2) cos( ∠ B (cid:48) /
2) cos( ∠ D (cid:48) / . In particular, it follows that t t + t t + t t − >
0. Hence,(14) 1 − t t > , − t t > , t t + t t + t t − > . Assume now that
ABCD is the axes-parallel square which is the desired pre-image of A (cid:48) B (cid:48) C (cid:48) D (cid:48) . Assume that this square has side l > A ( x, y ), B ( x + l, y ), C ( x + l, y + l ) , D ( x, y + l ), and x + y > x, y and l such that A (cid:48) = T ( A ), B (cid:48) = T ( B ), C (cid:48) = T ( C ), D (cid:48) = T ( D ). Denote a x = X , a y = Y and a l = L . Since a > x + y > l > XY >
L > ∠ A (cid:48) /
2) = t = X Y − X + Y , tan( ∠ B (cid:48) /
2) = t = X L + YXY L − , tan( ∠ C (cid:48) /
2) = t = XY L − L ( X + Y ) , tan( ∠ D (cid:48) /
2) = t = X + Y LXY L − . (15)After taking resultants of the expressions for t , t and t in (15) it followsthat X , Y and L satisfy the following quadratic equations X − t + t − t − t ( t + t )1 − t t X − , (16) Y − t − t + t − t ( t + t )1 − t t Y − , (17) L − ( t t + t t − + t + t + 1 t t + t t + t t − L + 1 = 0 . (18)These equations are well defined as we earlier proved that 1 − t t >
0, 1 − t t > t t + t t + t t − >
0. Let X , Y and L be the larger roots of (16), (17)and (18), respectively. n particular, X = 2 t + t − t − t ( t + t ) + (cid:112) (1 + t ) [( t t + t t − + ( t − t ) ]2(1 − t t ) Y = 2 t − t + t − t ( t + t ) + (cid:112) (1 + t ) [( t t + t t − + ( t − t ) ]2(1 − t t ) L = ( t t + t t − + t + t + 1 + ( t + t ) (cid:112) (1 + t ) [( t t + t t − + ( t − t ) ]2( t t + t t + t t − . Ensuring that these choices for X , Y and L satisfy the system (15) is just amatter of algebraic calculation; for the value of t one can use identity (13).Such verifications can be easily performed using any computer algebra software(for instance we checked them using MAPLE).It remains to prove that for these choices of X, Y and L we do have that XY >
L > a := 2 t + t − t − t ( t + t )1 − t t , (19) b := 2 t − t + t − t ( t + t )1 − t t , (20) c := ( t t + t ! t − + t + t + 1 t t + t t + t t − . (21)The numbers a , b and c above are the coefficients of the linear terms in thequadratics (16), (17) and (18). Hence, X , Y and L are the larger roots of theequations(22) X − aX − , Y − bY − , L − cL + 1 = 0 , that is,(23) X = a + √ a + 42 , Y = b + √ b + 42 , L = c + √ c − . Quick computations show that a + b = t (cid:0) (2 − t t − t t ) + ( t − t ) (cid:1) (1 − t t )(1 − t t ) , c = 2+ (2 − t t − t t ) + ( t − t ) ( t t + t t + t t − . Combining the above equalities with (14), it follows that a + b > c > a > − b and since the function h ( x ) = x + √ x + 4 is strictly increasingon R we have that h ( a ) > h ( − b ) from which a + (cid:112) a + 4 > − b + (cid:112) b + 4 , and after multiplying both sides by b + (cid:112) b + 4( a + (cid:112) a + 4)( b + (cid:112) b + 4) > −→ a + √ a + 42 · b + √ b + 42 > −→ XY > . On the other hand, since c > L = ( c + √ c − / > XY >
L > (cid:3)
It can be noticed that transformation T given by (11) proves slightly morethan that any given tangential quadrilateral has an n × n class preserving griddissection for every n ≥ orollary 6.3. For any given tangential quadrilateral Q and any dissection ofan axes-parallel square into smaller squares, there exists a topologically equivalentdissection of Q into smaller tangential quadrilaterals. Three nice properties of tangential quadrilaterals
There are three natural “centers” associated to every given tangential quadri-lateral A (cid:48) B (cid:48) C (cid:48) D (cid:48) . First, we have I , the incenter of A (cid:48) B (cid:48) C (cid:48) D (cid:48) . Second, we have S , the point of intersection of the diagonals A (cid:48) C (cid:48) and B (cid:48) D (cid:48) . Third, there is W ,the 2 × center of A (cid:48) B (cid:48) C (cid:48) D (cid:48) , the common point of the four smaller tangentialquadrilaterals which determine the 2 × A (cid:48) B (cid:48) C (cid:48) D (cid:48) - see figure7. One very interesting fact is that these points are always collinear. Theorem 7.1.
Let A (cid:48) B (cid:48) C (cid:48) D (cid:48) be a tangential quadrilateral placed such that A (cid:48) B (cid:48) ∩ C (cid:48) D (cid:48) = O (0 , and A (cid:48) D (cid:48) ∩ B (cid:48) C (cid:48) = P (1 , as shown in figure 7.a) Let I , S and W be the three centers defined above. Then these points arecollinear and the line determined by them is perpendicular to OP .b) Denote the radii of the inscribed circles of the four small tangential quadri-laterals A (cid:48) B (cid:48) C (cid:48) D (cid:48) is dissected into by r , r , r and r , respectively. Then (24) 1 r + 1 r = 1 r + 1 r . Figure 7.
The incenter I , the 2 × W , and the point of inter-section of the diagonals S , are collinear in any tangential quadrilateral Proof.
From theorems 6.1 and 6.2 it follows that there exists an axes-parallelsquare,
ABCD with A ( x, y ) , B ( x + l, y ) , C ( x + l, y + l ) , D ( x, y + l ), a > l > x + y > T given by (11) such that T ( ABCD ) = A (cid:48) B (cid:48) C (cid:48) D (cid:48) .As before, denote X := a x , Y := a y , L := a l . From our choices it follows that, X > , Y > , XY > L > e just have to compute the coordinates of the points I , S and W . Straight-forward calculations show that all these points have the same abscissa u I = u S = u W = X ( L Y − X + Y )( XY L − , which proves the first assertion . For the second part, similar computations give the expressions of r i , for 1 ≤ i ≤ r = XY ( L − X + Y )( XY L − , r = XY L ( L − XL + Y )( XY L − ,r = XY L ( L − X + Y L )( XY L − r = XY L ( L − X + Y ) L ( XY L − . Equality (24) follows. It is interesting to mention that a similar relation issatisfied by the inradii of triangles A (cid:48) SB (cid:48) , B (cid:48) SC (cid:48) , C (cid:48) SD (cid:48) and D (cid:48) SA (cid:48) . This wasproved in [2] to be a necessary and sufficient condition for a quadrilateral to betangential.(25) 1 r A (cid:48) SB (cid:48) + 1 r C (cid:48) SD (cid:48) = 1 r B (cid:48) SC (cid:48) + 1 r D (cid:48) SA (cid:48) . (cid:3) Another property an n × n class preserving grid dissection has is what we callthe triple-grid property . The situation is displayed below for the case n = 3. Figure 8.
The triple grid property: both incenters and the points ofintersection of the diagonals form an n × n grid. The green points represent the incenters of the nine smaller quadrilateralswhile the red points represent the intersections of the diagonals within each ofthese nine quadrilaterals. It can be easily seen that the incenters as well as theintersection points of the diagonals create a “grid-like” pattern. A third grid, notpictured in figure 8 in order to maintain clarity, is created by the 2 × eferences [1] D. Ismailescu and A. Vojdany, Class Preserving Dissections of Convex Quadrilaterals, Fo-rum Geometricorum (2009), pp. 195–211.[2] W. C. Wu and P. Simeonov, Problem 10698, American Mathematical Monthly , (1998)pp. 995; solution, (2000), pp. 657-658.[3] P. Yiu, Notes on Euclidean Geometry
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