A better solution to a difficult sangaku problem
J. Marshall Unger
Department of East Asian Languages & Literatures
The Ohio State University
PROBLEM. In ΔABC, AB = BC. If one chooses D on AB and J on CD such that
SOLUTION
The trick to solving the problem expeditiously is to clarify what is given and to prove the converse proposition first.
Notice that the only connection between
In the figure above, let a, b, d, and h be the lengths of CD,
1.  a  b  = 2d 
2.  2 r_{1}r_{2}  = ad 
Given h = 4r, square the equation for the inradius of a right triangle:
Now area( ΔACD) = ha/2 = 12r² = r_{1}s, where s is the semiperimter of ΔACD or 8r. Thus
From this, it is clear that solving the original problem comes down to proving, without knowing the value of h, that ΔACJ and ΔADJ are 3:4:5 right triangles.
It's easy to show that ΔDTU is similar to ΔDO_{2}U. Hence DU is the mean proportional between DT and
a/2  r + d  = [r² + (a/2  r)²] / (a/2  r)  
(a/2  r)² + d (a/2  r)  = r² + (a/2  r)²  
d (a/2  r)  = r²  
ad/2  dr  = r²  
r_{1}r  dr  = r²  
r_{1}  d  = r. 
(Lemma 2 takes the form 2r_{1}r = ad since we are assuming
Hence, by Lemma 2, 2rr_{1} = 2(2d)(3d) = 12d² = ad, so a = 12d = 6r. Since
The lemmas follow from the general case of any triangle ABC with incircle I. Draw a cevian CD and the incircles O_{1} and O_{2} of the resulting triangles ACD and BCD. Label the points of tangency of these incircles as shown. Then
PROOF: By equal tangents, CE = CF (1), and
AE = AG  BF = BG  CM = CL  DJ = DK  
AH = AK  BM = BN  CH = CJ  DN = DL. 
Subtracting the equations in second row from those in the first, we get
EH = GK (2)  FM = GN (3)  CM  CH = JL (4)  JL = DK  DN (5). 
Equating the left and right sides of (4) and (5),
CM + DN  = DK + CH  
(CF + FM) + DN  = DK + (CE + EH)  
(by 1) FM + DN  = DK + EH  
(by 3) GN + DN  = DK + EH  
(GN  DG) + DN  = (DK  DG) + EH  
2DN  = GK + EH  
(by 2) DN  = GK (6). 
Linking equations (2) and (6), EH = GK = DN.
Many more inferences can be drawn from this figure, but we need just two.
Lemma 1 follows immediately if E and J are the midpoints of AC and CD, respectively. Then
Lemma 2 in the general case is
(r_{1} + r_{2})² + (DJ  DL)² = (O_{1}O_{2})² = (r_{1}  r_{2})² + KN²,
but KN = DK + DN = DJ + DL by equal tangents. The rest is just algebra.
In introducing the solution of Kitagawa Moko (1764 1833), Fukagawa and Rothman remark, "It is somewhat complicated and we hope that readers can find a simpler one." I do not know whether the solution presented above is what Fukagawa and Rothman had in mind, but Kitagawa's solution is certainly complicated. It proves Lemma 1 relying in part on the given that ΔABC is isosceles; finds expressions for r_{1}, a, and d in terms of r and h; plugs these directly into Lemma 2; produces a thirddegree equation in r and h through more painstaking algebra; rejects the roots
References
 H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008
 R. Honsberger, Episodes in nineteenth and twentiethcentury Euclidean geometry, MAA, 1995
Sangaku

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