## 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 *r* = AJ/4,

In the figure above, let *a*, *b*, *d*, and *h* be the lengths of CD, *r*_{1} and *r*_{2} be the radii of the incircles of ΔACD and ΔBCD, respectively. We use two lemmas, which we will prove later:

1. | a - b | = 2d |

2. | 2 r_{1}r_{2} | = ad |

Given *h* = 4*r*, square the equation for the inradius of a right triangle: *b* = *h* + *a*/2 - 2*r* = 2*r* + *a*/2*b*² = 4*r*² + 2*ar* + *a*²/4,*b*² = (*a*/2)² + *h*² = *a*²/4 + 16*r*².*a* = 6*r*,*h* = 4*r*,*b* = 5*r*.*d* = *r*.

Now area( ΔACD) = *ha*/2 = 12*r*² = *r*_{1}*s*, where *s* is the semiperimter of ΔACD or 8*r*. Thus *r*_{1} = 3*r*/2 = 3*d*.*r*_{2} = *a*/6 = *r*.

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 _{2}U = DT + DN.

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 2*r*_{1}*r* = *ad* since we are assuming *r*_{2} = *r*.)*r*,_{1}S = *d*._{1} = *r* + *d* = 3*d* = r_{1}.

Hence, by Lemma 2, 2*rr*_{1} = 2(2*d*)(3*d*) = 12*d*² = *ad*, so *a* = 12*d* = 6*r*. Since *a* - *b* = 2*d**b* = 10*d* = 5*r*.*a*/2 = 3*r* and *b* = 5*r**h* = 4*r*.

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} = DJ·DL.

(*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 third-degree equation in *r* and *h* through more painstaking algebra; rejects the roots *h* = ± *r*√2*h* = 4*r*.

### References

- H. Fukagawa, A. Rothman,
*Sacred Mathematics: Japanese Temple Geometry*, Princeton University Press, 2008 - R. Honsberger,
*Episodes in nineteenth and twentieth-century Euclidean geometry*, MAA, 1995

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