# Sangaku in a Square

One of the simplest sangaku - geometric problems carved on colorful wooden tablets and offered in shinto shrines and buddhist temples during the Edo period (1603-1867) of self-imposed seclusion of Japan from the Western world - is presented by the following diagram: A triangle is formed by a line that joins the base of a square with the midpoint of the opposite side and a diagonal. Find the radius of the inscribed circle.

Solution  A triangle is formed by two lines that join the base of a square with the midpoint of the opposite side and a diagonal. Find the radius of the inscribed circle.

A solution could be observed on an extended diagram. Let A, B, C, D be the vertices of the square and M the midpoint of AD. BD and CM intersect in P. Let Q be the intersection of BM and CD and H the projection of P on BC. The task is to find the inradius of ΔBCP. Focus on ΔBCQ. MD is parallel to the base and is half as long which implies that this is a midline of the triangle. In other words, M and D are the midpoints of BQ and CQ, respectively. This means that BD and CM are two medians in the triangle and P its centroid. The centroid divides the medians in ratio 2:1 so that

(1)

CP = CM·2/3 and BP = BD·2/3.

Thus assuming BC = a, we can apply the Pythagorean theorem to first find BD (from ΔBCD) and CM (from ΔCDM) and subsequently, from (1) the sides of ΔBCP. By the same token, altitude PH = a·2/3. But in any triangle,

(2)

r·p = 2S,

where r is the inradius, p the perimeter, and S the area of the triangle. Putting everything together we see that

r·a·(1 + 5/2·2/3 + 2·2/3) = 2·a2/3,

from which r is easily found:

(3)

r = 2·a / (3 + 5 + 22).

The problem perhaps warrants a

### Remark

The problem is clearly computational: the question is to find the inradius of a triangle related to a given square. Students may be tempted to use calculators or even dynamic geometry software to avoid handling square roots and fractions. This is what they might get - more or less. Assume a = 1. Then

BD = 1.414213562,
BP = .9428090416,
CM = 1.118033989,
CP = .7453559925,
p = 2.688165034,
QH = .6666666667,
S = .3333333333,
2S = .6666666666,
r = .2480006466,

which is quite in agreement with the exact answer (3). The difference between the two is that the numeric value, however accurate, carries no information as to the manner in which it was obtained. In (3), the appearance of the square roots is suggestive and points to a possible application of the Pythagorean theorem to two right triangles, as we did above. Thus (3) exhibits a valuable pattern that may trigger a thought process that may lead to the recollection of properties of medians and centroids, whereas r = .2480006466 in itself is a naked approximate number that, for all we know, is at best close enough to the exact answer.

In short, this 300 year old problem serves an example where the use of calculators actually obscures relationships between numbers. More such examples can be found in an edifying book What The Numbers Say by D. Niederman and D. Boyum along with a multitude of observations on everyday number usage and the ability to see beyond the numbers. ### References

1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

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