Japanese Temple Problem From 1844

I recently learnt a Japanese geometry temple problem.

The problem is the following:

Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.

This is problem 6 in this article. I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.

• If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem. Commented Nov 24, 2018 at 14:03
• I checked it out. A great book indeed, thank you. Commented Nov 24, 2018 at 14:22
• The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $\pi/2$. Let $\theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $\pi-\theta$. As $\sin\theta=\sin(\pi-\theta)$, $T$ has the same area as $\triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $\triangle ABC$. Commented Nov 24, 2018 at 14:24
• ^ From J.G.'s answer. Commented Nov 24, 2018 at 14:26

We will, first of all, prove a very interesting property

$$\mathbf{Lemma\;1}$$

Given two squares PQRS and PTUV (as shown on the picture), the triangles $$\Delta STP$$ and $$\Delta PVQ$$ have equal area.

$$\mathbf {Proof}$$

Denote by $$\alpha$$ the angle SPT and by $$[...]$$ the area of the polygon "...". Hence $$[\Delta STP]=\frac{\overline {PS}\cdot\overline {PT}\cdot \sin(\alpha)}{2}$$ $$[\Delta PVQ]=\frac{\overline {QP}*\overline {PV}\cdot\sin\Bigl(360°-(90°+90+\alpha)\Bigr)}{2}=\frac{\overline {QP}\cdot\overline {PV}\cdot\sin\Bigl(180°-\alpha\Bigr)}{2}=\frac{\overline {QP}\cdot\overline {PV}\cdot\sin(\alpha)}{2}$$

Since $$\overline {PS}=\overline {PQ}$$ and $$\overline {PT}=\overline {PV}$$ $$[\Delta STP]=[\Delta PVQ]$$

Now, back to the problem

Let $$\overline {AB}=a$$ and $$\overline {IJ}=b$$. Note first of all that $$\Delta BEC \cong \Delta EIF$$ See why? $$\mathbf {Hint:}$$

It is obvious that $$\overline {CE}=\overline {EF}$$. Use the properties of right triangles in order to show that all angles are equal.

Thus $${(\overline{CE})^2}={a^2}+{b^2}=S$$

Note furthermore that $$[\Delta BEC]=[\Delta EIF]=\frac{ab}{2}$$ By Lemma 1: $$[\Delta DCG]=[\Delta BEC]=\frac{ab}{2}=[\Delta EIF]=[\Delta GFK]$$ The area of the polygon AJKGD is thus $$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[\Delta DCG]=2\Bigl({a^2}+{b^2}\Bigr)+2ab$$

The area of the trapezoid AJKD is moreover $$[AJKD]=\frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$

Finally $$T=[\Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S \Rightarrow S=T$$

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Nov 24, 2018 at 13:16

$$|\square P_1 P_2 P_3 P_4| = (a+b)^2 = \frac12(a+b)(2a+2b) = |\square Q_1 Q_2 Q_3 Q_4|\quad=:R$$

$$S \;=\; R - 4\cdot\frac12ab \;=\; T$$

(This space intentionally left blank.)

• I wonder what tools you use to create awesome graphs like this Commented Nov 23, 2018 at 14:24
• @Larry: I use GeoGebra.
– Blue
Commented Nov 23, 2018 at 19:47
• I see, thank you. Commented Nov 23, 2018 at 19:49
• @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$\square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$\triangle$" (in, say, "$\triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
– Blue
Commented Nov 25, 2018 at 11:40
• @Blue Thanks a lot for clarifying :)
– C.S.
Commented Nov 25, 2018 at 14:34

Because there are so many squares, coordinates are easy to compute.

The area of the shaded square is clearly $$u^2+v^2$$.

The area of the shaded triangle is one-half of the absolute value of the determinant of the array

$$\left[ \begin{array}{c} 1 & 1 & 1 \\ 2u-v & 3u & 2u \\ 3u+v & u+3v & u+v \end{array} \right]$$

which is also $$u^2+v^2$$.

I have a second solution.

$$\triangle GPN$$ is obtained by rotating $$\triangle GSD \ 90^\circ$$ clockwise. $$\triangle GQM$$ is obtained by rotating $$\triangle GRK \ 90^\circ$$ counterclockwise.

While the other solutions are obviously correct, they are also unnecessarily complicated.
Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.

*) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.
While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct

• That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
– Owen
Commented Nov 23, 2018 at 23:19
• Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"? Commented Nov 24, 2018 at 8:01
• I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird. Commented Nov 24, 2018 at 20:38
• If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math. Commented Nov 24, 2018 at 20:40
• The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
– Blue
Commented Dec 16, 2018 at 19:53

The four triangles adjacent to $$S$$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $$90^\circ$$ so that they are adjacent to $$T$$, as shown.

What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.

This is a long comment.

The shapes $$S,\,T$$ share a vertex $$A$$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $$\theta$$ be the angle in $$T$$, so the two other acute angles, and the right angle between them from $$S$$, sum to $$\pi-\theta$$. Since $$\sin\theta=\sin(\pi-\theta)$$, $$T$$ has the same area as $$\triangle ABC$$, where $$BA,\,AC$$ are sides of the upper squares.

The challenge, then, is to show $$S$$ has that much area too. Let $$D$$ denote the vertex of $$S$$ opposite $$A$$. I suspect we can show the quadrilateral $$BACD$$ has twice the area of $$S$$, and also twice the area of $$\triangle ABC$$.

• I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$. Commented Nov 22, 2018 at 21:35
• $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$. Commented Nov 22, 2018 at 21:41
• $BACD$ is not a rhombus. Commented Nov 22, 2018 at 21:42
• @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
– J.G.
Commented Nov 23, 2018 at 10:22
• I have converted your answer into a comment. Commented Nov 24, 2018 at 14:28