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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.

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
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$

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
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$

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

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

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

$$S ;=; R - 4cdotfrac12ab ;=; T$$

(This space intentionally left blank.)

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.

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 $u^2+v^2$.

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 $sintheta=sin(pi-theta)$, $T$ 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 $triangle ABC$.