Subfield: Geometry


Question: A large square $A B C D$ is drawn, with a second smaller square $P Q R S$ completely inside it so that the squares do not touch. Line segments $A P, B Q, C R$, and $D S$ are drawn, dividing the region between the squares into four nonoverlapping convex quadrilaterals, as shown. If the sides of $P Q R S$ are not parallel to the sides of $A B C D$, prove that the sum of the areas of quadrilaterals $A P S D$ and $B C R Q$ equals the sum of the areas of quadrilaterals $A B Q P$ and $C D S R$. (Note: A convex quadrilateral is a quadrilateral in which the measure of each of the four interior angles is less than $180^{\circ}$.)

![](https://cdn.mathpix.com/cropped/2023_12_21_67f99a025aa829c0a7f6g-1.jpg?height=466&width=461&top_left_y=968&top_left_x=1298)

Solution:
We begin by "boxing in" square $P Q R S$ by drawing horizontal and vertical lines through its vertices to form rectangle $W X Y Z$, as shown. (Because the four quadrilaterals $A B Q P$, $B C R Q, C D S R$, and $D A P S$ are convex, there will not be any configurations that look substantially different from this the diagram below.) We also label the various areas.

![](https://cdn.mathpix.com/cropped/2023_12_21_dc7a2c8c0b7b39460d9fg-1.jpg?height=651&width=678&top_left_y=962&top_left_x=821)

Since $W X$ is parallel to $A B$, then quadrilateral $A B X W$ is a trapezoid. Similarly, quadrilaterals $B C Y X, C D Z Y$, and $D A W Z$ are trapezoids.

We use the notation $|A B Q P|$ to denote the area of quadrilateral $A B Q P$, and similar notation for other areas.

Suppose that the side length of square $A B C D$ is $x$ and the side length of square $P Q R S$ is $y$.

Also, we let $\angle W P Q=\theta$.

Since each of $\triangle W P Q, \triangle X Q R, \triangle Y R S$, and $\triangle Z S P$ is right-angled and each of the four angles of square $P Q R S$ is $90^{\circ}$, then $\angle W P Q=\angle X Q R=\angle Y R S=\angle Z S P=\theta$. This is because, for example,

$\angle X Q R=180^{\circ}-\angle P Q R-\angle W Q P=90^{\circ}-\left(180^{\circ}-\angle P W Q-\angle W P Q\right)=90^{\circ}-\left(90^{\circ}-\theta\right)=\theta$

This fact, together with the fact that $P Q=Q R=R S=S P=y$, allows us to conclude that the four triangles $\triangle W P Q, \triangle X Q R, \triangle Y R S$, and $\triangle Z S P$ are congruent.

In particular, this tells us

* the four areas labelled $e, f, g$ and $h$ are equal (that is, $e=f=g=h$ ),
* $P Z=Q W=R X=S Y=y \sin \theta$, and

$* W P=X Q=Y R=Z S=y \cos \theta$.



Combining these last two facts tells us that $W Z=X W=Y X=Z Y$, since, for example, $W Z=W P+P Z=Z S+S Y=Z Y$. In other words, $W X Y Z$ is a square, with side length $z$, say.

Next, we show that $(a+r)+(c+n)$ is equal to $(b+m)+(d+s)$.

Note that the sum of these two quantities is the total area between square $A B C D$ and square $W X Y Z$, so equals $x^{2}-z^{2}$.

Thus, to show that the quantities are equal, it is enough to show that $(a+r)+(c+n)$ equals $\frac{1}{2}\left(x^{2}-z^{2}\right)$.

Let the height of trapezoid $A B X W$ be $k$ and the height of trapezoid $Z Y C D$ be $l$.

Then $|A B X W|=a+r=\frac{1}{2} k(A B+W X)=\frac{1}{2} k(x+z)$.

Also, $|Z Y C D|=c+n=\frac{1}{2} l(D C+Z Y)=\frac{1}{2} l(x+z)$.

Since $A B, W X, Z Y$, and $D C$ are parallel, then the sum of the heights of trapezoid $A B X W$, square $W X Y Z$, and trapezoid $Z Y C D$ equals the height of square $A B C D$, so $k+z+l=x$, or $k+l=x-z$.

Therefore,

$$
(a+r)+(c+n)=\frac{1}{2} k(x+z)+\frac{1}{2} l(x+z)=\frac{1}{2}(x+z)(k+l)=\frac{1}{2}(x+z)(x-z)=\frac{1}{2}\left(x^{2}-z^{2}\right)
$$

as required.

Therefore, $(a+r)+(c+n)=(b+m)+(d+s)$. We label this equation $(*)$.

Next, we show that $r+n=m+s$.

Note that $r=|\triangle Q X B|$. This triangle can be viewed as having base $Q X$ and height equal to the height of trapezoid $A B X W$, or $k$.

Thus, $r=\frac{1}{2}(y \cos \theta) k$.

Note that $n=|\triangle S Z D|$. This triangle can be viewed as having base $S Z$ and height equal to the height of trapezoid $Z Y C D$, or $l$.

Thus, $n=\frac{1}{2}(y \cos \theta) l$.

Combining these facts, we obtain

$$
n+r=\frac{1}{2}(y \cos \theta) k+\frac{1}{2}(y \cos \theta) l=\frac{1}{2} y \cos \theta(k+l)=\frac{1}{2} y \cos \theta(x-z)
$$

We note that this sum depends only on the side lengths of the squares and the angle of rotation of the inner square, so is independent of the position of the inner square within the outer square.

This means that we can repeat this analysis to obtain the same expression for $m+s$.

Therefore, $n+r=m+s$. We label this equation $(* *)$.

We subtract $(*)-(* *)$ to obtain $a+c=b+d$.

Finally, we can combine all of this information:

$$
\begin{aligned}
& (|A B Q P|+|C D S R|)-(|B C R Q|+|A P S D|) \\
& \quad=(a+e+s+c+g+m)-(b+f+r+d+h+n) \\
& \quad=((a+c)-(b+d))+((m+s)-(n+r))+((e+g)-(f+h)) \\
& \quad=0+0+0
\end{aligned}
$$

since $a+c=b+d$ and $n+r=m+s$ and $e=f=g=h$.

Therefore, $|A B Q P|+|C D S R|=|B C R Q|+|A P S D|$, as required.

Final answer: 证明题，略