โจทย์เรขา ช่วยหน่อยนะครับ ข้อ 2

[Yo WMU]

2. $\Delta ABC และ \Delta XYZ $ เท่ากันทุกประการ โดยมี $\propto จัตุรัส DEFB และ \propto จัตุรัส HIJK แนบใน \Delta ABC และ \Delta XYZ ตามลำดับ $

$\propto DEFB และ \propto HIJK มีพื้นที่ 441 และ 440 ตารางเมตร ตามลำดับ จงหาผลบวกของด้านประกอบมุมฉากของ \Delta ABC $

[Anton]

Problem. Let $ABC$ and $XYZ$ be triangle such that $AB=XY$, $\angle ABC=\angle XYZ=\dfrac{\pi}{2}$, and $BC=YZ$. Points $D$, $E$, and $F$ on $AB$, $CA$, and $BC$, respectively, are such that the quadrilateral $BDEF$ is a square. Points $H$, $I$, $J$, and $K$ on $ZX$, $YZ$, $XY$, and $ZX$, respectively, are such that the quadrilateral $HIJK$ is a square. If the areas of the squares $BDEF$ and $HIJK$ are $441$ and $440$ square meters, respectively, what is the sum of the squares of catheti of the triangle $ABC$?

Let $c:=AB$ and $a:=BC$. If $s$ is the side length of the square $BDEF$, then note that $\triangle BDE\sim \triangle EFC$, whence
$$\frac{c-s}{s}=\frac{AD}{DE}=\frac{EF}{FC}=\frac{s}{a-s}\,.$$
That is,
$$(c-s)(a-s)=s^2\,.$$
Thus,
$$s=\frac{ac}{a+c}\,.$$

Now, let $r$ be the side length of the square $HIJK$. Note that $YZ=a$ and $XY=c$. Observe that $\triangle JYI\sim \triangle XYZ$. Therefore,
$$\frac{JY}{r}=\frac{JY}{IJ}=\frac{XY}{ZX}=\frac{c}{\sqrt{a^2+c^2}}\,.$$
This means $JY=\dfrac{rc}{\sqrt{a^2+c^2}}$. Hence,
$$XJ=XY-JY=c-\frac{rc}{\sqrt{a^2+c^2}}=c\,\left(1-\frac{r}{\sqrt{a^2+c^2}}\right)\,.$$
Now, $\triangle XKJ\sim \triangle XYZ$. That is,
$$\frac{c}{\sqrt{a^2+c^2}}\,\left(1-\frac{r}{\sqrt{a^2+c^2}}\right)=\frac{XJ}{ZX}=\frac{JK}{YZ}=\frac{r}{a}\,.$$
Thus,
$$r=\frac{ac}{a^2+ac+c^2}\,\sqrt{a^2+c^2}\,.$$

Now, observe that
$$\frac{1}{s}=\frac{1}{a}+\frac{1}{c}\,,$$
making
$$\frac{1}{s^2}=\frac{1}{a^2}+\frac{2}{ac}+\frac{1}{c^2}\,.$$
Thus,
$$\frac{1}{s^2}=\frac{a^2c^2}{a^2+c^2}\,\left(\frac{1}{a^2}+\frac{1}{c^2}\right)\,\left(\frac{1}{a^2}+\frac{2}{ac}+\frac{1}{c^2} \right)\,,$$
whence
$$\frac{1}{s^2}=\frac{a^2c^2}{a^2+c^2}\,\left(\frac{1}{a^4}+\frac{2}{a^3c}+\frac{2}{a^2c^2}+\frac{2}{ac^3}+\frac{1}{c^4}\right) \,.$$
We also have
$$\frac{1}{r}=\frac{ac}{\sqrt{a^2+c^2}}\,\left(\frac{1}{a^2}+\frac{1}{ac}+\frac{1}{c^2}\right) \,,$$
so
$$\frac{1}{r^2}=\frac{a^2c^2}{a^2+c^2}\,\left(\frac{1}{a^4}+\frac{2}{a^3c}+\frac{3}{a^2c^2}+\frac{2}{ac^3}+\frac{1}{c^4}\right) \,.$$
Consequently,
$$\frac{1}{r^2}-\frac{1}{s^2}=\frac{a^2c^2}{a^2+c^2}\,\left(\frac{1}{a^2c^2}\right)=\frac{1}{a^2+c^2} \,.$$
That is,
$$a^2+c^2=\frac{1}{\frac{1}{r^2}-\frac{1}{s^2}}=\frac{s^2r^2}{s^2-r^2} \,.$$
With $s^2=441\texttt{ m}^2$ and $r^2=440\texttt{ m}^2$, we obtain
$$a^2+c^2=\frac{441\cdot 440}{441-440}\texttt{ m}^2=441\cdot 440\texttt{ m}^2=194040\texttt{ m}^2\,.$$
It can be shown that $(a,c)$ is a permutation of
$$\left(231-63\sqrt{11}\texttt{ m},231+63\sqrt{11}\texttt{ m}\right)\,.$$
In general, $t=a$ and $t=c$ are the roots of the quadratic polynomial
$$t^2-s\,\left(\frac{s+\sqrt{s^2-r^2}}{\sqrt{s^2-r^2}}\right)\,t+s^2\,\left(\frac{s+\sqrt{s^2-r^2}}{\sqrt{s^2-r^2}}\right)=0\,,$$
which are
$$\frac{s}{2}\,\left(\frac{s+\sqrt{s^2-r^2}}{\sqrt{s^2-r^2}}\right)\,\left(1\pm\sqrt{1-\frac{4\sqrt{s^2-r^2}}{s+\sqrt{s^2-r^2}}}\right)\,.$$
From this result, we can see that
$$r<s\leq \frac{3}{2\sqrt{2}}\,r\,.$$
The inequality on the right is an equality if and only if $a=c$.

If $r=\dfrac{(1+u^2)(1+2u-u^2)}{1+2u+2u^2-2u^3+u^4}\,s$ for some parameter $u\in (0,1)$, then up to swapping,
$$a=\frac{(2-u)(1+2u-u^2)}{2(1-u^2)}\,s$$
and
$$c=\frac{1+2u-u^2}{2u(1-u^2)}\,s\,.$$ This parametrization is useful if we want to find a tuple $(a,c,r,s)\in\mathbb{Q}_{>0}^4$. For example, $u=\dfrac12$ and $s=444$ yield $$(a,c,r,s)=(777,1036,420,444)\,.$$
However, if you want to make $b:=CA=\sqrt{a^2+c^2}$ also a rational multiple of $s$, then you will need to solve the following Diophantine equation
$$(2-u)^2u^2+1=v^2\,,$$
where $u$ and $v$ are rational numbers. So far, I only know that $(u,v)=\left(\dfrac12,\pm\dfrac54\right)$ and $(u,v)=\left(\dfrac{3}{2},\pm\dfrac54\right)$ are solutions to the equation above.