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everyone! I hope you're all doing well. I've been pondering over the following problem lately and I've hit a bit of a roadblock:

We know that point $P$ is inside square $ABCD$. Given that $PA = 5$, $PB = 8$, and $PC = 13$. Find the area of square $ABCD$.

This is my sol:

Draw $PE \perp AB$ through point $P$, with the foot of the perpendicular being $E$, and $PF \perp BC$, with the foot of the perpendicular being $F$. Let the side length of the square be $a$, $PE = x$, and $PF = y$. Then, by the Pythagorean theorem, we have: \begin{equation} a = AB = \sqrt{5^2 - x^2} + y\cdots \cdots (1) \end{equation} \begin{equation} a = BC = \sqrt{13^2 - y^2} + x \cdots \cdots (2) \end{equation} \begin{equation} x^2 + y^2 = 64 \cdots \cdots (3) \end{equation} From (1) and (2), we get: \begin{equation} (a - y)^2 = 25 - x^2 \end{equation} \begin{equation} (a - x)^2 = 169 - y^2 \end{equation} From (3), we get: \begin{equation} a^2 - 2ay = 25 - (x^2 + y^2) = -39 \end{equation} \begin{equation} a^2 - 2ax = 169 - (x^2 + y^2) = 105 \end{equation} Therefore, \begin{equation} 4a^2y^2 = (a^2 + 39)^2 \end{equation} \begin{equation} 4a^2x^2 = (a^2 - 105)^2 \end{equation} Adding the two equations and using (3), we obtain: \begin{equation} 4a^2 \cdot 64 = (a^2 + 39)^2 + (a^2 - 105)^2 \end{equation} Solving this, we get $a^4 - 194a^2 + 6273 = 0$, which gives $a^2 = 41$ or $153$. Since $PC < AC$, i.e., $13 < \sqrt{2}a$, $2a^2 > 169$, hence only $a^2 = 153$. Therefore, the area of square $ABCD$ is $153$.

However, I feel like my algebraic approach might not be the best way to tackle this. I've got a hunch that there should be a purely geometric solution to this problem. After quite a bit of searching and head-scratching, I'm afraid I haven't been able to find the right method. That's why I'm reaching out to all of you wonderful minds for help. I'd really appreciate it if someone could share an elementary, purely geometric proof. Your assistance would be greatly appreciated! Thanks a lot!

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  • $\begingroup$ Maybe british flag theorem would help en.wikipedia.org/wiki/British_flag_theorem $\endgroup$ Commented Mar 2 at 3:09
  • $\begingroup$ @user4372 OP essentially did that in their equations (1), (2), (3). $\endgroup$ Commented Mar 2 at 4:36
  • $\begingroup$ You can find a geometric answer here: math.stackexchange.com/a/5013553/715329. In fact, your question is a duplicate of this question: math.stackexchange.com/q/5013127/715329 $\endgroup$ Commented Mar 2 at 5:42
  • $\begingroup$ @above, that's a good idea, but it uses Heron's formula, which is not purely geometric and requires some computational effort. $\endgroup$ Commented Mar 2 at 6:29
  • $\begingroup$ You can also use $\cos^2\angle ABP+\cos^2\angle CBP=1$, in which less calculation is involved. $\endgroup$ Commented Mar 2 at 10:35

1 Answer 1

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enter image description here

Hint:In the picture we have:

$\triangle APB=\triangle BPT=\triangle BTE=\triangle BCE$

$S_{APB}=\frac{8\times 3}2=12$

$\triangle APD=\triangle ECD$

$S_{APD}=\frac{8\times 12}2=48$

$S_{PDT}=S_{ETD}=\frac {0.74\times 13}2=4.81$

$S_{ABCD}=4\times12+ 2\times 48+2\times 4.63=153.26$

You can also do this:

$S_{ABCD}=AK^2+KD^2=3^2+12^2=153$

Update: we have right angled triangle BPU . BP=5, suppose BU and PU must have integer magnitude. The only possible numbers are $RU=3$ and $PU=4$ , in this case we have:

$AU=8+4=12$

$BU=3$

which gives:

$S_{ABCD}=AB^2=12^2+3^2=153$

Here is a link https://www.geogebra.org/classic

To see the picture click on recover.

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  • $\begingroup$ Accurate drawing isn't a proof (but I can see why an Engineer might consider it so). Furthermore, with this lengths, we can calculate (EG via cosine rule) $\angle ABP \neq \frac{90^\circ}{4}$, so you actually don't have an accurate drawing. (And that would explain how you arrived at a contradictory result for $ S_{ABCD}$) $\endgroup$ Commented Mar 4 at 0:36

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