Let $A_1A_2A_3A_4$ be a rectangle and let $S_1,S_2,S_3,S_4$ four circumferences inside of the rectangle such that $S_k$ and $S_{k+1}$ are tangent to each other and tangent to the side $A_kA_{k+1}$ for $k=1,2,3,4$, where $A_5=A_1$ and $S_5=S_1$. Prove that $A_1A_2A_3A_4$ is a square.
Problem
Source: 2022 Centroamerican and Caribbean Mathematical Olympiad, P4
Tags: geometry, rectangle, square, tangent circles
Rinaldo
02.12.2022 21:59
This is my solution to this problem... apparently though there's something wrong about it but I don't' know what it is, can someone tell me?
soryn
02.12.2022 22:28
Ok,thanks.....
Supertinito
03.12.2022 22:30
Here is the official solution. Let $r_i$ be the radius of $S_i$. Let $X$ and $Y$ be the tangency points of $S_1$ and $S_2$ with $A_1A_2$ respectively. By Pythagoras' theorem theorem, we have that \[XY=\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}\]Thus we get that \[A_1A_2=A_1X+XY+YA_2=r_1+2\sqrt{r_1r_2}+r_2=(\sqrt{r_1}+\sqrt{r_2})^2\] Analogously, we get \[A_3A_4=(\sqrt{r_3}+\sqrt{r_4})^2\]And as $A_1A_2A_3A_4$ is a rectangle, \[\sqrt{r_1}+\sqrt{r_2}=\sqrt{r_3}+\sqrt{r_4}\]. Similarly \[\sqrt{r_1}+\sqrt{r_4}=\sqrt{r_2}+\sqrt{r_3}\]The last two equations imply that $r_1=r_3$ and $r_2=r_4$ Then \[A_1A_2=(\sqrt{r_1}+\sqrt{r_2})^2=(\sqrt{r_3}+\sqrt{r_2})^2=A_2A_3\]Thus $A_1A_2A_3A_4$ is a square
parmenides51
04.12.2022 12:01
I send this to friend Antreas Chrysikos yesterday
Let $r_i$ be the radius of $S_i$. Let $X$ and $Y$ be the tangency points of $S_1$ and $S_2$ with $A_1A_2$ respectively. By Pythagoras' theorem theorem, we have that
\[XY=\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}\]Thus we get that
\[A_1A_2=A_1X+XY+YA_2=r_1+2\sqrt{r_1r_2}+r_2=(\sqrt{r_1}+\sqrt{r_2})^2\]
$r_1+2\sqrt{r_1r_2}+r_2=r_3+2\sqrt{r_3r_4}+r_4$ (1)
$r_1+2\sqrt{r_1r_4}+r_4=r_2+2\sqrt{r_2r_3}+r_3$ (2)
(1) - (2) $\Rightarrow r_2- r_4+2\sqrt{r_1r_2}-2\sqrt{r_1r_4}=r_4- r_2+2\sqrt{r_3r_4}-2\sqrt{r_2r_3}$
$\Leftrightarrow (\sqrt{r_2}-\sqrt{r_4})( \sqrt{r_2}+\sqrt{r_4}+2\sqrt{r_1})=(\sqrt{r_4}-\sqrt{r_2})( \sqrt{r_4}+\sqrt{r_2}+2\sqrt{r_3})$ (3)
if $r_2<r_4$, looking at (3): $LHS > 0$ but $RHS <0$ , constradiction
if $r_4>r_2$ , looking at (3): $LHS < 0$ but $RHS >0$ , constradiction
therefore $r_2=r_4$
) At the photo is his figure and that lemma. The basic part of his solution is what I put into hide. I tried solving it without any progress
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parmenides51
04.12.2022 12:44
Rinaldo wrote: This is my solution to this problem... apparently though there's something wrong about it but I don't' know what it is, can someone tell me? I just checked, your attempt doesn't work,
the point $D$ you define as the intersection of the perpendiculars from those touchpoints point is the same as the center of circle $O_2$ by definition, you do not use anywhere that $D$ lies on diagonal $A_2A_4$.
If you use that point $D$ lies on diagonal $A_2A_4$ of the rectangle and drop the perpendiculars on sides of the rectangle, nothing justifies that those shall be the touchpoints of the circle ($O_4$) with the sides, in other words projection of point $D$ on diagonal $A_2A_4$ is in general different than point $B$ , which you define as touchpoint of circle ($O_4$) with side $A_1A_2$, the same applies for your point $C$
A different approach to understand what goes wrong, think of a random rectangle, from a point on it's diagonal draw the perpenicular on two consecutive sides, they generally are not equal , although they form a small rectangle, the only way that those perpendiculars were equal is if the starting big rectangle was a square