Let $ABC$ be a triangle with $\angle ABC=90^{\circ}$. The square $BDEF$ is inscribed in $\triangle ABC$, such that $D,E,F$ are in the sides $AB,CA,BC$ respectively. The inradius of $\triangle EFC$ and $\triangle EDA$ are $c$ and $b$, respectively. Four circles $\omega_1,\omega_2,\omega_3,\omega_4$ are drawn inside the square $BDEF$, such that the radius of $\omega_1$ and $\omega_3$ are both equal to $b$ and the radius of $\omega_2$ and $\omega_4$ are both equal to $a$. The circle $\omega_1$ is tangent to $ED$, the circle $\omega_3$ is tangent to $BF$, $\omega_2$ is tangent to $EF$ and $\omega_4$ is tangent to $BD$, each one of these circles are tangent to the two closest circles and the circles $\omega_1$ and $\omega_3$ are tangents. Determine the ratio $\frac{c}{a}$.
Problem
Source: Brazil National Olympiad Junior 2021 #7
Tags: geometry
16.02.2022 22:34
Call the centers of $\omega_1$, $\omega_2$, $\omega_3$, $\omega_4$, $O_1$, $O_2$, $O_3$ and $O_4$, respectively. Because of the tangencies we must have $O_1O_2=O_2O_3=O_3O_4=O_4O_1=a+b$. Hence $O_1O_2O_3O_4$ is a rhombus. Now, if $O_1O_3 \cap O_2O_4=T$, we must have that $T$ is the midpoint of $O_1O_3$, so it must be the tangency point of $\omega_1$ and $\omega_3$. Now, by pithagoras: $$TO_4^2+TO_1^2=O_1O_4^2 \iff \dfrac{O_2O_4^2}{4}=(a+b)^2-b^2=a^2+2ab \iff \boxed{O_2O_4=2\sqrt{a^2+2ab}}$$Now, if $l=BD$, because of the tangency between $\omega_2$, $\omega_3$, $\omega_4$ and the sides of the squares we have: $$l=2b+2b=4b$$$$l=a+2\sqrt{a^2+2ab}+a=2a+2\sqrt{a^2+2ab}$$So: $$2a+2\sqrt{a^2+2ab}=l=4b \iff (2b-a)^2=2a^2+4ab \iff 4b^2= 6ab \iff \boxed{ 2a=3b \text{ and } l=4b}$$Now let $\angle BAC =2\alpha$, $I$ is the incenter of $\triangle ADE$ and $X$, $Y$ are the contact points of the incircle of $\triangle ADE$ with $AD$ and $ED$, respectively. We have that $IXDY$ is a rectangle, so: $$b=IX=YD \iff EY=l-b=4b-b=3b$$Now in $\triangle EYI$: $\tan(45-\alpha)=\dfrac{IY}{EY}=\dfrac{b}{3b}=\dfrac{1}{3} \iff \dfrac{\tan(45)-\tan(\alpha)}{1+\tan(45)\tan(\alpha)} =\dfrac{1}{3} \iff \dfrac{1-\tan(\alpha)}{1+\tan(\alpha)} =\dfrac{1}{3} \iff \tan(\alpha)=\dfrac{1}{2} \iff \dfrac{BC}{AB}=\tan(\angle BAC)=\tan(2\alpha)=\dfrac{2\tan(\alpha)}{1-\tan^2(\alpha)}= \dfrac{4}{3}\iff \boxed{\dfrac{BC}{AB}=\dfrac{4}{3}}$ Now, notice that $E$ is equidistant from the sides $BC$ and $AB$, so $E$ is the intersection of the angle bissector of $\angle BAC$ with $AC$, so: $$\dfrac{AE}{EC}=\dfrac{AB}{BC}=\dfrac{3}{4} \iff \boxed{\dfrac{AE}{EC}=\dfrac{3}{4}}$$Clearly $\triangle CEF$ ~ $\triangle EAD$, so: $$\dfrac{b}{c}=\dfrac{AE}{CE}=\dfrac{3}{4} \iff \boxed{\dfrac{b}{c}=\dfrac{3}{4} }$$Finally: $$\dfrac{c}{a}=\dfrac{\frac{4b}{3}}{\frac{2b}{3}}=2$$
19.02.2022 05:05
This problem is false Consider the following figure, take any square $BDEF$ and construct circles $\omega_1, \omega_2$ of radius $b$ and $\omega_3, \omega_4$ of radius $a$ such that $\omega_1$ is tangent to ED, $\omega_2$ is tangent to FB and both are tangent to each other, and $\omega_3$ is tangent to $\omega_1, \omega_2$ and the side $FE$, $\omega_4$ are constructed analogously, so vary a line through $E $ so that, calling $A, C$ their intersections with $BD, FB$ we have the circle of $AED$ with radius $b$, then this figure obeys the statement and as we have infinity of these figures where $\dfrac{c}{a}$ changes, it's over, because the problem is false (just take a different $b$ or change the point of tangency of one of the circles of radius b).