First prove that the 3 triangles are similar, then some length bashing to prove $FG+EA'=GA$

did u write inmo @ tarzanjunior?

Yep, I wrote it

As the inradius of any right triangle is half the difference between the sum of its sides and its hypotenuse, it suffices to show that $$FD'+GD'-FG+A'B+BE-A'E = GC+CA'-A'G$$$$\iff FD'+GD'+A'B+BE+A'G = FG+A'E+GC+A'C$$$$\iff FD+A'D' + A'B + BE = FG+AE + GC + A'C$$$$\iff FD+AD+A'B+BE = AE+FC+CA'$$$$\iff 2(AE+FC+CA') = AB+BC+CD+DA$$WLOG $AD=1$ and let $A'B=x$. Let $X=AA'\cap EF$. Then by similar triangles $\dfrac{AE}{AX} = \dfrac{AA'}{AB}\implies AE=\frac{1}{2}AA'^2 = 2x^2+2$. By similar triangles again $AE-DF = x\implies DF = 2x^2-x+2$. Now $AE+FC+CA' = 2x^2+2 + 1-(2x^2-x+2) + (1-x) = 2$ which is indeed half the perimeter of the square, done. $\Box$ EDIT: sorry, I misread my own work. Fixed now.

bobthesmartypants wrote: As the inradius of any right triangle is half the difference between the sum of its sides and its hypotenuse, it suffices to show that $$FD'+GD'-FG+A'B+BE-A'E = GC+CA'-A'G$$$$\iff FD'+GD'+A'B+BE+A'G = FG+A'E+GC+A'C$$$$\iff FD+A'D' + A'B + BE = FG+AE + GC + A'C$$$\color{red}{\iff}$$A'E=AE$ true. Why?

What does INMO stand for?

http://www.artofproblemsolving.com/community/c260h1368546p7531805

Just show that FD + A'B = AE.

tarzanjunior wrote: First prove that the 3 triangles are similar, then some length bashing to prove $FG+EA'=GA$ How to prove it ?

The same question came in 8th Sharygin Geometry Olympiad Final for 8th grade.

Dear Mathlinkers, 1. E' the foot of the perpendicular to CD through E 2. E'F = A'B and this is the key of the solution... Sincerely Jean-Louis

This problem is lacking a nice synthetic solution. Here's one I found. Solution: It is equivalent to proving that $FG+A'E=A'G $. Mark $G'$ on $AB $ such that $EFGG'$ is a parallelogram. Let $AA'\cap GG'=H $. Claim 1: $A $ is the excenter of $\Delta A'CG $. Proof of claim 1: As $A'$ is the reflection of $A $ in $EF $, so, $\angle AA'G=\angle A'AD=\angle AA'B $. The proof is now obvious as $AC $ bisects $\angle BCD $. Claim 2: $AGA'G'$ is an isosceles trapezoid. Proof of claim 2: $AA'\perp EF\implies AA'\perp GG'\implies HA'BG'$ is cyclic $\implies \angle GA'A=\angle GG'A\implies AGA'G'$ is cyclic. Again, $\angle GAG'=\angle AGD=\angle AGA'\implies AGA'G'$ is an isosceles trapezoid. Main problem: From claim $2$, we have $AG'=A'G$. This completes our proof as $FG=EG'$.

Where can I find sharygin questions?

What is inradius? I can't find it in dictionary Can someone explain?

khbghvb wrote: What is inradius? I can't find it in dictionary Can someone explain? The radius of the incircle.

Wow, it is trig bashable after getting the three triangles similar.

Drunken_Master wrote: Wow, it is trig bashable after getting the three triangles similar. That's how I did it !

HighQXMoney wrote: What does INMO stand for? Indian national Olympiad

Sorry for late reply. Three triangles are similar. A'B/GC=BE/CA'=A'E/GA'=k1 FD'/GC=D'G/CA'=FG/GA'=k2 A'B/FD'=BE/D'G=A'E/FG=k1/k2 The inradius of a right triangle is half the difference between the sum of its sides and hypotenuse. So we get, r1=k1*r3 r2=k2*r3 r1=(k1/k2)*r2 After solving all the equations we get r3=r1+r2 Please verify my solution.

Here's another nice synthetic finish. After proving the three triangles ($FD'G, A'CG$ and $EBA'$) similar (no big deal), we only need to show that $A'C=FD'+EB$. Now let $F'$ be the projection of $F$ on $\overline{AB}$. Note that since $A'$ is the reflection of $A$ over $\overline{EF}$, we have $AA'\perp FE$. Also $\angle GCA'=90^{\circ}$. If we let $X=AA'\cap EF$, then $FCA'X$ is cyclic, so $\angle XEA=\angle XFC=\angle XA'B$. Thus $\triangle FF'E\cong\triangle AA'B$ (since $AB=FF'$). So $F'E=A'B$, implying $A'C=F'A+EB=FD+EB=FD'+EB$. $\square$ [asy][asy] size(5cm); pair A=(0,0),B=(1,0),C=(1,1),D=(0,1),Ap=(1,0.333),Dp,Ee,F,G; Ee=extension(A,B,(A+Ap)/2,bisectorpoint(A,Ap)); F=extension(C,D,(A+Ap)/2,bisectorpoint(A,Ap)); Dp=reflect(Ee,F)*D; G=extension(C,D,Ap,Dp); D(MP("A",A,W)--MP("E",Ee,S)--MP("B",B,E)--MP("A^{\prime}",Ap,E)--MP("C",C,E)--MP("G",G,NE)--MP("D^{\prime}",Dp,N)--MP("F",F,NNW)--MP("D",D,W)--cycle,black); draw(Ee--Ap--G--F); draw(A--Ap, green); pair X; X=((A+Ap)/2); dot(X); label("X", X, NW); pair Fp; pair I=(0,0.5), J=(1,0.5); Fp=reflect(I,J)*F; dot(Fp); draw(F--Fp, green); label("$F^{\prime}$", Fp, S); dot(A);dot(B);dot(C);dot(D);dot(Ap);dot(Dp);dot(Ee);dot(F);dot(G); draw(Ee--F,dashed); [/asy][/asy]

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20VI.pdf p. 16... Sincerely Jean-Louis

Hi all, this is a nice problem. I've created a beginner-friendly solution HERE using triangle similarity and a bit of algebra if anyone is interested. I'm also currently building a blog about Euclidean Geometry. Please kindly visit.

Note that we need to prove $D'F + A'B + BE = AD$ or $D'F + A'B = AE$. Let $AA'$ meet $FE$ at $K$ and Let $S$ be feet of $F$ on $AB$. Note that we need to prove $SE = A'B$. we have $FS = AB$ and $\angle FSE = \angle 90 = \angle ABA'$ and $\angle SFE = \angle SFK = \angle SAK = \angle BAA'$ so triangles $BAA'$ and $SFE$ are congruent so $SE = A'B$. we're Done.