I have a NASTY solution using only cosine theorem and brute force; you basically define the angles that are $\frac{1}{4}$ of the angles of $ABC$, express everything as a function of the trigonometric reasons of these angles, in particular wlog $XY^2$ and $YZ^2$, write the equality between the results, and operate, then operate some more, and after some additional operations, you reach some equality of the form $\left(\cos\frac{A}{4}-\cos\frac{C}{4}\right)\cdot\text{something}=0$, where this "something" cannot be zero, or $A=C$. Do it cyclically, you get the proposed result. If somebody is interested, I will post the trigonometric orgy of operations leading to this result, although I personally will wait for a more beautiful solution... What I found has no added value in terms of geometry, it is basically brute force and patience.

The same is true when it is isosceles.

This problem was tested in our last IMO TST last year, and I did the same solution like daniel73, which was essentially the same as the original solution. Seems weird that the only solution provided in ISL 2009 was that trigonometric brute force solution :p

I will prove the contrapositive. Assume that $\angle C> \angle B$. Lemma 1: If $C>B$ then $r_C>r_B$. Proof: Reflect $C$ over $IA$ to $C'$, we have that $C'\in AB$ so the incircle of $IC'A$ is inside $IBA$. Thus the incircle of $IBA$ is bigger than the incricle of $ICA$ so, $r_C>r_B$.$\boxed{}$ Lemma 2: If $C>B$ then $XY<XZ$. Proof: Using the law of cosines to find $XY$ and $XZ$ ($CX=\frac{r_A}{\sin C/4}$, etc.), it is sufficient to prove that \begin{align*}\sqrt{\frac{r_A^2+r_B^2-2r_Ar_B\cos C/2}{\sin^2 C/4}}&< \sqrt{\frac{r_A^2+r_C^2-2r_Ar_C\cos B/2}{\sin^2 B/4}}\\ \frac{r_A^2+r_B^2-2r_Ar_B\cos C/2}{1-\cos C/2}&< \frac{r_A^2+r_C^2-2r_Ar_C\cos B/2}{1-\cos B/2}\\ (r_A-r_B)^2(1-\cos B/2)+2r_Ar_B(1-\cos B/2)(1-\cos C/2)&< (r_A-r_C)^2(1-\cos C/2)+2r_Ar_C(1-\cos B/2)(1-\cos C/2)\end{align*} Which is true by Lemma 1 and the fact that $-\cos C/2<-\cos B/2$ if $C>B$. $\boxed{}$ Thus we are done.

Sorry, the last line of my proof is incorrect (thanks to ms1990). Here is the corrected line: \begin{align*}(r_{A}-r_{B})^{2}(1-\cos B/2)+2r_{A}r_{B}(1-\cos B/2)(1-\cos C/2)&< (r_{A}-r_{C})^{2}(1-\cos C/2)+2r_{A}r_{C}(1-\cos B/2)(1-\cos C/2)\end{align*} Notice that the inequality is still true in this case. moderator says: last line fixed. Edit: There are a couple more errors (but I believe are fixable) 1. $-\cos C/2>-\cos B/2$ if $C>B$. 2. We need to show that $(r_B-r_A)^2< (r_C-r_A)^2$ or $2r_A<r_B+r_C$. If $B\neq C>A$, then there is no problem. But if $B=C>A$, then there is no such pair of verticies that works, so I need to think of a different proof.

Is there a solution without trigonometry?

I think this problem is not very fit for the ISL.Just apply sine rule in $\triangle{ABZ},\triangle{BCX},\triangle{ACY}$ to obatin $AZ,BZ,BX,CX,CY,AY$ in terms of the sides and angles.Now apply the cosine rule in $\triangle{AYZ},\triangle{BZX},\triangle{CXY}$ to get $XZ^2,ZY^2,XY^2$.Now equate these three terms and obatin $a=b=c$.(Use of complex numbers is not too bad either).

sayantanchakraborty wrote: I think this problem is not very fit for the ISL.Just apply sine rule in $\triangle{ABZ},\triangle{BCX},\triangle{ACY}$ to obatin $AZ,BZ,BX,CX,CY,AY$ in terms of the sides and angles.Now apply the cosine rule in $\triangle{AYZ},\triangle{BZX},\triangle{CXY}$ to get $XZ^2,ZY^2,XY^2$.Now equate these three terms and obatin $a=b=c$.(Use of complex numbers is not too bad either). Can you please give a hint how to start using complex numbers?

The computation is actually pretty short, but since nobody's posted it yet I guess I'll do it for completeness. Let $\alpha = \frac{\angle A}{4}$ and define $\beta, \gamma$ similarly so $\alpha +\beta +\gamma=45^{\circ}$. Let $r_X$ be the inradius of $\triangle BIC$ and define $r_Y, r_Z$ similarly. Then clearly by projecting $X$ onto $BC$, we have $a = \dfrac{r_X}{\tan \beta} + \dfrac{r_X}{\tan \gamma}$, so solving gives $r_X = \dfrac{a \tan \beta\tan \gamma}{\tan \beta+\tan \gamma}= \dfrac{a\sin \beta\sin \gamma}{\sin (\beta + \gamma)} = \dfrac{a\sin\beta \sin \gamma}{\sin (45^{\circ} - \alpha)}$. It then follows that $IX = \frac{r_X}{\sin \frac{BIC}{2}} = \frac{r_X}{\sin (45^{\circ} + \alpha)} = \frac{a\sin \beta \sin \gamma}{\sin (45^{\circ}-\alpha )\sin (45^{\circ} + \alpha)} = \frac{a\sin \beta \sin \gamma}{\sin 2\alpha}$ by the product-to-sum identity. Since $a=2R\sin 4\alpha = 8R \cos 2\alpha \cos \alpha \sin \alpha$, we see that $IX = (8R\sin \alpha \sin \beta \sin \gamma) \cos \alpha$; by homogeneity we can assume the term in parenthesis is $1$ so $IX=\cos \alpha$ and similarly for $IY,IZ$. Next we'll show that $XY=XZ\implies b=c$. By squaring both sides and applying Law of Cosines, this is just $IX^2 + IY^2 -2IX\cdot IY \cos XIY = IX^2 + IZ^2 - 2IX\cdot IZ \cos XIZ$. Now $\cos XIY = \cos (135^{\circ} - \gamma) =-\sin (45^{\circ}-\gamma)$ and similarly for $\cos XIZ$, so this is just $IY^2 + 2IX\cdot IY \sin (45^{\circ} - \gamma) = IZ^2 + 2IX\cdot IZ \sin (45^{\circ} - \beta)$. We'll move terms around to get $\cos^2 \beta - \cos^2 \gamma = 2 \cos \alpha [ \cos \gamma \sin (45^{\circ} - \beta) - \cos \beta \sin (45^{\circ} -\gamma)]$. By well-known identities the left side equals $\sin^2 \gamma -\sin^2 \beta = \sin (\gamma - \beta) \sin (\gamma + \beta)$. Meanwhile, by product-to-sum we have $2\cos \gamma \sin (45^{\circ} - \beta)= \sin ( 45^{\circ} - \beta - \gamma) + \sin (45^{\circ} - \beta +\gamma )$ and $2\cos \beta \sin (45^{\circ} - \gamma) = \sin (45^{\circ} - \gamma-\beta ) + \sin (45^{\circ} - \gamma + \beta)$, so subtracting gives that the right side of the previous equation equals $\cos \alpha [ \sin (45^{\circ} - \beta + \gamma) - \sin (45^{\circ} - \gamma + \beta)] = \sqrt{2} \cos \alpha \sin (\gamma - \beta)$. Equating everything yields $\sin (\gamma - \beta) \sin (\gamma + \beta) = \sqrt{2} \cos \alpha \sin (\gamma - \beta)$, so either $\beta = \gamma$ or $\sin (\gamma + \beta) = \sqrt{2} \cos \alpha$. Since $\sqrt{2} \cos \alpha >1$ and $\sin (\gamma + \beta) <1$ this is not possible, so we must have $\beta=\gamma \implies \angle B=\angle C$. By symmetry $ABC$ is equilateral and we're done.

The trig bash isn't so nasty and it's actually quite brainless. Let $\alpha = \frac{\angle A}{4}$ and define $\beta, \gamma$ similarly. By law of sines in $\triangle AIZ$ and $\triangle AIY$ we get $IZ = \frac{AI \sin\alpha}{\cos\beta}$ and $IY = \frac{AI\sin\alpha}{\cos\gamma}$, so $\frac{IZ}{IY} = \frac{\cos\gamma}{\cos\beta}$. By scaling properly we can WLOG assume $IZ = \cos\gamma$ and so on. We use law of cosines to get $YZ^2 = \cos^2\beta + \cos^2\gamma - 2\cos\beta\cos\gamma\cos(135^{\circ} - \alpha)$. Equating $YZ^2$ and $XY^2$, for example, produces $$\cos^2\gamma - \cos^2\alpha = 2\cos\beta(\cos\gamma\cos(135^{\circ} - \alpha) - \cos\alpha\cos(135^{\circ} - \gamma)).$$Then because $\alpha+\beta+\gamma = 45^{\circ}$, the right hand side is equal to \begin{align*} &\ 2\cos\beta(\cos\gamma\cos(135^{\circ} - \alpha) - \cos\alpha\cos(135^{\circ} - \gamma))\\ =&\ 2\cos\beta(\cos\gamma\cos(90^{\circ} + \beta+\gamma) - \cos\alpha\cos(90^{\circ} + \beta + \alpha))\\ =&\ \cos\beta((\sin\beta-\sin(\beta+2\gamma))-(\sin\beta - \sin(\beta+2\alpha)))\\ =&\ \cos\beta(\sin(\beta+2\alpha)-\sin(\beta+2\gamma))\\ =&\ \sqrt{2}\cos\beta\sin(\alpha-\gamma). \end{align*}The left hand side is \begin{align*} &\ \cos^2\gamma - \cos^2\alpha\\ =&\ \sin(\alpha+\gamma)\sin(\alpha-\gamma)\\ =&\ \sin(45^{\circ} - \beta)\sin(\alpha-\gamma). \end{align*}Therefore if $\alpha\neq \gamma$ then we have $\sin(45^{\circ} - \beta)=\sqrt{2}\cos\beta$, which is clearly impossible because $\beta<45^{\circ}$. We conclude that $\alpha=\beta=\gamma$ and $\triangle ABC$ is equilateral.

In my opinion this problem is really unfit for a G7 position (a G7 should require a minimal amount of ingenuity atleast ) This was really just a straightforward calculation Let $R$ denote the circumradius of $\triangle ABC$ . Denote by $R_X$ and $r_X$ ,the circumradius and inradius of $\triangle BIC$ . Define $R_Y,r_Y,R_Z,r_Z$ similarly. Also set $\alpha = \frac {A}{4}$ , $\beta = \frac {B}{4}$ , $\gamma = \frac {C}{4}$ . Obviously $ \alpha + \beta + \gamma = 45^ {\circ}$ We will use the well known fact that $R_X= 2R \sin (2 \alpha )$ First we compute $r_X$ . We have $$r_X= 4R_X \sin {\frac {\angle IBC}{2}} \sin {\frac {\angle BIC}{2}} \sin {\frac {\angle ICB}{2}} = 4R_X \sin \alpha \sin \gamma \sin (45 + \alpha)=8R \sin {2\alpha} \sin \alpha \sin \gamma \sin (45 + \alpha)$$ Now note that $$IX = \frac {r_X}{\sin (45 + \alpha)}=8R \sin {2\alpha} \sin \alpha \sin \gamma = \mu \cos \alpha $$Where $\mu = 16R \sin \alpha \sin \beta \sin \gamma$ Similarly $IY= \mu \cos \beta$ and $IZ= \mu \cos \gamma$ Now we can WLOG assume that $\mu=1$ We now prove that $YZ=XZ \iff \alpha= \beta \implies CA=CB$ First note that by Law of cosines on $\triangle IYZ$ ,we have $$YZ^2= \cos^2 \beta +\cos^2 \gamma - 2\cos\beta \cos \gamma \cos (135-\alpha)= \cos^2 \beta +\cos^2 \gamma + 2\cos\beta \cos \gamma \sin (45-\alpha)$$Cyclically shifting the variables, we have $$ZX^2= \cos^2 \alpha +\cos^2 \gamma - 2\cos\alpha \cos \gamma \cos (135-\beta)= \cos^2 \alpha +\cos^2 \gamma + 2\cos\alpha \cos \gamma \sin (45-\beta)$$ Hence ,equating $ZX^2=YZ^2$ , we have $$ \cos^2 \beta - \cos^2 \alpha = 2 \cos \gamma [\cos \alpha \sin (45- \beta) - \cos \beta \sin (45-\alpha)]$$$$ \iff \sin (\alpha - \beta ) \sin (\alpha + \beta ) = \frac {2 \cos \gamma}{ \sqrt 2} [ \cos \alpha (\cos \beta -\sin \beta ) - \cos \beta ( \cos \alpha - \sin \alpha) ]$$$$\iff \sin (\alpha - \beta ) \sin (\alpha +\beta ) = \sqrt 2 \cos \gamma \sin (\alpha - \beta)$$ If $\alpha \neq \beta $ , this means $$ \sin (45- \gamma) = \sqrt 2 \cos \gamma$$ However this is impossible for $\gamma < 45^{\circ}$ and we win $\blacksquare$

I speak from the future and the idea fo bash it with complex numbers is the worst thing I ever tried

The world's first synthetic solution to this problem? Has no one come up with this in 25 years? Claime 1. Let $X$ be a point outside the acute-angled triangle $ABC$ such that $\angle BXC = 135^{\circ} - \frac{\angle A}{2}$ and $X$ lie on bisector $\angle A$; $O$ be point outside the triangle $ABC$ such that $BO = CO$ and $\angle BOC = 90^{\circ}$. Then $\angle AXB + \angle CXO = 180^{\circ}$.} Proof: Let $D$ is second intersection point $AX$ and $(ABC)$; $CD$ intersect $BO$ at $B_1, BD$ intersect $CO$ at $C_1$. Note that $\angle BB_1C = \angle BC_1C = 180^{\circ} - 45^{\circ} - \frac{\angle A}{2} = \angle BXC \Rightarrow{} B, C, B_1, C_1, X$ are concyclic. Let $BCX$ intersect $AX$ again at $D'$. Note that $\angle XCO = \angle XBC_1 = \angle XBC - \angle DBC = \angle XD'C - \angle DAC = \angle ACD'$. Similarly $\angle XBO = \angle ABD'$. Then $X$ and $D'$ are isogonally conjugate in $ABCO$. Then well known that $\angle AXB + \angle CXO = 180^{\circ}$. $\square$ Claime 2. Let $ABC$ be a triangle with incenter $I$ and let $I_a$, $I_b$ and $I_c$ be the incenters of the triangles $BIC$, $CIA$ and $AIB$, respectively. Let $I_bI_cEF$ be a square, construct inwardly on the side $I_bI_c$ of triangle $I_aI_bI_c$ and $x$ is center of $I_bI_cEF$. Then $I, X, I_a$ lie on one line. Proof: Note that $AI$ is bisector $\angle I_bAI_c$ and $\angle I_bII_c = \frac{360^{\circ} - 90^{\circ} - \angle A}{2} = 135^{\circ} - \frac{\angle I_bAI_c}{2}$. Then from Claime 1$: \angle XII_b + \angle AII_c = 180^{\circ}$ $\Rightarrow{} \angle AII_c + \angle I_bIC + \angle XIC = 180^{\circ} \Rightarrow{} \angle XIC = \frac{\angle BIC}{2} \Rightarrow{} IX -$ bisector $\angle BIC \Rightarrow{} I, X, I_a$ lie on one line. $\square$ Consequence: $I$ is Inner Vecten point(see $X_{486}$ in ETC) triangle $I_aI_bI_c$. Let's return to the original problem. Let $YZEF$ be a square, construct inwardly on the side $YZ$ of triangle $XYZ$ and $O$ is center of $XYEF$. $XY = XZ$ and $OY = OZ \Rightarrow{} OX$ - perpendicular bisector $YZ$. From lemma $2: X$ lie on $OX \Rightarrow{} IY = IZ$. There are 2 possible cases: a) $\angle II_bA + \angle II_cA = 180^{\circ};$ b) $\Delta AII_b = \Delta AII_c$. a) Note that $\angle II_bA + \angle II_cA = 180^{\circ} + \frac{\angle B + \angle C}{4} > 180^{\circ}$. Therefore this case is impossible. b) $\Delta AII_b = \Delta AII_c \Rightarrow{} \angle AII_b = \angle AII_c \Rightarrow{} 90^{\circ} + \frac{\angle B}{2} = 90^{\circ} + \frac{\angle C}{2} \Rightarrow{} \angle B = \angle C$. Similarity $\angle A = \angle B = \angle C$. It means that $ABC$ is equilateral too. $\square$ Consequence: If $XYZ$ is isosceles then $ABC$ is isosceles too.

home245 wrote: The world's first synthetic solution to this problem? Has no one come up with this in 25 years? Uhh.... 15 years. I think your solution is very neat!

Suppose $XYZ$ is equilateral. We can first prove the triangle with sidelengths $IX,IY,IZ$ has angles $75-\frac A4,75-\frac B4, 75-\frac C4$. This comes from constructing the point $X'$ such that $XIX'\overset{+}{\sim} XYZ$. Then $IX'=IX$ and $XX'Z\cong XIY$, so $X'Z=IY$. The triangle $IX'Z$ has sidelengths $IX,IY,IZ$ and $\angle ZIX'=\angle ZIX-60=\frac{90+A/2}{2}+\frac{90+C/2}{2}-60=75-B/4$, similarly you can compute the other angles of $IX'Z$ to be $75-\frac{A}{4}$ and $75-\frac{C}{4}$. Now note that by LoS in $IXC$ and $IYC$ we have $IX\sin \left(90+\frac{B}{4}\right)=IC\sin \left( \frac C4\right)=IY\sin \left(90+\frac{A}{4}\right)$. Hence $\frac{IX}{\sin \left(90+\frac{A}{4}\right)}$ is the same when you replace $(A,X)$ with $(B,Y)$ and $(C,Z)$. However $\frac{IX}{\sin \left( 75-\frac A4\right)}$ has the same property. Hence $\frac{\sin \left(75-\frac A4\right)}{\sin \left(90+\frac A4\right)}$ is the same when $A$ is replaced by $B$ or $C$. The derivative of $\frac{\sin (75-x)}{\sin (90+x)}$ is zero when $-\sin (90+x)\cos(75-x)=\cos(90+x)\sin (75-x)$, or $\tan (75-x)=\tan (90-x)$. This doesn't happen for $x$ in $(0,45)$, so $\frac{\sin (75-x)}{\sin (90+x)}$ is monotonic in $(0,45)$ and we must have $\frac{A}{4}=\frac{B}{4}=\frac{C}{4}$, as desired.