$AB=AC\implies\frac{AB}{AC}=1$.

YESMAths wrote: Let $ABC$ be a triangle in which $AB=AC$. Suppose the orthocentre of the triangle lies on the incircle. Find the ratio $AB/AC$. it must be $AB/BC$

Okay, so the answer was $3/4$. Here is my sketch of a solution: Let A1 be the foot of the perpendicular from A. It is shown that A1,A,I,H are collinear. Then, take angle B = 2θ and angle chase to get cos3θ = sinθsin2θ... then it's simple.. And yes, it was $AB/BC$

@YESMAths please remove the "edited thanks", it decreases the post's fitness to be added to the contests section. Edit: OP has been edited.

it was the easiest INMO geometry ever MY SOLUTION = as orthocentre $H$ lie on incircle. thus $AI=AH+HI=AH+r$ now since $AH=2RcosA,AI=Rcosec{A/2},r=4R(sin{A/2})(sin{B/2})^{2}$ and $\angle A=180-2\angle B$. where $R$ is circumradii and $r$ is inradii solving we get $cosB=\frac{2}{3}$ now by cosine law we easily get $AB/BC=3/4$ so we are done

This is essentially a problem from ARML 1983.

It was $5$ min computation, the last years' geo was pretty compared to this.....

Hmm, I still think this is a nice problem. The fact that a triangle with such a weird condition can be decomposed by very simple tools (as I wrote in my solution in the thread jlammy linked to) is actually somewhat interesting. Maybe a bit too easy for INMO, but putting an easy-ish problem in the beginning of a contest is pretty okay as long as the remaining problems aren't easier than normal as a result.

$\tan \hat{\frac{A}{2}} = \tan \widehat{HBD}=\frac{2DI}{BD} \ \text{ (where D is the midpont of BC) , but} \ \tan \hat{\frac{A}{2}} = \frac{r}{AB'}\text{ (where B' is the touchpoint of AB) } \implies \frac{2r}{\frac{BC}{2}}= \frac{r}{\frac{AB+AC-BC}{2}} \implies \frac{AB}{BC}=\frac{3}{4}$

I think it's not a bad problem, just a bit too computational, that's all. My solution: (Usual notations are followed) Notice that if $D$ is the intouch point then $HD=2r$. However, $HD=2R\cos B \cos C$. Now, let $\angle B=\angle C=y$ and $\angle A=180-2y$. Also, for the sake of simplicity, let $AB=AC=1$ and let $BC=x$. We want to compute $\frac{1}{x}$. Now, see that $\cos^2y=\frac{x^2}{4}$ and that $r=R\cos^2y$. Now, $R=\frac{abc}{4\Delta}$ and $r=\frac{\Delta}{s}$ so it is equivalent to, \begin{align*} \frac{x^2}{4}(1.1.x)(1+1+x)=8\Delta^2 \end{align*} Now, we see that $16\Delta^2=(1+1+x)(x)(x)(2-x)$. Solving this gives $x=\frac{4}{3}$. Thus, answer is $\frac{3}{4}$.

Did anyone notice that the touch point of the incircle with the side $BC$ is the Feuerbach's point.........

Viswanath wrote: Did anyone notice that the touch point of the incircle with the side $BC$ is the Feuerbach's point......... That's trivial isn't it?

anantmudgal09 wrote: Viswanath wrote: Did anyone notice that the touch point of the incircle with the side $BC$ is the Feuerbach's point......... That's trivial isn't it? Indeed that gives us a scope of generalizing the problem..........

How about this: In triangle $ABC$, if the Orthocenter, the incenter and the Feuerbach touchpoint are col-linear then the triangle is isosceles. EDIT:- (A conjecture, I don't know if this holds)

anantmudgal09 wrote: How about this: In triangle $ABC$, if the Orthocenter, the incenter and the Feuerbach touchpoint are col-linear then the triangle is isosceles. R us sure...........? since it sounds like a surprise to me..........I was thinking something like non isosceles triangle..........

Wow, I happen to generalize in an obvious manner and djmathman proves it with lightning speed! That's brilliant So we made a really nice problem now.

That's just how I roll *flips hair back dramatically*

This could have made a nice substitute for the INMO P1........

Would have been really hard! This reminds me of another problem #15 Also, another guess (I think it's false) If the orthocenter lies on the incircle then the triangle is isosceles. djmathman wrote: That's just how I roll *flips hair back dramatically* Lol

Well, another boring computations uses the fact that we do have a median............ Let $D$ be the touch point of $BC$ with the incircle, so we have $BI$ the median of $\triangle BHD$, so we have $\frac{sin(B/2)}{sin(90-3B/2)} = \frac{BH}{BD} = \frac{1}{sin(B)}$, this quickly gives us $cos(3B/2) = sin(B)sin(B/2)$, which in turn gives us $cos(B/2) = 3cos(3B/2)$, this yields $cos(B) = \frac{2}{3}$, now the result follows. Although I like the General version more than this...... This trivializes the problem..........I better say the problem is born trivial...........

Computation geo isn't my strength, so a messy solution: Let $D$ be the midpoint of $BC$ and $H$ the orthocenter of $\triangle ABC$. WLOG $AB=AC=1$ and let $BD=CD=x$; we want to find the value of $\dfrac{1}{2x}$. By Pythag $AD = \sqrt{1-x^2}$. Let the inradius be $r$. We easily calculate $r=\dfrac{A}{s} = \dfrac{x\sqrt{1-x^2}}{1+x}$. Note that $$\angle BAD = \angle BCH = \angle CBH \implies \triangle ABD \sim \triangle BHD\implies \dfrac{AD}{BD} = \dfrac{BD}{DH} =\dfrac{BD}{2r}$$$$\implies \dfrac{\sqrt{1-x^2}}{x}=\dfrac{x}{\dfrac{2x\sqrt{1-x^2}}{1+x}}\implies 2-2x^2 = x(1+x)\implies 2-2x=x\implies x=\dfrac{2}{3}$$ Thus, $\dfrac{AB}{BC} = \dfrac{1}{2\cdot \dfrac{2}{3}} = \boxed{\dfrac{4}{3}}$

$D$ intersection of incircle and $BC$. Similar triangles gives $\frac{BD}{DH} = \frac{DA}{CD}.$ Thus let $AB = b, BC = a$; then if $r$ is inradius, then $$\frac{a^2}{4} = 2r \cdot DA.$$But $2r \cdot DA = \frac{2[ABC]}{s} \cdot DA = \frac{2DA^2 \cdot a}{a + 2b}.$ Also $DA^2 = b^2 - \frac{a^2}{4}$. Thus $\frac{a^2}{4} = \frac{2(b^2 - \frac{a^2}{4}) \cdot a}{a + 2b}$. This gives $a(a+2b) = 8b^2 - 2a^2$. Or $3a^2 + 2ab - 8b^2 = 0$. $\frac{b}{a} = r$: $8r^2 - 2r - 3 = 0$. $(4r - 3)(2r + 1) = 0$. $r = \frac{3}{4}$.

bobthesmartypants wrote: Computation geo isn't my strength: Thus, $\dfrac{AB}{BC} = \dfrac{1}{2\cdot \dfrac{2}{3}} = \boxed{\dfrac{4}{3}}$ You are right

Crafty sol without any dirty trig.........I had say this is the most pretty of all...............

@bobthesmartypants, in case you didn't get it, what suli pointed out is that there is a mistake in the last line, it is $AB/BC = 3/4$

Answer is coming 3/4

TheOneYouWant wrote: @bobthesmartypants, in case you didn't get it, what suli pointed out is that there is a mistake in the last line, it is $AB/BC = 3/4$ bobthesmartypants in PM to me wrote: This message is regarding this post. suli wrote: bobthesmartypants wrote: Computation geo isn't my strength: Thus, $\dfrac{AB}{BC} = \dfrac{1}{2\cdot \dfrac{2}{3}} = \boxed{\dfrac{4}{3}}$ You are right I'll leave that there because it's too good to just edit out

Easy enough geometry for INMO. Although what was surprising was that we did the very same problem a day before in our Maths Class!!!

Askr wrote: Easy enough geometry for INMO. Although what was surprising was that we did the very same problem a day before in our Maths Class!!! That wouldn't surprise me......

Guys!! I have a pure geometric solution but the sad thing is that i don't know LATEX

DeepanshuPrasad wrote: Guys!! I have a pure geometric solution but the sad thing is that i don't know LATEX Just put the codes and lengths like AB, etc between dollar signs: $AB$ and for fractions: $x/y$. Waiting for your solution. PS.: Even I have a solution but don't know how many will consider that as pure geometric one!

Pretty amusingly: see here?

AMN300 wrote: Pretty amusingly: see here? This will not surprise us.

Suppose $E$, $D$ are the feet of altitude from $B$ to $AC$ and from $A$ to $BC$ respectively, and the incircle touches the side $AC$ at $K$. Now draw a line passing through $K$ and $I$ which meets $BC$ at $L$ then obviously $KL$ is parallel to $BE$. $I$ is the mid-point of $HD$ so by mid-point theorem $L$ is the mid-point of $BD$, also we have $DC$ equal to $KC$ and $\Delta AKI$ and $\Delta LID$ are congruent. hence, now we can compute and get the result as $AB/BC=3/4$

this question was also in Sharygin book Q.89

Can be bashed easily by trigonometry.

It is well known that $HI^2= 2r^2-4R^2 \cos(A)\cos(B)\cos(C)$ , where $r$ is inradius and $R$ is circumradius of $\triangle{ABC}$. since $H$ lies on incircle of $\triangle{ABC}$ we have $HI^2=r^2$ so we have $r^2=4R^2\cos(A)\cos(B)\cos(C)$ We squeeze our triangle $ABC$ in such a way that $BC=1$. denote $AB=AC=x$ so we have $\left(\frac{\triangle}{s}\right)^2=4\cdot \left(\frac{x^2}{4\triangle}\cos(A)\cdot \cos(B)^2\right)$ since $\cos(B)=\cos(C)$ and $\triangle=[ABC], s=\frac{2x+1}{2}$ now just plugging every expression value we get $(2x-1)^2=2(2x^2-1) \implies x=\frac{3}{4} \implies \frac{AB}{BC}=\boxed{\frac{3}{4}}$ $\blacksquare$

Another trigonometric solution... But, it wasn't bad. Cute problem? Solution: Let $D$ be the foot of perpendicular from $A$ to $BC$. Denote $H$ and $I$ as the orthocenter and incenter of $\triangle ABC$. Since we're dealing with isosceles triangle, $A-H-I-D$ are collinear and thus $H$ and $D$ are antipodes in the incircle. Finally let $b = \overline{AB} = \overline{AC}$ and set $\angle CAD = \angle DAB = \theta$ for acute $\theta$. [asy][asy] import olympiad; size(10cm); defaultpen(fontsize(12pt)); pair A = (0,sqrt(5)); pair B = (-2,0); pair C = (2,0); pair I = incenter(A,B,C); pair H = orthocenter(A,B,C); pair D = foot(H,B,C); markscalefactor = 0.03; draw(anglemark(D,A,C), deepgreen); label("$\theta$",(0,sqrt(5)), dir(295)*5); draw(A--B--C--A, red); draw(A--D, red); draw(H--C, fuchsia); draw(incircle(A,B,C), orange); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, E); dot("$H$", H, dir(150)); dot("$D$", D, S); [/asy][/asy] Alright, now it is well-known that $\overline{CH}= 2R \cos(90^\circ - \theta)$. So, from $\triangle CHD$, we have that \[r = R \sin^2(\theta).\]From here, on we will eliminate $r$ and $R$ and form a trigonometric equation to work with. By using Sine rule, replace $R$ by $\frac{b}{2\sin(90^\circ - \theta)}$. From $r = \Delta/s$, we can quickly get that \[r = \frac{\frac{1}{2} \cdot b^2 \sin(2\theta)}{\frac{2b+2b\sin(\theta)}{2}} = \frac{b\cdot\sin(2\theta)}{2+2\sin(\theta)}\]Putting everything back, we would get that \begin{align*} \frac{b\cdot\sin(2\theta)}{2+2\sin(\theta)} &= \frac{b}{2\cos(\theta)} \cdot \sin^2(\theta) \\ \iff \frac{\cos(\theta)}{1 + \sin(\theta)} &= \frac{\tan(\theta)}{2} \iff \sin(\theta) = \frac{2}{3} \end{align*}To finish it off observe that \[\frac{AB}{BC} = \frac{1}{2} \cdot \frac{AB}{BD} = \frac{1}{2} \csc(\theta) = \boxed{\frac{3}{4}}\]and thus the solution is complete. $\blacksquare$

Indian RMO 2024 P3 is similar.