Prove that IMO is isosceles

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Prove that IMO is isosceles

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Source: 2024 Brazil Ibero TST P2

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#1 h Oct 12, 2024, 6:10 PM

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Let $ABC$ be an acute-angled scalene triangle with circumcenter $O$ . Denote by $M$ , $N$ , and $P$ the midpoints of sides $BC$ , $CA$ , and $AB$ , respectively. Let $\omega$ be the circle passing through $A$ and tangent to $OM$ at $O$ . The circle $\omega$ intersects $AB$ and $AC$ at points $E$ and $F$ , respectively (where $E$ and $F$ are distinct from $A$ ). Let $I$ be the midpoint of segment $EF$ , and let $K$ be the intersection of lines $EF$ and $NP$ . Prove that $AO = 2IK$ and that triangle $IMO$ is isosceles.

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DottedCaculator

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DottedCaculator

#2 h Oct 13, 2024, 2:16 AM • 2 Y

Y by cursed_tangent1434, centslordm

Solved with some dihydrogen monoxide

Since $AFEO$ and $APNO$ are cyclic, $O$ is the Miquel point of $EFNP$ , so $PKOE$ and $FKON$ are cyclic. Now, $\angle EFN=\angle EFO+\angle OFN=90^\circ-\angle ACB+180^\circ-\angle AOM=90^\circ-2\angle ACB+\angle ABC$ . Therefore, $\angle EFN+\angle OAC=180^\circ-2\angle ACB$ , so the angle bisector of $EF$ and $AO$ is perpendicular to $BC$ . This means that if a homothety at $M$ with ratio $2$ maps $I$ to $I'$ and $K$ to $K'$ , $AK'\parallel BC$ , and if $IM=IO$ , then $OI'\parallel BC$ , so $AO=I'K'=2IK$ . Therefore, it suffices to show $IM=IO$ .

Let $H$ be the orthocenter of $APN$ , and let $D$ be the reflection of $O$ over $I$ . Then, since $EDFO$ and $PHNO$ are parallelograms, $\triangle DHO\sim\triangle FNO$ . Therefore, $\angle HDO=\angle NFO=\angle ABC-\angle ACB=\angle HAO=\angle HMO$ , so $DHMO$ is cyclic. Since $\angle DHO=\angle FNO=90^\circ$ , $D$ lies on $BC$ , so $I$ lies on the perpendicular bisector of $MO$ , which means $IM=IO$ .

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cursed_tangent1434

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cursed_tangent1434

#3 h Oct 14, 2024, 4:08 AM • 1 Y

Y by InterLoop

Very cool problem. It took me a while to realize how $M$ came into the picture. We let $H_A$ denote the reflection of the orthocenter of $\triangle ABC$ over side $BC$ (the intersection of the $A-$ altitude with $(ABC)$ ). We start off with the following minor observations.

Claim : Quadrilaterals $ONFK$ and $NKPE$ are cyclic.
Proof : This is an easy angle chase. Simply note that,
$\measuredangle OFK = \measuredangle OFE = \measuredangle OAE = \measuredangle OAP = \measuredangle ONP = \measuredangle ONK$ which implies that $ONFK$ is indeed cyclic. The proof that $NKPE$ is cyclic is entirely similar.

Now, we are in a position to attack the first part of the problem, via the following key claim.

Claim : Point $H_A$ lies on $(AFOE)$ and in particular, $FOH_AE$ is an isosceles trapezoid.
Proof : Note that,
$\measuredangle (\overline{OM},\overline{AO}) = \measuredangle AOM = 2\measuredangle ACB + \measuredangle BAC = \measuredangle ABC + \measuredangle ACB = \measuredangle OAH_A = \measuredangle AH_AO$ which implies that $H_A$ also lies on $\omega$ as claimed. Further note that,
$\measuredangle FH_AO = \measuredangle FAO = \measuredangle CAO = \measuredangle H_AAB = \measuredangle H_AAE = \measuredangle H_AFE$ which implies that $OH_A \parallel EF$ and $FOH_AE$ is an isosceles trapezoid as desired.

Now, note that since $I$ is the midpoint of $EF$ , it lies on the perpendicular bisector of segment $EF$ . Since we showed that $EFOH_A$ is an isoceles trapezoid this implies that $I$ also lies on the perpendicular bisector of $OH_A$ so $IO=IH_A$ . Now, note that,
$\measuredangle OKE = \measuredangle OPE = \frac{\pi}{2}$ So, $OK \perp EF$ . Now, consider the point $R$ such that $OKRH_A$ is a rectangle. $R$ must clearly lie on $\overline{EF}$ as a result of our previous observations. Since $I$ lies on the perpendicular bisector of segment $OH_A$ it must also lie on the perpendicular bisector of segment $KR$ . Thus,
$2IK = RK = OH_A=AO$ which finishes the first part of the problem.

Next, let $T$ be the intersection of the tangent to $(ABC)$ at $H_A$ and $\overline{BC}$ . We first prove the following claim.

Claim : Points $E$ and $F$ are respectively the centers of circles $(BH_AT)$ and $(CH_AT)$ .
Proof : We first note that,
$\measuredangle EH_AB = \measuredangle EH_AO + \measuredangle OH_AB = \measuredangle EAO + \measuredangle OH_AB = \measuredangle BAO + \measuredangle OH_AB = \frac{\pi}{2} + \measuredangle BCA + \measuredangle CBA = \measuredangle H_ABA = \measuredangle H_ABE$ which implies that $EB=EH_A$ . Further note that,
$\measuredangle BTH_A = \frac{\pi}{2}+\measuredangle AH_AT = \measuredangle ABC + \measuredangle ACB$ So, $2\measuredangle BTH_A = \measuredangle BEH_A$ which implies that $T$ also lies on the circle through $B$ and $H_A$ with center $E$ proving the claim. The proof that $F$ is the center of circle $(CH_AT)$ is similar.

Now, we simply need to piece everything together. Note that,
$TE = EH_A = OF \text{ and } TF = FH_A = OE$ which implies that $TEOF$ is a parallelogram. Thus, $T$ is the reflection of $O$ across $I$ , which is simply the $O$ -antipodal point in the circle through $H_A$ and $O$ with center $I$ . Thus, $T$ lies on this circle as well. To finish off simply notice that,
$\measuredangle MOH_A = \measuredangle ABC + \measuredangle ACB = \measuredangle MTH_A$ so $M$ also lies on $(OH_AT)$ which implies that $IO=IM$ and indeed $\triangle IMO$ is isosceles as required.

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SomeonesPenguin

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SomeonesPenguin

#4 h Apr 8, 2025, 7:53 AM

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This takes about 15 minutes to complex bash. Set $(ABC)$ as the unit circle. Let $X$ be the center of $\omega$ . We have $\frac{x}{b-c}\in\mathbb R\iff \overline{x}=-\frac{x}{bc}$ $|x-a|=|x|\iff a\overline{x}+\frac{1}{a}x=1\iff x=\frac{abc}{bc-a^2}$ $E\in AB\iff \overline{e}=\frac{a+b-e}{ab}$ $|x-e|=|x|\iff \left(e-\frac{abc}{bc-a^2}\right)\left(\frac{a+b-e}{ab}-\frac{a}{a^2-bc}\right)-\frac{abc}{bc-a}\cdot\frac{a}{a^2-bc}=0$ Note that $a$ is a root of the above equation so it easily factors as $\frac{(a-e)(a^2e-bce+abc+b^2c)}{ab(a^2-bc)}=0\iff e=\frac{bc(a+b)}{bc-a^2}$ Using symmetry we get $i=\frac{bc(2a+b+c)}{2bc-2a^2}$ Now, for $\triangle IMO$ to be isosceles, it would suffice to show $\frac{i-\frac{b+c}{4}}{b-c}\in\mathbb R\iff \frac{\frac{2bc(2a+b+c)}{bc-a^2}-(b+c)}{b-c}\in\mathbb R$ $\iff \frac{4abc+b^2c+bc^2+ba^2+ca^2}{(b-c)(bc-a^2)}=\frac{\frac{4}{abc}+\frac{1}{b^2c}+\frac{1}{bc^2}+\frac{1}{ba^2}+\frac{1}{ca^2}}{\left(\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{bc}-\frac{1}{a^2}\right)}$ This is clearly true (just multiply the numerator and denominator of the second fraction by $(abc)^2$ ). We now use the points $T=OX\cap EF$ , $S=BC\cap EF$ and $R\in EF$ such that $AR\parallel BC$ . Since $\triangle IMO$ is isosceles, we have that $I$ is the midpoint of $TS$ and its clear that $K$ is the midpoint of $SR$ , hence $IK=TR/2$ . Since $ATOR$ is a trapezoid, it suffices to prove it is a isosceles trapezoid. This is equivalent to $\measuredangle(AO,BC)=\measuredangle(FE,BC)$ , or $\frac{a}{b-c}\cdot\frac{\frac{bc(b-c)}{bc-a^2}}{b-c}\in\mathbb R\iff \frac{abc}{(b-c)(bc-a^2)}\in\mathbb R$ Which is clear. $\blacksquare$
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This post has been edited 2 times. Last edited by SomeonesPenguin, Apr 8, 2025, 4:26 PM

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