Let $ABC$ be an acute-angled, nonisosceles triangle. Altitudes $AA'$ and $BB' $meet at point $H$, and the medians of triangle $AHB$ meet at point $M$. Line $CM$ bisects segment $A'B'$. Find angle $C$. (D. Krekov)
Problem
Source: Sharygin Geometry Olympiad 2015 Final 9.7
Tags: geometry, angle, orthocenter, Centroid
hukilau17
01.08.2018 17:21
If we inscribe $ABC$ in the unit circle, we obtain
$$|a|=|b|=|c|=1$$$$a' = \frac12\left(a+b+c-\frac{bc}a\right)$$$$b' = \frac12\left(a+b+c-\frac{ac}b\right)$$$$h = a+b+c$$$$m = \frac{a+h+b}3 = \frac{2a+2b+c}3$$The midpoint of $A'B'$, which we call $N$, has coordinate
$$n = \frac{a'+b'}2 = \frac14\left(2a+2b+2c-\frac{bc}a-\frac{ac}b\right)$$We know that $CMN$ is collinear, that is,
$$\frac{n-c}{m-c}\in\mathbb{R}$$$$\frac{2a+2b+2c-\frac{bc}a-\frac{ac}b-4c}{2a+2b-2c}\in\mathbb{R}$$(Note that $a+b\neq c$, since otherwise we have $|a-c|=|b-c|$.)
$$\frac{(a+b)(ac+bc-2ab)}{ab(a+b-c)} = \overline{\left(\frac{(a+b)(ac+bc-2ab)}{ab(a+b-c)}\right)}$$$$\frac{(a+b)(ac+bc-2ab)}{ab(a+b-c)} = \frac{(a+b)(a+b-2c)}{ac+bc-ab}$$$$(ac+bc-2ab)(ac+bc-ab) = ab(a+b-2c)(a+b-c)$$$$a^2c^2+b^2c^2+2a^2b^2+2abc^2-3a^2bc-3ab^2c = a^3b+ab^3+2abc^2+2a^2b^2-3a^2bc-3ab^2c$$$$a^2c^2+b^2c^2 = a^3b+ab^3$$$$(a^2+b^2)(c^2-ab) = 0$$If $c^2=ab$, then $|a-c|=|c(a-c)|=|ac-ab|=|b-c|$, which is not allowed since $ABC$ is scalene. Thus $a^2+b^2=0$. Therefore, $\frac{b}{a}=\pm i$, and so arc $AB$ measures $90^{\circ}$. Inscribed angle $\angle{ACB}$ is half of that, or $\boxed{45^{\circ}}$. $\blacksquare$
math_pi_rate
07.08.2018 22:19
A synthetic solution: Let $O$ and $O_1$ be the centers of $\odot (ABC)$ and $\odot (AHB)$. Let $K,P,T$ be the midpoints of $AB,CH,A'B'$. Also, let $N$ be the nine point center of $\triangle ABC$. Note that $M$ lies on $CO_1$, as it is the Euler line of $\triangle AHB$. Also, it's easy to get that $K$ is also the midpoint of $OO_1$. Also, it is well-known that $T$ lies on $PK$. According to the question, $CO_1$ also passes through $T$. Now, As $CH \parallel OO_1$, by homothety, we get that $OH$ also passes through $T$. But, as $NT$ is perpendicular to $A'B'$, and both $N,T$ lie on $OH$, either $OH$ coincides with the perpendicular bisector of $A'B'$, or $N=T$. If $OH$ coincides with the perpendicular bisector of $A'B'$, then $A'B' \parallel AB$. But $A'B'$ is antiparallel to $AB$. This means that $\triangle ABC$ is isosceles, which is not true. If $N=T$, then $A'B'$ is a diameter of the nine point circle of $\triangle ABC$, which means that $\angle A'PB' = 90^{\circ} \Rightarrow \angle A'CB' = 45^{\circ} \Rightarrow \angle C = 45^{\circ}$.
w0w
22.11.2018 02:28
By the hypothesis, we can deduce line $CM$ contains the orthocenter $O$ of $\triangle HA'B'$. Since $\triangle HAB \sim \triangle HB'A'$, $OH$ contains the circumcenter $O'$ of $\triangle HAB$, which lies on $CM$ by the Euler Line. So $O=O'$. Noting that $A'OB'C$ is a parallelogram, we have $A'O\parallel AC \perp BH$ so $A'$ lies on the perpendicular bisector of $BH$. Thus $ \angle A'BH = \angle BHA'$ and $\angle HA'B=90^\circ$, so $\angle C = \angle BHA' =45^\circ$ as desired.