Triangle and Circumcircle Problem

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High School Math Grades 9-12, Ages 13-18

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Triangle and Circumcircle Problem

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#1 h Mar 12, 2025, 11:41 PM

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Let $\triangle{ABC}$ be an acute triangle with circumcenter $O$ and orthocenter $H$ . Let $M$ be the reflection of $O$ over $AC$ . Given that $\angle{BAH} = 30^{\circ}$ and the circumradius of $\triangle{ABC}$ is $10$ , the area of triangle $\triangle{AOM}$ can be written as $a\sqrt{b}$ . Find $a+b$ .

This post has been edited 1 time. Last edited by Eternica, Mar 13, 2025, 4:26 PM

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#3 h Mar 13, 2025, 8:26 PM

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Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$ .
Condition 1: $a=\sqrt{3}b$ .
Circumcenter $O(\frac{c-b}{2},\frac{a^{2}-bc}{2a})$ .
Condition 2: $AO=CO=10 \Rightarrow 3b^{2}+c^{2}=300$ .

Equation of the line $AC\ :\ \left|\frac{ax+cy-ac}{\sqrt{a^{2}+c^{2}}}\right|$ ,
distance of the point $O$ to this line $\left|\frac{a \cdot \frac{c-b}{2}+c \cdot \frac{a^{2}-bc}{2a}-ac}{\sqrt{a^{2}+c^{2}}}\right|$ ,
area of $\triangle AMO\ =\ \sqrt{a^{2}+c^{2}} \cdot \left|\frac{a \cdot \frac{c-b}{2}+c \cdot \frac{a^{2}-bc}{2a}-ac}{\sqrt{a^{2}+c^{2}}}\right|$ ,
area $= \frac{b(a^{2}+c^{2})}{2a}=\frac{3b^{2}+c^{2}}{4\sqrt{3}}=25\sqrt{3}$ .

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#4 h Yesterday at 5:33 AM

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Let D be the foot of the perpendicular from O to AC, i.e D is the midpoint of AC.
Since angle BAH is $30^\circ$ , it is easy to see that angle ABC is $60^\circ$ .
From the Inscribed Angle Theorem, angle $AOC$ is $120^\circ$ , which implies that $AOD=60^\circ$ (since triangles ADO and CDO are congruent)
Using the fact that $AO=10$ , it is easy to see that $OD=5$ and $AD=5\sqrt{3}$ which means that the area of AOM, which is twice the area of AOD, is $25\sqrt{3}$ . So our answer is 28.

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