[Inscribed and Circumscribed Circles]

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Points $A_{1}, B_{1}$, and $C_{1}$ are the feet of the altitudes of triangle $A B C$. It is known that $A_{1} B_{1}=13, B_{1} C_{1}=14, A_{1} C_{1}=15$. Find the area of triangle $A B C$.
Let $H$ be the orthocenter of triangle $ABC$; $A_{2}, B_{2}, C_{2}$ be the points of intersection of the extensions of the altitudes $AA_{1}$, $BB_{1}$, $CC_{1}$ with the circumcircle $\Omega$ of triangle $ABC$. Then $A_{1}, B_{1}, C_{1}$ are the midpoints of segments $HA_{2}, HB_{2}, HC_{2}$ (see problem 55463). Therefore, $A_{1}B_{1}, B_{1}C_{1}, A_{1}C_{1}$ are the midlines of triangles $A_{2}HB_{2}, B_{2}HC_{2}, A_{2}H_{2}$, so the sides of triangle $A_{2}B_{2}C_{2}$ are parallel to the sides of triangle $A_{1}B_{1}C_{1}$. Hence, triangle $A_{2}B_{2}C_{2}$ is similar to triangle $A_{1}B_{1}C_{1}$ with a similarity ratio of 2. Therefore, $A_{2}B_{2}=2A_{1}B_{1}=26$, $B_{2}C_{2}=2B_{1}C_{1}=28$,

$A_{2}C_{2}=2A_{1}C_{1}=30$

Let $O$ be the center of $\Omega$, and $R$ be its radius. Denote $\angle A_{2}B_{2}C_{2}=\angle A_{1}B_{1}C_{1}=\beta$. From triangle $A_{1}B_{1}C_{1}$, by the cosine rule, we find that $\cos \beta=\frac{13^{2}+14^{2}-15^{2}}{2 \cdot 13 \cdot 14}=\frac{5}{13}$.

Thus, $\sin \beta=\frac{12}{13}$, and $R=\frac{A_{2}C_{2}}{2 \sin \beta}=\frac{30}{2 \cdot \frac{12}{13}}=\frac{65}{4}$ (since $\Omega$ is the circumcircle of triangle $A_{2}B_{2}C_{2}$).

The radii $OA_{2}, OB_{2}, OC_{2}$ are perpendicular to the segments $B_{1}C_{1}, A_{1}C_{1}, A_{1}B_{1}$ respectively (see the solution to problem 116334). Therefore,

$S_{ABC}=S_{AB_{1}OC_{1}}+S_{BA_{1}OC_{1}}+S_{CA_{1}OB_{1}}=\frac{1}{2}B_{1}C_{1} \cdot OA+\frac{1}{2}A_{1}C_{1} \cdot OB+\frac{1}{2}A_{1}B_{1} \cdot OC=\frac{1}{2}(B_{1}C_{1}+A_{1}C_{1}+A_{1}B_{1})R=\frac{1}{2}(14+15+16) \cdot \frac{64}{4}=\frac{1365}{4}$

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Author: Bogoslov I.I.

Does there exist a convex $N$-sided polygon, all sides of which are equal, and all vertices of which lie on the parabola $y=x^{2}$, if

a) $N=2011$;

b) $N=2012$?

## Solution

a) Let $O$ be the vertex of the parabola. Lay off 1005 equal chords $OA_{1}, A_{1}A_{2}, A_{2}A_{3}, \ldots, A_{1004}A_{1005}$ of length $t$ on the right branch. Consider the broken line $OB_{1}B_{2} \ldots B_{1005}$, symmetric to $OA_{1}A_{2} \ldots A_{1005}$ with respect to the axis of the parabola. Clearly, the length $l(t)$ of the segment $B_{1005}A_{1005}$ continuously depends on $t$. For $t=\sqrt{2}$, $B_{1}A_{1}=2>t$, and thus $l(t)>t$. For $t=4020$, the ordinate of point $A_{1005}$ is less than $1005 \cdot 4020=2010^{2}$, so its abscissa is less than 2010, and $l(t)<t$. Therefore, there exists a value of $t$ such that $l(t)=t$. The polygon $A_{1}A_{2} \ldots A_{1005}B_{1005} \ldots B_{1}$ is the required one.

b) Suppose there exists a polygon $A_{1}A_{2} \ldots A_{2012}$ with the required properties. Let $A_{1006}A_{1007}$ be the side with the largest ordinate. Then the sides $A_{1}A_{2}, A_{2}A_{3}, \ldots, A_{1005}A_{1006}$ lie on the right branch of the parabola, and the sides $A_{1007}A_{1008}, A_{1008}A_{1009}, \ldots, A_{2011}A_{2012}$ lie on the left branch. The side $A_{1006}A_{1007}$ is the highest, so the sides $A_{1005}A_{1006}$ and $A_{1007}A_{1008}$ are lower. Therefore, $A_{1005}A_{1006}<A_{1006}A_{1007}$ and $A_{1007}A_{1008}<A_{1006}A_{1007}$. This contradicts the fact that all sides are equal.

## Corollary

Let quadrilateral $ABCD$ be inscribed in a parabola, $AD=BC$, and points $A$ and $B$ lie on the arc $CD$ of the parabola. Then $CD>AB$.

Proof. Clearly, quadrilateral $ABCD$ is part of the segment of the parabola cut off by the chord $CD$. Let the tangents to the parabola at points $C$ and $D$ intersect at point $M$. Triangle $CMD$ contains the mentioned segment, and thus the quadrilateral $ABCD$. Therefore,

$\angle BCD + \angle ADC > \angle BCD + \angle BCD = 2\angle BCD > 180^\circ$ (since $\angle BCD < 90^\circ$). This implies that $CD > AB$.

## Answer

a) Exists; b) does not exist.