Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.0 National olympiad Prove it Japan

Let $ABC$ be an isosceles triangle with $AB = AC = 5$ . Let $D$ be a point on side $AB$ satisfying $AD = 3$ , and let $E$ be a point on side $BC$ (excluding the endpoints $B$ and $C$ ). Let $\omega$ be the circle passing through $E$ and tangent to line $AB$ at $B$ . Suppose that $\omega$ is tangent to the circumcircle of triangle $ADE$ . Let $F$ be the intersection point of $\omega$ and line $AE$ , other than $E$ . When $CF = 10$ , find the length of side $BC$ .

Solution

$\frac{14\sqrt{65}}{13}$

By the alternate segment theorem and the assumption $AB = AC$ , we have $\angle EFB = \angle EBA = \angle ACB$ . Therefore, the four points $A$ , $B$ , $F$ , $C$ are concyclic.

Let $X$ be any point on the tangent to $\omega$ at $E$ , lying on the same side of line $AF$ as $B$ . Then, by the alternate segment theorem, we have
$\angle DEB = \angle DEX + \angle XEB = \angle DAE + \angle EFB = \angle DAE + \angle EBA = \angle CEA.$
Since $\angle DBE = \angle ACE$ , we conclude that triangles $DEB$ and $AEC$ are similar. Therefore, we have $EB : EC = DB : AC = 2 : 5$ . Hence, we may write $EB = 2x$ and $EC = 5x$ with some $x > 0$ .

Since $A$ , $B$ , $F$ , $C$ are concyclic, triangles $EBA$ and $EFC$ are similar. Therefore, we have
$EB : EF = EA : EC = AB : CF = 5 : 10 = 1 : 2,$
and hence, we have $EF = 4x$ and $EA = \frac{5}{2}x$ . By the power of a point theorem, we have
$25 = AB^2 = AE \cdot AF = \frac{5}{2} \cdot \frac{13}{2}x.$

Solving this equation with the condition $x > 0$ , we get $x = \frac{2\sqrt{65}}{13}$ . Thus, we conclude that

BC = BE + EC = 7x = $\frac{14\sqrt{65}}{13}.$

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