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2000 JBMO Problems/Problem 3

Problem 3

A half-circle of diameter $EF$ is placed on the side $BC$ of a triangle $ABC$ and it is tangent to the sides $AB$ and $AC$ in the points $Q$ and $P$ respectively. Prove that the intersection point $K$ between the lines $EP$ and $FQ$ lies on the altitude from $A$ of the triangle $ABC$ .

Solution

We begin by showing that $A$ is the circumcenter of $\triangle KPQ$ :

Consider the configuration where $\angle ACB$ is obtuse and diameter $EF$ lies on extended line $BC$ .

Let $O$ be the midpoint of diameter $EF$ . Let $KA$ meet $BC$ at $G$ .

Let us define $\angle KPA = \alpha$ and $\angle KQA = \beta$

By applying Tangent Chord Angle theorem, we get: $\angle POE = 2\alpha$ and $\angle QOF = 2\beta$

Now, $\angle POQ = 180 - 2(\alpha + \beta)$ , and since $PAQO$ is a cyclic quadrilateral, we have $\angle PAQ = 2(\alpha + \beta)$

Now $\angle POF = 180 - 2\alpha$ , so $\angle PEF = 90 - \alpha$

Similarly, we have $\angle QOE = 180 - 2\beta$ , so $\angle QFE = 90 - \beta$

From $\triangle EKF, \angle EKF = 180 - (\angle PEF + \angle QFE)$

$= 180 - (90 - \alpha + 90 - \beta) = (\alpha + \beta)$

Thus, we have $\angle PKQ = \angle EKF = \angle PAQ/2$

Also, $AP = AQ$ (Since $AP$ and $AQ$ are tangents to the same circle)

From the above 2 results, it readily follows that $A$ is the circumcenter of $\triangle KPQ$ .

Thus, we have $AK = AP$ , and so $\angle AKP = \angle KPA = \alpha$

So in $\triangle EKG, \angle KGE = 180 - (\angle AKP + \angle PEF)$

$= 180 - (\alpha + 90 - \alpha) = 90^{\circ}$

So $KA$ is perpendicular to $BC$ , hence $K$ lies on the altitude from $A$ of the triangle $ABC$ .

$Kris17$

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