2002 Indonesia MO Problems/Problem 4

Given a triangle $ABC$ with $AC > BC$ . On the circumcircle of triangle $ABC$ there exists point $D$ , which is the midpoint of arc $AB$ that contains $C$ . Let $E$ be a point on $AC$ such that $DE$ is perpendicular to $AC$ . Prove that $AE = EC + CB$ .

Solution

We use the method of phantom points.

Draw $AC$ and $BC$ , and extend line $AC$ past $C$ to a point $B'$ such that $BC = B'C$ . Draw point $E'$ at the midpoint of $AB'$ , and $D'$ at the intersection of the perpendicular to $AC$ from $E'$ and the perpendicular bisector of $AB$ .

Since $D'B = D'B', CB = CB', D'C = D'C$ , we have $\triangle D'BC \cong \triangle D'B'C$ by side-side-side similarity. Then $\angle D'AC = \angle D'AE' = \angle D'B'E' = \angle D'B'C = \angle D'BC$ , so $ADCB$ is cyclic.

In particular, since we have $AD' = B'D' = BD'$ , we know that $D'$ must be the midpoint of the arc of the circumcircle of $\triangle ABC$ that contains point $C$ , and since $D'$ was on the perpendicular to $AC$ from $E'$ , we must have that $E'$ is the foot of the perpendicular of $D'$ to $AC$ . But this uniquely identifies $D' = D, E' = E$ , and we are done.

2002 Indonesia MO (Problems)
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 5
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2002 Indonesia MO Problems/Problem 4

Solution

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