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Problem 573

AMC 10/12, early questions

Geometry Difficulty 3.5 Multiple choice Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO · Italy · 2007

In an isosceles triangle $ABC$ with $AC = BC \neq AB$ , fix a point $P$ on the base $AB$ . How many positions can a point $Q$ take in the plane if we want the points $A, P$ and $Q$ , taken in any order, to be the vertices of a triangle similar to $ABC$ ?

Pick one

Official solution

Solution:

The answer is $\mathbf{(E)}$ . The triangle $APQ$ can be constructed so that $AQ = QP$ , or so that $AP = QP$ , or so that $AQ = AP$ . For each of these 3 possibilities, there are two choices for the placement of the point $P$ in symmetric positions with respect to the line through $A$ and $B$ (to which a side must necessarily belong). In total we will therefore have $3 \times 2 = 6$ possible positions for the point $P$ .

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