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, 2015

Geometry Difficulty 5.1 AIME, harder Prove it Argentina

Rectangle $ABCD$ has sides $AB = 3$ , $BC = 2$ . Point $P$ on side $AB$ is such that the bisector of $C\hat{D}P$ passes through the midpoint of $BC$ . Find $BP$ .

Solution

Let $M$ be the midpoint of $BC$ , and let line $DM$ intersect

line $AB$ at $Q$ (it is exterior to the segment $AB$ ). Then $BQM = CDM$ as $AB \parallel CD$ . On the other hand $CDM = PDM$ by hypothesis (DM is the bisector of $CDP$ ). So $PQD = PDC$ and hence $PQ = PD$ . In addition $BQ = CD = 3$ because triangles $BQM$ and $CDM$ are congruent ( $BM = CM$ , $M\hat{B}Q = M\hat{C}D = 90^\circ$ , $BQM = CDM$ ).

Set $BP = x$ . Then $PQ = PB + BQ = x + 3$ and $PD = PQ = x + 3$ . Apply Pythagoras theorem to triangle $PDA$ in which $AP = 3 - x$ , $AD = 2$ , $PD = x + 3$ . This gives
$(3 - x)^2 + 2^2 = (3 + x)^2,$
and we find $x = \frac{1}{3}$ .

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