Isoceles Triangle in a Trapezium - BMO1 Q3

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Isoceles Triangle in a Trapezium - BMO1 Q3

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Source: British Mathematical Olympiad Round 1 2008 (BMO1 2008)

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#1 h Dec 6, 2008, 3:06 PM • 2 Y

Y by Adventure10, Mango247

Let $ABPC$ be a parallelogram such that $ABC$ is an acute-angled triangle. The circumcircle of triangle $ABC$ meets the line $CP$ again at $Q$ . Prove that $PQ = AC$ if, and only if, $BAC = 60^{\circ}$ . The circumcircle of a triangle is the circle which passes through its vertices.

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#2 h Dec 6, 2008, 3:41 PM • 2 Y

Y by Adventure10, Mango247

if $<CAB=60$ then $<BQP=60$ and we have also $<QPB=60$ so $PQ=PC=AC$ .

now if $PQ=AC=PB$ let $x=<PQB=<QBP$ ,we have $<ABP=x+<QBA=x+180-<ACQ=x+180-<ABP$ so $2<ABP=180+x$ .

and $<ABP+<QPB=180$ so $90+x/2+<QPB=180$ and we know that $2x+<QPB=180$ therefore $180-2x=90-x/2$ so $x=60$

so $<QPB=<CAB=60$ .

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#3 h Nov 29, 2010, 7:56 AM • 1 Y

Y by Adventure10

Clearly $AC=BP$ . Also, $\angle BAC=\angle BQP=\angle BPQ$ so triangle $QBP$ is isosceles. But then $BP=PQ$ if and only if $\triangle BQP$ is equilateral, that is, when $\angle BAC=180^\circ$ .

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