Yesterday I did a practice British Mathematical Olympiad (BMO) paper, under timed conditions, just to help me prepare for the real one next December. I think I got full, near perfect solutions to all of them, except one. Question 5, which was geometry, my weakest area, confused me.
Let $ABC$ be a triangle with $\angle A < \angle B < 90 ^{\circ} $ and let $ \Gamma$ be the circle through $A$, $B$ and $C$. The tangents to $\Gamma$ at $A$ and $C$ meet at $P$. The line segments $AB$ and $PC$ meet at $Q$. It is given that $[ACP] = [ABC]=[BQC]$ with $[XYZ]$ being the area. Prove that $\angle BCA = 90^{\circ}$
Now my problems understanding this come from the use of "line segment". Does this mean I should extend the lines? Also when I draw it, angle $B$ seems to be obtuse. I can't really understand how to draw it. I ran out of time by the time I'd finished the other five questions, so I only had time to sketch a drawing but I couldn't figure out what it looks like.
Please could you help me draw this and prove it?
Thank you.
Make the following.
Draw $\Delta ABC$ such that $\measuredangle ACB=90^{\circ}$ and $\alpha<\beta$.
Now, draw the circle (it's the circle with a diameter $AB$).
I am sure that you'll see the the line $AB$ and the tangent are intersected.
In the standard notation $S_{\Delta ABC}=S_{\Delta APC}$ gives $$S=\frac{\frac{b}{2}\tan\beta\cdot b}{2}$$ or $$4S=\frac{b^2\cdot\frac{2S}{ac}}{\frac{a^2+c^2-b^2}{2ac}}$$ or $$a^2+c^2=2b^2.$$ Now, $$BQ=AB=c$$ and from here $$QC^2=c\cdot2c,$$ which gives $$QC=\sqrt2c$$ and since $$S_{\Delta ABC}=S_{\Delta BCQ},$$ we obtain: $$2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=2(a^2c^2+2a^2c^2+2c^4)-a^4-c^4-4c^4$$ or $$2b^2(a^2+c^2)-b^4=4a^2c^2$$ or $$(a^2+c^2)^2-4a^2c^2=b^4$$ or $$(a^2-c^2)^2=b^4$$ and since $b>a$, we obtain: $$c^2-a^2=b^2$$ and $$\measuredangle ACB=90^{\circ}.$$