Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Find the answer

6. As shown in Figure $2, \odot O$ is the circumcircle of square $ABCD$ , with $O$ as the center. Point $P$ is on the minor arc $AB$ , and $DP$ intersects $AO$ at point $Q$ . If $PQ=QO$ , then $\frac{QC}{AQ}$ equals:

Pick one

Solution

6. B.

As shown in Figure 7, let's assume $A O=$ $O C=1$ , then $A D=C D=\sqrt{2}$ .
Let $Q O=Q P=x$ , then
$\begin{array}{l} A Q=1-x, \\ Q C=1+x . \end{array}$

By the intersecting chords theorem, we get
$\begin{array}{l} D Q=\frac{A Q \cdot Q C}{Q P} \\ =\frac{1-x^{2}}{x} . \end{array}$

Construct $Q H \perp A D, Q K \perp C D$ , with $H$ and $K$ being the feet of the perpendiculars, then $Q H=A H=\frac{1-x}{\sqrt{2}}, D H=Q K=C K=\frac{1+x}{\sqrt{2}}$ . By the Pythagorean theorem, $Q H^{2}+H D^{2}=D Q^{2}$ , i.e., $\left(\frac{1-x}{\sqrt{2}}\right)^{2}+\left(\frac{1+x}{\sqrt{2}}\right)^{2}=\left(\frac{1-x^{2}}{x}\right)^{2}$ .

Therefore, $x^{2}=\frac{1}{3} \Rightarrow x=\frac{1}{\sqrt{3}}$ .
Thus, $\frac{Q C}{A Q}=\frac{1+x}{1-x}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.