Track / Stage 4 / 265 of 340 #525 of 1964

Problem 525

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

14. As shown in Figure 6, it is known that quadrilateral $ABCD$ is inscribed in a circle $\odot O$ with a diameter of 3, diagonal $AC$ is the diameter, the intersection point of diagonals $AC$ and $BD$ is $P$ , $AB=BD$ , and $PC=0.6$ . Find the perimeter of quadrilateral $ABCD$ .

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

$1+$ Let the heart of the shadow be क”, so $\mathrm{BH} / \mathrm{C}$ .
From this, $\triangle O P B C \triangle C P D, \frac{C D}{B O}=\frac{C P}{P O}$ ,

which means $\frac{C D}{1.5}=\frac{0.6}{1.5-0.6}$ . Therefore, $C D=1$ .
Thus, $A D=\sqrt{A C^{2}-C D^{2}}=\sqrt{9-1}=2 \sqrt{2}$ .
Also, $O H=\frac{1}{2} C D=\frac{1}{2}$ , so,
$\begin{array}{l} A B=\sqrt{A H^{2}+B H^{2}}=\sqrt{2+4}=\sqrt{6}, \\ B C=\sqrt{A C^{2}-A B^{2}}=\sqrt{9-6}=\sqrt{3} . \end{array}$

Therefore, the perimeter of quadrilateral $A B C D$ is
$1+2 \sqrt{2}+\sqrt{3}+\sqrt{6} \text {. }$

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