Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Find the answer

1. As shown in Figure 2, in the right triangle $\triangle ABC$ , it is known that $\angle ACB=90^{\circ}, CA$ $=6, P$ is the midpoint of the semicircular arc $\overparen{AC}$ , and $BP$ is connected. The line segment $BP$ divides the figure $APCB$ into two parts. The absolute value of the difference in the areas of these two parts is $\qquad$

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

$=1.9$ .
As shown in Figure 6, let $AC$ intersect $BP$ at point $D$ , and the point symmetric to $D$ with respect to the circle center $O$ is denoted as $E$ . The line segment $BP$ divides the figure $APCB$ into two parts, and the absolute value of the difference in the areas of these two parts is the area of $\triangle BEP$ , which is twice the area of $\triangle BOP$ .
And $S_{\triangle BOP}=\frac{1}{2} PO \cdot CO=\frac{1}{2} \times 3 \times 3=\frac{9}{2}$ ,
Therefore, the absolute value of the difference in the areas of these two parts is 9.
Figure 6

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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.