Library / /996 of 1394

, 2023

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Acute triangle $ABC$ has circumcenter $O$ . The bisector of $\angle ABC$ and the altitude from $C$ to side $AB$ intersect at $X$ . Suppose that there is a circle passing through $B$ , $O$ , $X$ , and $C$ . If $\angle BAC = n^{\circ}$ , where $n$ is a positive integer, compute the largest possible value of $n$ .

Solution

Solution:

We have $\angle XBC = B/2$ and $\angle XCB = 90^{\circ} - B$ . Thus, $\angle BXC = 90^{\circ} + B/2$ . We have $\angle BOC = 2A$ , so
$90^{\circ} + B/2 = 2A$
This gives $B = 4A - 180^{\circ}$ , which gives $C = 360^{\circ} - 5A$ .

In order for $0^{\circ} < B < 90^{\circ}$ , we need $45^{\circ} < A < 67.5^{\circ}$ . In order for $0^{\circ} < C < 90^{\circ}$ , we require $54^{\circ} < A < 72^{\circ}$ . The largest integer value in degrees satisfying these inequalities is $A = 67^{\circ}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.