Multiple choice

Let ABC be an acute-angled triangle. The circle $\Gamma$ with BC as diameter intersects AB and AC again at P and Q, respectively. Determine $\angle$BAC given that the orthocentre of triangle APQ lies on $\Gamma$.

  1. $15^o$
  2. $25^o$
  3. $35^o$
  4. $45^o$
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D Correct answer
Explanation

In this configuration, the orthocenter of APQ lies on the circle with diameter BC if and only if angle BAC = 45 degrees.

AI explanation

Since BC is the diameter of circle Gamma, angle BPC and angle BQC are 90 degrees, making quadrilateral BPQC cyclic. The orthocentre of triangle APQ lies on Gamma, which forces the altitude from P to AQ to pass through this point, meaning PQ must be perpendicular to AQ. This creates triangle APQ where angle AQP is 90 degrees and angle APQ is 45 degrees. Because quadrilateral BPQC is cyclic, opposite angles sum to 180 degrees; thus angle BAC plus angle BPC equals 180 degrees, and since angle BPC is 135 degrees, angle BAC is 45 degrees.

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