In an equilateral triangle ABC, let P be the orthocenter, that is, the point where all three altitudes meet. What is the measure (in degrees) of angle ∠BPC?

Introduction / Context:
This question focuses on the orthocenter of a triangle and a special property relating the angles formed at that point. In an acute triangle, the orthocenter lies inside the triangle, and there is a well known relationship between angle ∠BHC, where H is the orthocenter, and the angle at the remaining vertex A. For an equilateral triangle all angles are equal, and the notable centres such as centroid, incenter, circumcenter and orthocenter all coincide. We use the general orthocenter angle property and then apply it to the equilateral case.

Given Data / Assumptions:
- ABC is an equilateral triangle, so ∠A = ∠B = ∠C = 60°. - P is the orthocenter of triangle ABC. - All altitudes meet at P. - We need to find ∠BPC in degrees.

Concept / Approach:
A key property in any acute triangle is that if H is the orthocenter, then angle ∠BHC equals 180° − ∠A, where A is the angle at vertex A opposite side BC. This comes from the cyclic quadrilateral formed by the triangle and its orthocenter. In an equilateral triangle, all interior angles are 60°, so applying this property directly yields ∠BHC = 180° − 60° = 120°. Because in an equilateral triangle the orthocenter coincides with the other centres and has full symmetry, the same relation holds with P as orthocenter, giving the desired angle ∠BPC.

Step-by-Step Solution:
Step 1: Recall that in an equilateral triangle, each interior angle is 60°. Step 2: Let P be the orthocenter of triangle ABC, so all three altitudes intersect at P. Step 3: In any acute triangle, a standard theorem states that ∠BHC = 180° − ∠A, where H is the orthocenter. Step 4: In our notation, H is replaced by P, so for triangle ABC we have ∠BPC = 180° − ∠A. Step 5: Since the triangle is equilateral, ∠A = 60°. Step 6: Substitute into the formula: ∠BPC = 180° − 60° = 120°. Step 7: Therefore angle ∠BPC is 120 degrees.

Verification / Alternative check:
To verify the result, consider that in an equilateral triangle, the orthocenter, centroid, incenter and circumcenter all coincide at the same central point. If we draw the altitudes, they also act as medians and angle bisectors. The segments from this central point to the vertices divide the triangle into three smaller congruent isosceles triangles, each having vertex angle 120° at the centre and base angles of 30° at the original vertices. This visual symmetry confirms that the angle between lines PB and PC must be 120°, matching the theoretical result from the orthocenter property.

Why Other Options Are Wrong:
Option 90° would correspond to a right angle at the orthocenter, which does not match the geometry of an equilateral triangle. Option 135° does not have a standard relationship with the interior angle of 60° in an equilateral triangle. Option 145° is a random value that is neither supplementary nor directly related to 60° in this context.

Common Pitfalls:
A common error is to forget the orthocenter angle theorem and treat P simply as the centre of the triangle without considering the specific angle relationships. Another mistake is confusing the orthocenter with the circumcenter or incenter and trying to apply circle theorems incorrectly. Remembering the identity ∠BHC = 180° − ∠A for an acute triangle and then substituting ∠A = 60° is the most efficient and reliable method here.

Final Answer:
The measure of angle ∠BPC is 120°.