$\triangle ABC$ is a scalene triangle and $\overline {AM}$ is the median relative to the side $\overline {BC}$. A circumference of diameter $\overline {AM}$ intersects by the second time sides $\overline {AB}$ and $\overline {AC}$ at points P and Q, respectively, both different from A. Assuming that $\overline {PQ}$ is parallel to $\overline {BC}$, find the measure of angle $\angle BAC$.

Background: I'm a 9th grader who has some experience in math contests. This is question 5 (level 2), from the 2013 Brazilian Math Olympic (OBM). The answer was not given, and I believe that it's impossible to find a solution, given that $\triangle ABC$ cannot be scalene in the conditions given by the question, but I'm not sure.

My attempt:

(1)Using the information given in the question and considering that the circumference intersects the sides of the triangle is at point A, I came up with enter image description here

(2)$\overline {CB}$ is tangent to the circle, so: $$\angle AMB=\angle AMC=90°$$

(3)Point M is the midpoint of $\overline {BC}$, so: $$\overline {CM}=\overline {BM}$$

Using (1), (2), and (3), I can conclude that $\triangle ABC$ is isosceles, which contradicts the statement in the question, in which it is stated that $\triangle ABC$ is scalene.

Question: Did I interpret something wrong? Is it impossible to find a scalene triangle that complies with the statement? Any help is appreciated.


Don't let the diagram fool you. It does not follow from the given information that the circle is tangent to side BC at M.

Here is another way to draw the diagram which should give you a hint about how to find the angle:


I have created a Geogebra web page where you may experiment with the triangle.

Move the vertices of the triangle here.

  • $\begingroup$ Initially, I had the wrong link to the Geogebra simulation, but it has been corrected. $\endgroup$ – John Wayland Bales Nov 17 '17 at 22:26

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