Problem
GEO-B1-M05-P035 A Median, an Altitude, and a Diameter Circle
In acute triangle \(ABC\), median \(AM\) and altitude \(BH\) are drawn. The line through \(M\) perpendicular to \(AM\) meets ray \(HB\) at point \(K\). Prove that if \(\angle MAC=30^\circ\), then \(AK=BC\).
C. Hint 1. Notice the circle with diameter \(AK\).
D. Hint 2. Compare chord \(HM\) of this circle with the median to the hypotenuse in right triangle \(BHC\).
E. Full solution.
Since \(BH\perp AC\) and \(K\) lies on ray \(HB\), we have \(KH\perp AH\). Also, by construction, \(KM\perp AM\).
Thus \(\angle AHK=\angle AMK=90^\circ\), so points \(A,H,M,K\) lie on one circle. This circle has diameter \(AK\).
Angle \(\angle MAH\) equals \(\angle MAC=30^\circ\), because \(H\in AC\). Hence chord \(HM\), subtending a \(30^\circ\) angle, equals the radius of the circle, so \(HM=\frac{AK}{2}\).
In right triangle \(BHC\), point \(M\) is the midpoint of hypotenuse \(BC\), because \(AM\) is a median of triangle \(ABC\). Therefore \(HM=\frac{BC}{2}\).
We get \(\frac{AK}{2}=HM=\frac{BC}{2}\), hence \(AK=BC\).
A. Source analysis. Main objects: a median, an altitude, a circle with a diameter, and a chord subtending \(30^\circ\). The obvious approach is to express \(AK\) directly, but the hidden circle \(A,H,M,K\) turns \(AK\) into twice chord \(HM\). Number of key ideas: 3.
F. Difficulty justification. This is Level 6: a regional-style problem with one hidden circle and a median to the hypotenuse.
G. Check. This is not a one-step exercise: one must see the cyclicity, use the \(30^\circ\) chord fact, and then move to right triangle \(BHC\).