+1348 Equilateral triangle $ABC$ is inscribed in a circle. Let $M$ and $N$ be the midpoints of sides $\overline{AB}$ and $\overline{AC},$ respectively. Line $MN$ intersects the circle at $P$ and $Q.$ Compute $\frac{AP}{AM}.$
+129771 Let the sides of the triangle = 2
Then AM = MB = 1
And MN = 1
Let PM = NQ = x
Angle AMP =120°
By the intersecting chord theorem
AM * MB = PM * MQ
AM * MB = PM ( MN + NQ)
1 * 1 = x ( 1 + x)
1 = x + x^2
x^2 + x - 1 = 0
x^2 + x = 1 complete the square on x
x^2 + x + 1/4 = 1 + 1/4
(x + 1/2) = 5/4 take the positive root of both sides
x + 1/2 = sqrt (5) / 2
x = [ sqrt (5) - 1 ] / 2 = PM = phi
By the Law of Cosines
AP^2 = PM^2 +AM^2 - 2 ( AM * PM) * cos (AMP) {cos AMP = cos 120° = -1/2 }
AP^2 = phi^2 + 1^2 - 2 (1 * phi) (-1/2)
AP^2 = phi^2 + 1 + phi { 1 + phi = Phi }
AP^2 = phi^2 + Phi { phi^2 + Phi = 2 }
AP^2 = 2
AP = sqrt (2)
AP / AM = sqrt (2) / 1 = sqrt (2)
