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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 $\frac{AP}{AM}.$

 

 Aug 19, 2023
 #1
avatar+128732 
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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)

 

 

cool cool cool

 Aug 19, 2023

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