Assume $ \triangle ABC$ is a right triangle where $\angle C= 90°$. Suppose that $ \overline {AX} $ bisects $\angle A$ and point $X$ lies on $BC$. Assume that the circumcircle of triangle $AXB$ intersects $AC$ on $Y$.

a) Show that if lengths $BC$ and $CY$ are intergers divisible by a prime $p$, then $AY$ is also an intenger divisible by $P$.

b) Show that if $CY= k$ and $BC= 3k$ ($k$ is an interger), then the lengths of sides of the triangle $ABC$ must be intergers.

This is what I have done:

Applying power of point of $C$, we get:

$$CY (CA) = (CX)(CB)$$

$$CY(CY+AY)= CB(CB -XB)$$

$$ (CY)^2+ CY(AY) = (CB)^2 - CB(XB)$$

$$CY(AY) = (CB)^2 - (CY)^2 - CB(XB)$$

$$ AY = \frac{(CB)^2 - (CY)^2 - CB(XB)}{CY}$$

But this doesn't prove that $AY$ is divisible by $p$ or that it is an interger.

  • $\begingroup$ Note, there are as many r's in intergers as there are in sherbert. $\endgroup$ – steven gregory May 31 '18 at 23:36
  • $\begingroup$ What work have you done? Where are you stuck? $\endgroup$ – bkarthik May 31 '18 at 23:45
  • $\begingroup$ @bkarthik I have also found some similar triangles but that doesn't help me I think. $\endgroup$ – Vmimi Jun 1 '18 at 21:25

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