$ \triangle ABC$ is a right triangle, $AX$ bisects $\angle A$ and ...

Question

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:

Aplying 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.

I did that considering this figure. I think I may be wrong because that might not be what a circumcircle is.

Answer 1

Since $\angle BAX = \angle XAY$ and $BXYA$ are concyclic, it means that $BX=XY$. Thus

$$ CX = BC - BX \land CX^2 = BX^2 - CY^2 $$ $$ (BC - BX)^2 = BX^2 - CY^2 $$ $$ BC^2 + BX^2 - 2BC\cdot BX = BX^2 - CY^2 $$ $$ BX = \frac{BC^2+CY^2}{2BC} $$ From the power of the point: $$ AC = \frac{BC\cdot BX}{CY} = \frac{BC^2 +CY^2}{2CY} $$ Why is it divisible by $p$?

Answer 2

Hint: Join the points $B$ and $Y$. Then study the triangles $BCY$ and $ACX$, which are similar. If we let $CB=m=pa, CY=n=pb, AY=M, CX=N$, then by we have that $$\frac{BC}{AC}=\frac{CY}{CX}=\frac{BY}{AX}.$$ From there you get two equations in the unknowns $M$ and $N$, which we can solve for $M$, which should tell us what we seek.

For the second part, we now know the sides of $\triangle ABC$, we can study them (after making appropriate substitutions) to see how they are integers given the constraints.

Answer 3

You might find it useful: The power of point equation you wrote is correct. $CY⋅CA=CX⋅CB$ Also, according to the internal bisector theorem: $\frac{CX}{CA}=\frac{BX}{AB}$