MHB Prove that the radius of the incircle of △ is rational

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The discussion centers on proving that the radius of the incircle of an isosceles triangle with rational side lengths is rational if and only if the two right triangles formed by the altitude to the base are similar to a right triangle with integer side lengths. Participants emphasize the importance of the triangle's properties and the relationship between rational dimensions and the incircle's radius. The proof hinges on the similarity of the right triangles, linking rationality to integer side lengths. The conversation includes acknowledgment of contributions and solutions presented. The overall focus is on the geometric and algebraic implications of the triangle's dimensions.
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Let $\bigtriangleup$ be an isosceles triangle for which the length of a side and the length of the base are rational. Prove that the radius of the incircle of $\bigtriangleup $ is rational if and only if the two right triangles formed by the altitude to the base are similar to a right triangle with integer side lengths
 
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lfdahl said:
Let $\bigtriangleup$ be an isosceles triangle for which the length of a side and the length of the base are rational. Prove that the radius of the incircle of $\bigtriangleup $ is rational if and only if the two right triangles formed by the altitude to the base are similar to a right triangle with integer side lengths
my solution:
given isosceles triangle $ABC$
$h^2=x^2-\dfrac {y^2}{4}$
$h=\sqrt{\dfrac {4x^2-y^2}{4}}$
$h$ must be a perfect square
$r=\dfrac {yh}{2x+y}$
for $x,y,r$ being rational ,if x,y and h are all integers then we are done
else we can enlarge the original triangle to make them (x,y,h) all integers
so the two right triangles formed by the altitude to the base are similar to a right triangle (the bigger one) with integer side lengths
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Albert said:
my solution:
given isosceles triangle $ABC$
$h^2=x^2-\dfrac {y^2}{4}$
$h=\sqrt{\dfrac {4x^2-y^2}{4}}$
$h$ must be a perfect square
$r=\dfrac {yh}{2x+y}$
for $x,y,r$ being rational ,if x,y and h are all integers then we are done
else we can enlarge the original triangle to make them (x,y,h) all integers
so the two right triangles formed by the altitude to the base are similar to a right triangle (the bigger one) with integer side lengths
Well done, Albert!Thankyou for your participation.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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