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Another geometry problem! Threeeee!

Prove that a triangle with rational sides and rational area can be formed by pasting together two right triangles with the same property.

Note by Vishnu C
1 year, 8 months ago

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Hint: If the area and the sides are rational, then the height is also going to be rational. Drop a perpendicular from A to BC, meeting the side at D. Now you only need to prove that the side is divided into two rational parts Vishnu C · 1 year, 8 months ago

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@Vishnu C If we call A,B,C the three sides of the traingle, and X,Y the two parts of the "devided" side of the triangle witch is C. We will have X^2 + H^2 = A^2 & Y^2 + H^2 = B^2 => X^2 - Y^2 = A^2 - B^2 => X - Y = (A-B)(A+B)/( X+Y) = (A-B)(A+B)/C Or A-B , A+B, C are rational witch means (A-B)*(A+B)/C = X-Y is rational. C = X+ Y is rational => C + X - Y is also rational : C+ X-Y = 2X, 2X rational => X rational : X is rational and C is rational : C - X = Y is also rational. Hence X & Y are both rational. You did the same ? Mohamed Laghlal · 1 year, 8 months ago

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@Vishnu C Nice use of contradiction. +1 Raghav Vaidyanathan · 1 year, 8 months ago

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