Area of Square – Singapore Mathematical Olympiad – 2013 – Problem No.17

This can the very first hint to start this sum:

Assume the length of the side is a.

Now from the given data we can apply Pythagoras’ Theorem :

Since, \(6^2+8^2 = 10^2\)

so \(\angle AXY = 90^\circ\).

From this, we can understand that \(\triangle ABX \) is similar to \(\triangle XCY\)

Try to do the rest of the sum……………………

From the previous hint we find that :

\(\triangle ABX \sim \triangle XCY\)

From this we can find \(\frac {AX}{XY} = \frac {AB}{XC} \)

\(\frac {8}{6} = \frac {a}{a – BX}\)

Can you now solve this equation ?????????????

This is the very last part of this sum :

Solving the equation from last hint we get :

a = 4BX and from this we can compute :

\(8^2 = {AB}^2 +{BX}^2 = {16BX}^2 + {BX}^2 \)

so , \( BX = \frac {8}{\sqrt {17}} and \(a^2 = 16 \times \frac {64}{17} = \frac {1024}{17}\)

Thus m + n = 1041 (Answer).