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Question Number 127788 by peter frank last updated on 02/Jan/21

	Answered by mr W last updated on 02/Jan/21

	$A B = [a b] = 10 a + b$ $C D = [b a] = a + 10 b$ $w i t h 1 ⩽ a, b ⩽ 9$ ${O E}^{2} = \frac{1}{4} ({A B}^{2} - {C D}^{2})$ $\Rightarrow O E = \frac{\sqrt{(A B + C D) (A B - C D)}}{2}$ $= \frac{3 \sqrt{11 (a + b) (a - b)}}{2} = r a t i o n a l$ $\Rightarrow (a + b) (a - b) = 11 n^{2}$ $O E = \frac{3 \times 11 n}{2} = \frac{33 n}{2}$ $c a s e 1 :$ $a + b = 11$ $a - b = n^{2}$ $\Rightarrow a = \frac{11 + n^{2}}{2}, b = \frac{11 - n^{2}}{2}$ $n = 1 : a = 6, b = 5 \Rightarrow O E = \frac{33}{2}$ $n = 3 : a = 10 > 9, b = 1 \Rightarrow b a d$ $c a s e 2 :$ $a + b = n^{2}$ $a - b = 11 \Rightarrow a > 11 > 9 \Rightarrow b a d$ $c a s e 3 :$ $a + b = 11 n$ $a - b = n$ $\Rightarrow a = 6 n, b = 5 n$ $n = 1 : a = 6, b = 5 \Rightarrow O E = \frac{33}{2}$ $n = 2 : a = 12 > 9, b = 10 > 9 \Rightarrow b a d$ $c a s e 4 :$ $a + b = n$ $a - b = 11 n > a + b \Rightarrow b a d$ $c a s e 5 :$ $a + b = 11 n^{2}$ $a - b = 1$ $\Rightarrow a = \frac{11 n^{2} + 1}{2}, b = \frac{11 n^{2} - 1}{2}$ $n = 1 : a = 6, b = 5 \Rightarrow O E = \frac{33}{2}$ $\Rightarrow t h e o n l y s o l u t i o n i s O E = \frac{33}{2} w i t h$ $A B = 65, C D = 56$
	Commented by peter frank last updated on 02/Jan/21

	$thank you . happy new year$
	Commented by mr W last updated on 02/Jan/21

	$t h e s a m e t o y o u!$
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