Question-127788

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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