The 3 Wisemen

Geometry Level 4

Once upon a time, Ptolemy drew 2 red perpendicular lines of length A C = 56 AC = 56 and B D = 63 BD = 63 and said, "This is a special intersection as the lengths from point E E to other vertices are all in integers."

Brahmagupta then drew 4 blue lines, connecting the vertices, and created a quadrilateral. "This is a unique quadrilateral," he claimed, "as the 4 side lengths are all in integers."

Finally, Parameshvara drew a circle, circumscribing that quadrilateral before declaring, "This is a wonderful circle, for its diameter also has the length in integer."

What is the length of this circumcircle's diameter according to these 3 wisemen?


The answer is 65.

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1 solution

Let A E = a AE = a and B E = b BE = b for some integers a , b a, b . Then C E = 56 a CE = 56 - a and D E = 63 b DE = 63 - b .

According to the chord theorem, A E C E = B E D E AE \cdot CE = BE \cdot DE .

Thus, a ( 56 a ) = b ( 63 b ) a(56 - a) = b(63 - b) __(1)

Then by Ptolemy's Theorem , A C B D = A B C D + A D B C AC \cdot BD = AB \cdot CD + AD \cdot BC

By Pythagorean Theorem, A B = a 2 + b 2 AB = \sqrt{a^2 + b^2} , B C = ( 56 a ) 2 + b 2 BC = \sqrt{(56-a)^2 + b^2} , C D = ( 56 a ) 2 + ( 63 b ) 2 CD = \sqrt{(56-a)^2 + (63-b)^2} , D A = a 2 + ( 63 b ) 2 DA = \sqrt{a^2 + (63-b)^2} .

Thus, 56 63 = ( a 2 + b 2 ) [ ( 56 a ) 2 + ( 63 b ) 2 ] + [ ( 56 a ) 2 + b 2 ] [ a 2 + ( 63 b ) 2 ] 56 \cdot 63 = \sqrt{(a^2 + b^2)[(56-a)^2 + (63-b)^2]} + \sqrt{[(56-a)^2 + b^2][a^2 + (63-b)^2]} _(2)

Solving for the 2 2 equations, we will obtain the side lengths of quadrilateral: 25 , 39 , 60 , 52 25, 39, 60, 52 .

The area of the quadrilateral is 1 2 ( 56 × 63 \dfrac{1}{2}(56\times63 = 1764)

Then by using Parameshvara's formula, we can compute the radius of the circumcircle as:

R = 1 4 × 1764 ( 25 × 39 + 60 × 52 ) + ( 25 × 60 + 39 × 52 ) + ( 25 × 52 + 39 × 60 ) = 65 2 R = \dfrac{1}{4\times 1764}\sqrt{(25\times39 + 60\times 52) +(25\times 60 + 39\times 52) +(25\times 52 + 39\times 60)} = \dfrac{65}{2} .

Therefore, the diameter of the circumcircle equals 65 \boxed{65} .

How stupid of me: didnt look further for that formula so i ended up clicking discuss solution. I came to the integer lengtes so i actually was close. I worked from multiples of the 3-4-5 triangle to find them.

Peter van der Linden - 4 years, 8 months ago

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