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 Two Digit Diameter Trial (Posted on 2016-07-11)
Consider a circle, where the length of diameter PQ is a 2-digit integer. Reversing the digits of PQ one obtains the length of a perpendicular chord RS.

PQ intersects RS at the point T. The center of the circle is denoted by O and it is known that the length of TO is a positive rational number.

Determine the length PQ.

 No Solution Yet Submitted by K Sengupta No Rating

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 analytic solution Comment 2 of 2 |
At the heart of this problem is the Pythagorean theorem:
(PQ/2)²+(OT)²=(OR)²
or since OR=PQ/2

PQ²-RS²=4OT²=[rational]²

Let PQ=10a+b so RS=10b+a
and
PQ²-RS²=99a²-99b²=11*9*(a²-b²)=[rational]²
So 11 must be a factor of (a²-b²) and any other factor must be a perfect square.
Checking the three possibilities:
(a²-b²)=99 not possible since a<10
(a²-b²)=44 not possible (easy to check)
(a²-b²)=11 yields a=6, b=5
so PQ=65

Checking:
(56/2)²+(33/2)²=(65/2)²
 Posted by Jer on 2016-07-11 17:24:07

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