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Today Puzzle #645

Pythagoras has drawn a three-four-five triangle. In it he inscribes a circle that just touches the three sides. What is the radius of the circle?

Today’s #PuzzleForToday has been set by Hugh Hunt, Reader in Engineering Dynamics and Vibration at Trinity College, Cambridge

One

AB is the hypotenuse of triangle ABC with sides a (opposite A), b opposite

(B) and c. The circle of radius r, centred at O, touches the sides at D on

AB, E on BC and F on CA.

Given C is a right angle, OECF is a square of side r.

OEB and ODB are both right angles so DOEB is a kite with sides r and a-r

and so we deduce that BD is of length a-r and AD is of length c-a+r.

Likewise AFOD is a kite of sides r and b-r from which we deduce that c-a+r

= b-r

or 2r = a+b-c [eq.1]

And with a,b,c = 3,4,5 we get r=1

It is interesting to note that for all Pythagorean triples (integer values)

the radius is always integer. This is because from a^2 + b^2 = c^2 we get

that either all of a, b and c are even or two of them are odd. This means

from [eq.1] that 2r is even, so r is always an integer.

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