# Going around incircles

“Did you know,” asked a student at third-hand ((a colleague of @reflectivemaths said so, as reported in Episode 39 of WBU)), “that the in-circle of a 3-4-5 triangle has a radius of 1?”

That’s the kind of thing I’d normally just fire up GeoGebra to check, but I was in the middle of a podcast! The best I could do was check to see if it violated anything I could think of, and I blurted out: it doesn’t violate the equal tangents theorem (if two tangents to the same circle meet at a point, the tangents are equal.)

That’s because the in-circle fits snugly into the right angle. If its radius is 1, the tangents on the ‘3’ side are of length 1 and 2, and those on the ‘4’ side are of length 1 and 3. The tangents on the ‘5’ side must be the (unmatched) 2 and 3, which sum to 5 - so an in-circle radius of 1 is consistent with the theorem.

Surprisingly to me, though, that theorem is enough to prove it! If the radius is $r$, the tangents on the ‘3’ side are $r$ and $3-r$; on the ‘4’ side, they’re $r$ and $4-r$. The unmatched tangents have to sum to 5, so $(3-r) + (4-r) = 5$; solving that gives $r=1$.

That leads to another result: given a right-angled triangle with legs of lengths $a$ and $b$ and hypotenuse $c$, the incircle radius must satisfy $(a-r) + (b-r) = c$, so $r = \frac{a + b - c}{2}$. (It doesn’t work for non-right-angled triangles, because one of the tangent lengths isn’t automatically $r$. I wonder - in an exercise-for-the-reader sort of way - how the result can be extended.)

## A selection of other posts

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