Which is absolutely correct! Well done. There’s a direct relationship between the radius of the circle and triangle, regardless of dimensions. r = (a + b + c) / 2

I have solved this to get r=20 by finding the equation of the radius crossing the hypotenuse at right angles in terms of r. Then finding where they intersect again in terms of r and finally using pythagoras to get a quadratic in terms of r. Is there a simpler method?

I found all 3 sides of the triangle first, then drew a square from the centre of the circle to the top tangent and right angled point of the triangle. You can then get two simple equations using a combination of r a b and c, then solve for r. Your answer is correct.

A solution that is mostly definitions. Label the points of intersection (chosen to suit the conclusion, above, that r=(a+b+c)/2 using the sides of the triangle):
* D, A, C: on the top horizontal line,
* C, B, E: on the right vertical line,
* F: the center, and
* G: the triangle’s tangent point with the circle
Define:
* a = BC (15 in the example)
* b = AC (8 in the example)
* c = BC (17 in the example)
* r = radius
By “tangents to a circle are congruent”, both:
* BG = BE = r-a
* AG = AD = r-b
Also:
* AG+BG = AB = c
So:
* (r-a)+(r-b)=c
Or 2r=a+b+c

Nice!

I’ve got to 400pi

Which is absolutely correct! Well done. There’s a direct relationship between the radius of the circle and triangle, regardless of dimensions. r = (a + b + c) / 2

Hi

I have solved this to get r=20 by finding the equation of the radius crossing the hypotenuse at right angles in terms of r. Then finding where they intersect again in terms of r and finally using pythagoras to get a quadratic in terms of r. Is there a simpler method?

I found all 3 sides of the triangle first, then drew a square from the centre of the circle to the top tangent and right angled point of the triangle. You can then get two simple equations using a combination of r a b and c, then solve for r. Your answer is correct.

Forgive my stupidity but I can’t visualise where the square is going

I’ll tweet you a pic.

Neat problem. Would you post a full work up of the solution?

Happy to tweet or email you??

A solution that is mostly definitions. Label the points of intersection (chosen to suit the conclusion, above, that r=(a+b+c)/2 using the sides of the triangle):

* D, A, C: on the top horizontal line,

* C, B, E: on the right vertical line,

* F: the center, and

* G: the triangle’s tangent point with the circle

Define:

* a = BC (15 in the example)

* b = AC (8 in the example)

* c = BC (17 in the example)

* r = radius

By “tangents to a circle are congruent”, both:

* BG = BE = r-a

* AG = AD = r-b

Also:

* AG+BG = AB = c

So:

* (r-a)+(r-b)=c

Or 2r=a+b+c

Pardon the typo: define “c = AB (17 in the example)”