Radius in a Triangle

[dtweah]

If a circle of radius r is inscribed in a right triangle whose sides x, y, and z take on only integral values, which one of the following can be a radius of the circle?


A. 1

B. 2^.5

C. (x+y+z)/2^.5

D. (x+y+z) 2^.5

E. It cannot be determined

[kanha81]

Is OA [spoiler][A][/spoiler]?

This is how I tried to solve:
Sides x, y, z are integral values:
x-y-z = (5-12-13) or (3-4-5) or (others)

a) 5-12-13:
radius of incircle of right angle triangle
= (sum of perp. sides - hyp) / 2
= (17 - 13) / 2
= 4/2
= 2 Not one of the answers

b) 3-4-5:
radius of incircle of right angle triangle
= (7 - 5) / 2
= 2/2
= 1. One of the answers

Choose [spoiler][A][/spoiler]

[dtweah]

Is OA [spoiler][A][/spoiler]?

This is how I tried to solve:
Sides x, y, z are integral values:
x-y-z = (5-12-13) or (3-4-5) or (others)

a) 5-12-13:
radius of incircle of right angle triangle
= (sum of perp. sides - hyp) / 2
= (17 - 13) / 2
= 4/2
= 2 Not one of the answers

b) 3-4-5:
radius of incircle of right angle triangle
= (7 - 5) / 2
= 2/2
= 1. One of the answers

Choose [spoiler][A][/spoiler]

Cool solution. Theorem: When a circle is inscribed in a pythagorean right triangle the radius of the circle must be an integer. Don't ask me for the proof.

[Uri]

interesting problem and interesting solution! i find two new things here.

kanha81, is the formula u used always valid? i've perhaps never seen this formula before
radius of incircle of right angle triangle
= (sum of perp. sides - hyp) / 2

dtweah, u've also given a new formula. i am afraid that if i keep on getting new formula at this rate, i'll never be even half prepared for the GMAT

[gmat740]

= (sum of perp. sides - hyp) / 2

I have never heard of any such formula.
where did you get this??

Moreover, this method is just not working,otherwise it would have given you the answer with any integral value.

5-12-13:
radius of incircle of right angle triangle
= (sum of perp. sides - hyp) / 2
= (17 - 13) / 2
= 4/2
= 2 Not one of the answers

[Ian Stewart]

I've never seen a real GMAT question where you'd need the formula kanha uses above, but it's certainly correct. You can prove it fairly easily, though it's a bit tricky to explain without a diagram:

-draw the circle inside the triangle, and draw a radius to all three points where the circle touches the triangle;
-notice the triangle's edges are tangent to the circle, so you have a 90 degree angle at each of the three points where the radius connects to a triangle's edge;

Now the stuff that's much easier to see with a diagram:

-when you've drawn the three radii, you've divided up the triangle into a square with sides of length r, and two 'kites' (quadrilaterals which have adjacent sides of equal length);
-If the base has length a, the side of the kite along the base has length a-r;
-if the height has length b, the side of the kite along the height has length b-r;
-notice now that the hypotenuse is made up of one side of the first kite, and one side of the second. So, if c is the length of the hypotenuse, we have:

c = (a - r) + (b - r)
c = a + b - 2r
r = (a + b - c)/2