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May 10th, 2014, 08:40 AM   #1
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A cirlce inscribed inside a right triangle...

This is a 3-part question in my Calculus textbook, one of the challenge problems. I don't even think the part I'm having trouble with even requires calculus, but it's driving me nuts because I can't solve it. Here's the question:

Quote:
Let ABC be a triangle with right angle A and hypotenuse a = BC. If the inscribed circle touches the hypotenuse at D, show that:

$\displaystyle \left | CD \right |=\frac{1}{2}(\left | BC \right |+\left | AC \right |-\left | AB \right |)$
(circle is tangent at D)
C
|\...D
|......\
|_______\B
A

Now matter how I set this thing up, I can't algebraically organize |AB|, |AC|, |BC|, and |CD| in such a way that the previous equation results. Perhaps I lack the foresight? There's no typo here; pretending the triangle in the image was of 3,4,5 proportion gave me |CD|=2 by other means, as well as in the above equation. I think I'm on to something by assuming it is necessary to utilize the formula for the inradius,
$\displaystyle r =\frac{1}{2}(\left | AB \right |+\left | AC \right |-\left | BC \right |)$

Can someone do this and explain what they noticed that allowed them to get the answer? Thanks.

Last edited by chameleojack; May 10th, 2014 at 08:45 AM.
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May 10th, 2014, 08:50 AM   #2
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Solutio....

It is easy.
Let touch points be D, E, F.

$\displaystyle BC=BD+DC. \; AC=CE+EF. \; AB= AF+FB.$

$\displaystyle And \; BD=BF=x, \; CD=CE=y, \; AE=AF=z.$

Therefore, $\displaystyle \frac{1}{2}(BC+AC-AB)=\frac{1}{2}[(x+y)+(z+y)-(x+z)]=\frac{1}{2} \cdot 2y = y.$
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May 10th, 2014, 08:59 AM   #3
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aww! of course! isosceles triangles! I'm embarassed. Thanks for the fresh perspective! I was trying to derive the equation from scratch using Areas and Perimeter. Now I can do the calculus bit =j

Last edited by chameleojack; May 10th, 2014 at 09:08 AM.
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May 10th, 2014, 09:13 AM   #4
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Quote:
Originally Posted by chameleojack View Post
aww! of course! isosceles triangles! I'm embarassed. Thanks for the fresh perspective! Now I can do the calculus bit =j
You should be embarassed more. It has nothing to do with isosceles. The shape does not matter.
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May 10th, 2014, 06:43 PM   #5
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Hey. Smart guy. Look at the picture! The points of tangency connect a chord that makes an isosceles triangle (that's how I was able to visualize the symmetry). You're bending over backward to make fun of me...and failing...for what? If you're only offering help so you can moan about how stupid the rest of us are, why don't go somewhere else?
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May 10th, 2014, 07:08 PM   #6
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Do you think we could call a halt to this discussion?
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May 11th, 2014, 07:29 PM   #7
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I don't moan and I don't mare fun, and I don't care whether you are backward or "forward".
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