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Old 09-30-2003, 06:59 PM   #1
Zlex
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This problem is killing my head. I can't figure it out, and I've been looking at it for like 20 minutes.

[i]Triangle PQR is a right angled at Q. PQ = 16cm and QR = 30cm. A circle is drawn inside the triangle, touching all sides. Find the radius of the cirlcle
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Old 09-30-2003, 07:16 PM   #2
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funny you should bring this up, I took the mensa test yesterday on jwa site and had to refresh my memory on finding the hypotenuse of a right triangle...



find the square root of PQ squared + QR squared and find that the hypotenus, or PR is 34

there's a start... if I had some graph paper I could help you better...
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Old 09-30-2003, 07:50 PM   #3
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Nevermind I solved it.

But here's a picture to help you solve it, if you still want to try.


Here's the solution, kinda messy and hard to understand.


Given: PQ = 16, QR = 30 PQ perp. QR, circle inside the triangle
RTF: Radius of that circle

Solution:

- Call points of intersection between circle and triangle: alpha, beta and gamma, intersecting at the lines PQ, QR, PR respectively

- Construct:
Radius from O(origin/centre of circle) to alpha (forms 90 degree angle, tangent to center theorm)
Radius from O to beta
Radius from O to gamma

-Construct:
Bisector of angle P which will fall on the line OP since O is equidistant from the two sides of the angle
Bisector of angle Q falls on line OQ
Bisector of angle R falls on line OR

Proof:
r^2 + (24-z)^2 = a^2 (pyth. theorm)
r^2 + (16 -y)^2 = a^2

r^2 + z^2 = b^2

r^2 + y^2 = c^2
Since <Q = 90 and the angle formed at alpha = 90 and the angle formed at beta = 90 That section must be a rectangle (sum int angles = 360)
Since the line segment from O to alpha and O to beta are equal AND opposite sides of a rectangle are equal we know that all the sides are equal
Therefore:
y = r

So!

If:
r^2 + (24-z)^2 = a^2 (pyth. theorm)
r^2 + (16 -y)^2 = a^2

AND:

z + r = 30
z = 30 - r

Then:

r^2 + (34 -z)^2 = r^2 + (16-y)^2
34^2 -68z + z^2 = 16^2 - 32y + y^2
34^2 - 2040 + 68R + 30^2 - 60R + R^2 = 16^2 - 32R + R^2
16 + 8R = 16^2 - 32R
40R = 240
R = 6

Therefore the radius is 6!

QED
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Old 09-30-2003, 07:53 PM   #4
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nevermind...
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Old 09-30-2003, 07:59 PM   #5
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ok now you have to tell me how you found the value of y and therefore c...


man it's been too long since I've done this stuff
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Old 10-04-2003, 09:26 PM   #6
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Sorry I didn't help, I haven't been on the site for a long time. That special circle is called the "Incircle" and the radius is the "Inradius". The circle has some special properties, and you can study them here:

http://mathworld.wolfram.com/Incircle.html
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Old 10-04-2003, 09:32 PM   #7
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i know the answer! wait...flashback to the math stuff i posted about a year ago... ahh!
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