# Isosceles triangle, angle, area of triangle inscribed in a circle!

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• Apr 26th 2008, 05:08 PM
rawrzjaja
Isosceles triangle, angle, area of triangle inscribed in a circle!

An isosceles triangle is inscribed in a circle of radius R. Determine the value of theta that maximizes the area of the triangle given that theta is the angle contained between the two equal sides.
• Apr 26th 2008, 05:13 PM
Mathstud28
Quote:

Originally Posted by rawrzjaja

An isosceles triangle is inscribed in a circle of radius R. Determine the value of theta that maximizes the area of the triangle give that theta is the angle contained between the two equal sides.

Try thinking that the height of the triangle si equal to 2R..where would you go from there
• Apr 26th 2008, 05:23 PM
rawrzjaja
Hmmmmmmmmmmmmm
well, ive drawn a diagram of the scenario
with the height 2r... i was thinking i split the triangles in symmetrically, making it a right angle triangle... with a height 2r... and that's as far as i've gone T_T
• Apr 26th 2008, 05:31 PM
icemanfan
I'm not sure what suggesting a height of 2r does for solving the problem, since no triangle inscribed in a circle of radius r could have that height. I would try writing the height and length of the base in terms of trigonometric functions and the radius r.
• Apr 26th 2008, 05:40 PM
icemanfan
I just followed my own suggestion. The area of the triangle is given by $(r sin \theta)(r + r cos \theta)$. Can you figure it out from there?
• Apr 26th 2008, 05:43 PM
rawrzjaja
Hmmmmmmmmmmmmm
Can you elaborate on how you got to that stage please... step by step please =o.. i want to understand >.<
• Apr 26th 2008, 05:51 PM
icemanfan
Quote:

Originally Posted by rawrzjaja
Can you elaborate on how you got to that stage please... step by step please =o.. i want to understand >.<

Ok, first let's establish a name for the triangle. I'll call the circumscribed triangle PQR, where Q has angle theta, and the sides PQ and QR are of the same length. The circle is O. So the lengths of OP, OQ, and OR are all equal to the radius r.

This means that OPQ, OQR, and OPR are all isosceles triangles, with OPQ and OQR congruent. The altitude of triangle PQR divides the angle PQR in half, so angles PQO, RQO, OPQ and ORQ are all equal to one-half theta. This means angles POQ and ROQ are equal to $\pi-\theta$.

With me so far?
• Apr 26th 2008, 05:57 PM
icemanfan

Call the point where the altitude of PQR from Q hits PR point S.
Then angles POS and ROS are both equal to theta. From this, the length of PS is $r sin \theta$ and the length of OS is $r cos \theta$. So the length of altitude QS is $r + r cos \theta$. That's where my area formula comes from. It's the length of PS times the length of QS.
• Apr 26th 2008, 06:03 PM
finch41
so when the altitude at Q to the base (line segment PR) is equal to the line segment PR?

where is the point S located? i am a little confused
• Apr 26th 2008, 06:06 PM
rawrzjaja
Hmmmmmmmmmmmmm
Im just having trouble trying to understand where the point O would be? would the point O be in the middle of the circle?
• Apr 26th 2008, 06:08 PM
icemanfan
Quote:

Originally Posted by rawrzjaja
Im just having trouble trying to understand where the point O would be? would the point O be in the middle of the circle?

Point O is the center of the circle and Point S is the end of the altitude from Q, so it is halfway between P and R on the line segment PR.
• Apr 26th 2008, 06:08 PM
finch41
yeah it should be there as far as i can tell
• Apr 26th 2008, 06:15 PM
rawrzjaja
Q contains one angle, and the other point R contains the other equal angle? is that what your saying? cause that's the only way i can say the radius being equal
• Apr 26th 2008, 06:15 PM
finch41
so are you getting the max area to be r^2? as setting theta to be 90? or...
• Apr 26th 2008, 06:16 PM
finch41
P, Q, and R are all on the circle
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