- anonymous

Is the triangle that this question is speaking of unique, or could there be multiple cases?
"Let O be the circumcenter of ΔABC. Suppose OA and OB are perpendicular to each other. What is the measure of ∠C?

- jamiebookeater

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- anonymous

Solving the problem is easy, but I just am worried that there could be multiple cases, one in which the triangle could be acute and one in which the triangle could be obtuse. I don't think this problem is possible were the triangle to be a right triangle.

- phi

|dw:1437855747306:dw|

- phi

the angle at C is always 1/2 of the intercepted arc
given OA and OB are perpendicular means the central angle is 90 degrees

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- anonymous

Yep, that would be the case for an acute. I was then worried if it were possible, based on how the problem had to be constructed, that you could have C be inside the angle .

- anonymous

|dw:1437855911900:dw|

- phi

oh, yes that is possible

- ikram002p

-.- i lost my graph with cancel xDDDDDDD

- anonymous

Oops, haha. I hate that x_X

- anonymous

And yeah, I thought it might be possible but I wasn't sure. If it's not possible then I don't want to claim multiple cases.

- ikram002p

|dw:1437856026860:dw|

- anonymous

I know the circumcenter is outside the triangle if the triangle is obtuse. But since the circumcenter is the intersection of the perpendicular bisectors, I couldn't conclude whether or not I could still have the perpendicular bisectors intersect while OA still remainded perpendicular to OB.

- phi

the other case has

- ikram002p

the thing is its not unique triangle but a unique angle :D xD

- anonymous

Ah. So there only is one case then?

- ikram002p

it need not to be a bisector though :O

- anonymous

So what are you saying then? xD

- phi

your picture with C forming an obtuse angles meets the given conditions.
O is the circumcenter
OA is 90 deg to OB
so it is valid... < C would be 135 in that case

- ikram002p

u already set two cases xD

- ikram002p

but 90+ obtuse noway

- anonymous

No, I agree, phi. But would O still be the intersection of the perpendicular bisectors under the given conditions?
|dw:1437856512007:dw|
I'm not sure if this picture is possible. I was trying to draw it out on my own and couldnt show that those perpendicular bisectors would still hit the circumcenter O. Sorry if my question isn't clear, lol.

- phi

yes, the perpendicular bisectors of the sides will meet at the circumcenter (but outside of the triangle)

- anonymous

Okay, so I do need two cases. Yeah, I couldnt prove it to myself that I need the answers of both 45 and 135 degrees, wanted to confirm :)

- anonymous

|dw:1437856799729:dw|
Two triangles one acute and one obtuse, but both with the ability to fit the given conditions. That's what I was unsure of. But either way, I'll make sure to mention both cases :3

- phi

yes

- anonymous

Awesome. Thanks

- phi

for what it is worth, here is a figure I made using Geogebra (free and useful tool)
where the radiuses are perpendicular, and thus the third angle is 135 degrees
It shows the perpendicular bisectors of the 3 legs as blue lines that intersect at the circumcenter.

##### 1 Attachment

- anonymous

Ah. Cool. I like that :) Yeah, our class has us using Geometer's Sketchpad, but that's not free :(
I'll check out Geogebra ^_^

- phi

each side of the triangle is a chord of the circumcircle, and the perpendicular bisectors of a chord always goes through the circle's center. in other words, it *has to work*

- anonymous

That's interesting. Yeah, when you say that it definitely clears things up. Thanks again :)

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