| Annie
Shepherd |
I'm not sure how to answer the following problem. A triangle has sides of lengths a, b, and c. It's circumcentre has radius R. Prove the triangle is right angled iff a2 +b2 +c2 =R2 . I am able to show if it is right angled the relationship holds, but not sure how to show it does not hold for a non right angled triangle. Thanks for your help. love annie x |
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| Marcos |
Hi Annie Did you mean a2+b2+c2=8R2 by any chance or is it me being stupid again? If you did mean that then (as you've proved right-angled Þ a2+b2+c2=8R2) let's prove a2+b2+c2=8R2Þ right-angled ... Firstly, let's assume WLOG that c > a, b and q is the angle opposite c. (i.e. if there's a right angle it's going to be q) Now, notice that it suffices to prove the result for when a=b (If you can't see why then fix c and note that q remains constant as you move the point opposite c around the circumcircle) Now, we need to get some equations relating everything together. We can get an expression (using the given equation and the new information we've added) in R, a and q (by using the cosine rule for c2). Now, to get another equation in R, a and q we can find the area of the triangle in two different ways. Then we can equate and use the previous result to eliminate R and a to (eventually but not very messily) give q = p/2. I hope that was clear, ask if anything wasn't... Marcos |
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| David
Loeffler |
Marcos, are you sure you can get away with assuming a = b? It's not obvious that moving point C around the circumcircle doesn't change the value of a2 + b2 + c2 ; indeed if c isn't a right angle, then this is false, so you need to assume the result you are trying to prove. This is a BMO round 1 problem from 2001, isn't it? My approach to it at the time was to use trigonometric addition laws. Start with the sine rule, which states that a/sin A = b/sin B = c/sin C = 2R. So the condition is equivalent to sin2 A + sin2 B + sin2 C = 2 Now, we know that sin2 A = 1/2 (1 - cos 2A) etc, so we get 3 - cos 2A - cos 2B - cos 2C = 4 cos 2A + cos 2B + cos 2C = -1. Now use some addition formulae. cos X + cos Y = 2 cos (X+Y)/2 cos (X-Y)/2, and cos 2C = 2 cos2 C - 1, so the statement is equivalent to 2 cos (A+B) cos (A-B) + (2 cos2 C - 1) = -1 Now cos (A + B) = cos (p - C) = -cos C, so this is -2 cos C ( cos (A-B) + cos (A+B) ) = 0 or -4 cos C cos A cos B = 0 Now we are done, since if cos A cos B cos C = 0, then one of the cosines must be zero, so one of the angles is right. And since this argument is perfectly reversible, we've done the if and the only if in one go. David |
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| Marcos |
David,
What I meant was (probably false, but I can't quite spot why) that given a triangle ABC (with AB the longest side) with its circumcircle then if we let the point C' vary around the circumcircle then the triangle ABC' has the same circumcircle and angle AC'B = angle ACB (for any C' in the same segment as C) Now, after that horrible explanation, if we want to prove a result about angle ACB we can change the problem a bit to proving the result for angle AC'B (where C' is chosen appropriately). In our problem, the "appropriate" C' is chosen so that AC' = C'B (Once the validity of C' being a right-angle is obtained we know that C is a right angle as it's in the same segment as C') Marcos P.S. Of course, in any case, David's argument is simpler on the whole (especially as it takes care of both directions of implication) |
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| David
Loeffler |
Yes, but we are trying to relate angle ACB to a2 + b2 + c2 , which is altered by moving C; in your notation it isn't the same as a'2 + b'2 + c'2 . (If C is a right angle, then wherever that right angle is, the quantity is unaffected; but this is precisely what we are trying to prove.) David |