Shadow Posted April 10, 2009 Share Posted April 10, 2009 Hey all, so, I was at this math contest last Tuesday. The contest consisted of 4 question, of which one was geometry. I had absolutely no clue on how to solve it, so I was wondering if anybody here could help me. We have a triangle [math]ABC[/math], which is not isosceles. Let [math]x[/math] be a line that bisects the angle [math]ACB[/math], [math]y[/math] be the perpendicular bisector of [math]AB[/math], [math]h_a[/math] be a line perpendicular to [math]BC[/math] and passing through [math]A[/math] and finally [math]h_b[/math] be a line perpendicular to [math]AC[/math] and passing through [math]B[/math]. Let [math]K \in x \cap y[/math], [math]P \in KC \cap h_a[/math] and [math]Q \in KC \cap h_b[/math]. Let [math]A_{AKP} = A_{BKQ}[/math], where [math]A_{XYZ}[/math] denotes the area of the triangle [math]XYZ[/math]. Determine the size of the angle [math]ACB[/math]. Since my knowledge of English mathematical terms is unsatisfactory at best, I attached a picture in case anyone was confused. Anyway, I already know the correct answer is 60°, but I haven't got the faintest idea why. I don't need the result for anything anymore, but I sure am curious, so thanks in advance for any light shed on this matter. Cheers, Gabe Link to comment Share on other sites More sharing options...
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