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发表于 25-1-2011 16:37:33|来自:新加坡
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大家好,我是磊磊。
这道题看起来是代数题,实际上有纯几何解法。以下我用英文作答。(如果懒得看的话,答案是(10乘以16)的平方根 :D)
restate the problem: given a right angled triangle ABC with right angle B and AB = 16. A point D lies between A and B such that AD = 6. Find the length of BC such that angle ACD is maximum.
suppose we could find this maximum angle, say beta. then we let the point C move around such that angle ACD is fixed. point C will trace a circle.
Now if this circle cuts the lines BC at two distinct points, say X and Y, then this angle is clearly not maximum; because we can just find a point Z between X and Y, and extend AZ to cut the circle at Z' then angle AZD > angle AZ'D= angle AXD =beta. In other words, we just construct a point Z on BC such that angle AZD is larger than maximum! impossible.
so the circle should just touch the line BC. then angle DCB = angle BAC, so we have triangle BCD similiar to triangle BAC. use ratio of similitude, we have BC^2=BD*BA=10*16=160, or BC = square root of 160=4 times (sqrt of 10). |
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