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Math challenge II-A Geometry 5.6

 
 
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Math challenge II-A Geometry 5.6
by Zeyin Wu - Tuesday, 19 October 2021, 5:15 AM
 

I don’t understand how is altitude BQ drawn. Doesn’t it overlap with H? Also, PD/PH is just 1, because in an equilateral triangle, a median is equally lengthy with an altitude, but AB/BQ=2/sqrt3, so why are they equivalent?

 
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Re: Math challenge II-A Geometry 5.6
by John Lensmire - Wednesday, 20 October 2021, 10:18 AM
 

Remember here in 5.6, P is an arbitrary point somewhere inside the triangle, so it is not necessarily the center. Therefore, the altitude from B to AC doesn't necessarily contain P or H.

I'm not sure about the ratio PD/PH, I think in the book we look at PD/BG which should be a ratio of an altitude to a side length.

The key idea in 5.6b is that in ANY equilateral triangle, the ratio of the altitude to the side length is sqrt3 : 2 (because an altitude of an equilateral triangle divides the triangle into two 30-60-90 triangles).

Adding the distances from P to the three sides is adding up the lengths of these three altitudes. But using the sqrt3 : 2 ratio we can write this in terms of the side lengths of the equilateral triangles. Then the result from part (a) helps us get the result we want.