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 Three heights in a triangle (Posted on 2017-03-11)
Given a triangle ABC.
Let Ha, Hb, and Hc be the altitudes to its sides a,b, c,respectively.
Prove:
Ha + Hb > Hc

Is it possible to construct a triangle with altitudes 7,11, and 20?
Justify your answer.

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Answer Comment 2 of 2 |
Start with the area formulas A = a*Ha/2 = b*Hb/2 = c*Hc/2 and the triangle inequatity a + b > c.

Rearrange the area formulas into a = 2*A/Ha, b = 2*A/Hb, c = 2*A/Hc

Then substitute into the triangle inequality to get 2*A/Ha + 2*A/Hb = 2*A/Hc, which simplifies to 1/Ha + 1/Hb > 1/Hc.

Test the given set of altitudes: 1/20 + 1/11 > 1/7.  Give each side a common denominator of 1540: 217/1540 > 220/1540.  This is false, so a trinagle cannot be constructed with altitudes 7, 11, and 20.

 Posted by Brian Smith on 2017-03-21 09:33:08
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