Given a triangle ABC.
Let Ha, Hb, and Hc
be the altitudes to its sides
Ha + Hb > Hc
Is it possible to construct a triangle with altitudes
7,11, and 20?
Justify your answer.
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.