Less than 30

Geometry Level 3

True or False?

In the acute triangle $ABC$ below, with a point $P$ inside of it, it is possible that all of the colored angles-- $\angle PAB, \angle PBC, \angle PCA$ --are each bigger than $30^\circ$ .

True False

1 solution

Steven Yuan
Sep 1, 2017

Suppose $\angle PAB, \angle PBC, \angle PCA > 30^{\circ}.$ Let the feet of the perpendiculars dropped from $P$ onto $AB, BC, CA$ be $X, Y, Z,$ respectively. Looking at right triangle $PAX$ we have $PX > PA \sin 30^{\circ} = \frac{1}{2} PA.$ Similarly, $PY > \frac{1}{2} PB$ and $PZ > \frac{1}{2} PC.$ Adding these inequalities gives

$PX + PY + PZ > \dfrac{1}{2}(PA + PB + PC).$

However, the Erdos-Mordell inequality states that $PX + PY + PZ \leq \frac{1}{2}(PA + PB + PC).$ Thus, we have reached a contradiction. At least one of $\angle PAB, \angle PBC, \angle PCA$ must be less than or equal to $30^{\circ}. \, \blacksquare$

Less than 30

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