## Wednesday, April 12, 2006

### Hidden Triangle. Topic: Geometry/Inequalities. Level: Olympiad.

Problem: (TJ USAMO) Show that on any tetrahedron we can find a vertex $V$ such that the lengths of the $3$ edges with $V$ as an endpoint can be the sides of a triangle.

Solution: Call the tetrahedron $ABCD$ and WLOG assume $AB$ is the longest edge.

From triangle $ABC$, we have $BC+CA > AB$. From triangle $ABD$, we have $BD+DA > AB$. Summing the two inequalities together, we have

$(BC+BD)+(CA+DA) > 2AB$.

It follows that one of $BC+BD$ and $CA+DA$ is greater than $AB$. WLOG, assume $CA+DA > AB$. But since $AB$ is the longest edge and we have $CA+DA > AB$, the edges $AB$, $AC$, and $AD$ must form a triangle, as desired. QED.

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Comment: This problem was not too difficult; using the extremal principle to pick the longest side works well with triangles because as long as $b+c > a$ when $a$ is the largest of the three, $a$, $b$, and $c$ can be the lengths of the sides of a triangle.

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Practice Problem: (TJ USAMO) Show that, for all positive reals $a$, $b$, and $c$ such that $a+b \ge c$; $b+c \ge a$; and $c+a \ge b$, we have

$2a^2(b + c) + 2b^2(c + a) + 2c^2(a + b) \ge a^3 + b^3 + c^3 + 9abc$.

When does equality hold?