## Friday, March 17, 2006

Problem: (360 Problems for Mathematical Contests - 3.1.59) Let $P$ be a point in the interior of a tetrahedron $ABCD$ such that its projections $A_1, B_1, C_1, D_1$ onto the planes $(BCD)$, $(CDA)$, $(DAB)$, $(ABC)$, respectively, are all situated in the interior of the faces. If $S$ is the total (surface) area and $r$ the inradius of the tetrahedron, prove that

$\frac{S_{BCD}}{PA_1} + \frac{S_{CDA}}{PB_1} + \frac{S_{DAB}}{PC_1} + \frac{S_{ABC}}{PD_1} \ge \frac{S}{r}$.

When does equality hold? (Note: $S_{ABC}$ represents the area of the triangle $ABC$)

Solution: We will use the following notation to simplify things.

$\displaystyle \sum (S_{ABC}) = S$ denotes the sum of all triangular face areas.

$\displaystyle \sum (PA_1 \cdot S_{BCD})$ denotes the sum of the product of the area of each triangular face with the corresponding altitude from $P$.

Let $V_{ABCP}$ denote the volume of the tetrahedron $ABCP$ and $V = V_{ABCD}$.

We have

$\displaystyle \sum (PA_1 \cdot S_{BCD}) = 3(V_{BCDP}+V_{CDAP}+V_{DABP}+V_{ABCP}) = 3V$, (1)

by the standard formula for the volume of a tetrahedron.

Similarly, we have

$\displaystyle r \cdot (\sum (S_{ABC})) = rS = 3V$. (2)

Combining (1) and (2), we get

$\displaystyle \sum (PA_1 \cdot S_{BCD}) = rS$. (3)

We apply Cauchy to get

$\displaystyle \left(\sum (PA_1 \cdot S_{BCD})\right)\left(\frac{S_{BCD}}{PA_1} + \frac{S_{CDA}}{PB_1} + \frac{S_{DAB}}{PC_1} + \frac{S_{ABC}}{PD_1}\right) \ge \left(\sum S_{ABC}\right)^2 = S^2$.

Substituting (3) and dividing through by $rS$, we have

$\frac{S_{BCD}}{PA_1} + \frac{S_{CDA}}{PB_1} + \frac{S_{DAB}}{PC_1} + \frac{S_{ABC}}{PD_1} \ge \frac{S}{r}$

as desired. Equality holds iff all the

$\frac{PA_1 \cdot S_{BCD}}{\frac{S_BCD}{PA_1}} = (PA_1)^2$

are equal, that is, $P$ is the incenter of $ABCD$. QED.

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Comment: Seeing the inequality with all the fractions should immediately make one think of Cauchy. After equating volumes in a few places, the result simply falls out.

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Practice Problem: Find the point in the tetrahedron $ABCD$ such that $PA_1+PB_1+PC_1+PD_1$ is minimized.