**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.

The Fermat point?

ReplyDeleteThe Fermat point I think would deal with PA+PB+PC+PD if you could generalize it to the third dimension (provide some sort of way of finding it).

ReplyDeleteConstruct regular tetrahedrons on each of the faces, and look at the intersection of the circumspheres? (Ugly, though.)

ReplyDelete=) *love your titles* cheerio says hi ^^

ReplyDelete