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    Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.
  
Where’s Your Paddle? Sure Does! Topic: Space Geometry/Inequalities. Level: Olympiad. March 18th, 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.

Posted in Geometry, Inequalities, Olympiad || 4 Comments »

4 Responses to “Where’s Your Paddle? Sure Does! Topic: Space Geometry/Inequalities. Level: Olympiad.”

  1. QC Says:
    March 18th, 2006 at 2:20 am

    The Fermat point?

  2. paladin8 Says:
    March 18th, 2006 at 5:17 am

    The 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).

  3. QC Says:
    March 18th, 2006 at 5:58 am

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

  4. LiZ Says:
    March 18th, 2006 at 6:12 am

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

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