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Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 3"

(solution by krsattack)
 
(Corrected absolute values.)
 
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== Solution ==
 
== Solution ==
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Let the altitude from <math>P</math> onto <math>AE</math> at <math>Q</math> have lengths <math>PQ = h</math> and <math>AQ = r</math>.  It is clear that, for a given <math>r</math> value, <math>AP</math>, <math>BP</math>, <math>CP</math>, <math>DP</math>, and <math>EP</math> are all minimized when <math>h = 0</math>.  So <math>P</math> is on <math>AE</math>, and therefore, <math>P = Q</math>.  Thus, <math>AP</math>=r, <math>BP = \abs{r - 1}</math>, <math>CP = \abs{r - 2}</math>, <math>DP = \abs{r - 4}</math>, and <math>EP = \abs{r - 13}</math>. Squaring each of these gives:
+
Let the altitude from <math>P</math> onto <math>AE</math> at <math>Q</math> have lengths <math>PQ = h</math> and <math>AQ = r</math>.  It is clear that, for a given <math>r</math> value, <math>AP</math>, <math>BP</math>, <math>CP</math>, <math>DP</math>, and <math>EP</math> are all minimized when <math>h = 0</math>.  So <math>P</math> is on <math>AE</math>, and therefore, <math>P = Q</math>.  Thus, <math>AP</math>=r, <math>BP = |r - 1|</math>, <math>CP = |r - 2|</math>, <math>DP = |r - 4|</math>, and <math>EP = |r - 13|.</math>  Squaring each of these gives:
  
 
<math>AP^2 + BP^2 + CP^2 + DP^2 + EP^2 = r^2 + (r - 1)^2 + (r - 2)^2 + (r - 4)^2 + (r - 13)^2 = 5r^2 - 40r + 190</math>
 
<math>AP^2 + BP^2 + CP^2 + DP^2 + EP^2 = r^2 + (r - 1)^2 + (r - 2)^2 + (r - 4)^2 + (r - 13)^2 = 5r^2 - 40r + 190</math>

Latest revision as of 05:18, 2 July 2015

Problem

$A, B, C, D,$ and $E$ are collinear in that order such that $AB = BC = 1, CD = 2,$ and $DE = 9$. If $P$ can be any point in space, what is the smallest possible value of $AP^2 + BP^2 + CP^2 + DP^2 + EP^2$?

Solution

Let the altitude from $P$ onto $AE$ at $Q$ have lengths $PQ = h$ and $AQ = r$. It is clear that, for a given $r$ value, $AP$, $BP$, $CP$, $DP$, and $EP$ are all minimized when $h = 0$. So $P$ is on $AE$, and therefore, $P = Q$. Thus, $AP$=r, $BP = |r - 1|$, $CP = |r - 2|$, $DP = |r - 4|$, and $EP = |r - 13|.$ Squaring each of these gives:

$AP^2 + BP^2 + CP^2 + DP^2 + EP^2 = r^2 + (r - 1)^2 + (r - 2)^2 + (r - 4)^2 + (r - 13)^2 = 5r^2 - 40r + 190$

This reaches its minimum at $r = \frac {40}{2\cdot 5} = 4$, at which point the sum of the squares of the distances is $\boxed{110}$.

See also

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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