# Mock AIME 1 Pre 2005 Problems/Problem 3

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## 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, $BP = \abs{r - 1}$ (Error compiling LaTeX. ! Undefined control sequence.), $CP = \abs{r - 2}$ (Error compiling LaTeX. ! Undefined control sequence.), $DP = \abs{r - 4}$ (Error compiling LaTeX. ! Undefined control sequence.), and $EP = \abs{r - 13}$ (Error compiling LaTeX. ! Undefined control sequence.). 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 $\boxed{110}$.