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

Revision as of 16:30, 21 March 2008

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. Unknown error_msg), $CP = \abs{r - 2}$ (Error compiling LaTeX. Unknown error_msg), $DP = \abs{r - 4}$ (Error compiling LaTeX. Unknown error_msg), and $EP = \abs{r - 13}$ (Error compiling LaTeX. Unknown error_msg). 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 $\boxed{110}$ .

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

Revision as of 16:30, 21 March 2008

Problem

Solution

See also