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Problem 693

AMC 10/12, early questions

Geometry Difficulty 3.8 Multiple choice CEMC Fermat · Canada · 2012

In the country of Nohills, each pair of cities is connected by a straight (and flat) road. The chart to the right shows the distances along the straight roads between some pairs of cities. The distance along the straight road between city $P$ and city $R$ is closest to

$P$
$Q$
$R$
$S$

$P$
0
25

$Q$
25
0
25
7

$R$

25
0
18

$S$
24
7
18
0

Pick one

Official solution

From the chart, we see that $QR=25$ , $QS=7$ and $SR=18$ .

Since $QR = QS+SR$ and $QR$ is the largest of these three lengths, then $S$ must be a point on line segment $QR$ .

This gives the following configuration so far:

[[IMAGE0]]

We have not yet used the fact that $PQ = 25$ or that $PS=24$ .

Note that $7^2 +24^2 = 49 + 576 = 625 = 25^2$ , so $QS^2 +PS^2 = PQ^2$ .

Since these lengths satisfy this property, then the points $P$ , $S$ and $Q$ form a triangle that is right-angled at $S$ .

This gives the following configuration so far:

[[IMAGE1]]

(We could have drawn $P$ “above" $QR$ .)

Since $\angle PSQ = 90^\circ$ , then $\angle PSR = 90^\circ$ .

Therefore, $PR^2 = PS^2 + SR^2 = 24^2 + 18^2 = 576 + 324 = 900$ .

Since $PR>0$ , then $PR = \sqrt{900}=30$ .

Thus, the distance between cities $P$ and $R$ is 30.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.