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, 2012

Geometry Difficulty 2.0 Junior Prove it Canada

Adam and Budan are playing a game of Bocce. Each wants their ball to land closest to the jack ball. The positions of Adam’s ball, $A$ , Budan’s ball, $B$ , and the jack ball, $J$ , are shown in the diagram.

What is the distance from A to J?
What is the distance from B to A?
Determine whose ball is closer to the jack ball, Adam’s or Budan’s.

Solution

In $\triangle JPA$ , $\angle JPA=90^{\circ}$ .

Using the Pythagorean Theorem, $AJ^2=15^2+20^2$ or $AJ^2=225+400=625$ and so $AJ=\sqrt{625}=25$ , since $AJ>0$ .

The distance from $A$ to $J$ is 25.
In $\triangle BAQ$ , $\angle BAQ=90^{\circ}$ .

Using the Pythagorean Theorem, $39^2=BA^2+15^2$ or $BA^2=1521-225=1296$ and so $BA=\sqrt{1296}=36$ , since $BA>0$ .

The distance from $B$ to $A$ is 36.
In $\triangle BJA$ , $\angle BJA=90^{\circ}$ .

Using the Pythagorean Theorem, $BA^2=BJ^2+AJ^2$ or $1296=BJ^2+625$ or $BJ^2=1296-625=671$ and so $BJ=\sqrt{671} \approx 25.904$ , since $BJ>0$ .

Thus, the distance from Budan’s ball to the jack ball is approximately 25.904.

In part (a), we determined the distance between Adam’s ball and the jack ball to be 25.

Therefore, Adam’s ball is closer to the jack ball.

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