# Problem of the Week Problem D and Solution Go Forth and Walk

## Problem

At noon three students, Abby, Ben, and Cassie, are standing so that Abby is $$100$$ m west of Ben and Cassie is $$160$$ m east of Ben. While Ben stays in his initial position, Abby begins walking south at a constant rate of $$20$$ m/min and Cassie begins walking north at a constant rate of $$41$$ m/min.

In how many minutes will the distance between Cassie and Ben be the twice the distance between Abby and Ben?

## Solution

Solution 1

Let $$t$$ represent the number of minutes until Cassie’s distance to Ben is twice that of Abby’s distance to Ben. In $$t$$ minutes Abby will walk $$20t$$ m and Cassie will walk $$41t$$ m. The following diagram contains the information showing Abby’s position, $$A$$, Ben’s position, $$B$$, and Cassie’s position, $$C$$, at time $$t>0$$.

Since both triangles in the diagram are right-angled triangles, we can use the Pythagorean Theorem to set up an equation.

\begin{aligned} CB&=2AB\\ (CB)^2&=\left(2AB\right)^2\\ (CB)^2&=4\left(AB\right)^2\\ (41t)^2+(160)^2&=4\left[(20t)^2+(100)^2\right]\\ 1681t^2+25600&=4\left[400t^2+10000\right]\\ 1681t^2+25600&=1600t^2+40000\\ 81t^2&=14400\\ t^2&=\frac{14400}{81}\\ t&=\frac{120}{9} & (\text{since }t>0)\\ t&=\frac{40}{3}\text{ min}\end{aligned}

Therefore, in $$13\frac{1}{3}$$ minutes ($$13$$ minutes $$20$$ seconds), Cassie’s distance to Ben will be twice that of Abby’s distance to Ben.

In Solution 2, an alternate solution that uses coordinate geometry is presented.

Solution 2

Represent Abby, Ben and Cassie’s respective positions at noon as points on the $$x$$-axis so that Ben is positioned at the origin $$B(0,0)$$, Abby is positioned $$100$$ units left of Ben at $$D(-100,0)$$ and Cassie is positioned $$160$$ units right of Ben at $$E(160,0)$$.

Let $$t$$ represent the number of minutes until Cassie’s distance to Ben is twice that of Abby’s distance to Ben.

In $$t$$ minutes Abby will walk south $$20t$$ m to the point $$A(-100,-20t)$$. In $$t$$ minutes Cassie will walk north $$41t$$ m to the point $$C(160,41t)$$.

The distance from a point $$P(x,y)$$ to the origin can be found using the formula $$d=\sqrt{x^2+y^2}$$.

Then $$AB=\sqrt{(-100)^2+(-20t)^2}=\sqrt{10000+400t^2}$$ and $$CB=\sqrt{(160)^2+(41t)^2}=\sqrt{25600+1681t^2}$$. \begin{aligned} CB&=2AB\\ \sqrt{25600+1681t^2}&=2\sqrt{10000+400t^2}\\ 25600+1681t^2&=4(10000+400t^2) & \text{(squaring both sides)}\\ 25600+1681t^2&=40000+1600t^2\\ 81t^2&=14400\\ t^2&=\frac{14400}{81}\\ t&=\frac{120}{9} & (\text{since } t>0)\\ t&=\frac{40}{3} \text{ min}\end{aligned} Therefore, in $$13\frac{1}{3}$$ minutes ($$13$$ minutes $$20$$ seconds), Cassie’s distance to Ben will be twice that of Abby’s distance to Ben.