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Problem of the Week
Problem D and Solution
Walking Away

Problem

Anna, Biljana, and Cees stand in a row in a large field. Anna is \(180\) m west of Biljana and Cees is \(450\) m east of Biljana. At the same time, Anna and Cees begin walking. Anna walks north at a constant rate of \(75\) m/min and Cees walks south at a constant rate of \(90\) m/min. Biljana does not move.

After how many minutes of walking will the distance between Biljana and Cees be twice the distance between Anna and Biljana?

Solution

Solution 1

Let \(t\) represent the number of minutes until the distance between Biljana and Cees is twice the distance between Anna and Biljana. In \(t\) minutes Anna will walk \(75t\) m north and Cees will walk \(90t\) m south. The following diagram shows Anna’s position, \(A\), Biljana’s position, \(B\), and Cees’s position, \(C\), in metres, at time \(t>0\).

Horizontal line segment DE is drawn with point B between D and E, point A above D and point C below E. Two right-angled triangles are formed. In triangle ADB, AD is 75t, DB is 180, and angle ADB is a right angle. In triangle BEC, BE is 450, EC is 90t, and BEC is a right angle.

Since both triangles in the diagram are right-angled triangles, we can use the Pythagorean Theorem to determine the value of \(t\) when \(BC=2AB\). \[\begin{aligned} BC &= 2AB\\ (BC)^2 &= (2AB)^2\\ BE^2+EC^2 &= 4(AD^2+DB^2)\\ 450^2+(90t)^2 &= 4((75t)^2+180^2)\\ 202\,500 + 8100t^2 &= 4(5625t^2 + 32\,400)\\ 202\,500 + 8100t^2 &= 22\,500t^2 + 129\,600\\ 72\,900 &= 14\,400t^2\\ \frac{81}{16} &= t^2 \end{aligned}\] Since \(t>0\), it follows that \(t=\frac{9}{4}=2 \frac{1}{4}\) min. Therefore, after \(2 \frac{1}{4}\) minutes of walking, the distance between Biljana and Cees will be twice the distance between Anna and Biljana.

Solution 2

In this solution, we represent Anna, Biljana and Cees’s original positions as points on the \(x\)-axis so that Biljana is positioned at the origin \(B(0,0)\), Anna is positioned \(180\) units left of Biljana at \(D(-180,0)\) and Cees is positioned \(450\) units right of Biljana at \(E(450,0)\).

Let \(t\) represent the number of minutes until the distance between Biljana and Cees is twice the distance between Anna and Biljana. In \(t\) minutes Anna will walk \(75t\) m north to the point \(A(-180,75t)\), and Cees will walk \(90t\) m south to the point \(C(450,-90t)\).

Points A, B, C, D, and E are plotted on a Cartesian plane. Lines AD, AB, BC, and EC are drawn to create right-angled triangles ADB and BEC.

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

Then \(AB=\sqrt{(-180)^2+(75t)^2}=\sqrt{32\,400+5625t^2}\) and \(BC=\sqrt{(450)^2+(-90t)^2}=\sqrt{202\,500+8100t^2}\). \[\begin{aligned} BC &= 2AB\\ \sqrt{202\,500+8100t^2} &= 2\sqrt{32\,400+5625t^2}\\ (\sqrt{202\,500+8100t^2})^2 &= (2\sqrt{32\,400+5625t^2})^2\\ 202\,500+8100t^2 &= 4(32\,400+5625t^2)\\ 202\,500+8100t^2 &= 129\,600 + 22\,500t^2\\ 72\,900 &= 14\,400t^2\\ \frac{81}{16} &= t^2 \end{aligned}\] Since \(t>0\), it follows that \(t=\frac{9}{4}=2 \frac{1}{4}\) min. Therefore, after \(2 \frac{1}{4}\) minutes of walking, the distance between Biljana and Cees will be twice the distance between Anna and Biljana.