# Problem of the Week Problem D and Solution Cartesian Geocaching

## Problem

Geocaching is a kind of outdoor treasure hunt where people use GPS devices to look for hidden objects, called caches. In Cartesian Geocaching, instead of using a GPS device, locations are described using Cartesian coordinates.

Hilde sets up a large field for Cartesian Geocaching, measuring the distances in kilometres so that the point $$(1,0)$$ lies $$1$$ km east of the point $$(0,0)$$, for example.

Hilde starts at point $$A(0,0)$$, then walks northwest in a straight line to some point $$B$$, where she hides a cache. Then, from $$B$$, she walks northeast in a straight line to point $$C(0,4)$$ where she hides another cache. Finally she walks straight back to point $$A$$.

How far does Hilde walk in total?

## Solution

We will show four different solutions to this problem.

Solution 1

If you travel northwest from $$A (0,0)$$, the line of travel will make a $$45^\circ$$ angle with the positive $$y$$-axis. Point $$B$$ is located somewhere on this line of travel. If you travel northeast from point $$B$$ to $$C(0,4)$$, the line will intersect the $$y$$-axis at a $$45^\circ$$ angle.

In $$\triangle ABC$$, $$\angle BAC=\angle BCA=45^\circ$$. It follows that $$\triangle ABC$$ is isosceles. Since two of the angles in $$\triangle ABC$$ are $$45^\circ$$, then the third angle, $$\angle ABC=90^\circ$$ and the triangle is right-angled.

The distance from point $$A$$ to point $$C$$ along the $$y$$-axis is $$AC=4$$ km. Let $$BC=AB=m$$, for some $$m>0$$. Using the Pythagorean Theorem, we can find the value of $$m$$. \begin{aligned} AC^2&=BC^2+AB^2\\ 4^2&=m^2+m^2\\ 16&=2m^2\\ 8&=m^2\end{aligned}

Then since $$m>0$$, we have $$m=\sqrt{8}$$.

Thus, the total distance walked by Hilde is $$AB+BC+AC=\sqrt{8}+\sqrt{8}+4=(2\sqrt{8}+4)$$ km.

Note that the answer $$\left(2\sqrt{8}+4\right)$$ is an exact answer. We can use a calculator to determine that this distance is approximately $$9.7$$ km.

The exact total distance travelled can be further simplified as follows: $2\sqrt{8}+4=2(\sqrt{4}\sqrt{2})+4=2(2\sqrt{2})+4=4\sqrt{2}+4$ This method of simplifying radicals is developed in later mathematics courses.

Solution 2

If you travel northwest from $$A (0,0)$$, the line of travel will make a $$45^\circ$$ angle with the positive $$y$$-axis. Point $$B$$ is located somewhere on this line of travel.

From $$B$$, draw a line segment perpendicular to the $$y$$-axis, meeting the $$y$$-axis at point $$D$$. A line of travel in a northeast direction from point $$B$$ to $$C$$ will make $$\angle DBC=45^\circ$$.

In $$\triangle ABD$$, $$\angle BAD=45^\circ$$ and $$\angle ADB=90^\circ$$. It follows that $$\angle ABD=45^\circ$$, $$\triangle ABD$$ is isosceles and $$BD=AD$$.

In $$\triangle CBD$$, $$\angle CBD=45^\circ$$ and $$\angle CDB=90^\circ$$. It follows that $$\angle BCD=45^\circ$$, $$\triangle CBD$$ is isosceles and $$CD=BD$$.

The distance from point $$A$$ to point $$C$$ along the $$y$$-axis is $$AC=4$$ km. Since $$CD=AD$$ and $$AC=CD+AD$$, then we know that $$CD=AD=2$$ km. But $$CD=BD$$ so $$CD=BD=AD=2$$ km.

Using the Pythagorean Theorem in right-angled $$\triangle ABD$$, we can calculate the length of $$AB$$. \begin{aligned} AB^2&=BD^2+AD^2\\ AB^2&=2^2+2^2\\ AB^2&=8\end{aligned}

Then since $$AB>0$$, we have $$AB=\sqrt{8}$$.

Using the same reasoning in $$\triangle CBD$$, we obtain $$BC=\sqrt{8}$$.

Thus, the total distance walked by Hilde is $$AB+BC+AC=\sqrt{8}+\sqrt{8}+4=(2\sqrt{8}+4)$$ km.

Note that the answer $$\left(2\sqrt{8}+4\right)$$ is an exact answer. We can use a calculator to determine that this distance is approximately $$9.7$$ km.

The exact total distance travelled can be further simplified as follows: $2\sqrt{8}+4=2(\sqrt{4}\sqrt{2})+4=2(2\sqrt{2})+4=4\sqrt{2}+4$ This method of simplifying radicals is developed in later mathematics courses.

Solution 3

If you travel northwest from $$A (0,0)$$, the line of travel will make a $$45^\circ$$ angle with the positive $$y$$-axis. It follows that this line has slope $$-1$$. Since this line passes through $$A(0,0)$$ and has slope $$-1$$, the equation of the line through $$A$$ and $$B$$ is $$y=-x$$.

Point $$B$$ is located somewhere on $$y=-x$$. A line drawn to the northeast would be perpendicular to a line drawn to the northwest. Since a line to the northwest has slope $$-1$$, it follows that a line to the northeast would have slope $$1$$. This second line passes through $$B$$ and $$C$$, so it has slope $$1$$ and $$y$$-intercept $$4$$, the $$y$$-coordinate of $$C$$. The equation of the second line is $$y=x+4$$.

Since point $$B$$ is located on both $$y=-x$$ and $$y=x+4$$, we can solve the system of equations to find the coordinates of $$B$$. Since $$y=y$$, \begin{aligned} -x&=x+4\\ -2x&=4\\ x&=-2\end{aligned} Substituting $$x=-2$$ into $$y=-x$$, we obtain $$y=2$$. The coordinates of $$B$$ are therefore $$(-2,2)$$.

Using the distance formula, we can find the lengths of $$AB$$ and $$BC$$. \begin{aligned} AB&=\sqrt{(-2-0)^2+(2-0)^2}=\sqrt{4+4}=\sqrt{8}\\ BC&=\sqrt{(0-(-2))^2+(4-2)^2}=\sqrt{4+4}=\sqrt{8}\end{aligned} The distance from point $$A$$ to point $$C$$ along the $$y$$-axis is $$AC=4$$ km. That is, $$AC=4$$.

Thus, the total distance walked by Hilde is $$AB+BC+AC=\sqrt{8}+\sqrt{8}+4=(2\sqrt{8}+4)$$ km.

Note that the answer $$\left(2\sqrt{8}+4\right)$$ is an exact answer. We can use a calculator to determine that this distance is approximately $$9.7$$ km.

The exact total distance travelled can be further simplified as follows: $2\sqrt{8}+4=2(\sqrt{4}\sqrt{2})+4=2(2\sqrt{2})+4=4\sqrt{2}+4$ This method of simplifying radicals is developed in later mathematics courses.

Solution 4

If you travel northwest from $$A (0,0)$$, the line of travel will make a $$45^\circ$$ angle with the positive $$y$$-axis. It follows that this line has slope $$-1$$. Since this line passes through $$A(0,0)$$ and has slope $$-1$$, the equation of the line through $$A$$ and $$B$$ is $$y=-x$$. A line drawn to the northeast would be perpendicular to a line drawn to the northwest, so $$AB$$ is perpendicular to $$BC$$, and thus $$\angle ABC=90^\circ$$.

Point $$B$$ is located somewhere on $$y=-x$$. Let the coordinates of $$B$$ be $$(-b,b)$$ for some $$b>0$$.

Using the distance formula, we can find expressions for the lengths of $$AB$$ and $$BC$$. \begin{aligned} AB&=\sqrt{(-b-0)^2+(b-0)^2}=\sqrt{b^2+b^2}=\sqrt{2b^2}\\ BC&=\sqrt{(0-(-b))^2+(4-b)^2}=\sqrt{b^2+16-8b+b^2}=\sqrt{2b^2-8b+16}\end{aligned} The distance from point $$A$$ to point $$C$$ along the $$y$$-axis is $$AC=4$$ km. That is, $$AC=4$$.

Using the Pythagorean Theorem, we can find the value of $$b$$. \begin{aligned} AC^2&=AB^2+BC^2\\ 4^2&=\left(\sqrt{2b^2}\right)^2+\left(\sqrt{2b^2-8b+16}\right)^2\\ 16&=2b^2+(2b^2-8b+16)\\ 16&=4b^2-8b+16\\ 0&=4b^2-8b\\ 0&=b^2-2b\\ 0&=b(b-2)\\ b&=0,2\end{aligned} Since $$b>0$$, it follows that $$b=2$$. We can substitute $$b=2$$ into our expressions for $$AB$$ and $$BC$$. \begin{aligned} AB&=\sqrt{2b^2}=\sqrt{2(2)^2}=\sqrt{8}\\ BC&=\sqrt{2b^2-8b+16}=\sqrt{2(2)^2-8(2)+16}=\sqrt{8}\end{aligned}

Thus, the total distance walked by Hilde is $$AB+BC+AC=\sqrt{8}+\sqrt{8}+4=(2\sqrt{8}+4)$$ km.

Note that the answer $$\left(2\sqrt{8}+4\right)$$ is an exact answer. We can use a calculator to determine that this distance is approximately $$9.7$$ km.

The exact total distance travelled can be further simplified as follows: $2\sqrt{8}+4=2(\sqrt{4}\sqrt{2})+4=2(2\sqrt{2})+4=4\sqrt{2}+4$ This method of simplifying radicals is developed in later mathematics courses.