# Angles in the Pentagon Here’s a geometry challenge. A plane intersects a cube in such a way as to form a pentagon. If AL, FJ, and CM are all one-fourth of the side length of the cube, what are the angles in the pentagon?

This problem relies on several strategies necessary to success in geometry, including being able to visualize at different perspectives, add lines as needed, move between shapes, and flatten out bent regions. So let’s begin.

First off, let’s say AL = 1 (and hence AB = 4). Because we’re looking at angles and not lengths, it doesn’t matter what the unit is, but having a base “unit” will make it easier to follow the math. I’m going to start with $$\angle JLM$$, and I’m going to find that measurement by finding each side of the triangle $$\Delta JML$$ and using the Law of Cosines. I’ve added two segments and gotten rid of the pentagon so I can focus just on the parts I need for now. The triangle $$\Delta QLJ$$ is a right triangle on one of the faces of the cube. Since JF = 1, we know that BQ = 1. Therefore, QL = 4 – 1 – 1 = 2. QJ is equal to the side length of the cube, that is, 4. By the Pythagorean Theorem, $$JL = \sqrt{QL^2 + QJ^2} = \sqrt{4 + 16} = 2\sqrt{5}$$.

We can see that QM = QJ and that therefore ML = LJ.

That leaves MJ. Note that FJ = CM, so $$MJ = FC = \sqrt{BF^2 + BC^2} = \sqrt{16 + 16} = 4\sqrt{2}$$.

So we have triangle $$\Delta LMJ$$ with sides $$(4\sqrt{2}, 2\sqrt{5}, 2\sqrt{5})$$, where $$\angle L$$ is opposite the long side. Arranging the Law of Cosines to solve for the cosine of the target angle gives us:

$\cos C = \frac{a^2 + b^2 – c^2}{2ab}$

that is,

$\cos{\angle L} = \frac{(2\sqrt{5})^2 + (2\sqrt{5})^2 – (4\sqrt{2})^2}{2\times 2\sqrt{5} \times 2\sqrt{5}} = \frac{20 + 20 – 32}{40} = \frac{8}{40} = 0.2$

Thus, $$m\angle L = \cos^{-1} 0.2 \approx 78.46^o$$. My next step is to prove that O and P represent the midpoints of their respective edges. Going back to the first diagram, imagine that the bottom and back faces are flattened out. We can do this because we’re focused on distances. That would give us the right triangle $$\Delta BLP$$. We know that $$BL = 3$$. Set $$CP = x$$. Then $$BP = 4 + x$$, $$MP = \sqrt{x^2 + 1}$$, and $$LP^2 = 3^2 + (4 + x)^2 =$$$$9 + 16 + 8x + x^2 = x^2 + 8x + 25$$. We already know that $$LM = \sqrt{20}$$, so $$LP = \sqrt{20} + \sqrt{x^2 + 1}$$ and $$LP^2 = (\sqrt{20} + \sqrt{x^2 + 1})^2 =$$$$20 + 2\sqrt{20}\sqrt{x^2 + 1} + x^2 + 1 = x^2 + 4\sqrt{5(x^2 + 1)} + 21$$.

We have two equations for $$LP^2$$, so this leads us to $$x^2 + 8x + 25 = x^2 + 4\sqrt{5(x^2 + 1)} + 21$$, and thus $$8x + 4 = 4\sqrt{5(x^2 + 1)}$$, that is, $$2x + 1 = \sqrt{5(x^2 + 1)}$$. Squaring both sides gives us $$4x^2 + 4x + 1 = 5x^2 + 5$$, that is, $$x^2 – 4x + 4 = 0$$, which has the real number solution $$x = 2$$. This means that CP = 2, which is what we were looking for. We can similarly show that O is the midpoint of edge segment FG. We can now proceed to figure out $$m\angle M$$. We will do this by again using the Law of Cosines. In the next diagram, I’ve drawn in two new segments as well as two sides of the pentagon. $$PM = \sqrt{1 + 2^2} = \sqrt{5}$$ and $$LM = 2\sqrt{5}$$. $$\Delta PBL$$ is a right triangle, so we can use the Pythagorean Theorem to calculate PL. BL = 3. $$\Delta CPB$$ is also a right triangle, so $$PB = \sqrt{4^2 + 2^2} = \sqrt{20}$$. Thus, $$PL^2 = 3^2 + \sqrt{20}^2 = 29$$ and $$PL = \sqrt{29}$$.

Applying the Law of Cosines this time gives us:

$\cos{\angle M} = \frac{(2\sqrt{5})^2 + \sqrt{5}^2 – \sqrt{29}^2}{2\times 2\sqrt{5} \times \sqrt{5}} = \frac{20 + 5 – 29}{20} = \frac{-4}{20} = -0.2$

So $$m\angle M = cos^{-1} -0.2 \approx 101.54^o$$. Let’s go back to the original image. $$\angle J \cong \angle M$$ and $$\angle P \cong \angle O$$. Furthermore, because this is a pentagon, all five measures add up to 540 degrees. So $$540^o = \cos^{-1} 0.2 + 2 \cos^{-1} -0.2 + 2 m\angle P$$ and thus $$m\angle P = \frac{540^o – \cos^{-1} 0.2 – 2 \cos^{-1} -0.2}{2}$$$$\approx 129.23^o$$.

We now have the measures of all five angles of the pentagon:

• $$m\angle L \approx 78.46^o$$
• $$m\angle M \approx 101.54^o$$
• $$m\angle P \approx 129.23^o$$
• $$m\angle O \approx 129.23^o$$
• $$m\angle J \approx 101.54^o$$ Clio Corvid

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