# 2021 AMC 12B Problems/Problem 15

The following problem is from both the 2021 AMC 10B #20 and 2021 AMC 12B #15, so both problems redirect to this page.

## Problem

The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$ $[asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]$

$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$

## Solution 1

$[asy] /* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-1.85,2); pair G=(-3.1,2); draw(A--G--A--F, lightgray); draw(B--F--C, lightgray); draw(E--G--D, lightgray); dot(F^^G, lightgray); draw(A--B--C--D--E--A); draw(A--C--A--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E); [/asy]$

Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$, they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$. For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$. Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$.

## Video Solution (🚀Under 3 min!🚀)

~Education, the Study of Everything

~ pi_is_3.14

~IceMatrix

~Interstigation

## Video Solution by The Power of Logic

~The Power of Logic

## Remark

This configuration of $11$ congruent line segments is known as the Moser Spindle https://en.wikipedia.org/wiki/Moser_spindle , and can be used to demonstrate that $3$ colors are not sufficient to color all of the points in the plane such that points that are $1$ unit apart have different colors. Finding the minimum such number of colors is a famous unsolved problem: the Nelson-Hadwiger problem. See: https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem

~hailstone