Difference between revisions of "2021 AMC 12B Problems/Problem 15"

Revision as of 19:37, 11 February 2021

Problem 20

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

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

Solution

Let $M$ be the midpoint of $CD$ . Noting that $AED$ and $ABC$ are $120-30-30$ triangles because of the equilateral triangles, $AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}$ . Also, $[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}$ and so $[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{(\textbf{D})23}$ .

~Lcz

@@ Line 26: / Line 26: @@
 ==Solution==
-Let <math>M</math> be the midpoint of <math>CD</math>.
+Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120-30-30</math> triangles because of the equilateral triangles, <math>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}</math>. Also, <math>[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}</math> and so <math>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{(\textbf{D})23}</math>.
+~Lcz

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Difference between revisions of "2021 AMC 12B Problems/Problem 15"

Revision as of 19:37, 11 February 2021

Problem 20

Solution