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Difference between revisions of "2021 AMC 12B Problems/Problem 15"
m (→‎Solution 2: Added a diagram)
Line 31: Line 31:
  
 
==Solution 2==
 
==Solution 2==
 +
 +
<asy>
 +
/* Made by samrocksnature, adapted by Tucker */
 +
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--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);
 +
dot(F^^G, gray);
 +
draw(A--G--A--F, gray);
 +
draw(B--F--C, gray);
 +
draw(E--G--D, gray);
 +
</asy>
  
 
Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>.
 
Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>.

Revision as of 21:05, 12 February 2021

Problem

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,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

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

Solution 2

[asy] /* Made by samrocksnature, adapted by Tucker */ 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--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); dot(F^^G, gray); draw(A--G--A--F, gray); draw(B--F--C, gray); draw(E--G--D, gray); [/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 by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)

https://youtu.be/QtSbAKUb1VE

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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