Difference between revisions of "2012 AMC 10A Problems/Problem 15"

Latest revision as of 22:41, 13 September 2023

Problem

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$ ?

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw((0,-2)--(1,-2)); draw(A--(0,-2)); draw(B--(1,-2)); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--(0,-2)); pair[] ps={A,B,C}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); [/asy]$

$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$

Solution 1

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--E); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$1$",(D--E),S); label("$1$",(A--B),N); label("$2$",(A--E),W); label("$\sqrt{5}$",(B--E),NW); [/asy]$

$AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\triangle ABC \sim \triangle BED$ . The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$ .

$\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC$

$\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}$

Since $AC=2BC$ , $BC=\frac{1}{\sqrt{5}}$ . $\triangle ABC$ is a right triangle so the area is just $\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}$

Solution 2 (coordbash)

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$F$",F,E); label("$G$",G,N); [/asy]$

Let $\text{E}$ be the origin. Then, $\text{D}=(1, 0)$ $\text{A}=(0, 2)$ $\text{B}=(1, 2)$ $\text{F}=(2, 1)$

${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\frac{1}{2}x+2$

Subtracting the second equation from the first gives us $\frac{5}{2}x-2=0$ . Thus, $x=\frac{4}{5}$ . Plugging this into the first equation gives us $y=\frac{8}{5}$ .

Since $\text{C} (0.8, 1.6)$ , $G$ is $(0.8, 2)$ ,

${AB}=1$ and ${CG}=0.4$ .

Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}$ . The answer is $\boxed{\textbf{(B)}\ \frac15}$ .

Solution 3

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); pair H=(0,-1), I=(0.5,-1); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); draw(H--I); pair[] ps={A,B,C,D,E,F,G, H, I}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$F$",F,E); label("$G$",G,N); label("$H$",H,W); label("$I$",I,E); [/asy]$

Triangle $EAB$ is similar to triangle $EHI$ ; line $HI = 1/2$

Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \frac{3}{2} = 2: 3$ .

Based on similarity the length of the height of $GC$ is thus $\frac{2}{5}\cdot1 = \frac{2}{5}$ .

Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}$ . The answer is $\boxed{\textbf{(B)}\ \frac15}$

Solution 4

Let $L$ be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let $K$ be the top right corner of the top right unit square, where segment $ABL$ is 2 units in length.

Because of the Pythagorean Theorem, since $AC = 2$ and $LK$ = 1, the diagonal of triangle $ALK$ is $\sqrt{5}$ .

Triangle $ALK$ is clearly a similar triangle to triangle $ABC$ . Segment $AB$ is the hypotenuse of triangle $ABC$ . So, we can write down:

$AK/AB = LK/BC$ , which is equal to: $\frac{\sqrt{5}}{1} = \frac{1}{BC}$ Solving this equation yields:

$BC = \frac{1}{\sqrt{5}}$

By Pythagorean theorem, we can now find segment $AC$ $(\frac{1}{\sqrt{5}})^2 + AC^2 = 1$ Solving this yields:

$AC^2 = \frac{4}{5}$ , so $AC = \frac{2}{\sqrt{5}}$

So then we can use $A = \frac{1}{2} * b * a.$ So $A = \frac{1}{2} * \frac{1}{\sqrt{5}} * \frac{2}{\sqrt{5}}$

$= \boxed{\textbf{(B)}\ \frac15}$

Video Solution

https://youtu.be/HVesU8cTjRU

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/4_x1sgcQCp4?t=1717

~ pi_is_3.14

@@ Line 22: / Line 22: @@
 <math> \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math>
+[[Category: Introductory Geometry Problems]]
 == Solution 1 ==
@@ Line 38: / Line 38: @@
 pair[] ps={A,B,C,D,E};
 dot(ps);
 label("$A$",A,N);
 label("$B$",B,N);
@@ Line 50: / Line 49: @@
 </asy></center>
-<math>AC</math> intersects <math>BC</math> at a right angle, so <math>\triangle ABC \sim \triangle BED</math>. The hypotenuse of right triangle BED is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>.
+<math>AC</math> intersects <math>BC</math> at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so <math>\triangle ABC \sim \triangle BED</math>. The hypotenuse of right triangle BED is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>.
 <cmath>\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC</cmath>
@@ Line 56: / Line 55: @@
 <cmath>\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}</cmath>
-Since AC=2BC, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math>
+Since <math>AC=2BC</math>, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math>
-== Solution 2 ==
+== Solution 2 (coordbash)==
 <center><asy>
@@ Line 89: / Line 88: @@
 <math>\text{F}=(2, 1)</math>
-<math>\widebar{EB}</math> can be represented by the line <math>y=2x</math>
+<math>{EB}</math> can be represented by the line <math>y=2x</math>
 Also, <math>{AF}</math> can be represented by the line <math>y=-\frac{1}{2}x+2</math>
@@ Line 100: / Line 99: @@
 <math>{AB}=1</math> and <math>{CG}=0.4</math>.
-Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}</math>. The answer is <math>(\text{B})</math>.
+Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}</math>. The answer is <math>\boxed{\textbf{(B)}\ \frac15}</math>.
+== Solution 3 ==
+<center><asy>
+unitsize(2cm);
+defaultpen(linewidth(.8pt)+fontsize(10pt));
+dotfactor=4;
+pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
+pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);
+pair H=(0,-1), I=(0.5,-1);
+draw(A--(2,0)); draw((0,-1)--F); draw(E--D);
+draw(A--E); draw(B--D); draw((2,0)--F);
+draw(A--F); draw(B--E); draw(C--G);
+draw(H--I);
+pair[] ps={A,B,C,D,E,F,G, H, I};
+dot(ps);
+label("$A$",A,N);
+label("$B$",B,N);
+label("$C$",C,W);
+label("$D$",D,S);
+label("$E$",E,S);
+label("$F$",F,E);
+label("$G$",G,N);
+label("$H$",H,W);
+label("$I$",I,E);
+</asy></center>
+Triangle <math>EAB</math> is similar to triangle <math>EHI</math>; line <math>HI = 1/2</math>
+Triangle <math>ACB</math> is similar to triangle <math>FCI</math> and the ratio of line <math>AB</math> to line <math>IF = 1 : \frac{3}{2} = 2: 3</math>.
+Based on similarity the length of the height of <math>GC</math> is thus <math>\frac{2}{5}\cdot1 = \frac{2}{5}</math>.
+Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}</math>.
+The answer is <math>\boxed{\textbf{(B)}\ \frac15}</math>
+==Solution 4==
+Let <math>L</math> be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let <math>K</math> be the top right corner of the top right unit square, where segment <math>ABL</math> is 2 units in length.
+Because of the Pythagorean Theorem, since <math>AC = 2</math> and <math>LK</math> = 1, the diagonal of triangle <math>ALK</math> is <math>\sqrt{5}</math>.
+Triangle <math>ALK</math> is clearly a similar triangle to triangle <math>ABC</math>. Segment <math>AB</math> is the hypotenuse of triangle <math>ABC</math>. So, we can write down:
+<cmath>AK/AB = LK/BC</cmath>, which is equal to:
+<cmath>\frac{\sqrt{5}}{1} = \frac{1}{BC}</cmath> Solving this equation yields:
+<cmath>BC = \frac{1}{\sqrt{5}}</cmath>
+By Pythagorean theorem, we can now find segment <math>AC</math>
+<cmath>(\frac{1}{\sqrt{5}})^2 + AC^2 = 1</cmath> Solving this yields:
+<cmath>AC^2 = \frac{4}{5}</cmath>, so <math>AC = \frac{2}{\sqrt{5}}</math>
+So then we can use <cmath>A = \frac{1}{2} * b * a.</cmath>
+So  <cmath>A = \frac{1}{2} * \frac{1}{\sqrt{5}} *  \frac{2}{\sqrt{5}}</cmath>
+<cmath>= \boxed{\textbf{(B)}\ \frac15}</cmath>
+==Video Solution==
+https://youtu.be/HVesU8cTjRU
+~savannahsolver
+== Video Solution by OmegaLearn ==
+https://youtu.be/4_x1sgcQCp4?t=1717
+~ pi_is_3.14
 == See Also ==

2012 AMC 10A (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 10 Problems and Solutions

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