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Difference between revisions of "2013 AMC 8 Problems/Problem 24"

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(→‎Solution 3)
 
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</asy>
 
</asy>
  
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==Video Solution for Problems 21-25=
+
==Solution 1 (shortcut)==
−
https://youtu.be/-mi3qziCuec
+
It can be proven that <math>\Delta ADX \cong \Delta JIX</math> (where <math>X</math> is the point where <math>\overline {AJ}</math> intersects <math>\overline {HC}</math>) which also means quadrilaterals <math>ABCX \cong JGHX</math> (due to the squares being equal in the area which means the squares are congruent, and since the triangles earlier mentioned are congruent). The area of the shaded region is equal to the area of one square since the quadrilaterals and triangles are congruent. The total area of the shape of three squares. Putting these two pieces of information together, the answer is <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math>.
  
−
==Video Solution==
 
−
https://youtu.be/sbBjMvq5GG4 ~savannahsolver
 
−
 
−
Easy Solution1-<br>
 
−
<img src="file:///Users/rajneesh/Desktop/Screenshot%202023-11-05%20at%203.06.53%20PM.png">
 
−
let side of 1 square be= 2 (doesn't matter if we are calculating ratios)<br>
 
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Then, total area= 3x2^2=12<br>
 
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Shaded area=Total area - Unshaded area<br>
 
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Unshaded area = [AJFEDA]<br>
 
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We can now extend AD to the centre point of FG. This can be called y.<br>
 
−
△AJY[{(2+2) x 2+(2/2)}/2]=6 + [DEFY]=2x1=2<br>
 
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Unshaded area=6+2=8<br>
 
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Shaded area = 12-8=4 <br>
 
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Hence, ratio= 4/12 = 1/3<br>
 
−
~aarnavSOni
 
−
 
−
==Easiest Solution==
 
−
 
−
We can see that the Pentagon is made of two congruent shapes. We can fit one triangle into the gap in the upper square. Therefore, the answer is just <math>\frac{1}{3}\implies\boxed{C}</math>
 
−
 
−
==Solution 1==
 
 
<asy>
 
<asy>
 
pair A,B,C,D,E,F,G,H,I,J,X;
 
pair A,B,C,D,E,F,G,H,I,J,X;
 +
 
A = (0.5,2);
 
A = (0.5,2);
 
B = (1.5,2);
 
B = (1.5,2);
Line 75: Line 55:
 
I = (2,1);
 
I = (2,1);
 
J = (2,0);  
 
J = (2,0);  
−
X= extension(I,J,A,B);
+
X = (1.25,1);
−
dot(X,red);
 
−
draw(I--X--B,red);
 
 
draw(A--B);  
 
draw(A--B);  
 
draw(C--B);  
 
draw(C--B);  
−
draw(D--A);  
+
draw(D--A);
 
draw(F--E);  
 
draw(F--E);  
 
draw(I--J);  
 
draw(I--J);  
Line 97: Line 75:
 
dot("$H$", H, N);
 
dot("$H$", H, N);
 
dot("$I$", I, NE);
 
dot("$I$", I, NE);
−
label("$X$", X,SE);
+
dot("$J$", J, SE);
−
dot("$J$", J, SE);</asy>
+
dot("$X$", X, NE);
 +
</asy>
  
  
−
First let <math>s=2</math> (where <math>s</math> is the side length of the squares) for simplicity. We can extend <math>\overline{IJ}</math> until it hits the extension of <math>\overline{AB}</math>. Call this point <math>X</math>. The area of triangle <math>AXJ</math> then is <math>\dfrac{3 \cdot 4}{2}</math> The area of rectangle <math>BXIC</math> is <math>2 \cdot 1 = 2</math>. Thus, our desired area is <math>6-2 = 4</math>. Now, the ratio of the shaded area to the combined area of the three squares is <math>\frac{4}{3\cdot 2^2} = \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
+
~ julia333
  
−
==Solution 2==
 
−
WARNING: Some errors
 
−
<asy>
 
−
pair A,B,C,D,E,F,G,H,I,J,X;
 
−
A = (0.5,2);
 
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B = (1.5,2);
 
−
C = (1.5,1);
 
−
D = (0.5,1);
 
−
E = (0,1);
 
−
F = (0,0);
 
−
G = (1,0);
 
−
H = (1,1);
 
−
I = (2,1);
 
−
J = (2,0);
 
−
X= (1.25,1);
 
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draw(A--B);
 
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draw(C--B);
 
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draw(D--A);
 
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draw(F--E);
 
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draw(I--J);
 
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draw(J--F);
 
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draw(G--H);
 
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draw(A--J);
 
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filldraw(A--B--C--I--J--cycle,grey);
 
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draw(E--I);
 
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dot(X,red);
 
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label("$A$", A, NW);
 
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label("$B$", B, NE);
 
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label("$C$", C, NE);
 
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label("$D$", D, NW);
 
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label("$E$", E, NW);
 
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label("$F$", F, SW);
 
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label("$G$", G, S);
 
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label("$H$", H, N);
 
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label("$I$", I, NE);
 
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label("$X$", X,SW,red);
 
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label("$J$", J, SE);</asy>
 
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Let the side length of each square be <math>1</math>.
 
−
 
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Let the intersection of <math>AJ</math> and <math>EI</math> be <math>X</math>.
 
−
 
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Since <math>[ABCD]=[GHIJ]</math>, <math>AD=IJ</math>. Since <math>\angle IXJ</math> and <math>\angle AXD</math> are vertical angles, they are congruent. We also have <math>\angle JIH\cong\angle ADC</math> by definition.
 
−
 
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So we have <math>\triangle ADX\cong\triangle JIX</math> by <math>\textit{AAS}</math> congruence. Therefore, <math>DX=JX</math>.
 
−
 
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Since <math>C</math> and <math>D</math> are midpoints of sides, <math>DH=CJ=\dfrac{1}{2}</math>. This combined with <math>DX=JX</math> yields <math>HX=CX=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}</math>.
 
−
 
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The area of trapezoid <math>ABCX</math> is <math>\dfrac{1}{2}(AB+CX)(BC)=\dfrac{1}{2}\times \dfrac{5}{4}\times 1=\dfrac{5}{8}</math>.
 
  
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The area of triangle <math>JIX</math> is <math>\dfrac{1}{2}\times XJ\times IJ=\dfrac{1}{2}\times \dfrac{3}{4}\times 1=\dfrac{3}{8}</math>.
 
  
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So the area of the pentagon <math>AJICB</math> is <math>\dfrac{3}{8}+\dfrac{5}{8}=1</math>.
 
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The area of the <math>3</math> squares is <math>1\times 3=3</math>.
 
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Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
 
  
 
==Solution 3==
 
==Solution 3==
  
−
<asy>
+
== Solution 3 ==
−
pair A,B,C,D,E,F,G,H,I,J,K;
+
We notice that ABCX is a trapezoid with the bases AB and CX. Assuming AB is 1, we find that CX is 0.25 since it is half of CH which is 0.5. Using the area of the trapezoid formula, we calculate the area of ABCX to be 0.625 and XCIJ to be 0.375. The combined areas equal 1. Therefore, the ratio of the area of the hexagon to the three squares is 1:3 because the area of the three squares is 3. The answer is <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math>-TheNerdwhoIsNerdy. (I don't know if this is correct, pls check).
−
A = (0.5,2);
 
−
B = (1.5,2);
 
−
C = (1.5,1);
 
−
D = (0.5,1);
 
−
E = (0,1);
 
−
F = (0,0);
 
−
G = (1,0);
 
−
H = (1,1);
 
−
I = (2,1);
 
−
J = (2,0);
 
−
K= (1.25,1);
 
−
draw(A--B);
 
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draw(C--B);
 
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draw(D--A);
 
−
draw(F--E);
 
−
draw(I--J);
 
−
draw(J--F);
 
−
draw(G--H);
 
−
draw(A--J);
 
−
filldraw(A--B--C--I--J--cycle,grey);
 
−
draw(E--I);
 
−
dot(K,red);
 
−
label("$A$", A, NW);
 
−
label("$B$", B, NE);
 
−
label("$C$", C, NE);
 
−
label("$D$", D, NW);
 
−
label("$E$", E, NW);
 
−
label("$F$", F, SW);
 
−
label("$G$", G, S);
 
−
label("$H$", H, N);
 
−
label("$I$", I, NE);
 
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label("$K$", K,SW,red);
 
−
label("$J$", J, SE);</asy>
 
−
 
 
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Let the intersection of <math>AJ</math> and <math>EI</math> be <math>K</math>.
 
−
 
 
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Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>.
 
−
 
 
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Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>.
 
−
 
 
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Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>.
 
−
 
 
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Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>.
 
−
 
 
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Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>.
 
−
 
 
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Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math>
 
−
 
 
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So the shaded area now completely covers the square <math>ABCD</math>
 
−
 
 
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Set the area of a square as <math>x</math>
 
−
 
 
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Therefore, <math>\frac{x}{3x}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>.
 
−
 
 
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==Solution 4==
 
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Given the information in the problem, we see that the ABCIJ is congruent to ADHGJ. Since it is half of 2/3, it takes up 1/3 of the area.
 
  
−
==Video Solution by OmegaLearn==
+
==Solution 2 (extremely simple)==
−
https://youtu.be/j3QSD5eDpzU?t=346
+
Extend line AD such that it intersects line FG. Call the intersection K. If we assign an arbitrary value to the side lengths of the squares, such as 2, the base of the now formed triangle AJK would have a length of 3, while the height would be 4. The area of triangle AJK would then be 6, and the rectangle EDKF would have an area of 2. 6 + 2 = 8, and since all of the squares' areas add up to a total of 12 (if each has side length 2), we obtain (12 - 8)/12 for the shaded part of the diagram which yields 4/12 = <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math>
  
 +
~ martianrunner
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=23|num-a=25}}
 
{{AMC8 box|year=2013|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:35, 15 September 2024

Problem

Squares $ABCD$, $EFGH$, and $GHIJ$ are equal in area. Points $C$ and $D$ are the midpoints of sides $IH$ and $HE$, respectively. What is the ratio of the area of the shaded pentagon $AJICB$ to the sum of the areas of the three squares?

$\textbf{(A)}\hspace{.05in}\frac{1}{4}\qquad\textbf{(B)}\hspace{.05in}\frac{7}{24}\qquad\textbf{(C)}\hspace{.05in}\frac{1}{3}\qquad\textbf{(D)}\hspace{.05in}\frac{3}{8}\qquad\textbf{(E)}\hspace{.05in}\frac{5}{12}$

[asy] pair A,B,C,D,E,F,G,H,I,J; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE); [/asy]

Solution 1 (shortcut)

It can be proven that $\Delta ADX \cong \Delta JIX$ (where $X$ is the point where $\overline {AJ}$ intersects $\overline {HC}$) which also means quadrilaterals $ABCX \cong JGHX$ (due to the squares being equal in the area which means the squares are congruent, and since the triangles earlier mentioned are congruent). The area of the shaded region is equal to the area of one square since the quadrilaterals and triangles are congruent. The total area of the shape of three squares. Putting these two pieces of information together, the answer is $\boxed{\textbf{(C)}\ \frac {1}{3}}$.

[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X = (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE); dot("$X$", X, NE); [/asy]


~ julia333



Solution 3

Solution 3

We notice that ABCX is a trapezoid with the bases AB and CX. Assuming AB is 1, we find that CX is 0.25 since it is half of CH which is 0.5. Using the area of the trapezoid formula, we calculate the area of ABCX to be 0.625 and XCIJ to be 0.375. The combined areas equal 1. Therefore, the ratio of the area of the hexagon to the three squares is 1:3 because the area of the three squares is 3. The answer is $\boxed{\textbf{(C)}\ \frac {1}{3}}$-TheNerdwhoIsNerdy. (I don't know if this is correct, pls check).

Solution 2 (extremely simple)

Extend line AD such that it intersects line FG. Call the intersection K. If we assign an arbitrary value to the side lengths of the squares, such as 2, the base of the now formed triangle AJK would have a length of 3, while the height would be 4. The area of triangle AJK would then be 6, and the rectangle EDKF would have an area of 2. 6 + 2 = 8, and since all of the squares' areas add up to a total of 12 (if each has side length 2), we obtain (12 - 8)/12 for the shaded part of the diagram which yields 4/12 = $\boxed{\textbf{(C)}\ \frac {1}{3}}$

~ martianrunner

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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