Difference between revisions of "2024 AMC 10A Problems/Problem 14"

Revision as of 21:28, 20 March 2025

Problem

All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is the length $AB$ ?

$\textbf{(A)}~4 + 4\sqrt{5}\qquad\textbf{(B)}~10\sqrt{2}\qquad\textbf{(C)}~5 + 5\sqrt{5}\qquad\textbf{(D)}~10\sqrt[4]{8}\qquad\textbf{(E)}~20$

Solution 1

Using the rectangle with area $1$ , let its short side be $x$ and the long side be $y$ . Observe that for every rectangle, since ratios of the side length of the rectangles are directly proportional to the ratios of the square roots of the areas (For example, each side of the rectangle with area $9$ is $\sqrt{9}=3$ times that of the rectangle with area $1$ ), as they are all similar to each other.

The side opposite $AB$ on the large rectangle is hence written as $6x + 4x + 2y\sqrt{2} + 3y\sqrt{2} = 10x+5y\sqrt{2}$ . However, $AB$ can be written as $4y\sqrt{2}+5x+7x = 4y\sqrt{2}+12x$ . Since the two lengths are equal, we can write $10x+5y\sqrt{2} = 4y\sqrt{2}+12x$ , or $y\sqrt{2} = 2x$ . Therefore, we can write $y=x\sqrt{2}$ .

Since $xy=1$ , we have $(x\sqrt{2})(x) = 1$ , which we can evaluate $x$ as $x=\frac{1}{\sqrt[4]{2}}$ . From this, we can plug back in to $xy=1$ to find $y=\sqrt[4]{2}$ . Substituting into $AB$ , we have $AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}}$ which can be evaluated to $\boxed{\textbf{(D) }10\sqrt[4]{8}}$ .

~i_am_suk_at_math_2

Remark

We know that the area is an integer, so after finding $y=x\sqrt{2}$ , AB must contain a fourth root. The only such option is $\boxed{\textbf{(D) }10\sqrt[4]{8}}$ .

Solution 2

Let the rectangle's height be $x,$ the length $AB=y.$ The entire rectangle has an area of $200.$ We will be using this fact for ratios.

Note that the short side of the rectangle with area 32 will have a height of $\sqrt{\frac{32}{200}}\cdot x = \frac{2}{5}x.$ We use $x$ because it is apparent that the height of the rectangle with area $32$ is the shorter side, corresponding with $x.$

Similarly, the long side of the rectangle with area 36 has a height of $\sqrt{\frac{36}{200}}\cdot y = \frac{3\sqrt{2}}{10}y.$

Noting that the total height of the big rectangle has height $x,$ we have the equation $\frac25 x + \frac{3\sqrt{2}}{10}y = x \Rightarrow x=\frac{y}{\sqrt{2}}.$

Since the area $xy=\frac{y^2}{\sqrt{2}}$ is equal to 200, we have:

$\begin{align*} y&=\sqrt{200\sqrt{2}} \\ &=\sqrt{100\sqrt{8}} \\ &=\boxed{\textbf{(D) }10\sqrt[4]{8}}. \end{align*}$

~mathboy282

Solution 3 - ruler (last effort)

Given that the diagram is drawn to scale, we can use a ruler to estimate the length of $AB$ .

We start by measuring the lengths of the rectangle with area $1$ , which may vary per viewing medium. For the sake of the solution, we use side lengths ~ $\frac{7}{10}$ cm and ~ $\frac{5}{10}$ cm.

To get the scale ratio from centimeters to the units in the problem, we need to find a ratio $x$ such that $\frac{7x}{10}\cdot\frac{5x}{10}=1$

Solving this equation, we get $x=\frac{10}{\sqrt{35}}$

We then measure the length of $AB$ (will vary), to get ~ $10.2$ cm. We multiply this length by our early ratio to get $\frac{10 \cdot 10.2}{\sqrt{35}} = \frac{102}{\sqrt{35}} \approx \frac{102}{6} \approx 17$

The answer choice closest to this would be $10\sqrt[4]{8}\approx16.8$ , so therefore the closest answer is $\boxed{\textbf{(D) }10\sqrt[4]{8}}$ .

~shreyan.chethan

Note: Never ever use this in an actual test

Remark

The specific numbers used in the solution above vary per test medium, but the method should still work.

Video Solution 1 by Pi Academy

https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE

Video Solution 2 by Power Solve

https://youtu.be/8abGnAJZ3AM

Video Solution 3 by Innovative Minds (Similar to Solution 1 above)

https://youtu.be/EepOGN0_Rgw

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

@@ Line 1: / Line 1: @@
-== Problem ==
+==Problem==
+All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is the length <math>AB</math>?
-One side of an equilateral triangle of height <math>24</math> lies on line <math>\ell</math>. A circle of radius <math>12</math> is tangent to line <math>\
+[[File:Screenshot 2024-11-08 2.08.49 PM.png|350px|center]]
-l</math> and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line <math>\ell</math> can be written as <math>a \sqrt{b} - c \pi</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers and <math>b</math> is not divisible by the square of any prime. What is <math>a + b + c</math>?
-<math>\textbf{(A)}~72\qquad\textbf{(B)}~73\qquad\textbf{(C)}~74\qquad\textbf{(D)}~75\qquad\textbf{(E)}~76</math>
+<math>\textbf{(A)}~4 + 4\sqrt{5}\qquad\textbf{(B)}~10\sqrt{2}\qquad\textbf{(C)}~5 + 5\sqrt{5}\qquad\textbf{(D)}~10\sqrt[4]{8}\qquad\textbf{(E)}~20</math>
-== Diagram ==
+==Solution 1==
-<asy>
+Using the rectangle with area <math>1</math>, let its short side be <math>x</math> and the long side be <math>y</math>. Observe that for every rectangle, since ratios of the side length of the rectangles are directly proportional to the ratios of the square roots of the areas (For example, each side of the rectangle with area <math>9</math> is <math>\sqrt{9}=3</math> times that of the rectangle with area <math>1</math>), as they are all similar to each other.
-/* Made by MRENTHUSIASM */
-size(250);
-pair A, B, C;
+The side opposite <math>AB</math> on the large rectangle is hence written as <math>6x + 4x + 2y\sqrt{2} + 3y\sqrt{2} = 10x+5y\sqrt{2}</math>. However, <math>AB</math> can be written as <math>4y\sqrt{2}+5x+7x = 4y\sqrt{2}+12x</math>. Since the two lengths are equal, we can write <math>10x+5y\sqrt{2} = 4y\sqrt{2}+12x</math>, or <math>y\sqrt{2} = 2x</math>. Therefore, we can write <math>y=x\sqrt{2}</math>.
-path p1, p2, p3;
-p1 = scale(16)*polygon(3);
-p2 = Circle((12*sqrt(3),4),12);
-A = intersectionpoint(p1,p2);
-B = (8*sqrt(3),-8);
-C = (12*sqrt(3),-8);
-Label L1 = Label("$24$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
-fill(A--Arc((12*sqrt(3),4),A,C)--B--cycle,yellow);
-draw(p1^^p2);
-draw((8*sqrt(3),-8)--(22+8*sqrt(3),-8));
-draw((-18,-8)--(-18,16), L=L1, arrow=Arrows(),bar=Bars(15));
-dot((12*sqrt(3),4),linewidth(4));
-draw((12*sqrt(3),4)--(12+12*sqrt(3),4));
-label("$12$",(6+12*sqrt(3),4),1.5S);
-dot(A^^B^^C,linewidth(4));
-</asy>
-~MRENTHUSIASM
+Since <math>xy=1</math>, we have <math>(x\sqrt{2})(x) = 1</math>, which we can evaluate <math>x</math> as <math>x=\frac{1}{\sqrt[4]{2}}</math>. From this, we can plug back in to <math>xy=1</math> to find <math>y=\sqrt[4]{2}</math>. Substituting into <math>AB</math>, we have <math>AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}}</math> which can be evaluated to <math>\boxed{\textbf{(D) }10\sqrt[4]{8}}</math>.
-== Solution 1 ==
+~i_am_suk_at_math_2
+===Remark===
+We know that the area is an integer, so after finding <math>y=x\sqrt{2}</math>, AB must contain a fourth root. The only such option is <math>\boxed{\textbf{(D) }10\sqrt[4]{8}}</math>.
-Call the bottom vertices of the triangle <math>B</math> and <math>C</math> (the one closer to the circle is <math>C</math>) and the top vertex <math>A</math>. The tangency point between the circle and the side of the triangle is <math>D</math>, and the tangency point on line <math>\ell</math> <math>E</math>, and the center of the circle is <math>O</math>.
+==Solution 2==
+Let the rectangle's height be <math>x,</math> the length <math>AB=y.</math> The entire rectangle has an area of <math>200.</math> We will be using this fact for ratios.
+Note that the short side of the rectangle with area 32 will have a height of <math>\sqrt{\frac{32}{200}}\cdot x = \frac{2}{5}x.</math> We use <math>x</math> because it is apparent that the height of the rectangle with area <math>32</math> is the shorter side, corresponding with <math>x.</math>
-Draw radii to the tangency points, the arc is <math>60</math> degrees because <math>\angle ACB</math> is <math>60</math>, and since <math>\angle DCE</math> is supplementary, it's <math>120^{\circ}</math>. The sum of the angles in a quadrilateral is <math>360</math>, which means <math>\angle COD</math> is <math>60^{\circ}</math>
+Similarly, the long side of the rectangle with area 36 has a height of <math>\sqrt{\frac{36}{200}}\cdot y = \frac{3\sqrt{2}}{10}y.</math>
+Noting that the total height of the big rectangle has height <math>x,</math> we have the equation <math>\frac25 x + \frac{3\sqrt{2}}{10}y = x \Rightarrow x=\frac{y}{\sqrt{2}}.</math>
-Triangle ODC is <math>30</math>-<math>60</math>-<math>90</math> triangle so CD is <math>4\sqrt{3}</math>.
+Since the area <math>xy=\frac{y^2}{\sqrt{2}}</math> is equal to 200, we have:
-Since we have <math>2</math> congruent triangles (<math>\triangle ODC</math> and <math>\triangle OEC</math>), the combined area of both is <math>48\sqrt{3}</math>.
-The area of the arc is <math>144 \cdot \frac{60}{360} \cdot \pi</math> which is <math>24\pi</math>, so the answer is <math>48\sqrt{3}-24\pi</math>
+<cmath>\begin{align*}
+y&=\sqrt{200\sqrt{2}} \\
+&=\sqrt{100\sqrt{8}} \\
+&=\boxed{\textbf{(D) }10\sqrt[4]{8}}.
+\end{align*}</cmath>
-<math>a+b+c</math> is <math>48+3+24</math> which is <math>\boxed{\textbf{(D)}~75}</math>
+~mathboy282
+==Solution 3 - ruler (last effort)==
-~ASPALAPATI75
+Given that the diagram is drawn to scale, we can use a ruler to estimate the length of <math>AB</math>.
-~andy_liu766 (latex)
-edits by KR
+We start by measuring the lengths of the rectangle with area <math>1</math>, which may vary per viewing medium. For the sake of the solution, we use side lengths ~<math>\frac{7}{10}</math> cm and ~<math>\frac{5}{10}</math> cm.
-==Note==
+To get the scale ratio from centimeters to the units in the problem, we need to find a ratio <math>x</math> such that <math>\frac{7x}{10}\cdot\frac{5x}{10}=1</math>
-There were two possible configurations from this problem; the one described in the solution above and the configuration in which the circle is tangent to the bottom of line <math>\ell</math> and the base of the equilateral triangle. However, since the area in this configuration is simply <math>0,</math> we can infer that the problem is talking about the configuration in Solution 1.
-~dbnl
+Solving this equation, we get <math>x=\frac{10}{\sqrt{35}}</math>
-==Solution 2 (Quick Guess)==
+We then measure the length of <math>AB</math>(will vary), to get ~<math>10.2</math> cm. We multiply this length by our early ratio to get <math>\frac{10 \cdot 10.2}{\sqrt{35}} = \frac{102}{\sqrt{35}} \approx \frac{102}{6} \approx 17</math>
-Since this problem involves equilateral triangles, the only possible number under the square root is <math>3</math>. Now subtracting all of the answer choices by <math>3</math>, we get:
+The answer choice closest to this would be <math>10\sqrt[4]{8}\approx16.8</math>, so therefore the closest answer is <math>\boxed{\textbf{(D) }10\sqrt[4]{8}}</math>.
-<cmath>\textbf{(A)}~72-3=69\qquad\textbf{(B)}~73-3=70\qquad\textbf{(C)}~74-3=71\qquad\textbf{(D)}~75-3=72\qquad\textbf{(E)}~76-3=73</cmath>
-Due to the even parity of the problem, we can safely assume that the answer is either <math>B</math> or <math>D</math>, but as <math>D</math> is a multiple of <math>12</math> and <math>24</math>, we get the answer of <math>\boxed{\textbf{(D)}~75}</math>.
+~shreyan.chethan
-~megaboy6679
+Note:
+Never ever use this in an actual test
-==Solution 3==
+==Remark==
-(pardon the diagrams :D)
+The specific numbers used in the solution above vary per test medium, but the method should still work.
-say the area we want to find is x.
+== Video Solution 1 by Pi Academy ==
-since the equilateral triangle has an internal angle of 60, the exterior angle formed by the triangle and the line is 120. simplifying the diagram you will get:
+https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
- \   #####
+== Video Solution 2 by Power Solve ==
-  \ ########
+https://youtu.be/8abGnAJZ3AM
-   \########
-    \#####___________
-make three of these that each circle is tangent to the other 2 circles
+== Video Solution 3 by Innovative Minds (Similar to Solution 1 above)==
+https://youtu.be/EepOGN0_Rgw
-    \
-     \    #####
-      \ ########
-  #####\########
- #######\_####__________
- #######/ #####
-  #####/########
-      / ########
-     /   ######
-Since they are 3 congruent triangles, you can make an equilateral triangle using their radius(12), with each vertex at the center of each circle. This will make an equilateral triangle of side length 24. if you look now, the area within the equilateral triangle consists of 3 <math>60/360</math> of a circle, and 3 of x.
-we first find the area of the triangle, which is <math>24 * 12\sqrt{3}</math>, we then find the area of <math>60/360</math> of a circle, which is <math>60/360 * 12^2\pi</math>, we subtract <math>24 * 12\sqrt{3}</math> by <math>60/360 * 12^2\pi</math>, and divide by 3, yielding the area of x.
- <math>24 * 12\sqrt{3}</math> - <math>60/360 * 12^2\pi</math>
-_________________________________________ = <math>48\sqrt{3}-24\pi</math>
-<math>a+b+c</math> is <math>48+3+24</math> which is <math>\boxed{\textbf{(D)}~75}</math>
-~Yiguo Zhang
-== Video Solution by Number Craft ==
-https://youtu.be/rIF5L_-zZYQ
-== Video Solution by Pi Academy ==
-https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
-== Video Solution 1 by Power Solve ==
-https://youtu.be/oCQ_QvXqV5s
 ==Video Solution by SpreadTheMathLove==
 https://www.youtube.com/watch?v=6SQ74nt3ynw
-==Video Solution by Just Math⚡==
-https://www.youtube.com/watch?v=fzXBMltyXjs&t=53s
 ==See also==
 {{AMC10 box|year=2024|ab=A|num-b=13|num-a=15}}
 {{MAA Notice}}

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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
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Difference between revisions of "2024 AMC 10A Problems/Problem 14"

Revision as of 21:28, 20 March 2025

Contents

Problem

Solution 1

Remark

Solution 2

Solution 3 - ruler (last effort)

Remark

Video Solution 1 by Pi Academy

Video Solution 2 by Power Solve

Video Solution 3 by Innovative Minds (Similar to Solution 1 above)

Video Solution by SpreadTheMathLove

See also