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Difference between revisions of "2021 AMC 12B Problems/Problem 17"

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==Problem 17==
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==Problem==
  
 
Let <math>ABCD</math> be an isoceles trapezoid having parallel bases <math>\overline{AB}</math> and <math>\overline{CD}</math> with <math>AB>CD.</math> Line segments from a point inside <math>ABCD</math> to the vertices divide the trapezoid into four triangles whose areas are <math>2, 3, 4,</math> and <math>5</math> starting with the triangle with base <math>\overline{CD}</math> and moving clockwise as shown in the diagram below. What is the ratio <math>\frac{AB}{CD}?</math>
 
Let <math>ABCD</math> be an isoceles trapezoid having parallel bases <math>\overline{AB}</math> and <math>\overline{CD}</math> with <math>AB>CD.</math> Line segments from a point inside <math>ABCD</math> to the vertices divide the trapezoid into four triangles whose areas are <math>2, 3, 4,</math> and <math>5</math> starting with the triangle with base <math>\overline{CD}</math> and moving clockwise as shown in the diagram below. What is the ratio <math>\frac{AB}{CD}?</math>
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<asy>
 
<asy>
 
unitsize(100);
 
unitsize(100);
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dot(A^^B^^C^^D^^P);
 
dot(A^^B^^C^^D^^P);
 
</asy>
 
</asy>
 +
 
<math>\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}</math>
 
<math>\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}</math>
  
 
==Solution==
 
==Solution==
  
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Without loss let <math>\mathcal T</math> have vertices <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, with <math>AB = r</math> and <math>CD = s</math>. Also denote by <math>P</math> the point in the interior of <math>\mathcal T</math>.
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Without the loss of generality, let <math>\mathcal T</math> have vertices <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, with <math>AB = r</math> and <math>CD = s</math>. Also denote by <math>P</math> the point in the interior of <math>\mathcal T</math>.
  
 
Let <math>X</math> and <math>Y</math> be the feet of the perpendiculars from <math>P</math> to <math>AB</math> and <math>CD</math>, respectively. Observe that <math>PX = \tfrac 8r</math> and <math>PY = \tfrac 4s</math>. Now using the formula for the area of a trapezoid yields
 
Let <math>X</math> and <math>Y</math> be the feet of the perpendiculars from <math>P</math> to <math>AB</math> and <math>CD</math>, respectively. Observe that <math>PX = \tfrac 8r</math> and <math>PY = \tfrac 4s</math>. Now using the formula for the area of a trapezoid yields
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<cmath>14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr.</cmath>
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<cmath>14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 2\cdot\frac rs + 4\cdot\frac sr.</cmath>
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Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>2\rho + \rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{2+\sqrt 2\textbf{ (B)}}</math>.
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Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>\rho + 2\rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{\textbf{(B)}\: 2+\sqrt{2}}</math>.
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(Observe that the given areas of <math>3</math> and <math>5</math> are irrelevant to the ratio <math>\frac{AB}{CD}</math>.)
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 +
==Solution 2==
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Let <math>b_1</math> be the bottom base, <math>b_2</math> be the top base, <math>h_1</math> be the height of the bottom triangle, <math>h_2</math> be the height of the top triangle.
 +
Thus, <math>b_1h_1 = 8, b_2h_2 = 4, (b_1+b_2)(h_1+h_2) = 28,</math> so <math>b_1h_2 + b_2h_1 = 16.</math> Let <math>b_2 = 1, h_2 = 4,</math> so we get <math>b_1h_1 = 8, 4b_1+h_1 = 16.</math>
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This gives us a quadratic in <math>b_1,</math> ie. <math>4b_1^2+8=16b_1,</math> so <math>b_1 = \boxed{\textbf{(B)}\: 2+\sqrt{2}}.</math>
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 +
- Solution by MathAwesome123, added by ccx09
 +
 
 +
 
 +
==Video Solution by Challenge 25==
 +
https://www.youtube.com/watch?v=0IB_1K8Dta8
 +
 
 +
==Video Solution (🚀 Just 3 min 🚀)==
 +
https://youtu.be/XrIM3cdEk2k
 +
 
 +
<i>~Education, the Study of Everything </i>
 +
 
 +
== Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area) ==
 +
https://youtu.be/MpMdRI9wC54
 +
 
 +
~ pi_is_3.14
 +
 
 +
==See Also==
 +
 
 +
{{AMC12 box|year=2021|ab=B|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Latest revision as of 11:58, 10 September 2023

Problem

Let $ABCD$ be an isoceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB>CD.$ Line segments from a point inside $ABCD$ to the vertices divide the trapezoid into four triangles whose areas are $2, 3, 4,$ and $5$ starting with the triangle with base $\overline{CD}$ and moving clockwise as shown in the diagram below. What is the ratio $\frac{AB}{CD}?$

[asy] unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.3, 0.9), D=(-0.3, 0.9), P=(0.2, 0.5), E=(0.1, 0.75), F=(0.4, 0.5), G=(0.15, 0.2), H=(-0.3, 0.5); draw(A--B--C--D--cycle, black); draw(A--P, black); draw(B--P, black); draw(C--P, black); draw(D--P, black); label("$A$",A,(-1,0)); label("$B$",B,(1,0)); label("$C$",C,(1,-0)); label("$D$",D,(-1,0)); label("$2$",E,(0,0)); label("$3$",F,(0,0)); label("$4$",G,(0,0)); label("$5$",H,(0,0)); dot(A^^B^^C^^D^^P); [/asy]

$\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}$

Solution

Without the loss of generality, let $\mathcal T$ have vertices $A$, $B$, $C$, and $D$, with $AB = r$ and $CD = s$. Also denote by $P$ the point in the interior of $\mathcal T$.

Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Observe that $PX = \tfrac 8r$ and $PY = \tfrac 4s$. Now using the formula for the area of a trapezoid yields \[14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 2\cdot\frac rs + 4\cdot\frac sr.\] Thus, the ratio $\rho := \tfrac rs$ satisfies $\rho + 2\rho^{-1} = 4$; solving yields $\rho = \boxed{\textbf{(B)}\: 2+\sqrt{2}}$.

(Observe that the given areas of $3$ and $5$ are irrelevant to the ratio $\frac{AB}{CD}$.)

Solution 2

Let $b_1$ be the bottom base, $b_2$ be the top base, $h_1$ be the height of the bottom triangle, $h_2$ be the height of the top triangle. Thus, $b_1h_1 = 8, b_2h_2 = 4, (b_1+b_2)(h_1+h_2) = 28,$ so $b_1h_2 + b_2h_1 = 16.$ Let $b_2 = 1, h_2 = 4,$ so we get $b_1h_1 = 8, 4b_1+h_1 = 16.$ This gives us a quadratic in $b_1,$ ie. $4b_1^2+8=16b_1,$ so $b_1 = \boxed{\textbf{(B)}\: 2+\sqrt{2}}.$

- Solution by MathAwesome123, added by ccx09


Video Solution by Challenge 25

https://www.youtube.com/watch?v=0IB_1K8Dta8

Video Solution (🚀 Just 3 min 🚀)

https://youtu.be/XrIM3cdEk2k

~Education, the Study of Everything

Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area)

https://youtu.be/MpMdRI9wC54

~ pi_is_3.14

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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