Difference between revisions of "2020 AMC 10A Problems/Problem 20"

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('Pythagorean' Theorem had a typo, and Latex update on last solution.)
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It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. By dropping this altitude, we can also see two similar triangles, <math>BFE</math> and <math>DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math>. Now expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, <math>[ABC]=60</math>. So <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}</math>
 
It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. By dropping this altitude, we can also see two similar triangles, <math>BFE</math> and <math>DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math>. Now expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, <math>[ABC]=60</math>. So <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}</math>
  
(I'm very sorry if you're a visual learner)
+
(I'm very sorry if you're a visual learner but now you have a diagram by ciceronii)
  
~Ultraman, diagram by ciceronii
+
~ Solution by Ultraman
  
==Solution 2 (Pro Guessing Strats)==
+
~ Diagram by ciceronii
We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to <math>\sqrt{20a}</math> and <math>\sqrt{20b}</math>, and because the hypotenuse is 20 we get <math>a+b=20</math>. Testing small numbers, we get that when <math>a=2</math> and <math>b=18</math>, <math>ab</math> is indeed a square. The area of the triangle is thus 60, so the answer is <math>\boxed {\textbf{(D) }360}</math>.
 
  
~tigershark22
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==Solution 2 (coordinates)==
~(edited by HappyHuman)
 
 
 
==Solution 3 (coordinates)==
 
 
<asy>
 
<asy>
 
size(10cm,0);
 
size(10cm,0);
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Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>.
 
Let the points be <math>A(-10,0)</math>, <math>\:B(x,y)</math>, <math>\:C(10,0)</math>, <math>\:D(10,30)</math>,and <math>\:E(-5,0)</math>, respectively. Since <math>B</math> lies on line <math>DE</math>, we know that <math>y=2x+10</math>. Furthermore, since <math>\angle{ABC}=90^\circ</math>, <math>B</math> lies on the circle with diameter <math>AC</math>, so <math>x^2+y^2=100</math>. Solving for <math>x</math> and <math>y</math> with these equations, we get the solutions <math>(0,10)</math> and <math>(-8,-6)</math>. We immediately discard the <math>(0,10)</math> solution as <math>y</math> should be negative. Thus, we conclude that <math>[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}</math>.
  
==Solution 4 (Trigonometry)==
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==Solution 3 (Trigonometry)==
 
Let <math>\angle C = \angle{ACB}</math> and <math>\angle{B} = \angle{CBE}.</math> Using Law of Sines on <math>\triangle{BCE}</math> we get <cmath>\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}</cmath> and LoS on <math>\triangle{ABE}</math> yields <cmath>\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.</cmath> Divide the two to get <math>\tan{B} = 3 \tan{C}.</math> Now, <cmath>\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}</cmath> and solve the quadratic, taking the positive solution (C is acute) to get <math>\tan{C} = \frac{1}{3}.</math> So if <math>AB = a,</math> then <math>BC = 3a</math> and <math>[ABC] = \frac{3a^2}{2}.</math> By Pythagorean Theorem, <math>10a^2 = 400 \iff \frac{3a^2}{2} = 60</math> and the answer is <math>300 + 60 \iff \boxed{\textbf{(D)}}.</math>
 
Let <math>\angle C = \angle{ACB}</math> and <math>\angle{B} = \angle{CBE}.</math> Using Law of Sines on <math>\triangle{BCE}</math> we get <cmath>\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}</cmath> and LoS on <math>\triangle{ABE}</math> yields <cmath>\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.</cmath> Divide the two to get <math>\tan{B} = 3 \tan{C}.</math> Now, <cmath>\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}</cmath> and solve the quadratic, taking the positive solution (C is acute) to get <math>\tan{C} = \frac{1}{3}.</math> So if <math>AB = a,</math> then <math>BC = 3a</math> and <math>[ABC] = \frac{3a^2}{2}.</math> By Pythagorean Theorem, <math>10a^2 = 400 \iff \frac{3a^2}{2} = 60</math> and the answer is <math>300 + 60 \iff \boxed{\textbf{(D)}}.</math>
  
 
(This solution is incomplete, can someone complete it please-Lingjun) ok
 
(This solution is incomplete, can someone complete it please-Lingjun) ok
 
Latex edited by kc5170
 
Latex edited by kc5170
 +
 +
We could use the famous m-n rule in trigonometry in <math>\triangle ABC</math> with Point <math>E</math>
 +
[Unable to write it here.Could anybody write the expression]
 +
. We will find that <math>\overrightarrow{BD}</math> is an angle bisector of <math>\triangle ABC</math> (because we will get <math>\tan(x) = 1</math>).   
 +
Therefore by converse of angle bisector theorem <math>AB:BC = 1:3</math>. By using Pythagorean theorem, we have values of <math>AB</math> and <math>AC</math>.
 +
Computing <math>AB \cdot AC = 120</math>. Adding the areas of <math>ABC</math> and <math>ACD</math>, hence the answer is <math>\boxed{\textbf{(D)}\:360}</math>.
 +
 +
By: Math-Amaze
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Latex: Catoptrics.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 22:22, 18 March 2020

The following problem is from both the 2020 AMC 12A #18 and 2020 AMC 10A #20, so both problems redirect to this page.

Problem

Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$

$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$

Solution 1 (Just Drop An Altitude)

[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1,1.5), N); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((0,0)--(1,1.5), dashed); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (1,1.5)--(1.714,1.143), NE); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); draw(rightanglemark((0,0),(1,1.5),(0,2))); [/asy]

It's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$, we get $[ACD]=300$. Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$. Let $FE=x$. Since $AE=5$, then $AF=5-x$. By dropping this altitude, we can also see two similar triangles, $BFE$ and $DCE$. Since $EC$ is $20-5=15$, and $DC=30$, we get that $BF=2x$. Now, if we redraw another diagram just of $ABC$, we get that $(2x)^2=(5-x)(15+x)$. Now expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$. This factors to $(x+5)(x-3)$. Since lengths cannot be negative, $x=3$. Since $x=3$, $[ABC]=60$. So $[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}$

(I'm very sorry if you're a visual learner but now you have a diagram by ciceronii)

~ Solution by Ultraman

~ Diagram by ciceronii

Solution 2 (coordinates)

[asy] size(10cm,0); draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); draw((-20,0)--(20,0)); draw((0,-15)--(0,35)); draw((10,30)--(-8,-6)); draw(circle((0,0),10)); label("E",(-4.05,-.25),S); label("D",(10,30),NE); label("C",(10,0),NE); label("B",(-8,-6),SW); label("A",(-10,0),NW); label("5",(-10,0)--(-5,0), NE); label("15",(-5,0)--(10,0), N); label("30",(10,0)--(10,30), E); dot((-5,0)); dot((-10,0)); dot((-8,-6)); dot((10,0)); dot((10,30)); [/asy] Let the points be $A(-10,0)$, $\:B(x,y)$, $\:C(10,0)$, $\:D(10,30)$,and $\:E(-5,0)$, respectively. Since $B$ lies on line $DE$, we know that $y=2x+10$. Furthermore, since $\angle{ABC}=90^\circ$, $B$ lies on the circle with diameter $AC$, so $x^2+y^2=100$. Solving for $x$ and $y$ with these equations, we get the solutions $(0,10)$ and $(-8,-6)$. We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}$.

Solution 3 (Trigonometry)

Let $\angle C = \angle{ACB}$ and $\angle{B} = \angle{CBE}.$ Using Law of Sines on $\triangle{BCE}$ we get \[\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}\] and LoS on $\triangle{ABE}$ yields \[\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.\] Divide the two to get $\tan{B} = 3 \tan{C}.$ Now, \[\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}\] and solve the quadratic, taking the positive solution (C is acute) to get $\tan{C} = \frac{1}{3}.$ So if $AB = a,$ then $BC = 3a$ and $[ABC] = \frac{3a^2}{2}.$ By Pythagorean Theorem, $10a^2 = 400 \iff \frac{3a^2}{2} = 60$ and the answer is $300 + 60 \iff \boxed{\textbf{(D)}}.$

(This solution is incomplete, can someone complete it please-Lingjun) ok Latex edited by kc5170

We could use the famous m-n rule in trigonometry in $\triangle ABC$ with Point $E$ [Unable to write it here.Could anybody write the expression] . We will find that $\overrightarrow{BD}$ is an angle bisector of $\triangle ABC$ (because we will get $\tan(x) = 1$). Therefore by converse of angle bisector theorem $AB:BC = 1:3$. By using Pythagorean theorem, we have values of $AB$ and $AC$. Computing $AB \cdot AC = 120$. Adding the areas of $ABC$ and $ACD$, hence the answer is $\boxed{\textbf{(D)}\:360}$.

By: Math-Amaze Latex: Catoptrics.

Video Solution

On The Spot STEM https://www.youtube.com/watch?v=hIdNde2Vln4

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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
2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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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