Difference between revisions of "2020 AMC 10A Problems/Problem 20"
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+ | {{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #18]] and [[2020 AMC 10A Problems|2020 AMC 10A #20]]}} | ||
+ | |||
+ | == Problem == | ||
+ | Quadrilateral ABCD satisfies <math>\angle ABC = \angle ACD = 90^{\circ}, AC=20,</math> and <math>CD=30.</math> Diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at point <math>E,</math> and <math>AE=5.</math> What is the area of quadrilateral <math>ABCD?</math> | ||
+ | |||
+ | <math>\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370</math> | ||
+ | |||
+ | == 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 <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> because of the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into. 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> | ||
+ | |||
+ | |||
+ | |||
+ | ~ 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 <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 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> | ||
+ | |||
+ | (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 <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 | ||
+ | Latex: Catoptrics. | ||
+ | |||
+ | ==Solution 4 (Answer Choices)== | ||
+ | We know that the big triangle has area 300. Using the answer choices 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. We 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 <math>60</math>, so the answer is <math>\boxed {\textbf{(D) }360}</math>. | ||
+ | |||
+ | ~tigershark22 | ||
+ | |||
+ | ==Solution 5 (LoC)== | ||
+ | |||
+ | <asy> | ||
+ | import olympiad; | ||
+ | pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798); | ||
+ | dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N); | ||
+ | draw(A--B--C--D--A); | ||
+ | draw(A--C, dotted); draw(B--D, dotted); | ||
+ | </asy> | ||
+ | |||
+ | Denote <math>EB</math> as <math>x</math>. By the Law of Cosine: | ||
+ | <cmath>AB^2 = 25 + x^2 - 10x\cos(\angle DEC)</cmath> | ||
+ | <cmath>BC^2 = 225 + x^2 + 30x\cos(\angle DEC)</cmath> | ||
+ | |||
+ | Adding these up yields: | ||
+ | <cmath>400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0</cmath> | ||
+ | By the quadratic formula, <math>x = 3\sqrt5</math>. | ||
+ | |||
+ | Observe: | ||
+ | <cmath>[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60</cmath>. | ||
+ | |||
+ | Thus the desired area is <math>\frac{1}{2}(30)(20) + 60 = \boxed{\textbf{(D) } 360}</math> | ||
+ | |||
+ | ~qwertysri987 | ||
+ | |||
+ | ==Solution 6 (basic vectors/coordinates)== | ||
+ | |||
+ | Let <math>C = (0, 0)</math> and <math>D = (0, 30)</math>. Then <math>E = (-15, 0), A = (-20, 0),</math> and <math>B</math> lies on the line <math>y=2x+30.</math> So the coordinates of <math>B</math> are <cmath>(x, 2x+30).</cmath> | ||
+ | |||
+ | We can make this a vector problem. | ||
+ | <math>\overrightarrow{\mathbf{B}} = \begin{pmatrix} | ||
+ | x \\ | ||
+ | 2x+30 | ||
+ | \end{pmatrix}.</math> We notice that point <math>B</math> forms a right angle, meaning vectors <math>\overrightarrow{\mathbf{BC}}</math> and <math>\overrightarrow{\mathbf{BA}}</math> are orthogonal, and their dot-product is <math>0</math>. | ||
+ | |||
+ | We determine <math>\overrightarrow{\mathbf{BC}}</math> and <math>\overrightarrow{\mathbf{BA}}</math> to be <math>\begin{pmatrix} | ||
+ | -x \\ | ||
+ | -2x-30 | ||
+ | \end{pmatrix}</math> and <math>\begin{pmatrix} | ||
+ | -20-x \\ | ||
+ | -2x-30 | ||
+ | \end{pmatrix}</math> , respectively. (To get this, we use the fact that <math>\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}</math> and similarly, <math>\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.</math> ) | ||
+ | |||
+ | Equating the cross-product to <math>0</math> gets us the quadratic <math>-x(-20-x)+(-2x-30)(-2x-30)=0.</math> The solutions are <math>x=-18, -10.</math> Since <math>B</math> clearly has a more negative x-coordinate than <math>E</math>, we take <math>x=-18</math>. So <math>B = (-18, -6).</math> | ||
+ | |||
+ | From here, there are multiple ways to get the area of <math>\Delta{ABC}</math> to be <math>60</math>, and since the area of <math>\Delta{ACD}</math> is <math>300</math>, we get our final answer to be <cmath>60 + 300 = \boxed{\text{(D) } 360}.</cmath> | ||
+ | |||
+ | -PureSwag | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | On The Spot STEM | ||
+ | |||
+ | https://www.youtube.com/watch?v=hIdNde2Vln4 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://www.youtube.com/watch?v=sHrjx968ZaM&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=2 ~ MathEx | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | Education, The Study of Everything | ||
+ | |||
+ | https://youtu.be/5lb8kk1qbaA | ||
+ | |||
+ | ==Video Solution 4== | ||
+ | The Beauty Of Math | ||
+ | https://www.youtube.com/watch?v=RKlG6oZq9so&ab_channel=TheBeautyofMath | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}} | ||
+ | {{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:19, 17 January 2021
- The following problem is from both the 2020 AMC 12A #18 and 2020 AMC 10A #20, so both problems redirect to this page.
Contents
Problem
Quadrilateral ABCD satisfies and Diagonals and intersect at point and What is the area of quadrilateral
Solution 1 (Just Drop An Altitude)
It's crucial to draw a good diagram for this one. Since and , we get . Now we need to find to get the area of the whole quadrilateral. Drop an altitude from to and call the point of intersection . Let . Since , then . By dropping this altitude, we can also see two similar triangles, and . Since is , and , we get that . Now, if we redraw another diagram just of , we get that because of the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into. Now expanding, simplifying, and dividing by the GCF, we get . This factors to . Since lengths cannot be negative, . Since , . So
~ Solution by Ultraman
~ Diagram by ciceronii
Solution 2 (Coordinates)
Let the points be , , , ,and , respectively. Since lies on line , we know that . Furthermore, since , lies on the circle with diameter , so . Solving for and with these equations, we get the solutions and . We immediately discard the solution as should be negative. Thus, we conclude that .
Solution 3 (Trigonometry)
Let and Using Law of Sines on we get and LoS on yields Divide the two to get Now, and solve the quadratic, taking the positive solution (C is acute) to get So if then and By Pythagorean Theorem, and the answer is
(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 with Point [Unable to write it here.Could anybody write the expression] . We will find that is an angle bisector of (because we will get ). Therefore by converse of angle bisector theorem . By using Pythagorean theorem, we have values of and . Computing . Adding the areas of and , hence the answer is .
By: Math-Amaze Latex: Catoptrics.
Solution 4 (Answer Choices)
We know that the big triangle has area 300. Using the answer choices 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. We guess that the legs are equal to and , and because the hypotenuse is 20 we get . Testing small numbers, we get that when and , is indeed a square. The area of the triangle is thus , so the answer is .
~tigershark22
Solution 5 (LoC)
Denote as . By the Law of Cosine:
Adding these up yields: By the quadratic formula, .
Observe: .
Thus the desired area is
~qwertysri987
Solution 6 (basic vectors/coordinates)
Let and . Then and lies on the line So the coordinates of are
We can make this a vector problem. We notice that point forms a right angle, meaning vectors and are orthogonal, and their dot-product is .
We determine and to be and , respectively. (To get this, we use the fact that and similarly, )
Equating the cross-product to gets us the quadratic The solutions are Since clearly has a more negative x-coordinate than , we take . So
From here, there are multiple ways to get the area of to be , and since the area of is , we get our final answer to be
-PureSwag
Video Solution 1
On The Spot STEM
https://www.youtube.com/watch?v=hIdNde2Vln4
Video Solution 2
https://www.youtube.com/watch?v=sHrjx968ZaM&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=2 ~ MathEx
Video Solution 3
Education, The Study of Everything
Video Solution 4
The Beauty Of Math https://www.youtube.com/watch?v=RKlG6oZq9so&ab_channel=TheBeautyofMath
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
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 (Problems • Answer Key • Resources) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.