2020 AMC 10A Problems/Problem 20

Revision as of 09:51, 19 April 2021 by MRENTHUSIASM (talk | contribs) (Solution 1 (Just Drop An Altitude): Deleted extra spaces.)
The following problem is from both the 2020 AMC 12A #18 and 2020 AMC 10A #20, so both problems redirect to this page.


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 $(2x)^2=(5-x)(15+x)$ 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 (for any right triangle). 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 $[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}$

~ 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.

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 $\sqrt{20a}$ and $\sqrt{20b}$, and because the hypotenuse is 20 we get $a+b=20$. Testing small numbers, we get that when $a=2$ and $b=18$, $ab$ is indeed a square. The area of the triangle is thus $60$, so the answer is $\boxed {\textbf{(D) }360}$.


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 $EB$ as $x$. By the Law of Cosine: \[AB^2 = 25 + x^2 - 10x\cos(\angle DEC)\] \[BC^2 = 225 + x^2 + 30x\cos(\angle DEC)\]

Adding these up yields: \[400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0\] By the quadratic formula, $x = 3\sqrt5$.

Observe: \[[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60\].

Thus the desired area is $\frac{1}{2}(30)(20) + 60 = \boxed{\textbf{(D) } 360}$


Solution 6 (basic vectors/coordinates)

Let $C = (0, 0)$ and $D = (0, 30)$. Then $E = (-15, 0), A = (-20, 0),$ and $B$ lies on the line $y=2x+30.$ So the coordinates of $B$ are \[(x, 2x+30).\]

We can make this a vector problem. $\overrightarrow{\mathbf{B}} = \begin{pmatrix} x \\ 2x+30 \end{pmatrix}.$ We notice that point $B$ forms a right angle, meaning vectors $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ are orthogonal, and their dot-product is $0$.

We determine $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ to be $\begin{pmatrix} -x \\ -2x-30 \end{pmatrix}$ and $\begin{pmatrix} -20-x \\ -2x-30 \end{pmatrix}$ , respectively. (To get this, we use the fact that $\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}$ and similarly, $\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.$ )

Equating the cross-product to $0$ gets us the quadratic $-x(-20-x)+(-2x-30)(-2x-30)=0.$ The solutions are $x=-18, -10.$ Since $B$ clearly has a more negative x-coordinate than $E$, we take $x=-18$. So $B = (-18, -6).$

From here, there are multiple ways to get the area of $\Delta{ABC}$ to be $60$, and since the area of $\Delta{ACD}$ is $300$, we get our final answer to be \[60 + 300 = \boxed{\text{(D) } 360}.\]


Video Solutions

Video Solution 1

On The Spot STEM


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

Video Solution 5



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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