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 == | == Problem == | ||
− | Quadrilateral <math>ABCD</math> satisfies <math>\angle ABC = \angle ACD = 90^{\circ}, AC=20,</math> and <math>CD=30.</math> Diagonals <math>\overline{AC}</math> and | + | Quadrilateral <math>ABCD</math> 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 == | ||
+ | |||
+ | <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); | ||
+ | label("5$-$$x$", (1,1.5)--(0,2), 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>\triangle BFE \sim \triangle 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 in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into. | ||
+ | |||
+ | 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>, which has roots of <math>x=-5, 3</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, that means the altitude <math>BF=2\cdot3=6</math>, or <math>[ABC]=60</math>. Thus <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 (Law of Cosines)== | ||
+ | |||
+ | <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 Cosines: | ||
+ | <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 5 (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 | ||
+ | |||
+ | == Solution 6 (Power of a Point)== | ||
+ | |||
+ | <asy> | ||
+ | import olympiad; | ||
+ | pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798),F=(285,94.5),G=(361.2,361.2); | ||
+ | dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);dot("$F$",F,N);dot("$G$",G,N); | ||
+ | draw(A--B--C--D--A); | ||
+ | draw(A--C, dotted); draw(B--D, dotted); draw(F--G, dotted); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>F</math> be the midpoint of <math>AC</math>, and draw <math>FG // CD</math> where <math>G</math> is on <math>BD</math>. We have <math>EF=5,FC=10</math>. | ||
+ | |||
+ | <math>\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC</math>. Therefore <math>ABCG</math> is a cyclic quadrilateral. | ||
+ | |||
+ | Notice that <math>\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}</math> via Power of a Point. | ||
+ | |||
+ | The altitude from <math>B</math> to <math>AC</math> is then equal to <math>GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6</math>. | ||
+ | |||
+ | Finally, the total area of <math>ABCD</math> is equal to <math>\frac 12 \cdot 20 \left(30+6 \right) =\boxed{\text{(D) } 360}.</math> | ||
+ | |||
+ | ~asops | ||
+ | |||
+ | ==Solution 7 (Solving Equations)== | ||
+ | |||
+ | <asy> | ||
+ | size(15cm,0); | ||
+ | import olympiad; | ||
+ | draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); | ||
+ | label("A", (0,2), NW); | ||
+ | label("B", (0,0), SW); | ||
+ | label("C", (4,0), SE); | ||
+ | label("D", (6,4), NE); | ||
+ | label("E", (1.714,1.143), N); | ||
+ | label("F", (1.714,0), SE); | ||
+ | draw((0,2)--(4,0), dashed); | ||
+ | draw((0,0)--(6,4), dashed); | ||
+ | draw((1.714,1.143)--(1.714,0), dashed); | ||
+ | label("20", (0,2)--(4,0), SW); | ||
+ | label("30", (4,0)--(6,4), SE); | ||
+ | label("$x$", (-0.3,2)--(-0.3,0), N); | ||
+ | label("$y$", (0,-0.3)--(4,-0.3), E); | ||
+ | draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); | ||
+ | draw(rightanglemark((0,2),(0,0),(4,0))); | ||
+ | draw(rightanglemark((0,2),(4,0),(6,4))); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>AB = x</math>, <math>BC = y</math> | ||
+ | |||
+ | Looking at the diagram we have <math>x^2 + y^2 = 20^2</math>, | ||
+ | <math>DE = \sqrt{30^2+15^2} = 15\sqrt{5}</math>, <math>[ACD] = \frac{1}{2} \cdot 20 \cdot 30 = 300</math> | ||
+ | |||
+ | Because <math>\triangle CEF \sim \triangle CAB</math>, <math>EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}</math> | ||
+ | |||
+ | <math>BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = \frac{1}{4} \cdot BC = \frac{y}{4}</math> | ||
+ | |||
+ | <math>BE = \sqrt{ \left( \frac{3x}{4} \right) ^2 + \left( \frac{y}{4} \right) ^2 } = \frac{ \sqrt{9x^2 + y^2} }{4}</math> , substituting <math>x^2 + y^2 = 400</math>, we get <math>BE = \frac{ \sqrt{8x^2 + 400} }{4} = \frac{ \sqrt{2x^2 + 100} }{2}</math> | ||
+ | |||
+ | <math>[ABC] = \frac{1}{2} \cdot x \cdot y</math> | ||
+ | |||
+ | Because <math>\triangle ABC</math> and <math>\triangle ACD</math> share the same base, <math>\frac{[ABC]}{[ACD]} = \frac{BE}{DE}</math> | ||
+ | |||
+ | <math>[ABC] = [ACD] \cdot \frac{BE}{DE} = 300 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ 15 \sqrt{5} }</math> | ||
+ | |||
+ | <math>\frac{1}{2} \cdot x \cdot y = 20 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ \sqrt{5} }</math> | ||
+ | |||
+ | <math>xy = 4 \sqrt{10x^2 + 500}</math> | ||
+ | |||
+ | By <math>x^2 + y^2 = 400</math>, <math>y = \sqrt{400 - x^2}</math>. So, <math>x \cdot \sqrt{400 - x^2} = 4 \sqrt{10x^2 + 500}</math> | ||
+ | |||
+ | <math>x^2 (400 - x^2) = 16 (10x^2 + 500)</math> | ||
+ | |||
+ | Let <math>x^2 = a</math>, <math>a (400 - a) = 16 (10a + 500)</math>, <math>400a - a^2 = 160a + 8000</math>, <math>a^2 - 240a + 8000 = 0</math>, <math>(a-200)(a-40) = 0</math> | ||
+ | |||
+ | Because <math>x < 20</math>, <math>a</math> can only equal 40. <math>a = 40</math>, <math>x = 2 \sqrt{10}</math>, <math>y = 6 \sqrt{10}</math> | ||
+ | |||
+ | <math>[ABC] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} = 60</math> | ||
+ | |||
+ | <math>[ABCD] = [ABC] + [ACD] = 60 + 300 = \boxed{\text{(D) } 360}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 8== | ||
+ | Drop perpendiculars <math>\overline{AF}</math> and <math>\overline{CG}</math> to <math>\overline{BD}.</math> Notice that since <math>\angle AEF=\angle CEG</math> (since they are vertical angles) and <math>\angle AFE=\angle CGE=90^\circ,</math> triangles <math>AEF</math> and <math>CEG</math> are similar. Therefore, we have | ||
+ | |||
+ | <cmath>x/EF=CE/AE=15/5=3,</cmath> | ||
+ | |||
+ | where <math>EG=x.</math> Therefore, <math>EF=x/3.</math> | ||
+ | |||
+ | Additionally, angle chasing shows that triangles <math>CEG</math> and <math>DCG</math> are also similar. This gives <math>CG/x=DC/CE=30/15=2,</math> so <math>CG=2x.</math> Thus, applying the Pythagorean Theorem to triangle <math>CEG</math> gives | ||
+ | |||
+ | <cmath>x^2+(2x)^2=15^2,</cmath> | ||
+ | |||
+ | so <math>EG=x=3\sqrt 5.</math> Our pairs of similar triangles then allow us to fill in the following lengths (in this order): | ||
+ | |||
+ | <cmath>EF=x/3=\sqrt 5, CG=2x=6\sqrt 5, AF=CG/3=2\sqrt 5, DG=2\cdot CG=12\sqrt 5.</cmath> | ||
+ | |||
+ | Now, let <math>BF=y.</math> Angle chasing shows that triangle <math>ABF</math> and <math>BCG</math> are similar, so <math>BG/AF=CG/BF.</math> Plugging in known lengths gives | ||
+ | |||
+ | <cmath>\dfrac{y+4\sqrt 5}{2\sqrt 5}=\dfrac{6\sqrt 5}{y}.</cmath> | ||
+ | |||
+ | This gives <math>y=2\sqrt 5.</math> Now we know all the lengths that make up <math>BD,</math> which allows us to find | ||
+ | |||
+ | <cmath>BD=2\sqrt 5+\sqrt 5+3\sqrt 5+12\sqrt 5=18\sqrt 5.</cmath> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | [ABCD] &= [ABD]+[CBD] \\ | ||
+ | &= (BD)(AF)/2+(BD)(CG)/2 \\ | ||
+ | &= (18\sqrt 5)(2\sqrt 5)/2+(18\sqrt 5)(6\sqrt 5)/2 \\ | ||
+ | &= \boxed{\text{(D) } 360}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | --vaporwave | ||
+ | |||
+ | ==Solution 9 Trigonometry == | ||
+ | [[Image:2020_AMC_10A_Problem_20.png|thumb|center|600px]] | ||
+ | |||
+ | set <cmath>\angle ACB = \theta , C= \cos(\theta), S = \sin(\theta) </cmath> | ||
+ | |||
+ | <cmath>\dfrac{E_y}{E_x} = \dfrac{30C} { 20C+30S} = \dfrac{15S} {20C-15C} </cmath> | ||
+ | |||
+ | <cmath>2SC = \dfrac35</cmath> | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | [ABCD] &= [ABD]+[CBD] \\ | ||
+ | &= \dfrac12\cdot 20C\cdot 30C + \dfrac12 \cdot 20S (20C+30S) \\ | ||
+ | &= 100\cdot 2SC + 300 \\ | ||
+ | &= \boxed{\text{(D) } 360}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Video Solution by Education, The Study of Everything== | ||
+ | https://youtu.be/5lb8kk1qbaA | ||
+ | |||
+ | ==Video Solution by On The Spot STEM== | ||
+ | https://www.youtube.com/watch?v=hIdNde2Vln4 | ||
+ | |||
+ | ==Video Solution by MathEx== | ||
+ | https://www.youtube.com/watch?v=sHrjx968ZaM | ||
+ | |||
+ | ==Video Solution by TheBeautyOfMath== | ||
+ | https://www.youtube.com/watch?v=RKlG6oZq9so | ||
− | == Solution == | + | ==Video Solution by Triviality== |
+ | https://youtu.be/R220vbM_my8?t=658 | ||
+ | (amritvignesh0719062.0) | ||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/hDsoyvFWYxc?t=1224 | ||
+ | ~ pi_is_3.14 | ||
==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 13:36, 25 August 2024
- The following problem is from both the 2020 AMC 12A #18 and 2020 AMC 10A #20, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Coordinates)
- 4 Solution 3 (Trigonometry)
- 5 Solution 4 (Law of Cosines)
- 6 Solution 5 (Vectors / Coordinates)
- 7 Solution 6 (Power of a Point)
- 8 Solution 7 (Solving Equations)
- 9 Solution 8
- 10 Solution 9 Trigonometry
- 11 Video Solution by Education, The Study of Everything
- 12 Video Solution by On The Spot STEM
- 13 Video Solution by MathEx
- 14 Video Solution by TheBeautyOfMath
- 15 Video Solution by Triviality
- 16 Video Solution by OmegaLearn
- 17 See Also
Problem
Quadrilateral satisfies
and
Diagonals
and
intersect at point
and
What is the area of quadrilateral
Solution 1
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, . 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 in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into.
Expanding, simplifying, and dividing by the GCF, we get . This factors to
, which has roots of
. Since lengths cannot be negative,
. Since
, that means the altitude
, or
. Thus
~ 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 (Law of Cosines)
Denote as
. By the Law of Cosines:
Adding these up yields:
By the quadratic formula,
.
Observe:
.
Thus the desired area is
~qwertysri987
Solution 5 (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
Solution 6 (Power of a Point)
Let be the midpoint of
, and draw
where
is on
. We have
.
. Therefore
is a cyclic quadrilateral.
Notice that via Power of a Point.
The altitude from to
is then equal to
.
Finally, the total area of is equal to
~asops
Solution 7 (Solving Equations)
Let ,
Looking at the diagram we have ,
,
Because ,
, substituting
, we get
Because and
share the same base,
By ,
. So,
Let ,
,
,
,
Because ,
can only equal 40.
,
,
Solution 8
Drop perpendiculars and
to
Notice that since
(since they are vertical angles) and
triangles
and
are similar. Therefore, we have
where Therefore,
Additionally, angle chasing shows that triangles and
are also similar. This gives
so
Thus, applying the Pythagorean Theorem to triangle
gives
so Our pairs of similar triangles then allow us to fill in the following lengths (in this order):
Now, let Angle chasing shows that triangle
and
are similar, so
Plugging in known lengths gives
This gives Now we know all the lengths that make up
which allows us to find
Therefore,
--vaporwave
Solution 9 Trigonometry
set
Video Solution by Education, The Study of Everything
Video Solution by On The Spot STEM
https://www.youtube.com/watch?v=hIdNde2Vln4
Video Solution by MathEx
https://www.youtube.com/watch?v=sHrjx968ZaM
Video Solution by TheBeautyOfMath
https://www.youtube.com/watch?v=RKlG6oZq9so
Video Solution by Triviality
https://youtu.be/R220vbM_my8?t=658 (amritvignesh0719062.0)
Video Solution by OmegaLearn
https://youtu.be/hDsoyvFWYxc?t=1224 ~ pi_is_3.14
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 |
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