Difference between revisions of "2019 AMC 8 Problems/Problem 24"
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==Problem 24== | ==Problem 24== | ||
− | In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> | + | In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> so that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and let <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>? |
<asy> | <asy> | ||
Line 33: | Line 30: | ||
==Solution 1== | ==Solution 1== | ||
− | Draw <math>X</math> | + | |
+ | Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}</math> | ||
==Solution 2 (Mass Points)== | ==Solution 2 (Mass Points)== | ||
Line 82: | Line 80: | ||
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. | First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. | ||
− | + | The triangle we will consider is <math>\triangle ABC</math> (obviously), and we will let <math>E</math> be the center of mass, so that <math>D</math> balances <math>A</math> and <math>C</math> (this is true since <math>E</math> balances <math>B</math> and <math>D</math>, but <math>E</math> also balances <math>A</math> and <math>B</math> and <math>C</math> so <math>D</math> balances <math>A</math> and <math>C</math>), and <math>F</math> balances <math>B</math> and <math>C</math>. | |
+ | |||
+ | We know that <math>AD:CD=1:2</math> and <math>D</math> balances <math>A</math> and <math>C</math> so we assign <math>2</math> to <math>A</math> and <math>1</math> to <math>C</math>. Then, since <math>D</math> balances <math>A</math> and <math>C</math>, we get <math>D = A + C = 2 + 1 = 3</math> (by mass points addition). | ||
+ | |||
+ | Next, since <math>E</math> balances <math>B</math> and <math>D</math> in a ratio of <math>BE:DE=1:1</math>, we know that <math>B=D=3</math>. Similarly, by mass points addition, <math>E=B+D=3+3=6</math>. | ||
+ | |||
+ | Finally, <math>F</math> balances <math>B</math> and <math>C</math> so <math>F=B+C=3+1=4</math>. We can confirm we have done everything right by noting that <math>E</math> balances <math>A</math> and <math>F</math>, so <math>E</math> should equal <math>A+F</math>, which it does. | ||
− | Now, we | + | Now that our points have weights, we can solve the problem. <math>BF:FC=1:3</math> so <math>BF:BC=1:4</math> so <math>[ABF]=\frac{1}{4}[ABC]=90</math>. Also, <math>EF:EA=2:4=1:2</math> so <math>EF:AF=1:3</math> so <math>[EBF]=\frac{1}{3}[ABF]=30</math>. |
− | + | So we get the area of <math>EBF</math> as <math>\boxed{\textbf{(B) }30}</math> | |
+ | -Firebolt360 and Brudder | ||
− | + | Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{\textbf{(B) }30}</math> | |
− | |||
− | Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{B}</math> | ||
-Brudder | -Brudder | ||
==Solution 3== | ==Solution 3== | ||
− | <math>\frac{BF}{FC}</math> is equal to <math>\frac{The area of triangle ABE}{The area of triangle ACE}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{(B) 30}</math>. ~~SmileKat32 | + | <math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{\textbf{(B) }30}</math>. ~~SmileKat32 |
+ | |||
==Solution 4 (Similar Triangles)== | ==Solution 4 (Similar Triangles)== | ||
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown: | Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown: | ||
Line 118: | Line 122: | ||
label("$G$", G, ENE); | label("$G$", G, ENE); | ||
</asy> | </asy> | ||
− | Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math> | + | Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>. |
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>. | Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>. | ||
Line 148: | Line 152: | ||
</asy> | </asy> | ||
(Credit to MP8148 for the idea) | (Credit to MP8148 for the idea) | ||
+ | |||
+ | ==Solution 5 (Area Ratios)== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A, B, C, D, E, F; | ||
+ | B = (0,0); | ||
+ | A = (2, 3); | ||
+ | C = (5, 0); | ||
+ | D = (3, 2); | ||
+ | E = (1.5, 1); | ||
+ | F = (1.25, 0); | ||
+ | |||
+ | draw(A--B--C--A--D--B); | ||
+ | draw(A--F); | ||
+ | draw(E--C); | ||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, WSW); | ||
+ | label("$C$", C, ESE); | ||
+ | label("$D$", D, dir(0)*1.5); | ||
+ | label("$E$", E, SSE); | ||
+ | label("$F$", F, S); | ||
+ | label("$60$", (A+E+D)/3); | ||
+ | label("$60$", (A+E+B)/3); | ||
+ | label("$120$", (D+E+C)/3); | ||
+ | label("$x$", (B+E+F)/3); | ||
+ | label("$120-x$", (F+E+C)/3); | ||
+ | </asy> | ||
+ | As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. | ||
+ | (Credit to scrabbler94 for the idea) | ||
+ | |||
+ | ==Solution 6 (Coordinate Bashing)== | ||
+ | Let <math>ADB</math> be a right triangle, and <math>BD=CD</math> | ||
+ | |||
+ | Let <math>A=(-2\sqrt{30}, 0)</math> | ||
+ | |||
+ | <math>B=(0, 4\sqrt{30})</math> | ||
+ | |||
+ | <math>C=(4\sqrt{30}, 0)</math> | ||
+ | |||
+ | <math>D=(0, 0)</math> | ||
+ | |||
+ | <math>E=(0, 2\sqrt{30})</math> | ||
+ | |||
+ | <math>F=(\sqrt{30}, 3\sqrt{30})</math> | ||
+ | |||
+ | The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math> | ||
+ | |||
+ | The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math> | ||
+ | |||
+ | Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math> | ||
+ | |||
+ | Now we can find <math>EF=BF=2\sqrt{15}</math> | ||
+ | |||
+ | <math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare</math> | ||
+ | |||
+ | -Trex4days | ||
+ | |||
+ | == Solution 7 == | ||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
+ | import graph; size(15cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */ | ||
+ | |||
+ | /* draw figures */ | ||
+ | draw(circle((0,0), 5), linewidth(2)); | ||
+ | draw((-4,-3)--(4,3), linewidth(2)); | ||
+ | draw((-4,-3)--(0,5), linewidth(2)); | ||
+ | draw((0,5)--(4,3), linewidth(2)); | ||
+ | draw((12,-1)--(-4,-3), linewidth(2)); | ||
+ | draw((0,5)--(0,-5), linewidth(2)); | ||
+ | draw((-4,-3)--(0,-5), linewidth(2)); | ||
+ | draw((4,3)--(0,2.48), linewidth(2)); | ||
+ | draw((4,3)--(12,-1), linewidth(2)); | ||
+ | draw((-4,-3)--(4,3), linewidth(2)); | ||
+ | /* dots and labels */ | ||
+ | dot((0,0),dotstyle); | ||
+ | label("E", (0.27,-0.24), NE * labelscalefactor); | ||
+ | dot((-5,0),dotstyle); | ||
+ | dot((-4,-3),dotstyle); | ||
+ | label("B", (-4.45,-3.38), NE * labelscalefactor); | ||
+ | dot((4,3),dotstyle); | ||
+ | label("$D$", (4.15,3.2), NE * labelscalefactor); | ||
+ | dot((0,5),dotstyle); | ||
+ | label("A", (-0.09,5.26), NE * labelscalefactor); | ||
+ | dot((12,-1),dotstyle); | ||
+ | label("C", (12.23,-1.24), NE * labelscalefactor); | ||
+ | dot((0,-5),dotstyle); | ||
+ | label("$G$", (0.19,-4.82), NE * labelscalefactor); | ||
+ | dot((0,2.48),dotstyle); | ||
+ | label("I", (-0.33,2.2), NE * labelscalefactor); | ||
+ | dot((0,0),dotstyle); | ||
+ | label("E", (0.27,-0.24), NE * labelscalefactor); | ||
+ | dot((0,-2.5),dotstyle); | ||
+ | label("F", (0.23,-2.2), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | Let <math>A[\Delta XYZ]</math> = <math>\text{Area of Triangle XYZ}</math> | ||
+ | |||
+ | |||
+ | <math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math> | ||
+ | |||
+ | |||
+ | <math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts) | ||
+ | |||
+ | |||
+ | Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>. | ||
+ | |||
+ | |||
+ | <math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic) | ||
+ | |||
+ | |||
+ | But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>. | ||
+ | |||
+ | \therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem). | ||
+ | |||
+ | In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>. | ||
+ | |||
+ | Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math> | ||
+ | |||
+ | |||
+ | Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>. | ||
+ | |||
+ | <math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>. | ||
+ | |||
+ | Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math> | ||
+ | |||
+ | Thus <math>A[\Delta BEF] = \boxed{\textbf{(B) }30}</math> | ||
+ | |||
+ | |||
+ | ~phoenixfire & flamewavelight | ||
+ | |||
+ | == Solution 8 == | ||
+ | <asy> | ||
+ | import geometry; | ||
+ | unitsize(2cm); | ||
+ | pair A,B,C,DD,EE,FF, M; | ||
+ | B = (0,0); C = (3,0); M = (1.45,0); | ||
+ | A = (1.2,1.7); | ||
+ | DD = (2/3)*A+(1/3)*C; | ||
+ | EE = (B+DD)/2; | ||
+ | FF = intersectionpoint(B--C,A--A+2*(EE-A)); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--FF); | ||
+ | draw(B--DD);dot(A); | ||
+ | label("$A$",A,N); | ||
+ | dot(B); | ||
+ | label("$B$", | ||
+ | B,SW);dot(C); | ||
+ | label("$C$",C,SE); | ||
+ | dot(DD); | ||
+ | label("$D$",DD,NE); | ||
+ | dot(EE); | ||
+ | label("$E$",EE,NW); | ||
+ | dot(FF); | ||
+ | label("$F$",FF,S); | ||
+ | draw(EE--M,StickIntervalMarker(1,1)); | ||
+ | label("$M$",M,S); | ||
+ | draw(A--DD,invisible,StickIntervalMarker(1,1)); | ||
+ | dot((DD+C)/2); | ||
+ | draw(DD--C,invisible,StickIntervalMarker(2,1)); | ||
+ | </asy> | ||
+ | Using the ratio of <math>\overline{AD}</math> and <math>\overline{CD}</math>, we find the area of <math>\triangle ADB</math> is <math>120</math> and the area of <math>\triangle BDC</math> is <math>240</math>. Also using the fact that <math>E</math> is the midpoint of <math>\overline{BD}</math>, we know <math>\triangle ADE = \triangle ABE = 60</math>. | ||
+ | Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. We immediatley know that <math>\triangle BEM \sim BDC</math> by <math>2</math>. Using that we can conclude <math>EM</math> has ratio <math>1</math>. Using <math>\triangle EFM \sim \triangle AFC</math>, we get <math>EF:AE = 1:2</math>. Therefore using the fact that <math>\triangle EBF</math> is in <math>\triangle ABF</math>, the area has ratio <math>\triangle BEF : \triangle ABE=1:2</math> and we know <math>\triangle ABE</math> has area <math>60</math> so <math>\triangle BEF</math> is <math>\boxed{\textbf{(B) }30}</math>. - fath2012 | ||
+ | |||
+ | ==Solution 9 (Menelaus's Theorem)== | ||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair A,B,C,DD,EE,FF; | ||
+ | B = (0,0); C = (3,0); | ||
+ | A = (1.2,1.7); | ||
+ | DD = (2/3)*A+(1/3)*C; | ||
+ | EE = (B+DD)/2; | ||
+ | FF = intersectionpoint(B--C,A--A+2*(EE-A)); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--FF); | ||
+ | draw(B--DD);dot(A); | ||
+ | label("$A$",A,N); | ||
+ | dot(B); | ||
+ | label("$B$", | ||
+ | B,SW);dot(C); | ||
+ | label("$C$",C,SE); | ||
+ | dot(DD); | ||
+ | label("$D$",DD,NE); | ||
+ | dot(EE); | ||
+ | label("$E$",EE,NW); | ||
+ | dot(FF); | ||
+ | label("$F$",FF,S); | ||
+ | </asy> | ||
+ | By Menelaus's Theorem on triangle <math>BCD</math>, we have <cmath>\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.</cmath> Therefore, <cmath>[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.</cmath> | ||
+ | |||
+ | ==Solution 10 (Graph Paper)== | ||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair A,B,C,D,E,F,a,b,c,d,e,f; | ||
+ | A = (2,3); | ||
+ | B = (0,2); | ||
+ | C = (2,0); | ||
+ | D = (2/3)*A+(1/3)*C; | ||
+ | E = (B+D)/2; | ||
+ | F = intersectionpoint(B--C,A--A+2*(E-A)); | ||
+ | a = (0,0); | ||
+ | b = (1,0); | ||
+ | c = (2,1); | ||
+ | d = (1,3); | ||
+ | e = (0,3); | ||
+ | f = (0,1); | ||
+ | draw(a--C,dashed); | ||
+ | draw(f--c,dashed); | ||
+ | draw(e--A,dashed); | ||
+ | draw(a--e,dashed); | ||
+ | draw(b--d,dashed); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--F); | ||
+ | draw(B--D); | ||
+ | dot(A); | ||
+ | label("$A$",A,NE); | ||
+ | dot(B); | ||
+ | label("$B$",B,dir(180)); | ||
+ | dot(C); | ||
+ | label("$C$",C,SE); | ||
+ | dot(D); | ||
+ | label("$D$",D,dir(0)); | ||
+ | dot(E); | ||
+ | label("$E$",E,SE); | ||
+ | dot(F); | ||
+ | label("$F$",F,SW); | ||
+ | </asy> | ||
+ | <b>Note:</b> If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.<br> | ||
+ | <br> | ||
+ | As triangle <math>ABC</math> is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.<br> | ||
+ | <br> | ||
+ | As point <math>D</math> splits line segment <math>\overline{AC}</math> in a <math>1:2</math> ratio, we draw <math>\overline{AC}</math> as a vertical line segment <math>3</math> units long. Point <math>D</math> is thus <math>1</math> unit below point <math>A</math> and <math>2</math> units above point <math>C</math>. By definition, Point <math>E</math> splits line segment <math>\overline{BD}</math> in a <math>1:1</math> ratio, so we draw <math>\overline{BD}</math> <math>2</math> units long directly left of <math>D</math> and draw <math>E</math> directly between <math>B</math> and <math>D</math>, <math>1</math> unit away from both.<br> | ||
+ | <br> | ||
+ | We then draw line segments <math>\overline{AB}</math> and <math>\overline{BC}</math>. We can easily tell that triangle <math>ABC</math> occupies <math>3</math> square units of space. Constructing line <math>AE</math> and drawing <math>F</math> at the intersection of <math>AE</math> and <math>BC</math>, we can easily see that triangle <math>EBF</math> forms a right triangle occupying <math>\frac{1}{4}</math> of a square unit of space.<br> | ||
+ | <br> | ||
+ | The ratio of the areas of triangle <math>EBF</math> and triangle <math>ABC</math> is thus <math>\frac{1}{4}\div3=\frac{1}{12}</math>, and since the area of triangle <math>ABC</math> is <math>360</math>, this means that the area of triangle <math>EBF</math> is <math>\frac{1}{12}\times360=\boxed{\textbf{(B) }30}</math>. ~[[User:emerald_block|emerald_block]]<br> | ||
+ | <br> | ||
+ | <b>Additional note:</b> There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|tahiti121]] | ||
+ | |||
+ | ==Solution 11== | ||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair A,B,C,DD,EE,FF,G; | ||
+ | B = (0,0); C = (3,0); | ||
+ | A = (1.2,1.7); | ||
+ | DD = (2/3)*A+(1/3)*C; | ||
+ | EE = (B+DD)/2; | ||
+ | FF = intersectionpoint(B--C,A--A+2*(EE-A)); | ||
+ | G = (1.5,0); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--FF); | ||
+ | draw(B--DD); | ||
+ | draw(G--DD); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$", | ||
+ | B,SW); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",DD,NE); | ||
+ | label("$E$",EE,NW); | ||
+ | label("$F$",FF,S); | ||
+ | label("$G$",G,S); | ||
+ | </asy> | ||
+ | We know that <math>AD = \dfrac{1}{3} AC</math>, so <math>[ABD] = \dfrac{1}{3} [ABC] = 120</math>. Using the same method, since <math>BE = \dfrac{1}{2} BD</math>, <math>[ABE] = \dfrac{1}{2} [ABD] = 60</math>. Next, we draw <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{DG}</math> is parallel to <math>\overline{AF}</math> and create segment <math>DG</math>. We then observe that <math>\triangle AFC \sim \triangle DGC</math>, and since <math>AD:DC = 1:2</math>, <math>FG:GC</math> is also equal to <math>1:2</math>. Similarly (no pun intended), <math>\triangle DBG \sim \triangle EBF</math>, and since <math>BE:ED = 1:1</math>, <math>BF:FG</math> is also equal to <math>1:1</math>. Combining the information in these two ratios, we find that <math>BF:FG:GC = 1:1:2</math>, or equivalently, <math>BF = \dfrac{1}{4} BC</math>. Thus, <math>[BFA] = \dfrac{1}{4} [BCA] = 90</math>. We already know that <math>[ABE] = 60</math>, so the area of <math>\triangle EBF</math> is <math>[BFA] - [ABE] = \boxed{\textbf{(B) }30}</math>. ~[[User:tahiti121|tahiti121]] | ||
+ | |||
+ | ==Solution 12 (Fastest Solution if you have no time)== | ||
+ | The picture is misleading. Assume that the triangle ABC is right. | ||
+ | |||
+ | Then find two factors of <math>720</math> that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near <math>720</math> to use difference of squares, we find <math>24</math> and <math>30</math> as our numbers. Then the coordinates of D are <math>(10,16)</math>(note, A=0,0). E is then <math>(5,8)</math>. Then the equation of the line AE is <math>-16x/5+24=y</math>. Plugging in <math>y=0</math>, we have <math>x=\dfrac{15}{2}</math>. Now notice that we have both the height and the base of EBF. | ||
+ | |||
+ | Solving for the area, we have <math>(8)(15/2)(1/2)=30</math>. | ||
+ | |||
+ | == Solution 13 == | ||
+ | <math>AD : DC = 1:2</math>, so <math>ADB</math> has area <math>120</math> and <math>CDB</math> has area <math>240</math>. <math>BE = ED</math> so the area of <math>ABE</math> is equal to the area of <math>ADE = 60</math>. | ||
+ | Draw <math>\overline{DG}</math> parallel to <math>\overline{AF}</math>.<br> | ||
+ | Set area of BEF = <math>x</math>. BEF is similar to BDG in ratio of 1:2<br> | ||
+ | so area of BDG = <math>4x</math>, area of EFDG=<math>3x</math>, and area of CDG<math>=240-4x</math>.<br> | ||
+ | CDG is similar to CAF in ratio of 2:3 so area CDG = <math>4/9</math> area CAF, and area AFDG=<math>5/4</math> area CDG.<br> | ||
+ | Thus <math>60+3x=5/4(240-4x)</math> and <math>x=30</math>. | ||
+ | ~EFrame | ||
+ | |||
+ | == Solution 14 - Geometry & Algebra== | ||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair A,B,C,DD,EE,FF; | ||
+ | B = (0,0); C = (3,0); | ||
+ | A = (1.2,1.7); | ||
+ | DD = (2/3)*A+(1/3)*C; | ||
+ | EE = (B+DD)/2; | ||
+ | FF = intersectionpoint(B--C,A--A+2*(EE-A)); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--FF); | ||
+ | draw(DD--FF,blue); | ||
+ | draw(B--DD);dot(A); | ||
+ | label("$A$",A,N); | ||
+ | dot(B); | ||
+ | label("$B$", | ||
+ | B,SW);dot(C); | ||
+ | label("$C$",C,SE); | ||
+ | dot(DD); | ||
+ | label("$D$",DD,NE); | ||
+ | dot(EE); | ||
+ | label("$E$",EE,NW); | ||
+ | dot(FF); | ||
+ | label("$F$",FF,S); | ||
+ | </asy> | ||
+ | |||
+ | We draw line <math>FD</math> so that we can define a variable <math>x</math> for the area of <math> \triangle BEF = \triangle DEF</math>. Knowing that <math> \triangle ABE</math> and <math> \triangle ADE</math> share both their height and base, we get that <math>ABE = ADE = 60</math>. | ||
+ | |||
+ | Since we have a rule where 2 triangles, (<math>\triangle A</math> which has base <math>a</math> and vertex <math>c</math>), and (<math>\triangle B</math> which has Base <math>b</math> and vertex <math>c</math>)who share the same vertex (which is vertex <math>c</math> in this case), and share a common height, their relationship is : Area of <math>A : B = a : b</math> (the length of the two bases), we can list the equation where <math>\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}</math>. Substituting <math>x</math> into the equation we get: | ||
+ | |||
+ | <cmath>\frac{x+60}{300-x} = \frac{2x}{240-2x}</cmath>. <cmath>(2x)(300-x) = (60+x)(240-x).</cmath> <cmath>600-2x^2 = 14400 - 120x + 240x - 2x^2.</cmath> <cmath>480x = 14400.</cmath> and we now have that <math> \triangle BEF=30.</math> | ||
+ | ~<math>\bold{\color{blue}{onionheadjr}}</math> | ||
+ | |||
+ | ==Video Solutions== | ||
+ | Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo | ||
+ | |||
+ | https://youtu.be/Ns34Jiq9ofc —DSA_Catachu | ||
+ | |||
+ | https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) | ||
+ | |||
+ | https://www.youtube.com/watch?v=nyevg9w-CCI&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=6 ~ MathEx | ||
+ | |||
+ | https://www.youtube.com/watch?v=m04K0Q2SNXY&t=1s | ||
+ | |||
+ | ==Solution 15 (Straightfoward & Simple Solution)== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A, B, C, D, E, F; | ||
+ | B = (0,0); | ||
+ | A = (2, 3); | ||
+ | C = (5, 0); | ||
+ | D = (3, 2); | ||
+ | E = (1.5, 1); | ||
+ | F = (1.25, 0); | ||
+ | |||
+ | draw(A--B--C--A--D--B); | ||
+ | draw(A--F); | ||
+ | draw(E--C); | ||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, WSW); | ||
+ | label("$C$", C, ESE); | ||
+ | label("$D$", D, dir(0)*1.5); | ||
+ | label("$E$", E, SSE); | ||
+ | label("$F$", F, S); | ||
+ | label("$60$", (A+E+D)/3); | ||
+ | label("$60$", (A+E+B)/3); | ||
+ | label("$120$", (D+E+C)/3); | ||
+ | label("$x$", (B+E+F)/3); | ||
+ | label("$120-x$", (F+E+C)/3); | ||
+ | </asy> | ||
+ | Since <math>AD:DC=1:2</math> thus <math>\triangle ABD=\frac{1}{3} \cdot 360 = 120.</math> | ||
+ | |||
+ | Similarly, <math>\triangle DBC = \frac{2}{3} \cdot 360 = 240.</math> | ||
+ | |||
+ | Now, since <math>E</math> is a midpoint of <math>BD</math>, <math>\triangle ABE = \triangle AED = 120 \div 2 = 60.</math> | ||
+ | |||
+ | We can use the fact that <math>E</math> is a midpoint of <math>BD</math> even further. Connect lines <math>E</math> and <math>C</math> so that <math>\triangle BEC</math> and <math>\triangle DEC</math> share 2 sides. | ||
+ | |||
+ | We know that <math>\triangle BEC=\triangle DEC=240 \div 2 = 120</math> since <math>E</math> is a midpoint of <math>BD.</math> | ||
+ | |||
+ | Let's label <math>\triangle BEF</math> <math>x</math>. We know that <math>\triangle EFC</math> is <math>120-x</math> since <math>\triangle BEC = 120.</math> | ||
+ | |||
+ | Note that with this information now, we can deduct more things that are needed to finish the solution. | ||
+ | |||
+ | Note that <math>\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.</math> because of triangles <math>EBF, ABE, AEC,</math> and <math>EFC.</math> | ||
+ | |||
+ | We want to find <math>x.</math> | ||
+ | |||
+ | This is a simple equation, and solving we get <math>x=\boxed{\textbf{(B)}30}.</math> | ||
+ | |||
+ | ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. | ||
+ | |||
+ | ==Solution 16== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | pair A, B, C, D, E, F; | ||
+ | B = (0,0); | ||
+ | A = (2, 3); | ||
+ | C = (5, 0); | ||
+ | D = (3, 2); | ||
+ | E = (1.5, 1); | ||
+ | F = (1.25, 0); | ||
+ | |||
+ | draw(A--B--C--A--D--B); | ||
+ | draw(A--F); | ||
+ | draw(E--C); | ||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, WSW); | ||
+ | label("$C$", C, ESE); | ||
+ | label("$D$", D, dir(0)*1.5); | ||
+ | label("$E$", E, SSE); | ||
+ | label("$F$", F, S); | ||
+ | label("$60$", (A+E+D)/3); | ||
+ | label("$60$", (A+E+B)/3); | ||
+ | label("$120$", (D+E+C)/3); | ||
+ | </asy> | ||
+ | |||
+ | Because <math>AD:DC=1:2</math> and <math>E</math> is the midpoint of <math>BD</math>, we know that the areas of <math>ABE</math> and <math>AED</math> are <math>60</math> and the areas of <math>DEC</math> and <math>EBC</math> are <math>120</math>. | ||
+ | <cmath>\frac{[EBF]}{[EFC]} = \frac{[ABF]}{[AFC]} = \frac{ [ABE]}{[AEC]} = \frac{60}{180}</cmath> | ||
+ | <math>[EBF] = \frac{120}{4} = \boxed{\textbf{(B) }30}</math> | ||
+ | |||
+ | ==Note== | ||
+ | This question is extremely similar to 1971 AHSME Problem 26. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2019|num-b=23|num-a=25}} | {{AMC8 box|year=2019|num-b=23|num-a=25}} | ||
+ | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:05, 17 January 2022
Contents
- 1 Problem 24
- 2 Solution 1
- 3 Solution 2 (Mass Points)
- 4 Solution 3
- 5 Solution 4 (Similar Triangles)
- 6 Solution 5 (Area Ratios)
- 7 Solution 6 (Coordinate Bashing)
- 8 Solution 7
- 9 Solution 8
- 10 Solution 9 (Menelaus's Theorem)
- 11 Solution 10 (Graph Paper)
- 12 Solution 11
- 13 Solution 12 (Fastest Solution if you have no time)
- 14 Solution 13
- 15 Solution 14 - Geometry & Algebra
- 16 Video Solutions
- 17 Solution 15 (Straightfoward & Simple Solution)
- 18 Solution 16
- 19 Note
- 20 See Also
Problem 24
In triangle , point divides side so that . Let be the midpoint of and let be the point of intersection of line and line . Given that the area of is , what is the area of ?
Solution 1
Draw on such that is parallel to . That makes triangles and congruent since . so . Since ( and , so ), the altitude of triangle is equal to of the altitude of . The area of is , so the area of
Solution 2 (Mass Points)
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us.
The triangle we will consider is (obviously), and we will let be the center of mass, so that balances and (this is true since balances and , but also balances and and so balances and ), and balances and .
We know that and balances and so we assign to and to . Then, since balances and , we get (by mass points addition).
Next, since balances and in a ratio of , we know that . Similarly, by mass points addition, .
Finally, balances and so . We can confirm we have done everything right by noting that balances and , so should equal , which it does.
Now that our points have weights, we can solve the problem. so so . Also, so so .
So we get the area of as -Firebolt360 and Brudder
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of over the product of the mass points of which is which also yields -Brudder
Solution 3
is equal to . The area of triangle is equal to because it is equal to on half of the area of triangle , which is equal to one third of the area of triangle , which is . The area of triangle is the sum of the areas of triangles and , which is respectively and . So, is equal to =, so the area of triangle is . That minus the area of triangle is . ~~SmileKat32
Solution 4 (Similar Triangles)
Extend to such that as shown: Then and . Since , triangle has four times the area of triangle . Since , we get .
Since is also , we have because triangles and have the same height and same areas and so their bases must be the congruent. Thus triangle has twice the side lengths and therefore four times the area of triangle , giving .
(Credit to MP8148 for the idea)
Solution 5 (Area Ratios)
As before we figure out the areas labeled in the diagram. Then we note that Solving gives . (Credit to scrabbler94 for the idea)
Solution 6 (Coordinate Bashing)
Let be a right triangle, and
Let
The line can be described with the equation
The line can be described with
Solving, we get and
Now we can find
-Trex4days
Solution 7
Let =
(the median divides the area of the triangle into two equal parts)
Construction: Draw a circumcircle around with as is diameter. Extend to such that it meets the circle at . Draw line .
(Since is cyclic)
But is common in both with an area of 60. So, .
\therefore (SAS Congruency Theorem).
In , let be the median of .
Which means
Rotate to meet at and at . will fit exactly in (both are radii of the circle). From the above solutions, .
is a radius and is half of it implies = .
Which means
Thus
~phoenixfire & flamewavelight
Solution 8
Using the ratio of and , we find the area of is and the area of is . Also using the fact that is the midpoint of , we know . Let be a point such is parellel to . We immediatley know that by . Using that we can conclude has ratio . Using , we get . Therefore using the fact that is in , the area has ratio and we know has area so is . - fath2012
Solution 9 (Menelaus's Theorem)
By Menelaus's Theorem on triangle , we have Therefore,
Solution 10 (Graph Paper)
Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.
As triangle is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.
As point splits line segment in a ratio, we draw as a vertical line segment units long. Point is thus unit below point and units above point . By definition, Point splits line segment in a ratio, so we draw units long directly left of and draw directly between and , unit away from both.
We then draw line segments and . We can easily tell that triangle occupies square units of space. Constructing line and drawing at the intersection of and , we can easily see that triangle forms a right triangle occupying of a square unit of space.
The ratio of the areas of triangle and triangle is thus , and since the area of triangle is , this means that the area of triangle is . ~emerald_block
Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~tahiti121
Solution 11
We know that , so . Using the same method, since , . Next, we draw on such that is parallel to and create segment . We then observe that , and since , is also equal to . Similarly (no pun intended), , and since , is also equal to . Combining the information in these two ratios, we find that , or equivalently, . Thus, . We already know that , so the area of is . ~tahiti121
Solution 12 (Fastest Solution if you have no time)
The picture is misleading. Assume that the triangle ABC is right.
Then find two factors of that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near to use difference of squares, we find and as our numbers. Then the coordinates of D are (note, A=0,0). E is then . Then the equation of the line AE is . Plugging in , we have . Now notice that we have both the height and the base of EBF.
Solving for the area, we have .
Solution 13
, so has area and has area . so the area of is equal to the area of .
Draw parallel to .
Set area of BEF = . BEF is similar to BDG in ratio of 1:2
so area of BDG = , area of EFDG=, and area of CDG.
CDG is similar to CAF in ratio of 2:3 so area CDG = area CAF, and area AFDG= area CDG.
Thus and .
~EFrame
Solution 14 - Geometry & Algebra
We draw line so that we can define a variable for the area of . Knowing that and share both their height and base, we get that .
Since we have a rule where 2 triangles, ( which has base and vertex ), and ( which has Base and vertex )who share the same vertex (which is vertex in this case), and share a common height, their relationship is : Area of (the length of the two bases), we can list the equation where . Substituting into the equation we get:
. and we now have that ~
Video Solutions
Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo
https://youtu.be/Ns34Jiq9ofc —DSA_Catachu
https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution)
https://www.youtube.com/watch?v=nyevg9w-CCI&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=6 ~ MathEx
https://www.youtube.com/watch?v=m04K0Q2SNXY&t=1s
Solution 15 (Straightfoward & Simple Solution)
Since thus
Similarly,
Now, since is a midpoint of ,
We can use the fact that is a midpoint of even further. Connect lines and so that and share 2 sides.
We know that since is a midpoint of
Let's label . We know that is since
Note that with this information now, we can deduct more things that are needed to finish the solution.
Note that because of triangles and
We want to find
This is a simple equation, and solving we get
~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.
Solution 16
Because and is the midpoint of , we know that the areas of and are and the areas of and are .
Note
This question is extremely similar to 1971 AHSME Problem 26.
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.