Difference between revisions of "2019 AMC 8 Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
− | 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> | + | 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)== | ||
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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. | |
− | -Brudder | + | |
+ | 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{\textbf{(B) }30}</math> | ||
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</asy> | </asy> | ||
− | Let <math>A[\Delta XYZ]</math> = <math>Area | + | Let <math>A[\Delta XYZ]</math> = <math>\text{Area of Triangle XYZ}</math> |
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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> | 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> | <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| | + | <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== | ==Solution 11== | ||
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label("$G$",G,S); | label("$G$",G,S); | ||
</asy> | </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: | + | 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)== | ==Solution 12 (Fastest Solution if you have no time)== | ||
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== Solution 13 == | == Solution 13 == | ||
− | AD : DC = 1:2, so ADB has area 120 and CDB has area 240. BE=ED so area of ABE | + | <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> | 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> | Set area of BEF = <math>x</math>. BEF is similar to BDG in ratio of 1:2<br> | ||
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Thus <math>60+3x=5/4(240-4x)</math> and <math>x=30</math>. | Thus <math>60+3x=5/4(240-4x)</math> and <math>x=30</math>. | ||
~EFrame | ~EFrame | ||
− | |||
== Solution 14 - Geometry & Algebra== | == Solution 14 - Geometry & Algebra== | ||
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~<math>\bold{\color{blue}{onionheadjr}}</math> | ~<math>\bold{\color{blue}{onionheadjr}}</math> | ||
− | == | + | ==Video Solutions== |
− | https://youtu.be/Ns34Jiq9ofc | + | 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 | ||
+ | |||
+ | https://youtu.be/vZjPUW_ZupA | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==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== | ==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== |
Latest revision as of 10:59, 18 February 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
~savannahsolver
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.