Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39 </math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the [[ratio]] of the area of triangle <math>ADE</math> to the area of the [[quadrilateral]] <math>BCED</math>? | + | In <math>ABC</math> we have <math> AB = 25 </math>, <math> BC = 39</math>, and <math>AC=42</math>. Points <math>D</math> and <math>E</math> are on <math>AB</math> and <math>AC</math> respectively, with <math> AD = 19 </math> and <math> AE = 14 </math>. What is the [[ratio]] of the area of triangle <math>ADE</math> to the area of the [[quadrilateral]] <math>BCED</math>? |
− | <math> \ | + | <math> \textbf{(A) } \frac{266}{1521}\qquad \textbf{(B) } \frac{19}{75}\qquad \textbf{(C) }\frac{1}{3}\qquad \textbf{(D) } \frac{19}{56}\qquad \textbf{(E) } 1 </math> |
− | ==Solution 1 | + | ==Solution 1== |
− | We have | + | We have |
<cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.</cmath> | <cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.</cmath> | ||
+ | |||
+ | (Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.). | ||
<asy> | <asy> | ||
Line 33: | Line 35: | ||
</asy> | </asy> | ||
− | + | <math>[BCED] = [ABC] - [ADE]</math>, so | |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ | \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ | ||
− | &= \frac{1}{ | + | &= \frac{1}{\frac{ABC}{ADE} - 1} \\ |
− | &= \frac{1}{75 | + | &= \frac{1}{\frac{75}{19} - 1} \\ |
− | &= \boxed{\frac{19}{56} | + | &= \boxed{\textbf{(D) }\frac{19}{56}}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | ==Solution 2( | + | |
+ | Note: If it is hard to understand why <cmath>\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}</cmath>, you can use the fact that the area of a triangle equals <math>\frac{1}{2} \cdot ab \cdot \sin(C)</math>. If angle <math>DAE = Z</math>, we have that <cmath>\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}</cmath>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/VXyOJWcpi00 | ||
+ | |||
+ | ==Solution 2 == | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.15 cm); | ||
+ | |||
+ | pair A, B, C, D, E; | ||
+ | |||
+ | A = (191/39,28*sqrt(1166)/39); | ||
+ | B = (0,0); | ||
+ | C = (39,0); | ||
+ | D = (6*A + 19*B)/25; | ||
+ | E = (28*A + 14*C)/42; | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(D--E); | ||
+ | |||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, SW); | ||
+ | label("$C$", C, SE); | ||
+ | label("$D$", D, W); | ||
+ | label("$E$", E, NE); | ||
+ | label("$19$", (A + D)/2, W); | ||
+ | label("$6$", (B + D)/2, W); | ||
+ | label("$14$", (A + E)/2, NE); | ||
+ | label("$28$", (C + E)/2, NE); | ||
+ | </asy> | ||
+ | |||
We can let <math>[ADE]=x</math>. | We can let <math>[ADE]=x</math>. | ||
− | Since <math>EC=2 | + | Since <math>EC=2 \cdot EA</math>, <math>[DEC]=2x</math>. |
So, <math>[ADC]=3x</math>. | So, <math>[ADC]=3x</math>. | ||
− | This means | + | This means that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>. |
− | Thus, < | + | Thus, <math>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.</math> |
-Conantwiz2023 | -Conantwiz2023 | ||
− | ==Solution 3(trig)== | + | ==Solution 3 (trig)== |
− | The [[area]] of a [[triangle]] is <math>\frac{1}{2}bc\sin A</math>. | + | The [[area]] of a [[triangle]] is <math>\frac{1}{2} bc\sin A</math>. |
Using this formula: | Using this formula: | ||
Line 67: | Line 102: | ||
<math>[BCED] = 525\sin A - 133\sin A = 392\sin A</math>. | <math>[BCED] = 525\sin A - 133\sin A = 392\sin A</math>. | ||
− | Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56} | + | Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\boxed{\textbf{(D) }\frac{19}{56}}</math> |
− | |||
− | |||
+ | Note: <math>BC=39</math> was not used in this problem. | ||
== Solution 4 == | == Solution 4 == | ||
Let <math>F</math> be on <math>AC</math> such that <math>DE\parallel BF</math> then we have | Let <math>F</math> be on <math>AC</math> such that <math>DE\parallel BF</math> then we have | ||
− | <cmath>\frac{[ADE]}{[ABF]}=(\frac{AD}{AB})^2=(\frac{19}{25})^2=\frac{361}{625}</cmath> | + | <cmath>\frac{[ADE]}{[ABF]}=\left(\frac{AD}{AB}\right)^2=\left(\frac{19}{25}\right)^2=\frac{361}{625}</cmath> |
<cmath>\frac{[ADE]}{[DEFB]}=\frac{361}{625-361}=\frac{361}{364}</cmath> | <cmath>\frac{[ADE]}{[DEFB]}=\frac{361}{625-361}=\frac{361}{364}</cmath> | ||
Since <math>\bigtriangleup ADE\sim\bigtriangleup ABF</math> we have | Since <math>\bigtriangleup ADE\sim\bigtriangleup ABF</math> we have | ||
Line 82: | Line 116: | ||
Thus <math>FC=EC-EF=\frac{448}{19}</math> and | Thus <math>FC=EC-EF=\frac{448}{19}</math> and | ||
<cmath>\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}</cmath> | <cmath>\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}</cmath> | ||
− | <cmath>\frac{[ADE]}{[BFC]}=(\frac{361}{625})(\frac{350}{448})=\frac{126350}{280000}</cmath> | + | <cmath>\frac{[ADE]}{[BFC]}=\left(\frac{[ADE]}{[ABF]}\right)\left(\frac{[ABF]}{[BFC]}\right)=\left(\frac{361}{625}\right)\left(\frac{350}{448}\right)=\frac{126350}{280000}</cmath> |
− | Finally, | + | Finally, after some calculations, |
− | <math></math>\frac{[ | + | <cmath>\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}</cmath>. |
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | ~ LaTeX changes by tkfun | ||
+ | |||
+ | |||
+ | == Solution 5 == | ||
+ | Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD]. | ||
+ | |||
+ | <cmath>\text{Diagram:}</cmath> | ||
+ | <asy> | ||
+ | unitsize(0.15 cm); | ||
+ | |||
+ | pair A, B, C, D, E; | ||
+ | |||
+ | A = (191/39,28*sqrt(1166)/39); | ||
+ | B = (0,0); | ||
+ | C = (39,0); | ||
+ | D = (6*A + 19*B)/25; | ||
+ | E = (28*A + 14*C)/42; | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(D--E); | ||
+ | |||
+ | label("$A$", A, N); | ||
+ | label("$B$", B, SW); | ||
+ | label("$C$", C, SE); | ||
+ | label("$D$", D, W); | ||
+ | label("$E$", E, NE); | ||
+ | label("$19$", (A + D)/2, W); | ||
+ | label("$6$", (B + D)/2, W); | ||
+ | label("$14$", (A + E)/2, NE); | ||
+ | label("$28$", (C + E)/2, NE); | ||
+ | </asy> | ||
+ | |||
+ | Let the area of <math>\triangle ABC</math> be <math>x</math>. <math>\triangle ABE</math> and <math>\triangle BEC</math> share a height, and the ratio of their bases are <math>1:2</math>, so the area of <math>\triangle ABE</math> is <math>\frac{x}{3}</math>. | ||
+ | |||
+ | Similarly, <math>\triangle AED</math> and <math>\triangle DEB</math> share a height, and the ratio of their bases is <math>19:6</math>, so the ratio of <math>\frac{[AED]}{[AEB]}=\frac{19}{25}</math>. Therefore, | ||
+ | <cmath>[AED]=\frac{19}{25}\cdot\left[AEB\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot\left[ABC\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot x=\frac{19}{75}x</cmath> | ||
+ | <cmath>[DECB]=[ABC]-[AED]=x-\frac{19}{75}x=\frac{56}{75}x</cmath> | ||
+ | The ratio <math>\frac{[AED]}{[DECB]}=\frac{\frac{19}{75}}{\frac{56}{75}}=\frac{19}{56}</math> which is answer choice <math>\boxed{\textbf{(D) } \frac{19}{56}}</math>. | ||
+ | |||
+ | |||
+ | ~JH. L | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}} | {{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}} | ||
− | + | [[Category:Triangle Area Ratio Problems]] | |
− | [[Category: | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg |
Latest revision as of 13:31, 13 October 2024
Contents
Problem
In we have , , and . Points and are on and respectively, with and . What is the ratio of the area of triangle to the area of the quadrilateral ?
Solution 1
We have
(Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.).
, so
Note: If it is hard to understand why , you can use the fact that the area of a triangle equals . If angle , we have that .
Video Solution
Solution 2
We can let .
Since , .
So, .
This means that .
Thus,
-Conantwiz2023
Solution 3 (trig)
Using this formula:
Since the area of is equal to the area of minus the area of ,
.
Therefore, the desired ratio is
Note: was not used in this problem.
Solution 4
Let be on such that then we have Since we have Thus and Finally, after some calculations, .
~ Nafer
~ LaTeX changes by tkfun
Solution 5
Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD].
Let the area of be . and share a height, and the ratio of their bases are , so the area of is .
Similarly, and share a height, and the ratio of their bases is , so the ratio of . Therefore, The ratio which is answer choice .
~JH. L
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg