Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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<math> \mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1 </math> | <math> \mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1 </math> | ||
− | ==Solution 1 | + | ==Solution 1== |
We have that | We have that | ||
<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> | ||
Line 43: | Line 43: | ||
</cmath> | </cmath> | ||
+ | |||
+ | |||
+ | 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>. - SuperJJ | ||
+ | |||
+ | ==Video Solution== | ||
+ | CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 | ||
==Solution 2(no trig)== | ==Solution 2(no trig)== | ||
+ | |||
+ | <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>. | ||
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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{<cmath>ABF</cmath>}{<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> | ||
+ | Since <math>\bigtriangleup ADE\sim\bigtriangleup ABF</math> we have | ||
+ | <cmath>\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}</cmath> | ||
+ | Thus <math>FC=EC-EF=\frac{448}{19}</math> and | ||
+ | <cmath>\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}</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, | ||
+ | <cmath>\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}</cmath> | ||
+ | after some calculations. | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | ~ LaTeX changes by tkfun | ||
==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}} | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | <asy>https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg</asy> |
Revision as of 21:29, 26 April 2021
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 that
But , 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 . - SuperJJ
Video Solution
CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00
Solution 2(no trig)
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
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 (Error making remote request. Unknown error_msg)