During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

Difference between revisions of "2005 AMC 10A Problems/Problem 25"

(Solution 2(no trig))
(Solution 4)
 
(58 intermediate revisions by 13 users not shown)
Line 1: Line 1:
 
==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> \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> \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(no trig)==
+
==Solution 1==
We have that
+
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>
  
Line 37: Line 37:
 
\begin{align*}
 
\begin{align*}
 
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\
 
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\
&= \frac{1}{[ABC]/[ADE] - 1} \\
+
&= \frac{1}{\frac{ABC}{ADE} - 1} \\
&= \frac{1}{75/19 - 1} \\
+
&= \frac{1}{\frac{75}{19} - 1} \\
&= \boxed{\frac{19}{56}\Longrightarrow D}.
+
&= \boxed{\textbf{(D) }\frac{19}{56}}.
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
  
  
==Solution 2(no trig)==
 
  
We can let <math>[ADE]=x</math>. Since <math>EC=2*EA</math>, <math>[DEC]=2x</math>. So, <math>[ADC]=3x</math>. This means  that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>. Thus, <cmath>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\frac{19}{56}\Longrightarrow D}.</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
  
==Solution (trig)==
+
==Video Solution==
The [[area]] of a [[triangle]] is <math>\frac{1}{2}bc\sin A</math>.  
+
CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00
 +
 
 +
==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>.
 +
Since <math>EC=2 \cdot EA</math>, <math>[DEC]=2x</math>.
 +
So, <math>[ADC]=3x</math>.
 +
This means that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>.
 +
Thus, <math>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.</math>
 +
 
 +
-Conantwiz2023
 +
 
 +
==Solution 3 (trig)==
 +
The [[area]] of a [[triangle]] is <math>\frac{1}{2} bc\sin A</math>.  
  
 
Using this formula:
 
Using this formula:
Line 61: Line 99:
 
<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}\Longrightarrow \mathrm{(D)}</math>
+
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 ==
 +
 
 +
Let <math>F</math> be on <math>AC</math> such that <math>DE\parallel BF</math> then we have
 +
<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, after some calculations,
 +
<cmath>\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}</cmath>.
  
 +
~ Nafer
  
Note: <math>BC=39</math> was not used in this problem
+
~ 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}}
  
[[Category:Introductory Number Theory Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg

Latest revision as of 12:50, 14 December 2021

Problem

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?

$\textbf{(A) } \frac{266}{1521}\qquad \textbf{(B) } \frac{19}{75}\qquad \textbf{(C) }\frac{1}{3}\qquad \textbf{(D) } \frac{19}{56}\qquad \textbf{(E) } 1$

Solution 1

We have \[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.\]

[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]

But $[BCED] = [ABC] - [ADE]$, so \begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{\frac{ABC}{ADE} - 1} \\ &= \frac{1}{\frac{75}{19} - 1} \\ &= \boxed{\textbf{(D) }\frac{19}{56}}. \end{align*}


Note: If it is hard to understand why \[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}\], you can use the fact that the area of a triangle equals $\frac{1}{2} \cdot ab \cdot \sin(C)$. If angle $DAE = Z$, we have that \[\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}\]. - SuperJJ

Video Solution

CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00

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 $[ADE]=x$. Since $EC=2 \cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}$. Thus, $\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.$

-Conantwiz2023

Solution 3 (trig)

The area of a triangle is $\frac{1}{2} bc\sin A$.

Using this formula:

$[ADE]=\frac{1}{2}\cdot19\cdot14\cdot\sin A = 133\sin A$

$[ABC]=\frac{1}{2}\cdot25\cdot42\cdot\sin A = 525\sin A$

Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$,

$[BCED] = 525\sin A - 133\sin A = 392\sin A$.

Therefore, the desired ratio is $\frac{133\sin A}{392\sin A}=\boxed{\textbf{(D) }\frac{19}{56}}$


Note: $BC=39$ was not used in this problem.

Solution 4

Let $F$ be on $AC$ such that $DE\parallel BF$ then we have \[\frac{[ADE]}{[ABF]}=\left(\frac{AD}{AB}\right)^2=\left(\frac{19}{25}\right)^2=\frac{361}{625}\] \[\frac{[ADE]}{[DEFB]}=\frac{361}{625-361}=\frac{361}{364}\] Since $\bigtriangleup ADE\sim\bigtriangleup ABF$ we have \[\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}\] Thus $FC=EC-EF=\frac{448}{19}$ and \[\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}\] \[\frac{[ADE]}{[BFC]}=\left(\frac{[ADE]}{[ABF]}\right)\left(\frac{[ABF]}{[BFC]}\right)=\left(\frac{361}{625}\right)\left(\frac{350}{448}\right)=\frac{126350}{280000}\] Finally, after some calculations, \[\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}\].

~ Nafer

~ LaTeX changes by tkfun

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png

https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg

Invalid username
Login to AoPS