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

Latest revision as of 13:31, 13 October 2024

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

(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.).

$[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$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]$

$[BCED] = [ABC] - [ADE]$ , 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 $\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}$ .

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 $[ADE]=x$ . 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

Solution 5

Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD].

$\text{Diagram:}$ $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$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 $\triangle ABC$ be $x$ . $\triangle ABE$ and $\triangle BEC$ share a height, and the ratio of their bases are $1:2$ , so the area of $\triangle ABE$ is $\frac{x}{3}$ .

Similarly, $\triangle AED$ and $\triangle DEB$ share a height, and the ratio of their bases is $19:6$ , so the ratio of $\frac{[AED]}{[AEB]}=\frac{19}{25}$ . Therefore, $[AED]=\frac{19}{25}\cdot\left[AEB\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot\left[ABC\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot x=\frac{19}{75}x$ $[DECB]=[ABC]-[AED]=x-\frac{19}{75}x=\frac{56}{75}x$ The ratio $\frac{[AED]}{[DECB]}=\frac{\frac{19}{75}}{\frac{56}{75}}=\frac{19}{56}$ which is answer choice $\boxed{\textbf{(D) } \frac{19}{56}}$ .

~JH. L

@@ Line 1: / Line 1: @@
 ==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>
+(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>
@@ Line 33: / Line 35: @@
 </asy>
-But <math>[BCED] = [ABC] - [ADE]</math>, so
+<math>[BCED] = [ABC] - [ADE]</math>, so
 <cmath>
 \begin{align*}
 \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*}
 </cmath>
-==Solution 2(no trig)==
+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>.
-Since <math>EC=2*EA</math>, <math>[DEC]=2x</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>.
+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>
+Thus, <math>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.</math>
 -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:
@@ Line 67: / Line 102: @@
 <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
+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{<cmath>ABF</cmath>}{\[ADE]\}=(\frac{AD}{AE})^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>
+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
+~ 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==
 {{AMC10 box|year=2005|ab=A|num-b=24|after=Last Problem}}
+[[Category:Triangle Area Ratio Problems]]
-[[Category:Introductory Number Theory Problems]]
 {{MAA Notice}}
+https://ivyleaguecenter.files.wordpress.com/2017/11/amc-10-picture.jpg

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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