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Difference between revisions of "2006 AIME I Problems/Problem 7"

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== Problem ==
 
== Problem ==
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An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] <math> \mathcal{C} </math> to the area of shaded region <math> \mathcal{B} </math> is 11/5. Find the ratio of shaded region <math> \mathcal{D} </math> to the area of shaded region <math> \mathcal{A}. </math>
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An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] <math> C </math> to the area of shaded region <math> B </math> is 11/5. Find the ratio of shaded region <math> D </math> to the area of shaded region <math> A. </math>
  
−
[[Image:2006AimeA7.PNG]]
+
<asy>
 +
defaultpen(linewidth(0.7)+fontsize(10));
 +
for(int i=0; i<4; i=i+1) {
 +
fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray);
 +
}
 +
pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);
 +
draw(B--A--C);
 +
fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white);
 +
clip(B--A--C--cycle);
 +
for(int i=0; i<9; i=i+1) {
 +
draw((i,1)--(i,6));
 +
}
 +
label("$\mathcal{A}$", A+0.2*dir(-17), S);
 +
label("$\mathcal{B}$", A+2.3*dir(-17), S);
 +
label("$\mathcal{C}$", A+4.4*dir(-17), S);
 +
label("$\mathcal{D}$", A+6.5*dir(-17), S);</asy>
  
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== Solution ==
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== Solution 1 ==
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Apex of the angle is not on the parallel lines.
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Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]].
  
−
Let...
+
Let the set of parallel lines be [[perpendicular]] to the [[x-axis]], such that they cross it at <math>0, 1, 2 \ldots</math>. The base of region <math>\mathcal{A}</math> is on the line <math>x = 1</math>. The bigger base of region <math>\mathcal{D}</math> is on the line <math>x = 7</math>.
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*The set of parallel lines be
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Let the top side of the angle be <math>y = x - s</math> and the bottom side be x-axis, as dividing the angle doesn't change the problem.
−
:perpendicular to x-axis
 
−
:& cross x-axis at 0, 1, 2...
 
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*Base of Region <math>\mathcal{A}</math> be at <math>x = 1</math>; right base of Region <math>\mathcal{D}</math> at <math>x = 7</math>
 
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*One side of the angle be x-axis.
 
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*The other side be <math>y = x - h</math>
 
−
<br>
 
−
Then...
 
−
<br><br>
 
−
As area of triangle =.5 base x height...
 
−
<br><br>
 
−
<math>
 
−
\frac{Region \mathcal{C}}{Region \mathcal{B}} = \frac{11}{5}
 
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= \frac{.5(5-h)^2 - .5(4-h)^2}{.5(3-h)^2 - .5(2-h)^2}
 
−
</math>
 
−
<br><br>
 
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<math>h = \frac{5}{6}</math>
 
  
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By similar method, <math>\frac{Region \mathcal{D}}{Region \mathcal{A}}</math> seems to be 408.
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Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,
 +
 
 +
<cmath>
 +
\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5}
 +
= \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}
 +
</cmath>
 +
 
 +
Solve this to find that <math>s = \frac{5}{6}</math>.
 +
 
 +
Using the same reasoning as above, we get <math>\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}</math>, which is <math>\boxed{408}</math>.
 +
 
 +
== Solution 2 ==
 +
 
 +
Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be <math>x</math> and the area of it be <math>x^2</math>. Also, let all sections of the line on the same side as the side with length <math>x</math> on a trapezoid be equal to <math>1</math>.
 +
 
 +
Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is <math>{\left(\frac{x+1}{x}\right)}^2</math>. Multiplying, we get <math>(x+1)^2</math> as the area of the triangle, so the area of the trapezoid is <math>2x+1</math>. Repeating this process, we get that the area of B is <math>2x+3</math>, the area of C is <math>2x+7</math>, and the area of D is <math>2x+11</math>.
 +
 
 +
We can now use the given condition that the ratio of C and B is <math>\frac{11}{5}</math>.
 +
 
 +
<math>\frac{11}{5} = \frac{2x+7}{2x+3}</math> gives us <math>x = \frac{1}{6}</math>
 +
 
 +
So now we compute the ratio of D and A, which is <math>\frac{(2)(\frac{1}{6}) + 11}{(\frac{1}{6})^2} = \boxed{408.}</math>
 +
 
 +
Edit: fixed misplaced brackets
 +
 
 +
== Solution 3 (Bash) ==
 +
 
 +
Let the distances from the apex to the parallel lines be <math>x</math> and <math>y</math> and the distance between the intersections be <math>a,b.</math> We know the area ratio means <math>\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}</math> which simplifying yields <math>ab = 3ay+3bx.</math> The ratio we seek is <math>\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\frac{ay+yx+11ab}{xy}.</math> We know that <math>ab = 3ay+3bx</math> so the ratio we seed is <math>\frac{33(ay+yx)}{11xy}.</math> Finally note that by similar triangles <math>\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.</math> Therefore the ratio we seek is <math>\frac{66(ay)}{11xy} =\frac{66a}{11x}.</math> Finally note that <math>ab=3ay+3bx \implies ab = 6bx \implies a = 6x</math> so the final ratio is <math>6 \cdot 68 = \boxed{408}.</math>
  
 
== See also ==
 
== See also ==
−
* [[2006 AIME I Problems]]
+
{{AIME box|year=2006|n=I|num-b=6|num-a=8}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:50, 10 January 2022

Problem

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$

[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]

Solution 1

Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof.

Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at $0, 1, 2 \ldots$. The base of region $\mathcal{A}$ is on the line $x = 1$. The bigger base of region $\mathcal{D}$ is on the line $x = 7$. Let the top side of the angle be $y = x - s$ and the bottom side be x-axis, as dividing the angle doesn't change the problem.

Since the area of the triangle is equal to $\frac{1}{2}bh$,

\[\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}\]

Solve this to find that $s = \frac{5}{6}$.

Using the same reasoning as above, we get $\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$, which is $\boxed{408}$.

Solution 2

Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be $x$ and the area of it be $x^2$. Also, let all sections of the line on the same side as the side with length $x$ on a trapezoid be equal to $1$.

Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is ${\left(\frac{x+1}{x}\right)}^2$. Multiplying, we get $(x+1)^2$ as the area of the triangle, so the area of the trapezoid is $2x+1$. Repeating this process, we get that the area of B is $2x+3$, the area of C is $2x+7$, and the area of D is $2x+11$.

We can now use the given condition that the ratio of C and B is $\frac{11}{5}$.

$\frac{11}{5} = \frac{2x+7}{2x+3}$ gives us $x = \frac{1}{6}$

So now we compute the ratio of D and A, which is $\frac{(2)(\frac{1}{6}) + 11}{(\frac{1}{6})^2} = \boxed{408.}$

Edit: fixed misplaced brackets

Solution 3 (Bash)

Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.$ Therefore the ratio we seek is $\frac{66(ay)}{11xy} =\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \implies ab = 6bx \implies a = 6x$ so the final ratio is $6 \cdot 68 = \boxed{408}.$

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AIME Problems and Solutions

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