Difference between revisions of "2006 AIME I Problems/Problem 7"
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== Problem == | == Problem == | ||
− | An [[angle]] is drawn on a set of equally spaced [[parallel]] [[line]]s as shown. The [[ratio]] of the [[area]] of shaded [[region]] <math> | + | 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> |
− | + | <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 == | + | == Solution 1 == |
Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]]. | Note that the apex of the angle is not on the parallel lines. Set up a [[coordinate proof]]. | ||
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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>. | 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 == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 01:50, 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 to the area of shaded region
is 11/5. Find the ratio of shaded region
to the area of shaded region
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 . The base of region
is on the line
. The bigger base of region
is on the line
.
Let the top side of the angle be
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 ,
Solve this to find that .
Using the same reasoning as above, we get , which is
.
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 and the area of it be
. Also, let all sections of the line on the same side as the side with length
on a trapezoid be equal to
.
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 . Multiplying, we get
as the area of the triangle, so the area of the trapezoid is
. Repeating this process, we get that the area of B is
, the area of C is
, and the area of D is
.
We can now use the given condition that the ratio of C and B is .
gives us
So now we compute the ratio of D and A, which is
Edit: fixed misplaced brackets
Solution 3 (Bash)
Let the distances from the apex to the parallel lines be and
and the distance between the intersections be
We know the area ratio means
which simplifying yields
The ratio we seek is
We know that
so the ratio we seed is
Finally note that by similar triangles
Therefore the ratio we seek is
Finally note that
so the final ratio is
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.