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Difference between revisions of "2009 AMC 12A Problems/Problem 13"

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Revision as of 21:43, 3 July 2013

Problem

A ship sails $10$ miles in a straight line from $A$ to $B$, turns through an angle between $45^{\circ}$ and $60^{\circ}$, and then sails another $20$ miles to $C$. Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$? [asy] unitsize(2mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C=20*dir(50); draw(A--B--C); draw(A--C,linetype("4 4")); dot(A); dot(B); dot(C); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); [/asy]

$\textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]$ $\textbf{(E)}\ [800,900]$

Solution

Answering the question

To answer the question we are asked, it is enough to compute $AC^2$ for two different angles, preferably for both extremes ($45$ and $60$ degrees). You can use the law of cosines to do so.

Alternately, it is enough to compute $AC^2$ for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our $AC^2$ is the minimal or the maximal possible value of $AC^2$.

Below we show a complete solution in which we also show that all possible values of $AC^2$ do indeed lie in the given interval.

Complete solution

Let $C_1$ be the point the ship would reach if it turned $45^\circ$, and $C_2$ the point it would reach if it turned $60^\circ$. Obviously, $C_1$ is the furthest possible point from $A$, and $C_2$ is the closest possible point to $A$.

Hence the interval of possible values for $AC^2$ is $[AC_2^2,AC_1^2]$.

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60); draw(A--B--C1); draw(A--C1,linetype("4 4")); draw(A--B--C2); draw(A--C2,linetype("4 4")); draw( arc(A, length(C1-A), 0, 55 ), dotted ); draw( arc(A, length(C2-A), 0, 55 ), dotted ); draw( arc(B, C1, C2) ); dot(A); dot(B); dot(C1); dot(C2); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C1),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C_1$",C1,NE); label("$C_2$",C2,N); [/asy]

We can find $AC_1^2$ and $AC_2^2$ as follows:

Let $D_1$ and $D_2$ be the feet of the heights from $C_1$ and $C_2$ onto $AB$. The angles in the triangle $BD_1C_1$ are $45^\circ$, $45^\circ$, and $90^\circ$, hence $BD_1 = D_1C_1 = BC_1 / \sqrt 2$. Similarly, the angles in the triangle $BD_2C_2$ are $30^\circ$, $60^\circ$, and $90^\circ$, hence $BD_2 = BC_2 / 2$ and $D_2C_2 = BC_2 \sqrt 3 / 2$.

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60); draw(A--B--C1); draw(A--C1,linetype("4 4")); dot(A); dot(B); dot(C1); pair D1 = (C1.x,0); dot(D1); draw(B--D1--C1); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C1),SE); label("$20/\sqrt 2$",midpoint(B--D1),S); label("$20/\sqrt 2$",midpoint(D1--C1),E); label("$A$",A,SW); label("$B$",B,SE); label("$C_1$",C1,NE); label("$D_1$",D1,S); [/asy]

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60); draw(A--B--C2); draw(A--C2,linetype("4 4")); dot(A); dot(B); dot(C2); pair D2 = (C2.x,0); dot(D2); draw(B--D2--C2); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C2),SE); label("$10$",midpoint(B--D2),S); label("$10\sqrt 3$",midpoint(D2--C2),E); label("$A$",A,SW); label("$B$",B,SE); label("$C_2$",C2,NE); label("$D_2$",D2,S); [/asy]

Hence we get:

\[AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\sqrt 3)^2 = 400 + 300 = 700\]

\[AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 = 100 + 400/\sqrt 2 + 200 + 200 = 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800\]

Therefore for any valid $C$ the value $AC^2$ is surely in the interval $\boxed{ [700,800] }$.

See Also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 12 Problems and Solutions

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