Difference between revisions of "2012 AMC 12B Problems/Problem 17"
Flamedragon (talk | contribs) |
|||
Line 5: | Line 5: | ||
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 </math> | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 </math> | ||
− | ==Solution 1== | + | ==Solutions== |
+ | ===Solution 1=== | ||
Let the four points be labeled <math>P_1</math>, <math>P_2</math>, <math>P_3</math>, and <math>P_4</math>, respectively. Let the lines that go through each point be labeled <math>L_1</math>, <math>L_2</math>, <math>L_3</math>, and <math>L_4</math>, respectively. Since <math>L_1</math> and <math>L_2</math> go through <math>SP</math> and <math>RQ</math>, respectively, and <math>SP</math> and <math>RQ</math> are opposite sides of the square, we can say that <math>L_1</math> and <math>L_2</math> are parallel with slope <math>m</math>. Similarly, <math>L_3</math> and <math>L_4</math> have slope <math>-\frac{1}{m}</math>. Also, note that since square <math>PQRS</math> lies in the first quadrant, <math>L_1</math> and <math>L_2</math> must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: <math>L_1: y = m(x-3)</math>, <math>L_2: y = m(x-5)</math>, <math>L_3: y = -\frac{1}{m}(x-7)</math>, <math>L_4: y = -\frac{1}{m}(x-13)</math>. | Let the four points be labeled <math>P_1</math>, <math>P_2</math>, <math>P_3</math>, and <math>P_4</math>, respectively. Let the lines that go through each point be labeled <math>L_1</math>, <math>L_2</math>, <math>L_3</math>, and <math>L_4</math>, respectively. Since <math>L_1</math> and <math>L_2</math> go through <math>SP</math> and <math>RQ</math>, respectively, and <math>SP</math> and <math>RQ</math> are opposite sides of the square, we can say that <math>L_1</math> and <math>L_2</math> are parallel with slope <math>m</math>. Similarly, <math>L_3</math> and <math>L_4</math> have slope <math>-\frac{1}{m}</math>. Also, note that since square <math>PQRS</math> lies in the first quadrant, <math>L_1</math> and <math>L_2</math> must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: <math>L_1: y = m(x-3)</math>, <math>L_2: y = m(x-5)</math>, <math>L_3: y = -\frac{1}{m}(x-7)</math>, <math>L_4: y = -\frac{1}{m}(x-13)</math>. | ||
Line 12: | Line 13: | ||
Since <math>PQRS</math> is a square, it follows that <math>\Delta x</math> between points <math>P</math> and <math>Q</math> is equal to <math>\Delta y</math> between points <math>Q</math> and <math>R</math>. Our approach will be to find <math>\Delta x</math> and <math>\Delta y</math> in terms of <math>m</math> and equate the two to solve for <math>m</math>. <math>L_1</math> and <math>L_3</math> intersect at point <math>P</math>. Setting the equations for <math>L_1</math> and <math>L_3</math> equal to each other and solving for <math>x</math>, we find that they intersect at <math>x = \frac{3m^2 + 7}{m^2 + 1}</math>. <math>L_2</math> and <math>L_3</math> intersect at point <math>Q</math>. Intersecting the two equations, the <math>x</math>-coordinate of point <math>Q</math> is found to be <math>x = \frac{5m^2 + 7}{m^2 + 1}</math>. Subtracting the two, we get <math>\Delta x = \frac{2m^2}{m^2 + 1}</math>. Substituting the <math>x</math>-coordinate for point <math>Q</math> found above into the equation for <math>L_2</math>, we find that the <math>y</math>-coordinate of point <math>Q</math> is <math>y = \frac{2m}{m^2+1}</math>. <math>L_2</math> and <math>L_4</math> intersect at point <math>R</math>. Intersecting the two equations, the <math>y</math>-coordinate of point <math>R</math> is found to be <math>y = \frac{8m}{m^2 + 1}</math>. Subtracting the two, we get <math>\Delta y = \frac{6m}{m^2 + 1}</math>. Equating <math>\Delta x</math> and <math>\Delta y</math>, we get <math>2m^2 = 6m</math> which gives us <math>m = 3</math>. Finally, note that the line which goes though the midpoint of <math>P_1</math> and <math>P_2</math> with slope <math>3</math> and the line which goes through the midpoint of <math>P_3</math> and <math>P_4</math> with slope <math>-\frac{1}{3}</math> must intersect at at the center of the square. The equation of the line going through <math>(4,0)</math> is given by <math>y = 3(x-4)</math> and the equation of the line going through <math>(10,0)</math> is <math>y = -\frac{1}{3}(x-10)</math>. Equating the two, we find that they intersect at <math>(4.6, 1.8)</math>. Adding the <math>x</math> and <math>y</math>-coordinates, we get <math>6.4</math>. Thus, answer choice <math>\boxed{\textbf{(C)}}</math> is correct. | Since <math>PQRS</math> is a square, it follows that <math>\Delta x</math> between points <math>P</math> and <math>Q</math> is equal to <math>\Delta y</math> between points <math>Q</math> and <math>R</math>. Our approach will be to find <math>\Delta x</math> and <math>\Delta y</math> in terms of <math>m</math> and equate the two to solve for <math>m</math>. <math>L_1</math> and <math>L_3</math> intersect at point <math>P</math>. Setting the equations for <math>L_1</math> and <math>L_3</math> equal to each other and solving for <math>x</math>, we find that they intersect at <math>x = \frac{3m^2 + 7}{m^2 + 1}</math>. <math>L_2</math> and <math>L_3</math> intersect at point <math>Q</math>. Intersecting the two equations, the <math>x</math>-coordinate of point <math>Q</math> is found to be <math>x = \frac{5m^2 + 7}{m^2 + 1}</math>. Subtracting the two, we get <math>\Delta x = \frac{2m^2}{m^2 + 1}</math>. Substituting the <math>x</math>-coordinate for point <math>Q</math> found above into the equation for <math>L_2</math>, we find that the <math>y</math>-coordinate of point <math>Q</math> is <math>y = \frac{2m}{m^2+1}</math>. <math>L_2</math> and <math>L_4</math> intersect at point <math>R</math>. Intersecting the two equations, the <math>y</math>-coordinate of point <math>R</math> is found to be <math>y = \frac{8m}{m^2 + 1}</math>. Subtracting the two, we get <math>\Delta y = \frac{6m}{m^2 + 1}</math>. Equating <math>\Delta x</math> and <math>\Delta y</math>, we get <math>2m^2 = 6m</math> which gives us <math>m = 3</math>. Finally, note that the line which goes though the midpoint of <math>P_1</math> and <math>P_2</math> with slope <math>3</math> and the line which goes through the midpoint of <math>P_3</math> and <math>P_4</math> with slope <math>-\frac{1}{3}</math> must intersect at at the center of the square. The equation of the line going through <math>(4,0)</math> is given by <math>y = 3(x-4)</math> and the equation of the line going through <math>(10,0)</math> is <math>y = -\frac{1}{3}(x-10)</math>. Equating the two, we find that they intersect at <math>(4.6, 1.8)</math>. Adding the <math>x</math> and <math>y</math>-coordinates, we get <math>6.4</math>. Thus, answer choice <math>\boxed{\textbf{(C)}}</math> is correct. | ||
− | ==Solution 2== | + | ===Solution 2=== |
Note that the center of the square lies along a line that has an <math>x-</math>intercept of <math>\frac{3+5}{2}=4</math>, and also along another line with <math>x-</math>intercept <math>\frac{7+13}{2}=10</math>. Since these 2 lines are parallel to the sides of the square, they are perpindicular (since the sides of a square are). Let <math>m</math> be the slope of the first line. Then <math>-\frac{1}{m}</math> is the slope of the second line. We may use the point-slope form for the equation of a line to write <math>l_1:y=m(x-4)</math> and <math>l_2:y=-\frac{1}{m}(x-10)</math>. We easily calculate the intersection of these lines using substitution or elimination to obtain <math>\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)</math> as the center or the square. Let <math>\theta</math> denote the (acute) angle formed by <math>l_1</math> and the <math>x-</math>axis. Note that <math>\tan\theta=m</math>. Let <math>s</math> denote the side length of the square. Then <math>\sin\theta=s/2</math>. On the other hand the acute angle formed by <math>l_2</math> and the <math>x-</math>axis is <math>90-\theta</math> so that <math>\cos\theta=\sin(90-\theta)=s/6</math>. Using <math>\cos\theta=\sqrt{1-\sin^2\theta}</math> (for acute <math>\theta</math>) we have <math>\frac{s}{6}=\sqrt{1-\left(\frac{s}{2}\right)^2}</math> where upon <math>s=\frac{3\sqrt{10}}{5}</math>. Then <math>m=\tan\theta=3</math>. Substituting into <math>\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)</math> we obtain <math>\left(\frac{23}{5},\frac{9}{5}\right)</math> so that the sum of the coordinates is <math>\frac{32}{5}=6.4</math>. Hence the answer is <math>\framebox{C}</math>. | Note that the center of the square lies along a line that has an <math>x-</math>intercept of <math>\frac{3+5}{2}=4</math>, and also along another line with <math>x-</math>intercept <math>\frac{7+13}{2}=10</math>. Since these 2 lines are parallel to the sides of the square, they are perpindicular (since the sides of a square are). Let <math>m</math> be the slope of the first line. Then <math>-\frac{1}{m}</math> is the slope of the second line. We may use the point-slope form for the equation of a line to write <math>l_1:y=m(x-4)</math> and <math>l_2:y=-\frac{1}{m}(x-10)</math>. We easily calculate the intersection of these lines using substitution or elimination to obtain <math>\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)</math> as the center or the square. Let <math>\theta</math> denote the (acute) angle formed by <math>l_1</math> and the <math>x-</math>axis. Note that <math>\tan\theta=m</math>. Let <math>s</math> denote the side length of the square. Then <math>\sin\theta=s/2</math>. On the other hand the acute angle formed by <math>l_2</math> and the <math>x-</math>axis is <math>90-\theta</math> so that <math>\cos\theta=\sin(90-\theta)=s/6</math>. Using <math>\cos\theta=\sqrt{1-\sin^2\theta}</math> (for acute <math>\theta</math>) we have <math>\frac{s}{6}=\sqrt{1-\left(\frac{s}{2}\right)^2}</math> where upon <math>s=\frac{3\sqrt{10}}{5}</math>. Then <math>m=\tan\theta=3</math>. Substituting into <math>\left(\frac{4m^2+10}{m^2+1},\frac{6m}{m^2+1}\right)</math> we obtain <math>\left(\frac{23}{5},\frac{9}{5}\right)</math> so that the sum of the coordinates is <math>\frac{32}{5}=6.4</math>. Hence the answer is <math>\framebox{C}</math>. | ||
− | ==Solution 3 (Fast)== | + | ===Solution 3 (Fast)=== |
Suppose | Suppose | ||
Revision as of 13:20, 15 February 2014
Problem
Square lies in the first quadrant. Points and lie on lines , and , respectively. What is the sum of the coordinates of the center of the square ?
Solutions
Solution 1
Let the four points be labeled , , , and , respectively. Let the lines that go through each point be labeled , , , and , respectively. Since and go through and , respectively, and and are opposite sides of the square, we can say that and are parallel with slope . Similarly, and have slope . Also, note that since square lies in the first quadrant, and must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: , , , .
Since is a square, it follows that between points and is equal to between points and . Our approach will be to find and in terms of and equate the two to solve for . and intersect at point . Setting the equations for and equal to each other and solving for , we find that they intersect at . and intersect at point . Intersecting the two equations, the -coordinate of point is found to be . Subtracting the two, we get . Substituting the -coordinate for point found above into the equation for , we find that the -coordinate of point is . and intersect at point . Intersecting the two equations, the -coordinate of point is found to be . Subtracting the two, we get . Equating and , we get which gives us . Finally, note that the line which goes though the midpoint of and with slope and the line which goes through the midpoint of and with slope must intersect at at the center of the square. The equation of the line going through is given by and the equation of the line going through is . Equating the two, we find that they intersect at . Adding the and -coordinates, we get . Thus, answer choice is correct.
Solution 2
Note that the center of the square lies along a line that has an intercept of , and also along another line with intercept . Since these 2 lines are parallel to the sides of the square, they are perpindicular (since the sides of a square are). Let be the slope of the first line. Then is the slope of the second line. We may use the point-slope form for the equation of a line to write and . We easily calculate the intersection of these lines using substitution or elimination to obtain as the center or the square. Let denote the (acute) angle formed by and the axis. Note that . Let denote the side length of the square. Then . On the other hand the acute angle formed by and the axis is so that . Using (for acute ) we have where upon . Then . Substituting into we obtain so that the sum of the coordinates is . Hence the answer is .
Solution 3 (Fast)
Suppose
where .
Recall that the distance between two parallel lines and is , we have distance between and equals to , and the distance between and equals to . Equating them, we get .
Then, the center of the square is just the intersection between the following two "mid" lines:
The solution is , so we get the answer . .
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.