Difference between revisions of "2012 AMC 12B Problems/Problem 17"
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Square <math>PQRS</math> lies in the first quadrant. Points <math>(3,0), (5,0), (7,0),</math> and <math>(13,0)</math> lie on lines <math>SP, RQ, PQ</math>, and <math>SR</math>, respectively. What is the sum of the coordinates of the center of the square <math>PQRS</math>? | Square <math>PQRS</math> lies in the first quadrant. Points <math>(3,0), (5,0), (7,0),</math> and <math>(13,0)</math> lie on lines <math>SP, RQ, PQ</math>, and <math>SR</math>, respectively. What is the sum of the coordinates of the center of the square <math>PQRS</math>? | ||
− | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ | + | <math> \textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5 </math> |
+ | ==Solutions== | ||
+ | |||
+ | <asy> size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);</asy> | ||
+ | |||
+ | (diagram by MSTang) | ||
− | |||
===Solution 1=== | ===Solution 1=== | ||
+ | |||
+ | <asy> size(14cm); | ||
+ | pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8); | ||
+ | |||
+ | dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H); | ||
+ | draw(A--SS--D--cycle); | ||
+ | draw(P--Q--R^^B--Q--C); | ||
+ | draw(EE--M--F^^G--B^^C--H,dotted); | ||
+ | |||
+ | label("A",A,SW); | ||
+ | label("B",B,S); | ||
+ | label("C",C,S); | ||
+ | label("D",D,SE); | ||
+ | label("E",EE,S); | ||
+ | label("F",F,S); | ||
+ | label("P",P,W); | ||
+ | label("Q",Q,NW); | ||
+ | label("R",R,NE); | ||
+ | label("S",SS,N); | ||
+ | label("M",M,S); | ||
+ | label("G",G,W); | ||
+ | label("H",H,NE);</asy> | ||
+ | |||
+ | Construct the midpoints <math>E=(4,0)</math> and <math>F=(10,0)</math> and triangle <math>\triangle EMF</math> as in the diagram, where <math>M</math> is the center of square <math>PQRS</math>. Also construct points <math>G</math> and <math>H</math> as in the diagram so that <math>BG\parallel PQ</math> and <math>CH\parallel QR</math>. | ||
+ | |||
+ | Observe that <math>\triangle AGB\sim\triangle CHD</math> while <math>PQRS</math> being a square implies that <math>GB=CH</math>. Furthermore, <math>CD=6=3\cdot AB</math>, so <math>\triangle CHD</math> is 3 times bigger than <math>\triangle AGB</math>. Therefore, <math>HD=3\cdot GB=3\cdot HC</math>. In other words, the longer leg is 3 times the shorter leg in any triangle similar to <math>\triangle AGB</math>. | ||
+ | |||
+ | Let <math>K</math> be the foot of the perpendicular from <math>M</math> to <math>EF</math>, and let <math>x=EK</math>. Triangles <math>\triangle EKM</math> and <math>\triangle MKF</math>, being similar to <math>\triangle AGB</math>, also have legs in a 1:3 ratio, therefore, <math>MK=3x</math> and <math>KF=9x</math>, so <math>10x=EF=6</math>. It follows that <math>EK=0.6</math> and <math>MK=1.8</math>, so the coordinates of <math>M</math> are <math>(4+0.6,1.8)=(4.6,1.8)</math> and so our answer is <math>4.6+1.8 = 6.4 =</math> <math>\boxed{\mathbf{(C)}\ 32/5}</math>. | ||
+ | |||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | <asy> size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);</asy> | ||
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>. | ||
− | 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 = 32/5</math>. Thus, answer choice <math>\boxed{\textbf{(C)}}</math> is correct. |
− | ===Solution | + | ===Solution 3=== |
− | 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 | + | 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 perpendicular (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>. 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 | + | ===Solution 4 (Fast)=== |
Suppose | Suppose | ||
Line 35: | Line 72: | ||
The solution is <math>(4.6,1.8)</math>, so we get the answer <math>4.6+1.8=6.4</math>. <math>\framebox{C}</math>. | The solution is <math>(4.6,1.8)</math>, so we get the answer <math>4.6+1.8=6.4</math>. <math>\framebox{C}</math>. | ||
+ | |||
+ | ===Solution 5 (Trigonometry)=== | ||
+ | |||
+ | Using the diagram shown in Solution 1, we can set angle <math>BCQ</math> as <math>\theta</math>. We know that <math>AB=2</math> and <math>BC=2</math>. Now using <math>AA</math> | ||
+ | |||
+ | similarity, we know that <math>\triangle BCQ\sim\triangle ACP</math> in a <math>1:2</math> ratio. Now we can see that <math>CQ=-2</math><math>\cos\theta</math>, therefore, | ||
+ | |||
+ | meaning that <math>PQ=-2</math><math>\cos\theta</math>. <math>PQRS</math> is a square, so <math>QR=-2</math><math>\cos\theta</math>. We also know that <math>QCHR</math> is also a square since its | ||
+ | |||
+ | angles are <math>90^\circ</math> and all of its sides are equal. Because squares <math>PQRS</math> and <math>QCHR</math> have equal side lengths, they are | ||
+ | |||
+ | congruent leading to the conclusion that side <math>CH=-2</math><math>\cos\theta</math>. Since <math>PQRS</math> is a square, lines <math>PQ</math> and <math>SR</math> are parallel | ||
+ | |||
+ | meaning that angle <math>CDH</math> and angle <math>BCQ</math> are congruent. We can easily calculate that the length of <math>CD=6</math> and furthermore that | ||
+ | |||
+ | <math>CH=6</math><math>\sin\theta</math>. Setting <math>6\sin\theta=-2\cos\theta</math>, we get that <math>\tan\theta=-1/3</math>. This means <math>-1/3</math> is the slope of line <math>PQ</math> | ||
+ | |||
+ | and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of | ||
+ | |||
+ | points <math>E</math> and <math>F</math> because they are the midpoints. This will make solving for the center of square <math>PQRS</math> easier. <math>E=(4,0)</math> and | ||
+ | |||
+ | <math>F=(10,0)</math>. We know the slopes of lines <math>MF</math> and <math>ME</math>, which are <math>-1/3</math> and <math>3</math> respectively. Now we can get the two equations. | ||
+ | |||
+ | <cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath> | ||
+ | |||
+ | By solving: <center><math> -1/3x+10/3=3x-12, </math></center>we find that <math>x=4.6</math>. Then plugging <math>x</math> back into one of the first equations, we can find <math>y</math> and the final coordinate turns out to be <math>(4.6,1.8)</math>. Summing up the values of <math>x</math> and <math>y</math>, you get <math>4.6+1.8=6.4=32/5</math>. <math>\boxed{\mathbf{(C)}\ 32/5}</math>. | ||
+ | |||
+ | |||
+ | ~Hithere22702 | ||
== See Also == | == See Also == |
Latest revision as of 18:54, 10 August 2020
Contents
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
(diagram by MSTang)
Solution 1
Construct the midpoints and and triangle as in the diagram, where is the center of square . Also construct points and as in the diagram so that and .
Observe that while being a square implies that . Furthermore, , so is 3 times bigger than . Therefore, . In other words, the longer leg is 3 times the shorter leg in any triangle similar to .
Let be the foot of the perpendicular from to , and let . Triangles and , being similar to , also have legs in a 1:3 ratio, therefore, and , so . It follows that and , so the coordinates of are and so our answer is .
Solution 2
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 3
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 perpendicular (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 . Then . Substituting into we obtain so that the sum of the coordinates is . Hence the answer is .
Solution 4 (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 . .
Solution 5 (Trigonometry)
Using the diagram shown in Solution 1, we can set angle as . We know that and . Now using
similarity, we know that in a ratio. Now we can see that , therefore,
meaning that . is a square, so . We also know that is also a square since its
angles are and all of its sides are equal. Because squares and have equal side lengths, they are
congruent leading to the conclusion that side . Since is a square, lines and are parallel
meaning that angle and angle are congruent. We can easily calculate that the length of and furthermore that
. Setting , we get that . This means is the slope of line
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of
points and because they are the midpoints. This will make solving for the center of square easier. and
. We know the slopes of lines and , which are and respectively. Now we can get the two equations.
By solving:
we find that . Then plugging back into one of the first equations, we can find and the final coordinate turns out to be . Summing up the values of and , you get . .
~Hithere22702
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.