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> |
+ | |||
+ | ==Diagram== | ||
+ | |||
+ | <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 2== | + | ==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 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 4 (Fast)== | ||
+ | Suppose | ||
+ | |||
+ | <cmath>SP: y=m(x-3)</cmath> | ||
+ | <cmath>RQ: y=m(x-5)</cmath> | ||
+ | <cmath>PQ: -my=x-7</cmath> | ||
+ | <cmath>SR: -my=x-13</cmath> | ||
+ | |||
+ | where <math>m >0</math>. | ||
+ | |||
+ | Recall that the distance between two parallel lines <math>Ax+By+C=0</math> and <math>Ax+By+C_1=0</math> is <math>|C-C_1|/\sqrt{A^2+B^2}</math>, we have distance between <math>SP</math> and <math>RQ</math> equals to <math>2m/\sqrt{1+m^2}</math>, and the distance between <math>PQ</math> and <math>SR</math> equals to <math>6/\sqrt{1+m^2}</math>. Equating them, we get <math>m=3</math>. | ||
+ | |||
+ | Then, the center of the square is just the intersection between the following two "mid" lines: | ||
+ | |||
+ | <cmath>L_1: y=3(x-4)</cmath> | ||
+ | <cmath>L_2: -3y = x-10</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | ~kempwood | ||
+ | |||
+ | == Solution 6 == | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <math>SP: y = mx - 3m</math>, <math>RQ: y = mx-5m</math>, <math>PQ: y = -\frac{1}{m}x + \frac{7}{m}</math>, <math>SR: y = -\frac{1}{m}x + \frac{13}{m}</math> | ||
+ | |||
+ | Let <math>SP = RP = PQ = SR = a</math>, <math>\angle GAB = \angle HCD = \theta</math>, and the slope of <math>SP</math> be <math>m</math>. | ||
+ | |||
+ | When the slope of <math>SP</math> is <math>m</math>, the slope of <math>SR</math> is <math>-\frac{1}{m}</math>, <math>\tan \theta = m</math>, <math>\cot \theta = -\frac{1}{m}</math> | ||
+ | |||
+ | <math>\sin \theta = \frac{GB}{AB} = \frac{a}{2}</math>, <math>\cos \theta = \frac{HC}{CD} = \frac{a}{6}</math> | ||
+ | |||
+ | As <math>\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{a}{2}}{\frac{a}{6}} = 3</math>, <math>m = 3</math> | ||
+ | |||
+ | <math>SP: y = 3x - 9</math>, <math>RQ: y = 3x-15</math>, <math>PQ: y = -\frac{1}{3}x + \frac{7}{3}</math>, <math>SR: y = -\frac{1}{3}x + \frac{13}{3}</math> | ||
+ | |||
+ | <math>3x - 9 = -\frac{1}{3}x + \frac{13}{3}</math>, <math>x = 4</math>, <math>y = 3</math>, <math>S = (4, 3)</math> | ||
+ | |||
+ | <math>3x-15 = -\frac{1}{3}x + \frac{7}{3}</math>, <math>x = 5.2</math>, <math>y = 0.6</math>, <math>Q = (5.2, 0.6)</math> | ||
+ | |||
+ | <math>M = (4.6, 1.8)</math>, <math>4.6 + 1.8 = \boxed{\mathbf{(C)}\ 32/5}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 21:05, 23 December 2023
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
?
Diagram
(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:
![$-1/3x+10/3=3x-12,$](http://latex.artofproblemsolving.com/a/6/0/a605b25ccab64192549a405301271cde75a1f644.png)
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
.
.
~kempwood
Solution 6
,
,
,
Let ,
, and the slope of
be
.
When the slope of is
, the slope of
is
,
,
,
As ,
,
,
,
,
,
,
,
,
,
,
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 |
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