https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hithere22702&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:43:27ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_3&diff=2082642019 AIME II Problems/Problem 32023-12-24T01:05:31Z<p>Hithere22702: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Find the number of <math>7</math>-tuples of positive integers <math>(a,b,c,d,e,f,g)</math> that satisfy the following systems of equations:<br />
<cmath>\begin{align*}<br />
abc&=70,\\<br />
cde&=71,\\<br />
efg&=72.<br />
\end{align*}</cmath><br />
<br />
==Solution 1==<br />
As 71 is prime, <math>c</math>, <math>d</math>, and <math>e</math> must be 1, 1, and 71 (in some order). However, since <math>c</math> and <math>e</math> are divisors of 70 and 72 respectively, the only possibility is <math>(c,d,e) = (1,71,1)</math>. Now we are left with finding the number of solutions <math>(a,b,f,g)</math> satisfying <math>ab = 70</math> and <math>fg = 72</math>, which separates easily into two subproblems. The number of positive integer solutions to <math>ab = 70</math> simply equals the number of divisors of 70 (as we can choose a divisor for <math>a</math>, which uniquely determines <math>b</math>). As <math>70 = 2^1 \cdot 5^1 \cdot 7^1</math>, we have <math>d(70) = (1+1)(1+1)(1+1) = 8</math> solutions. Similarly, <math>72 = 2^3 \cdot 3^2</math>, so <math>d(72) = 4 \times 3 = 12</math>.<br />
<br />
Then the answer is simply <math>8 \times 12 = \boxed{096}</math>.<br />
<br />
-scrabbler94<br />
<br />
==Solution 2==<br />
We know that any two consecutive numbers are coprime. Using this, we can figure out that <math>c=1</math> and <math>e=1</math>. <math>d</math> then has to be <math>71</math>. Now we have two equations left. <math>ab=70</math> and <math>fg=72</math>. To solve these we just need to figure out all of the factors. Doing the prime factorization of <math>70</math> and <math>72</math>, we find that they have <math>8</math> and <math>12</math> factors, respectively. The answer is <math>8 \times 12=\boxed{096}</math><br />
<br />
~kempwood<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=II|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=2082632012 AMC 12B Problems/Problem 172023-12-24T01:05:19Z<p>Hithere22702: /* Solution 5 (Trigonometry) */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<br />
<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><br />
<br />
==Diagram==<br />
<br />
<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><br />
<br />
(diagram by MSTang)<br />
<br />
==Solution 1==<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
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>.<br />
<br />
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>.<br />
<br />
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>.<br />
<br />
<br />
==Solution 2==<br />
<br />
<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><br />
<br />
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>.<br />
<br />
<br />
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.<br />
<br />
==Solution 3==<br />
<br />
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>.<br />
<br />
==Solution 4 (Fast)==<br />
Suppose<br />
<br />
<cmath>SP: y=m(x-3)</cmath><br />
<cmath>RQ: y=m(x-5)</cmath><br />
<cmath>PQ: -my=x-7</cmath><br />
<cmath>SR: -my=x-13</cmath><br />
<br />
where <math>m >0</math>.<br />
<br />
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>.<br />
<br />
Then, the center of the square is just the intersection between the following two "mid" lines:<br />
<br />
<cmath>L_1: y=3(x-4)</cmath><br />
<cmath>L_2: -3y = x-10</cmath><br />
<br />
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>.<br />
<br />
==Solution 5 (Trigonometry)==<br />
<br />
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><br />
<br />
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, <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
<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><br />
<br />
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of <br />
<br />
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 <br />
<br />
<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.<br />
<br />
<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath><br />
<br />
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>.<br />
<br />
<br />
~kempwood<br />
<br />
== Solution 6 ==<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
<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><br />
<br />
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>.<br />
<br />
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><br />
<br />
<math>\sin \theta = \frac{GB}{AB} = \frac{a}{2}</math>, <math>\cos \theta = \frac{HC}{CD} = \frac{a}{6}</math><br />
<br />
As <math>\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{a}{2}}{\frac{a}{6}} = 3</math>, <math>m = 3</math><br />
<br />
<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><br />
<br />
<math>3x - 9 = -\frac{1}{3}x + \frac{13}{3}</math>, <math>x = 4</math>, <math>y = 3</math>, <math>S = (4, 3)</math><br />
<br />
<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><br />
<br />
<math>M = (4.6, 1.8)</math>, <math>4.6 + 1.8 = \boxed{\mathbf{(C)}\ 32/5}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_11&diff=2082622012 AMC 12B Problems/Problem 112023-12-24T01:05:02Z<p>Hithere22702: /* Solution 2 (Answer Choices) */</p>
<hr />
<div>==Problem==<br />
<br />
In the equation below, <math>A</math> and <math>B</math> are consecutive positive integers, and <math>A</math>, <math>B</math>, and <math>A+B</math> represent number bases: <cmath>132_A+43_B=69_{A+B}.</cmath><br />
What is <math>A+B</math>?<br />
<br />
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17 </math><br />
<br />
==Solution 1==<br />
<br />
Change the equation to base 10: <cmath>A^2 + 3A +2 + 4B +3= 6A + 6B + 9</cmath> <cmath> A^2 - 3A - 2B - 4=0</cmath> <br />
<br />
Either <math>B = A + 1</math> or <math>B = A - 1</math>, so either <math>A^2 - 5A - 6, B = A + 1</math> or <math>A^2 - 5A - 2, B = A - 1</math>. The second case has no integer roots, and the first can be re-expressed as <math>(A-6)(A+1) = 0, B = A + 1</math>. Since A must be positive, <math>A = 6, B = 7</math> and <math>A+B = \boxed{\textbf{(C)}\ 13}</math><br />
<br />
==Solution 2 (Answer Choices)==<br />
<br />
We can eliminate answer choice <math>\textbf{(A)}</math> because you can't have a <math>9</math> in base <math>9</math>. Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most <math>8</math> cases. Eventually, after testing a few cases, you will find that <math>A=6</math> and <math>B=7</math>. The solution is <math>\boxed{\mathbf{(C)}\ 13}</math>.<br />
<br />
~kempwood<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=10|num-a=12}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=2082612010 AMC 8 Problems/Problem 172023-12-24T01:04:44Z<p>Hithere22702: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>.<br />
<br />
That means that the area of the trapezoid is <math>5+1=6</math>. <br />
<br />
<math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. <br />
<br />
Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>.<br />
<br />
Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
~kempwood<br />
<br />
==Solution 3==<br />
We can move the cube on the bottom right to the top left, thus creating a rectangle. We thus know that now this has been separated to 2 triangle, with equal area, although, obviously, one is not a perfect triangle. From this we can subtract 1 from one of the 5's (the area of one of the triangle-shapes-polygons) and add it to the other, since we have just moved the block, original in the triangle-shaped-polygon on the bottom, and thus, we get 4/6, or <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
-RealityWrites<br />
<br />
==Video Solution by OmegaLearn==<br />
https://youtu.be/j3QSD5eDpzU?t=937<br />
<br />
<br />
==Video by MathTalks== <br />
<br />
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be<br />
<br />
==Video Solution by WhyMath==<br />
https://youtu.be/N7Yu9-bLqls<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_21&diff=2082602004 AMC 10B Problems/Problem 212023-12-24T01:04:22Z<p>Hithere22702: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>1</math>; <math>4</math>; <math>\ldots</math> and <math>9</math>; <math>16</math>; <math>\ldots</math> be two arithmetic progressions. The set <math>S</math> is the union of the first <math>2004</math> terms of each sequence. How many distinct numbers are in <math>S</math>?<br />
<br />
<math> \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 </math><br />
<br />
==Solution 1==<br />
The two sets of terms are <math>A=\{ 3k+1 : 0\leq k < 2004 \}</math> and <math>B=\{ 7l+9 : 0\leq l<2004\}</math>. <br />
<br />
Now <math>S=A\cup B</math>. We can compute <math>|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|</math>. We will now find <math>|A\cap B|</math>. <br />
<br />
Consider the numbers in <math>B</math>. We want to find out how many of them lie in <math>A</math>. In other words, we need to find out the number of valid values of <math>l</math> for which <math>7l+9\in A</math>.<br />
<br />
The fact "<math>7l+9\in A</math>" can be rewritten as "<math>1\leq 7l+9 \leq 3\cdot 2003 + 1</math>, and <math>7l+9\equiv 1\pmod 3</math>".<br />
<br />
The first condition gives <math>0\leq l\leq 857</math>, the second one gives <math>l\equiv 1\pmod 3</math>.<br />
<br />
Thus the good values of <math>l</math> are <math>\{1,4,7,\dots,856\}</math>, and their count is <math>858/3 = 286</math>.<br />
<br />
Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{(A) 3722}</math>.<br />
<br />
==Solution 2==<br />
We can start by finding the first non-distinct term from both sequences. We find that that number is <math>16</math>. Now, to find every <br />
<br />
other non-distinct terms, we can just keep adding <math>21</math>. We know that the last terms of both sequences are <math>1+3\cdot 2003</math> and <br />
<br />
<math>9+7\cdot 2003</math>. Clearly, <math>1+3\cdot 2003</math> is smaller and that is the last possible common term of both sequences. Now, we can <br />
<br />
create the inequality <math>16+21k \leq 1+3\cdot 2003</math>. Using the inequality, we find that there are <math>286</math> common terms. There are 4008 <br />
<br />
terms in total. <math>4008-286=\boxed{(A) 3722}</math><br />
<br />
~kempwood<br />
<br />
== See also ==<br />
<br />
{{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_25&diff=2082592000 AMC 12 Problems/Problem 252023-12-24T01:04:08Z<p>Hithere22702: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
Eight congruent [[equilateral triangle]]s, each of a different color, are used to construct a regular [[octahedron]]. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) <br />
<br />
<center><asy><br />
import three;<br />
import math;<br />
unitsize(1.5cm);<br />
currentprojection=orthographic(2,0.2,1);<br />
<br />
triple A=(0,0,1);<br />
triple B=(sqrt(2)/2,sqrt(2)/2,0);<br />
triple C=(sqrt(2)/2,-sqrt(2)/2,0);<br />
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);<br />
triple E=(-sqrt(2)/2,sqrt(2)/2,0);<br />
triple F=(0,0,-1);<br />
draw(A--B--E--cycle);<br />
draw(A--C--D--cycle);<br />
draw(F--C--B--cycle);<br />
draw(F--D--E--cycle,dotted+linewidth(0.7));<br />
</asy></center><br />
<br />
<math>\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680</math><br />
<br />
== Solution 1 ==<br />
Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.<br />
<center><asy><br />
size(8cm);<br />
defaultpen(0.5);<br />
import three;<br />
import math;<br />
currentprojection=orthographic(2,0.2,1);<br />
triple A=(0,0,1);<br />
triple B=(sqrt(2)/2,sqrt(2)/2,0);<br />
triple C=(sqrt(2)/2,-sqrt(2)/2,0);<br />
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);<br />
triple E=(-sqrt(2)/2,sqrt(2)/2,0);<br />
triple F=(0,0,-1);<br />
draw(A--B--E--cycle);<br />
draw(A--C--D--cycle);<br />
draw(F--C--B--cycle);<br />
draw(F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);</asy></center><br />
There are <math>7!</math> ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is <math>7!/3 = 1680</math>.<br />
<center><asy><br />
size(8cm);<br />
defaultpen(0.5);<br />
import three;<br />
import math;<br />
currentprojection=orthographic(2,0,1);<br />
triple A=(0,0,1);<br />
triple B=(sqrt(2)/2,sqrt(2)/2,0);<br />
triple C=(sqrt(2)/2,-sqrt(2)/2,0);<br />
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);<br />
triple E=(-sqrt(2)/2,sqrt(2)/2,0);<br />
triple F=(0,0,-1);<br />
triple right=(0,1,0);<br />
picture p = new picture, r = new picture, s = new picture;<br />
draw(p,A--B--E--cycle);<br />
draw(p,A--C--D--cycle);<br />
draw(p,F--C--B--cycle);<br />
draw(p,F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);<br />
draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight);<br />
add(scale3(2.2)*p);<br />
draw(r,A--B--E--cycle);<br />
draw(r,A--C--D--cycle);<br />
draw(r,F--C--B--cycle);<br />
draw(r,F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);<br />
draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight);<br />
add(scale3(2.2)*shift(2*right)*r);<br />
draw(s,A--B--E--cycle);<br />
draw(s,A--C--D--cycle);<br />
draw(s,F--C--B--cycle);<br />
draw(s,F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);<br />
draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight);<br />
add(scale3(2.2)*shift(4*right)*s);<br />
</asy></center><br />
<br />
== Solution 2 ==<br />
We consider the dual of the octahedron, the [[cube (geometry)|cube]]; a cube can be inscribed in an octahedron with each of its [[vertex|vertices]] at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube.<br />
<br />
Select any vertex and call it <math>A</math>; there are <math>8</math> color choices for this vertex, but this vertex can be rotated to any of <math>8</math> locations. After fixing <math>A</math>, we pick another vertex <math>B</math> adjacent to <math>A</math>. There are seven color choices for <math>B</math>, but there are only three locations to which <math>B</math> can be rotated to (since there are three edges from <math>A</math>). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is <math>\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}</math>.<br />
<br />
== Solution 3 ==<br />
There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and <math>8!/24 = 1680 \Rightarrow \mathrm{(E)}</math><br />
<br />
== Solution 4 ==<br />
If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0<math>^{\circ}</math> rotation, 120<math>^{\circ}</math> rotation, and 240<math>^{\circ}</math> rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is <math>8 \cdot 3 = 24</math>. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be <math>8!/24 \Rightarrow \mathrm{(E)}</math><br />
<br />
~kempwood<br />
<br />
==Solution 5 (Graph Theory)==<br />
<br />
[[File:2000AMC12P25.png|500px|center]]<br />
<br />
This problem can be approached by [https://en.wikipedia.org/wiki/Graph_theory Graph Theory]. Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.<br />
<br />
Now the problem becomes how many distinguishable ways to color the <math>8</math> vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other.<br />
<br />
Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs.<br />
<br />
There are <math>8</math> colors and <math>4</math> outer vertices, therefore there are <math>\binom{8}{4}</math> ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are <math>4</math> colors left, therefore there are <math>4!</math> ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is <math>\binom{8}{4} \cdot 4! = \boxed{\textbf{(E) } 1680 }</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
==Solution 6==<br />
Let the colors be <math>1</math> to <math>8</math> inclusive, then rotate the octahedron such that color <math>1</math> is on top. You have <math>7</math> choices of what color is on the bottom, WLOG <math>2</math>. Then, there's two rings of each <math>3</math> colors on the top and bottom. For the top ring, you can choose any <math>3</math> out of the <math>6</math> remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have <math>3!</math> ways to put the three remaining colors in three spaces.<br />
In total this is <math>7 \cdot \binom{6}{3} \cdot 2 \cdot 3!=\boxed{\textbf{(E) } 1680 }</math><br />
<br />
-mathfan2020<br />
<br />
== Video Solution ==<br />
https://youtu.be/8WRpCVwQNBo<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2000|num-b=24|after=Last question}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_14&diff=2082582022 AMC 12A Problems/Problem 142023-12-24T01:03:50Z<p>Hithere22702: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
What is the value of <cmath>(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)</cmath> where <math>\log</math> denotes the base-ten logarithm?<br />
<br />
<math>\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3</math><br />
<br />
==Solution 1==<br />
Let <math>\text{log } 2 = x</math>. The expression then becomes <cmath>(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.</cmath><br />
<br />
-bluelinfish<br />
<br />
==Solution 2==<br />
Using sum of cubes<br />
<cmath>(\log 5)^{3}+(\log 20)^{3}</cmath><br />
<cmath>= (\log 5 + \log 20)((\log 5)^{2}-(\log 5)(\log 20) + (\log 20)^{2})</cmath><br />
<cmath>= 2((\log 5)^{2}-(\log 5)(2\log 2 + \log 5) + (2\log 2 + \log 5)^{2})</cmath><br />
Let x = <math>\log 5</math> and y = <math>\log 2</math>, so <math>x+y=1</math><br />
<br />
The entire expression becomes<br />
<cmath>2(x^2-x(2y+x)+(2y+x)^2)-6y^2</cmath><br />
<cmath>=2(x^2+2xy+4y^2-3y^2)</cmath><br />
<cmath>=2(x+y)^2 = \boxed{2}</cmath><br />
<br />
~kempwood<br />
<br />
==Solution 3==<br />
<br />
We can estimate the solution. Using <math>\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) = 1 + 0.3 \approx 1.3, \log(8) \approx 0.9</math> and <math>\log(.25) = \log(1)-\log(4)= 0 - 0.6\approx -0.6,</math> we have<br />
<br />
<cmath>0.7^3 + 1.7^3 + .9\cdot(-0.6) = \boxed{2}</cmath><br />
~kxiang<br />
<br />
==Solution 4(log bash)==<br />
<br />
Using log properties, we combine the terms to make our expression equal to <math>\log {( (5^{\log^2{5}}) \cdot (20^{\log^2{20}}) \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. <br />
By exponent properties, we separate the part with base <math>20</math> to become <math>20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}}</math>. Then, we substitute this into the original expression to get <math>\log {( (5^{\log^2{5}}) \cdot 20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) } = \log {( (100^{\log^2{5}}) \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. <br />
Because <math>100^{\log^2{5}} = 25^{\log{5}}</math>, and <math>\log^2{20}-\log^2{5} = (\log{20}+\log{5})(\log{20}-\log{5}) = \log{100}\cdot\log{4} = 2\log{4}</math>, this expression is equal to <math> \log {( 25 ^ {\log{5}} \cdot 400^{\log{4}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. <br />
We perform the step with the base combining on <math>25</math> and <math>400</math> to get <math>25 ^ {\log{5}} \cdot 400^{\log{4}} = 25 ^ {\log{5}-\log{4}} \cdot 10000^{\log{4}} = 25^{\log{\frac{5}{4}}}\cdot 256</math>. Putting this back into the whole equation gives <math>\log{( 25^{\log{\frac{5}{4}}}\cdot 256 \cdot 8^{\log{\frac{1}{4}}})}</math>. <br />
One last base merge remains - but <math>25\cdot 8</math> isn't a power of 10. We can rectify this by converting <math>8^{\log{\frac{1}{4}}}</math> to <math>(4^\frac{3}{2})^{\log{\frac{1}{4}}} = 4^{\log{ \frac{1}{8} }}</math>. <br />
Finally, we complete this arduous process by performing the base merge on <math>\log{( 25^{\log{\frac{5}{4}}}\cdot 256 \cdot 4^{\log{\frac{1}{8}}})}</math>. <br />
We get <math>25^{\log{\frac{5}{4}}} \cdot 4^{\log{\frac{1}{8}}} = 25^{\log{\frac{5}{4}}-\log{\frac{1}{8}}} \cdot 100^{\log{\frac{1}{8}}} = 25^{\log{10}} \cdot \frac{1}{64} = \frac{25}{64}</math>. <br />
Putting this back into that original equation one last time, we get <math>\log(256 \cdot \frac{25}{64}) = \log{100} = \boxed{2}</math>.<br />
~aop2014<br />
<br />
==Video Solution (Speedy)==<br />
https://www.youtube.com/watch?v=pai2A9FXI9U<br />
<br />
~Education, the Study of Everything<br />
<br />
==Video Solution (Simple)==<br />
https://youtu.be/7yAh4MtJ8a8?si=9vbP5erdxlCLlG82&t=2957<br />
<br />
~Math-x<br />
<br />
==See Also==<br />
{{AMC12 box|year=2022|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_1&diff=2082571984 AIME Problems/Problem 12023-12-24T01:03:31Z<p>Hithere22702: /* Solution 5 */</p>
<hr />
<div>== Problem ==<br />
Find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math> if <math>a_1</math>, <math>a_2</math>, <math>a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>a_1+a_2+a_3+\ldots+a_{98}=137</math>.<br />
<br />
== Solution 1 ==<br />
<br />
One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to find the value of <math>a_1</math>, then use that to calculate <math>a_2</math> and sum another arithmetic series to get our answer.<br />
<br />
A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substitute this into our given equation to get <math>(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137</math>. The left-hand side of this equation is simply <math>2(a_2 + a_4 + \ldots + a_{98}) - 49</math>, so our desired value is <math>\frac{137 + 49}{2} = \boxed{093}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
If <math> a_1 </math> is the first term, then <math> a_1+a_2+a_3 + \cdots + a_{98} = 137 </math> can be rewritten as:<br />
<br />
<math> 98a_1 + 1+2+3+ \cdots + 97 = 137 </math> <math>\Leftrightarrow</math><br />
<math> 98a_1 + \frac{97 \cdot 98}{2} = 137 </math><br />
<br />
Our desired value is <math> a_2+a_4+a_6+ \cdots + a_{98} </math> so this is:<br />
<br />
<math> 49a_1 + 1+3+5+ \cdots + 97 </math><br />
<br />
which is <math> 49a_1+ 49^2 </math>. So, from the first equation, we know <math> 49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2} </math>. So, the final answer is:<br />
<br />
<math> \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{093} </math>.<br />
<br />
== Solution 3 ==<br />
A better approach to this problem is to notice that from <math>a_{1}+a_{2}+\cdots a_{98}=137</math> that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be <math>\frac{137-49}{2}</math>. Thus, if we want to find the sum of all of the even elements we simply add <math>49</math> common differences to this giving us <math>\frac{137-49}{2}+49=\fbox{093}</math>.<br />
<br />
Or, since the sum of the odd elements is 44, then the sum of the even terms must be <math>\fbox{093}</math>.<br />
<br />
== Solution 4 ==<br />
We want to find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math>, which can be rewritten as <math>a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}</math>.<br />
We can split <math>a_1+a_2+a_3+\ldots+a_{98}</math> into two parts:<br />
<cmath>a_1+a_2+a_3+\ldots+a_{49}</cmath> and <cmath>a_{50}+a_{51}+a_{52}+\ldots+a_{98}</cmath><br />
Note that each term in the second expression is <math>49</math> greater than the corresponding term, so, letting the first equation be equal to <math>x</math>, we get <math>a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}</math>. Calculating <math>49^2</math> by sheer multiplication is not difficult, but you can also do <math>(50-1)(50-1)=2500-100+1=2401</math>. We want to find the value of <math>x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225</math>. Since <math>x=\frac{137-2401}{2}</math>, we find <math>x=-1132</math>. <math>-1132+1225=\boxed{93}</math>.<br />
<br />
- PhunsukhWangdu<br />
<br />
== Solution 5 ==<br />
<br />
Since we are dealing with an arithmetic sequence, <br />
<cmath>a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}</cmath><br />
We can also figure out that <br />
<cmath>a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137</cmath><br />
<cmath>a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137</cmath><br />
Thus, <math>49a_{50} = \frac{137 + 49}{2} = \boxed{093}</math><br />
<br />
~kempwood<br />
<br />
== Video Solution by OmegaLearn ==<br />
https://youtu.be/re8DTg-Lbu0?t=234<br />
<br />
~ pi_is_3.14<br />
<br />
== See also ==<br />
{{AIME box|year=1984|before=First Question|num-a=2}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category: Intermediate Algebra Problems]]</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_1&diff=1829021984 AIME Problems/Problem 12022-11-23T23:50:26Z<p>Hithere22702: </p>
<hr />
<div>== Problem ==<br />
Find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math> if <math>a_1</math>, <math>a_2</math>, <math>a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>a_1+a_2+a_3+\ldots+a_{98}=137</math>.<br />
<br />
== Solution 1 ==<br />
<br />
One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to find the value of <math>a_1</math>, then use that to calculate <math>a_2</math> and sum another arithmetic series to get our answer.<br />
<br />
A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substitute this into our given equation to get <math>(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137</math>. The left-hand side of this equation is simply <math>2(a_2 + a_4 + \ldots + a_{98}) - 49</math>, so our desired value is <math>\frac{137 + 49}{2} = \boxed{093}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
If <math> a_1 </math> is the first term, then <math> a_1+a_2+a_3 + \cdots + a_{98} = 137 </math> can be rewritten as:<br />
<br />
<math> 98a_1 + 1+2+3+ \cdots + 97 = 137 </math> <math>\Leftrightarrow</math><br />
<math> 98a_1 + \frac{97 \cdot 98}{2} = 137 </math><br />
<br />
Our desired value is <math> a_2+a_4+a_6+ \cdots + a_{98} </math> so this is:<br />
<br />
<math> 49a_1 + 1+3+5+ \cdots + 97 </math><br />
<br />
which is <math> 49a_1+ 49^2 </math>. So, from the first equation, we know <math> 49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2} </math>. So, the final answer is:<br />
<br />
<math> \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{093} </math>.<br />
<br />
== Solution 3 ==<br />
A better approach to this problem is to notice that from <math>a_{1}+a_{2}+\cdots a_{98}=137</math> that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be <math>\frac{137-49}{2}</math>. Thus, if we want to find the sum of all of the even elements we simply add <math>49</math> common differences to this giving us <math>\frac{137-49}{2}+49=\fbox{093}</math>.<br />
<br />
Or, since the sum of the odd elements is 44, then the sum of the even terms must be <math>\fbox{093}</math>.<br />
<br />
== Solution 4 ==<br />
We want to find the value of <math>a_2+a_4+a_6+a_8+\ldots+a_{98}</math>, which can be rewritten as <math>a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}</math>.<br />
We can split <math>a_1+a_2+a_3+\ldots+a_{98}</math> into two parts:<br />
<cmath>a_1+a_2+a_3+\ldots+a_{49}</cmath> and <cmath>a_{50}+a_{51}+a_{52}+\ldots+a_{98}</cmath><br />
Note that each term in the second expression is <math>49</math> greater than the corresponding term, so, letting the first equation be equal to <math>x</math>, we get <math>a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}</math>. Calculating <math>49^2</math> by sheer multiplication is not difficult, but you can also do <math>(50-1)(50-1)=2500-100+1=2401</math>. We want to find the value of <math>x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225</math>. Since <math>x=\frac{137-2401}{2}</math>, we find <math>x=-1132</math>. <math>-1132+1225=\boxed{93}</math>.<br />
<br />
- PhunsukhWangdu<br />
<br />
== Solution 5 ==<br />
<br />
Since we are dealing with an arithmetic sequence, <br />
<cmath>a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}</cmath><br />
We can also figure out that <br />
<cmath>a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137</cmath><br />
<cmath>a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137</cmath><br />
Thus, <math>49a_{50} = \frac{137 + 49}{2} = \boxed{093}</math><br />
<br />
~Hithere22702<br />
<br />
== See also ==<br />
{{AIME box|year=1984|before=First Question|num-a=2}}<br />
* [[AIME Problems and Solutions]]<br />
* [[American Invitational Mathematics Examination]]<br />
* [[Mathematics competition resources]]<br />
<br />
[[Category: Intermediate Algebra Problems]]</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_14&diff=1814332022 AMC 12A Problems/Problem 142022-11-14T19:06:04Z<p>Hithere22702: </p>
<hr />
<div>==Problem==<br />
What is the value of <cmath>(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)</cmath> where <math>\log</math> denotes the base-ten logarithm?<br />
<br />
<math>\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3</math><br />
<br />
==Solution 1==<br />
Let <math>\text{log } 2 = x</math>. The expression then becomes <cmath>(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.</cmath><br />
<br />
-bluelinfish<br />
<br />
==Solution 2==<br />
Using sum of cubes<br />
<cmath>(\log 5)^{3}+(\log 20)^{3}</cmath><br />
<cmath>= (\log 5 + \log 20)((\log 5)^{2}-(\log 5)(\log 20) + (\log 20)^{2})</cmath><br />
<cmath>= 2((\log 5)^{2}-(\log 5)(2\log 2 + \log 5) + (2\log 2 + \log 5)^{2})</cmath><br />
Let x = <math>\log 5</math> and y = <math>\log 2</math>, so <math>x+y=1</math><br />
<br />
The entire expression becomes<br />
<cmath>2(x^2-x(2y+x)+(2y+x)^2)-6y^2</cmath><br />
<cmath>=2(x^2+2xy+4y^2-3y^2)</cmath><br />
<cmath>=2(x+y)^2 = \boxed{2}</cmath><br />
<br />
~Hithere22702<br />
<br />
==See Also==<br />
{{AMC12 box|year=2022|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_25&diff=1768602000 AMC 12 Problems/Problem 252022-08-11T02:55:19Z<p>Hithere22702: /* Solutions */</p>
<hr />
<div>== Problem ==<br />
Eight congruent [[equilateral triangle]]s, each of a different color, are used to construct a regular [[octahedron]]. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.) <br />
<br />
<center><asy><br />
import three;<br />
import math;<br />
unitsize(1.5cm);<br />
currentprojection=orthographic(2,0.2,1);<br />
<br />
triple A=(0,0,1);<br />
triple B=(sqrt(2)/2,sqrt(2)/2,0);<br />
triple C=(sqrt(2)/2,-sqrt(2)/2,0);<br />
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);<br />
triple E=(-sqrt(2)/2,sqrt(2)/2,0);<br />
triple F=(0,0,-1);<br />
draw(A--B--E--cycle);<br />
draw(A--C--D--cycle);<br />
draw(F--C--B--cycle);<br />
draw(F--D--E--cycle,dotted+linewidth(0.7));<br />
</asy></center><br />
<br />
<math>\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.<br />
<center><asy><br />
size(8cm);<br />
defaultpen(0.5);<br />
import three;<br />
import math;<br />
currentprojection=orthographic(2,0.2,1);<br />
triple A=(0,0,1);<br />
triple B=(sqrt(2)/2,sqrt(2)/2,0);<br />
triple C=(sqrt(2)/2,-sqrt(2)/2,0);<br />
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);<br />
triple E=(-sqrt(2)/2,sqrt(2)/2,0);<br />
triple F=(0,0,-1);<br />
draw(A--B--E--cycle);<br />
draw(A--C--D--cycle);<br />
draw(F--C--B--cycle);<br />
draw(F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);</asy></center><br />
There are <math>7!</math> ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is <math>7!/3 = 1680</math>.<br />
<center><asy><br />
size(8cm);<br />
defaultpen(0.5);<br />
import three;<br />
import math;<br />
currentprojection=orthographic(2,0,1);<br />
triple A=(0,0,1);<br />
triple B=(sqrt(2)/2,sqrt(2)/2,0);<br />
triple C=(sqrt(2)/2,-sqrt(2)/2,0);<br />
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);<br />
triple E=(-sqrt(2)/2,sqrt(2)/2,0);<br />
triple F=(0,0,-1);<br />
triple right=(0,1,0);<br />
picture p = new picture, r = new picture, s = new picture;<br />
draw(p,A--B--E--cycle);<br />
draw(p,A--C--D--cycle);<br />
draw(p,F--C--B--cycle);<br />
draw(p,F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);<br />
draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight);<br />
add(scale3(2.2)*p);<br />
draw(r,A--B--E--cycle);<br />
draw(r,A--C--D--cycle);<br />
draw(r,F--C--B--cycle);<br />
draw(r,F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);<br />
draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight);<br />
add(scale3(2.2)*shift(2*right)*r);<br />
draw(s,A--B--E--cycle);<br />
draw(s,A--C--D--cycle);<br />
draw(s,F--C--B--cycle);<br />
draw(s,F--D--E--cycle,dotted+linewidth(0.7));<br />
draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight);<br />
draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight);<br />
add(scale3(2.2)*shift(4*right)*s);<br />
</asy></center><br />
<br />
=== Solution 2 ===<br />
We consider the dual of the octahedron, the [[cube (geometry)|cube]]; a cube can be inscribed in an octahedron with each of its [[vertex|vertices]] at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube.<br />
<br />
Select any vertex and call it <math>A</math>; there are <math>8</math> color choices for this vertex, but this vertex can be rotated to any of <math>8</math> locations. After fixing <math>A</math>, we pick another vertex <math>B</math> adjacent to <math>A</math>. There are seven color choices for <math>B</math>, but there are only three locations to which <math>B</math> can be rotated to (since there are three edges from <math>A</math>). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is <math>\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}</math>.<br />
<br />
=== Solution 3 ===<br />
There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and <math>8!/24 = 1680 \Rightarrow \mathrm{(E)}</math><br />
<br />
=== Solution 4 ===<br />
If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0<math>^{\circ}</math> rotation, 120<math>^{\circ}</math> rotation, and 240<math>^{\circ}</math> rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is <math>8 \cdot 3 = 24</math>. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be <math>8!/24 \Rightarrow \mathrm{(E)}</math><br />
<br />
~Hithere22702<br />
<br />
== Video Solution ==<br />
https://youtu.be/8WRpCVwQNBo<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2000|num-b=24|after=Last question}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=1228412010 AMC 8 Problems/Problem 172020-05-22T20:39:59Z<p>Hithere22702: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>.<br />
<br />
That means that the area of the trapezoid is <math>5+1=6</math>. <br />
<br />
<math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. <br />
<br />
Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>.<br />
<br />
Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
~Hithere22702<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=1228402010 AMC 8 Problems/Problem 172020-05-22T20:38:20Z<p>Hithere22702: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <br />
<br />
<math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <br />
<br />
<math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
~Hithere22702<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_21&diff=1218672004 AMC 10B Problems/Problem 212020-05-01T02:12:47Z<p>Hithere22702: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>1</math>; <math>4</math>; <math>\ldots</math> and <math>9</math>; <math>16</math>; <math>\ldots</math> be two arithmetic progressions. The set <math>S</math> is the union of the first <math>2004</math> terms of each sequence. How many distinct numbers are in <math>S</math>?<br />
<br />
<math> \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 </math><br />
==Solution 1==<br />
The two sets of terms are <math>A=\{ 3k+1 : 0\leq k < 2004 \}</math> and <math>B=\{ 7l+9 : 0\leq l<2004\}</math>. <br />
<br />
Now <math>S=A\cup B</math>. We can compute <math>|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|</math>. We will now find <math>|A\cap B|</math>. <br />
<br />
Consider the numbers in <math>B</math>. We want to find out how many of them lie in <math>A</math>. In other words, we need to find out the number of valid values of <math>l</math> for which <math>7l+9\in A</math>.<br />
<br />
The fact "<math>7l+9\in A</math>" can be rewritten as "<math>1\leq 7l+9 \leq 3\cdot 2003 + 1</math>, and <math>7l+9\equiv 1\pmod 3</math>".<br />
<br />
The first condition gives <math>0\leq l\leq 857</math>, the second one gives <math>l\equiv 1\pmod 3</math>.<br />
<br />
Thus the good values of <math>l</math> are <math>\{1,4,7,\dots,856\}</math>, and their count is <math>858/3 = 286</math>.<br />
<br />
Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{(A) 3722}</math>.<br />
<br />
==Solution 2==<br />
We can start by finding the first non-distinct term from both sequences. We find that that number is <math>16</math>. Now, to find every <br />
<br />
other non-distinct terms, we can just keep adding <math>21</math>. We know that the last terms of both sequences are <math>1+3\cdot 2003</math> and <br />
<br />
<math>9+7\cdot 2003</math>. Clearly, <math>1+3\cdot 2003</math> is smaller and that is the last possible common term of both sequences. Now, we can <br />
<br />
create the inequality <math>16+21k \leq 1+3\cdot 2003</math>. Using the inequality, we find that there are <math>286</math> common terms. There are 4008 <br />
<br />
terms in total. <math>4008-286=\boxed{(A) 3722}</math><br />
<br />
~Hithere22702<br />
== See also ==<br />
<br />
{{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_21&diff=1218662004 AMC 10B Problems/Problem 212020-05-01T02:12:05Z<p>Hithere22702: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>1</math>; <math>4</math>; <math>\ldots</math> and <math>9</math>; <math>16</math>; <math>\ldots</math> be two arithmetic progressions. The set <math>S</math> is the union of the first <math>2004</math> terms of each sequence. How many distinct numbers are in <math>S</math>?<br />
<br />
<math> \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 </math><br />
==Solution 1==<br />
The two sets of terms are <math>A=\{ 3k+1 : 0\leq k < 2004 \}</math> and <math>B=\{ 7l+9 : 0\leq l<2004\}</math>. <br />
<br />
Now <math>S=A\cup B</math>. We can compute <math>|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|</math>. We will now find <math>|A\cap B|</math>. <br />
<br />
Consider the numbers in <math>B</math>. We want to find out how many of them lie in <math>A</math>. In other words, we need to find out the number of valid values of <math>l</math> for which <math>7l+9\in A</math>.<br />
<br />
The fact "<math>7l+9\in A</math>" can be rewritten as "<math>1\leq 7l+9 \leq 3\cdot 2003 + 1</math>, and <math>7l+9\equiv 1\pmod 3</math>".<br />
<br />
The first condition gives <math>0\leq l\leq 857</math>, the second one gives <math>l\equiv 1\pmod 3</math>.<br />
<br />
Thus the good values of <math>l</math> are <math>\{1,4,7,\dots,856\}</math>, and their count is <math>858/3 = 286</math>.<br />
<br />
Therefore <math>|A\cap B|=286</math>, and thus <math>|S|=4008-|A\cap B|=\boxed{(A) 3722}</math>.<br />
<br />
==Solution 2==<br />
We can start by finding the first non-distinct term from both sequences. We find that that number is <math>16</math>. Now, to find every other <br />
<br />
non-distinct terms, we can just keep adding 21. We know that the last terms of both sequences are <math>1+3\cdot 2003</math> and <math>9+7\cdot <br />
<br />
2003</math>. Clearly, <math>1+3\cdot 2003</math> is smaller and that is the last possible common term of both sequences. Now, we can create the <br />
<br />
inequality <math>16+21k \leq 1+3\cdot 2003</math>. Using the inequality, we find that there are <math>286</math> common terms. There are 4008 terms in <br />
<br />
total. <math>4008-286=\boxed{(A) 3722}</math><br />
<br />
~Hithere22702<br />
== See also ==<br />
<br />
{{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=1218332010 AMC 8 Problems/Problem 172020-04-29T22:54:50Z<p>Hithere22702: /* Solution 2(More rigorous) */</p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
~Hithere22702<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=1218012010 AMC 8 Problems/Problem 172020-04-29T03:48:11Z<p>Hithere22702: /* Solution 2(More rigorous) */</p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2(More rigorous)==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <math>5(XQ+2)/2=6</math>. Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
~Hithere22702<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=1218002010 AMC 8 Problems/Problem 172020-04-29T03:46:50Z<p>Hithere22702: /* Solution 2(More rigorous) */</p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2(More rigorous)==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <math>5(XQ+2)/2=6</math> Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
~Hithere22702<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=1217992010 AMC 8 Problems/Problem 172020-04-29T03:44:01Z<p>Hithere22702: /* Solution 2(More rigorous) */</p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2(More rigorous)==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <math>(XQ+2)/2/times5=6</math> Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
~Hithere22702<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_17&diff=1217982010 AMC 8 Problems/Problem 172020-04-29T03:41:35Z<p>Hithere22702: </p>
<hr />
<div>==Problem==<br />
The diagram shows an octagon consisting of <math>10</math> unit squares. The portion below <math>\overline{PQ}</math> is a unit square and a triangle with base <math>5</math>. If <math>\overline{PQ}</math> bisects the area of the octagon, what is the ratio <math>\dfrac{XQ}{QY}</math>?<br />
<br />
<asy><br />
import graph; size(300); <br />
real lsf = 0.5; <br />
pen dp = linewidth(0.7) + fontsize(10); <br />
defaultpen(dp); <br />
pen ds = black; <br />
pen xdxdff = rgb(0.49,0.49,1);<br />
draw((0,0)--(6,0),linewidth(1.2pt)); <br />
draw((0,0)--(0,1),linewidth(1.2pt)); <br />
draw((0,1)--(1,1),linewidth(1.2pt)); <br />
draw((1,1)--(1,2),linewidth(1.2pt)); <br />
draw((1,2)--(5,2),linewidth(1.2pt)); <br />
draw((5,2)--(5,1),linewidth(1.2pt)); <br />
draw((5,1)--(6,1),linewidth(1.2pt)); <br />
draw((6,1)--(6,0),linewidth(1.2pt)); <br />
draw((1,1)--(5,1),linewidth(1.2pt));<br />
draw((1,1)--(1,0),linewidth(1.2pt)); <br />
draw((2,2)--(2,0),linewidth(1.2pt)); <br />
draw((3,2)--(3,0),linewidth(1.2pt)); <br />
draw((4,2)--(4,0),linewidth(1.2pt)); <br />
draw((5,1)--(5,0),linewidth(1.2pt));<br />
draw((0,0)--(5,1.5),linewidth(1.2pt));<br />
dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); <br />
dot((0,1),ds); <br />
dot((1,1),ds); <br />
dot((1,2),ds); <br />
dot((5,2),ds); <br />
label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); <br />
label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); <br />
dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); <br />
dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); <br />
dot((4,2),ds); dot((5,1.5),ds); <br />
label("$Q$", (5.14,1.51),NE*lsf); <br />
clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle);<br />
</asy><br />
<br />
<math> \textbf{(A)}\ \frac{2}{5}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4} </math><br />
<br />
==Solution 1==<br />
We see that half the area of the octagon is <math>5</math>. We see that the triangle area is <math>5-1 = 4</math>. That means that <math>\frac{5h}{2} = 4 \rightarrow h=\frac{8}{5}</math>.<br />
<cmath>\text{QY}=\frac{8}{5} - 1 = \frac{3}{5}</cmath><br />
Meaning, <math>\frac{\frac{2}{5}}{\frac{3}{5}} = \boxed{\textbf{(D) }\frac{2}{3}}</math><br />
<br />
==Solution 2(More rigorous)==<br />
Like stated in solution 1, we know that half the area of the octagon is <math>5</math>. That means that the area of the trapezoid is <math>5+1=6</math>. <math>(XQ+2)/2/times5=6</math> Solving for <math>XQ</math>, we get <math>XQ=2/5</math>. Subtracting <math>2/5</math> from <math>1</math>, we get <math>QY=3/5</math>. Therefore, the answer comes out to <math>\boxed{\textbf{(D) }\frac{2}{3}}</math><br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_3&diff=1217892019 AIME II Problems/Problem 32020-04-29T00:21:41Z<p>Hithere22702: </p>
<hr />
<div>==Problem 3==<br />
Find the number of <math>7</math>-tuples of positive integers <math>(a,b,c,d,e,f,g)</math> that satisfy the following systems of equations:<br />
<cmath>\begin{align*}<br />
abc&=70,\\<br />
cde&=71,\\<br />
efg&=72.<br />
\end{align*}</cmath><br />
<br />
==Solution 1==<br />
As 71 is prime, <math>c</math>, <math>d</math>, and <math>e</math> must be 1, 1, and 71 (in some order). However, since <math>c</math> and <math>e</math> are divisors of 70 and 72 respectively, the only possibility is <math>(c,d,e) = (1,71,1)</math>. Now we are left with finding the number of solutions <math>(a,b,f,g)</math> satisfying <math>ab = 70</math> and <math>fg = 72</math>, which separates easily into two subproblems. The number of positive integer solutions to <math>ab = 70</math> simply equals the number of divisors of 70 (as we can choose a divisor for <math>a</math>, which uniquely determines <math>b</math>). As <math>70 = 2^1 \cdot 5^1 \cdot 7^1</math>, we have <math>d(70) = (1+1)(1+1)(1+1) = 8</math> solutions. Similarly, <math>72 = 2^3 \cdot 3^2</math>, so <math>d(72) = 4 \times 3 = 12</math>.<br />
<br />
Then the answer is simply <math>8 \times 12 = \boxed{096}</math>.<br />
<br />
-scrabbler94<br />
<br />
==Solution 2==<br />
We know that any two consecutive numbers are coprime. Using this, we can figure out that <math>c=1</math> and <math>e=1</math>. <math>d</math> then has to be 71. Now we have two equations left. <math>ab=70</math> and <math>fg=72</math>. To solve these we just need to figure out all of the factors. Doing the prime factorization of <math>70</math> and <math>72</math>, we find that they have <math>8</math> and <math>12</math> factors, respectively. The answer is <math>8 \times 12=\boxed{096}</math><br />
<br />
~Hithere22702<br />
==See Also==<br />
{{AIME box|year=2019|n=II|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_3&diff=1217882019 AIME II Problems/Problem 32020-04-29T00:20:50Z<p>Hithere22702: </p>
<hr />
<div>==Problem 3==<br />
Find the number of <math>7</math>-tuples of positive integers <math>(a,b,c,d,e,f,g)</math> that satisfy the following systems of equations:<br />
<cmath>\begin{align*}<br />
abc&=70,\\<br />
cde&=71,\\<br />
efg&=72.<br />
\end{align*}</cmath><br />
<br />
==Solution 1==<br />
As 71 is prime, <math>c</math>, <math>d</math>, and <math>e</math> must be 1, 1, and 71 (in some order). However, since <math>c</math> and <math>e</math> are divisors of 70 and 72 respectively, the only possibility is <math>(c,d,e) = (1,71,1)</math>. Now we are left with finding the number of solutions <math>(a,b,f,g)</math> satisfying <math>ab = 70</math> and <math>fg = 72</math>, which separates easily into two subproblems. The number of positive integer solutions to <math>ab = 70</math> simply equals the number of divisors of 70 (as we can choose a divisor for <math>a</math>, which uniquely determines <math>b</math>). As <math>70 = 2^1 \cdot 5^1 \cdot 7^1</math>, we have <math>d(70) = (1+1)(1+1)(1+1) = 8</math> solutions. Similarly, <math>72 = 2^3 \cdot 3^2</math>, so <math>d(72) = 4 \times 3 = 12</math>.<br />
<br />
Then the answer is simply <math>8 \times 12 = \boxed{096}</math>.<br />
<br />
-scrabbler94<br />
<br />
==Solution 2==<br />
We know that any two consecutive numbers are coprime. Using this, we can figure out that <math>c=1</math> and <math>e=1</math>. <math>d</math> then has to be 71. Now we have two equations left. <math>ab=70</math> and <math>fg=72</math>. To solve these we just need to figure out all of the factors. Doing the prime factorization of <math>70</math> and <math>72</math>, we find that they have <math>8</math> and <math>12</math> factors, respectively. The answer is <math>8 \times 12=\boxed{96}</math><br />
<br />
~Hithere22702<br />
==See Also==<br />
{{AIME box|year=2019|n=II|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_3&diff=1217862019 AIME II Problems/Problem 32020-04-29T00:17:42Z<p>Hithere22702: </p>
<hr />
<div>==Problem 3==<br />
Find the number of <math>7</math>-tuples of positive integers <math>(a,b,c,d,e,f,g)</math> that satisfy the following systems of equations:<br />
<cmath>\begin{align*}<br />
abc&=70,\\<br />
cde&=71,\\<br />
efg&=72.<br />
\end{align*}</cmath><br />
<br />
==Solution 1==<br />
As 71 is prime, <math>c</math>, <math>d</math>, and <math>e</math> must be 1, 1, and 71 (in some order). However, since <math>c</math> and <math>e</math> are divisors of 70 and 72 respectively, the only possibility is <math>(c,d,e) = (1,71,1)</math>. Now we are left with finding the number of solutions <math>(a,b,f,g)</math> satisfying <math>ab = 70</math> and <math>fg = 72</math>, which separates easily into two subproblems. The number of positive integer solutions to <math>ab = 70</math> simply equals the number of divisors of 70 (as we can choose a divisor for <math>a</math>, which uniquely determines <math>b</math>). As <math>70 = 2^1 \cdot 5^1 \cdot 7^1</math>, we have <math>d(70) = (1+1)(1+1)(1+1) = 8</math> solutions. Similarly, <math>72 = 2^3 \cdot 3^2</math>, so <math>d(72) = 4 \times 3 = 12</math>.<br />
<br />
Then the answer is simply <math>8 \times 12 = \boxed{096}</math>.<br />
<br />
-scrabbler94<br />
<br />
==Solution 2==<br />
We know that any two consecutive numbers are coprime. Using this, we can figure out that <math>c=1</math> and <math>e=1</math>. <math>d</math> then has to be 71. Now we have two equations left. <math>ab=70</math> and <math>fg=72</math>. To solve these we just need to figure out all of the factors. Doing the prime factorization of <math>70</math> and <math>72</math>, we find that they have <math>8</math> and <math>12</math> factors, respectively. The answer is <math>8 /times 12=\boxed{96}</math><br />
<br />
~Hithere22702<br />
==See Also==<br />
{{AIME box|year=2019|n=II|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=1216762012 AMC 12B Problems/Problem 172020-04-26T02:08:11Z<p>Hithere22702: /* Solution 5 (Trigonometry) */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<br />
<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><br />
<br />
==Solutions==<br />
<br />
<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><br />
<br />
(diagram by MSTang)<br />
<br />
===Solution 1===<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
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>.<br />
<br />
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>.<br />
<br />
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>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
<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><br />
<br />
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>.<br />
<br />
<br />
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.<br />
<br />
===Solution 3===<br />
<br />
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>.<br />
<br />
===Solution 4 (Fast)===<br />
Suppose<br />
<br />
<cmath>SP: y=m(x-3)</cmath><br />
<cmath>RQ: y=m(x-5)</cmath><br />
<cmath>PQ: -my=x-7</cmath><br />
<cmath>SR: -my=x-13</cmath><br />
<br />
where <math>m >0</math>.<br />
<br />
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>.<br />
<br />
Then, the center of the square is just the intersection between the following two "mid" lines:<br />
<br />
<cmath>L_1: y=3(x-4)</cmath><br />
<cmath>L_2: -3y = x-10</cmath><br />
<br />
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>.<br />
<br />
===Solution 5 (Trigonometry)===<br />
<br />
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><br />
<br />
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, <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
<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><br />
<br />
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of <br />
<br />
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 <br />
<br />
<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.<br />
<br />
<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath><br />
<br />
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</math>. <math>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
~Bluebell<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_11&diff=1216452012 AMC 12B Problems/Problem 112020-04-25T16:35:36Z<p>Hithere22702: </p>
<hr />
<div>==Problem==<br />
<br />
In the equation below, <math>A</math> and <math>B</math> are consecutive positive integers, and <math>A</math>, <math>B</math>, and <math>A+B</math> represent number bases: <cmath>132_A+43_B=69_{A+B}.</cmath><br />
What is <math>A+B</math>?<br />
<br />
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17 </math><br />
<br />
==Solution 1==<br />
<br />
Change the equation to base 10: <cmath>A^2 + 3A +2 + 4B +3= 6A + 6B + 9</cmath> <cmath> A^2 - 3A - 2B - 4=0</cmath> <br />
<br />
Either <math>B = A + 1</math> or <math>B = A - 1</math>, so either <math>A^2 - 5A - 6, B = A + 1</math> or <math>A^2 - 5A - 2, B = A - 1</math>. The second case has no integer roots, and the first can be re-expressed as <math>(A-6)(A+1) = 0, B = A + 1</math>. Since A must be positive, <math>A = 6, B = 7</math> and <math>A+B = 13</math> <math> \textrm{ (C) } </math>.<br />
<br />
==Solution 2 (Answer Choices)==<br />
<br />
We can eliminate answer choice <math>\textbf{(A)}</math> because you can't have a <math>9</math> in base <math>9</math>. Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most <math>8</math> cases. Eventually, after testing a few cases, you will find that <math>A=6</math> and <math>B=7</math>. The solution is <math>\boxed{\mathbf{(C)}\ 13}</math>.<br />
<br />
~Hithere22702<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=10|num-a=12}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=1216082012 AMC 12B Problems/Problem 172020-04-24T03:16:18Z<p>Hithere22702: /* Solution 5 (Trigonometry) */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<br />
<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><br />
<br />
==Solutions==<br />
<br />
<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><br />
<br />
(diagram by MSTang)<br />
<br />
===Solution 1===<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
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>.<br />
<br />
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>.<br />
<br />
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>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
<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><br />
<br />
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>.<br />
<br />
<br />
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.<br />
<br />
===Solution 3===<br />
<br />
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>.<br />
<br />
===Solution 4 (Fast)===<br />
Suppose<br />
<br />
<cmath>SP: y=m(x-3)</cmath><br />
<cmath>RQ: y=m(x-5)</cmath><br />
<cmath>PQ: -my=x-7</cmath><br />
<cmath>SR: -my=x-13</cmath><br />
<br />
where <math>m >0</math>.<br />
<br />
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>.<br />
<br />
Then, the center of the square is just the intersection between the following two "mid" lines:<br />
<br />
<cmath>L_1: y=3(x-4)</cmath><br />
<cmath>L_2: -3y = x-10</cmath><br />
<br />
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>.<br />
<br />
===Solution 5 (Trigonometry)===<br />
<br />
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><br />
<br />
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, <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
<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><br />
<br />
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of <br />
<br />
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 <br />
<br />
<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.<br />
<br />
<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath><br />
<br />
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</math>. <math>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
~Hithere22702<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=1216072012 AMC 12B Problems/Problem 172020-04-24T03:13:46Z<p>Hithere22702: /* Solution 5 (Trigonometry) */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<br />
<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><br />
<br />
==Solutions==<br />
<br />
<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><br />
<br />
(diagram by MSTang)<br />
<br />
===Solution 1===<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
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>.<br />
<br />
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>.<br />
<br />
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>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
<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><br />
<br />
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>.<br />
<br />
<br />
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.<br />
<br />
===Solution 3===<br />
<br />
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>.<br />
<br />
===Solution 4 (Fast)===<br />
Suppose<br />
<br />
<cmath>SP: y=m(x-3)</cmath><br />
<cmath>RQ: y=m(x-5)</cmath><br />
<cmath>PQ: -my=x-7</cmath><br />
<cmath>SR: -my=x-13</cmath><br />
<br />
where <math>m >0</math>.<br />
<br />
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>.<br />
<br />
Then, the center of the square is just the intersection between the following two "mid" lines:<br />
<br />
<cmath>L_1: y=3(x-4)</cmath><br />
<cmath>L_2: -3y = x-10</cmath><br />
<br />
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>.<br />
<br />
===Solution 5 (Trigonometry)===<br />
<br />
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><br />
<br />
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, <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
<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><br />
<br />
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of <br />
<br />
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 <br />
<br />
<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.<br />
<br />
<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath><br />
<br />
Solving <center><math> -1/3x+10/3=3x-12, </math></center>and plugging <math>x</math> back into the equation, we get the coordinate <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</math>. <math>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
~Hithere22702<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=1216062012 AMC 12B Problems/Problem 172020-04-24T03:12:33Z<p>Hithere22702: /* Solution 5 (Trigonometry) */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<br />
<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><br />
<br />
==Solutions==<br />
<br />
<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><br />
<br />
(diagram by MSTang)<br />
<br />
===Solution 1===<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
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>.<br />
<br />
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>.<br />
<br />
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>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
<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><br />
<br />
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>.<br />
<br />
<br />
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.<br />
<br />
===Solution 3===<br />
<br />
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>.<br />
<br />
===Solution 4 (Fast)===<br />
Suppose<br />
<br />
<cmath>SP: y=m(x-3)</cmath><br />
<cmath>RQ: y=m(x-5)</cmath><br />
<cmath>PQ: -my=x-7</cmath><br />
<cmath>SR: -my=x-13</cmath><br />
<br />
where <math>m >0</math>.<br />
<br />
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>.<br />
<br />
Then, the center of the square is just the intersection between the following two "mid" lines:<br />
<br />
<cmath>L_1: y=3(x-4)</cmath><br />
<cmath>L_2: -3y = x-10</cmath><br />
<br />
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>.<br />
<br />
===Solution 5 (Trigonometry)===<br />
<br />
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><br />
<br />
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, <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
<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><br />
<br />
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of <br />
<br />
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 <br />
<br />
<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.<br />
<br />
<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath><br />
<br />
Solving <center><math> -1/3x+10/3=3x-12, </math></center>we get the coordinate <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</math>. <math>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
~Hithere22702<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=1216052012 AMC 12B Problems/Problem 172020-04-24T03:11:41Z<p>Hithere22702: /* Solution 5(Trig) */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<br />
<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><br />
<br />
==Solutions==<br />
<br />
<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><br />
<br />
(diagram by MSTang)<br />
<br />
===Solution 1===<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
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>.<br />
<br />
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>.<br />
<br />
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>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
<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><br />
<br />
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>.<br />
<br />
<br />
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.<br />
<br />
===Solution 3===<br />
<br />
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>.<br />
<br />
===Solution 4 (Fast)===<br />
Suppose<br />
<br />
<cmath>SP: y=m(x-3)</cmath><br />
<cmath>RQ: y=m(x-5)</cmath><br />
<cmath>PQ: -my=x-7</cmath><br />
<cmath>SR: -my=x-13</cmath><br />
<br />
where <math>m >0</math>.<br />
<br />
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>.<br />
<br />
Then, the center of the square is just the intersection between the following two "mid" lines:<br />
<br />
<cmath>L_1: y=3(x-4)</cmath><br />
<cmath>L_2: -3y = x-10</cmath><br />
<br />
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>.<br />
<br />
===Solution 5 (Trigonometry)===<br />
<br />
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><br />
<br />
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, <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
<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><br />
<br />
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of <br />
<br />
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 <br />
<br />
<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.<br />
<br />
<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath><br />
<br />
Solving <center><math> -1/3x+10/3=3x-12, </math></center>we get the coordinate <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</math>. <math>\framebox{C} 6.4</math>.<br />
<br />
<br />
~Hithere22702<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hithere22702https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_17&diff=1215992012 AMC 12B Problems/Problem 172020-04-24T02:27:10Z<p>Hithere22702: </p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<br />
<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><br />
<br />
==Solutions==<br />
<br />
<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><br />
<br />
(diagram by MSTang)<br />
<br />
===Solution 1===<br />
<br />
<asy> size(14cm);<br />
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);<br />
<br />
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);<br />
draw(A--SS--D--cycle);<br />
draw(P--Q--R^^B--Q--C);<br />
draw(EE--M--F^^G--B^^C--H,dotted);<br />
<br />
label("A",A,SW);<br />
label("B",B,S);<br />
label("C",C,S);<br />
label("D",D,SE);<br />
label("E",EE,S);<br />
label("F",F,S);<br />
label("P",P,W);<br />
label("Q",Q,NW);<br />
label("R",R,NE);<br />
label("S",SS,N);<br />
label("M",M,S);<br />
label("G",G,W);<br />
label("H",H,NE);</asy><br />
<br />
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>.<br />
<br />
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>.<br />
<br />
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>\boxed{\mathbf{(C)}\ 6.4}</math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
<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><br />
<br />
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>.<br />
<br />
<br />
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.<br />
<br />
===Solution 3===<br />
<br />
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>.<br />
<br />
===Solution 4 (Fast)===<br />
Suppose<br />
<br />
<cmath>SP: y=m(x-3)</cmath><br />
<cmath>RQ: y=m(x-5)</cmath><br />
<cmath>PQ: -my=x-7</cmath><br />
<cmath>SR: -my=x-13</cmath><br />
<br />
where <math>m >0</math>.<br />
<br />
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>.<br />
<br />
Then, the center of the square is just the intersection between the following two "mid" lines:<br />
<br />
<cmath>L_1: y=3(x-4)</cmath><br />
<cmath>L_2: -3y = x-10</cmath><br />
<br />
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>.<br />
<br />
===Solution 5(Trig)===<br />
<br />
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><br />
<br />
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, <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
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 <br />
<br />
<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><br />
<br />
and the lines parallel to it. This is good news because we are dealing with easy numbers. We can solve for the coordinates of <br />
<br />
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 <br />
<br />
<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.<br />
<br />
<cmath>\left\{\begin{array}{l}y=-1/3x+10/3\\y=3x-12\end{array}\right.</cmath><br />
<br />
Solving <center><math> -1/3x+10/3=3x-12, </math></center>we get the coordinate <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</math>. <math>\framebox{C}</math>.<br />
<br />
<br />
-Hithere22702<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hithere22702