https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Kagrat&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:06:33ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_21&diff=2027712023 AMC 12A Problems/Problem 212023-11-12T23:22:24Z<p>Kagrat: /* Solution 3 */</p>
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<div>{{duplicate|[[2023 AMC 10A Problems/Problem 25|2023 AMC 10A #25]] and [[2023 AMC 12A Problems/Problem 21|2023 AMC 12A #21]]}}<br />
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==Problem==<br />
If <math>A</math> and <math>B</math> are vertices of a polyhedron, define the distance <math>d(A,B)</math> to be the minimum number of edges of the polyhedron one must traverse in order to connect <math>A</math> and <math>B</math>. For example, if <math>\overline{AB}</math> is an edge of the polyhedron, then <math>d(A, B) = 1</math>, but if <math>\overline{AC}</math> and <math>\overline{CB}</math> are edges and <math>\overline{AB}</math> is not an edge, then <math>d(A, B) = 2</math>. Let <math>Q</math>, <math>R</math>, and <math>S</math> be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that <math>d(Q, R) > d(R, S)</math>?<br />
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<math>\textbf{(A) } \frac{7}{22} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{3}{8} \qquad \textbf{(D) } \frac{5}{12} \qquad \textbf{(E) } \frac{1}{2}</math><br />
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==Solution 1==<br />
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To find the total amount of vertices we first find the amount of edges, and that is <math>\frac{20 \times 3}{2}</math>. Next, to find the amount of vertices we can use Euler's characteristic, <math>V - E + F = 2</math>, and therefore the amount of vertices is <math>12</math> <br />
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So there are <math>P(12,3) = 1320</math> ways to choose 3 distinct points.<br />
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Now, the furthest distance we can get from one point to another point in a icosahedron is 3. Which gives us a range of <math>1 \leq d(Q, R), d(R, S) \leq 3</math><br />
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With some case work, we get two cases:<br />
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Case 1: <math>d(Q, R) = 3; d(R, S) = 1, 2</math><br />
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Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S.<br />
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Then, we get <math>12 \times 1 \times 10 = 120</math> (ways to choose R × ways to choose Q × ways to choose S)<br />
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Case 2: <math>d(Q, R) = 2; d(R, S) = 1</math><br />
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We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row.<br />
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Therefore, we have <math>12 \times 5 \times 5 = 300</math> (ways to choose R × ways to choose Q × ways to choose S)<br />
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Hence, <math>P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \boxed{\textbf{(A) } \frac{7}{22}}</math><br />
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~lptoggled, edited by ESAOPS<br />
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==Solution 2 (Cheese + Actual way)==<br />
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In total, there are <math>\binom{12}{3}=220</math> ways to select the points. However, if we look at the denominators of <math>B,C,D</math>, they are <math>3,8,12</math> which are not divisors of <math>220</math>. Also <math>\frac{1}{2}</math> is impossible as cases like <math>d(Q, R) = d(R, S)</math> exist. The only answer choice left is <math>\boxed{\textbf{(A) } \frac{7}{22}}</math><br />
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Note: this cheese is actually wrong because the total number of ways to select the points is actually <math>12 \times 11 \times 10 = 1320</math> as order matters, so all denominators are possible. Rather, you can arrive at the same conclusion by fixing R WLOG, leading to <math>11 \times 10 = 110</math> ways in total, which works for the original cheese. ~awesomeguy856<br />
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(Actual way)<br />
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Fix an arbitrary point, to select the rest <math>2</math> points, there are <math>\binom{11}{2}=55</math> ways. To make <math>d(Q, R)=d(R, S), d=1/2</math>. Which means there are in total <math>2\cdot \binom{5}{2}=20</math> ways to make the distance the same. <math>\frac{1}{2}(1-\frac{20}{55})= \boxed{\textbf{(A) } \frac{7}{22}}</math><br />
~bluesoul<br />
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== Solution 3 ==<br />
We can imagine the icosahedron as having 4 layers. 1 vertex at the top, 5 vertices below connected to the top vertex, 5 vertices below that which are 2 edges away from the top vertex, and one vertex at the bottom that is 3 edges away. WLOG because the icosahedron is symmetric around all vertices, we can say that R is the vertex at the top. So now, we just need to find the probability that S is on a layer closer to the top than Q. We can do casework on the layer S is on to get <br />
<cmath>\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}</cmath><br />
So the answer is <math>\boxed{\textbf{(A) }\frac{7}{22}}</math>. -awesomeparrot<br />
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== Solution 4 ==<br />
We can actually see that the probability that <math>d(Q, R) > d(R, S)</math> is the exact same as <math>d(Q, R) < d(R, S)</math> because <math>d(Q, R)</math> and <math>d(R, S)</math> have no difference. (In other words, we can just swap Q and S, meaning that can be called the same probability-wise.) Therefore, we want to find the probability that <math>d(Q, R) = d(R, S)</math>.<br />
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WLOG, we can rotate the icosahedron so that R is the top of the icosahedron. Then we can divide this into 2 cases:<br />
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1. They are on the second layer<br />
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There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 1</math>. So, there are <math>5 \cdot 4 = 20</math> ways to put them on the second layer.<br />
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2. They are on the third layer<br />
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There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 2</math>. So, there are <math>5 \cdot 4 = 20</math> ways to put them on the third layer.<br />
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The total number of ways to choose P and S are <math>11 \cdot 10 = 110</math> (because there are 12 vertices), so the probability that <math>d(Q, R) = d(R, S)</math> is <math>\frac{20+20}{110} = \frac{4}{11}</math>.<br />
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Therefore, the probability that <math>d(Q, R) > d(R, S)</math> is <math>\frac{1 - \frac{4}{11}}{2} = \boxed{\textbf{(A) }\frac{7}{22}}</math><br />
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~Ethanzhang1001<br />
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==Solution 5==<br />
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We know that there are <math>20</math> faces. Each of those faces has <math>3</math> borders (since each is a triangle), and each edge is used as a border twice (for each face on either side). Thus, there are <math>\dfrac{20\cdot3}2=30</math> edges. <br />
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By Euler's formula, which states that <math>v-e+f=2</math> for all convex polyhedra, we know that there are <math>2-f+e=12</math> vertices. <br />
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The answer can be counted by first counting the number of possible paths that will yield <math>d(Q, R) > d(R, S)</math> and dividing it by <math>12\cdot11\cdot10</math> (or <math>\dbinom{12}3</math>, depending on the approach). In either case, one will end up dividing by <math>11</math> somewhere in the denominator. We can then hope that there will be no factor of <math>11</math> in the numerator (which would cancel the <math>11</math> in the denominator out), and answer the only option that has an <math>11</math> in the denominator: <math>\boxed{\textbf{(A) }\frac{7}{22}}</math>. <br />
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~Technodoggo<br />
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Additional note by "Fruitz": Note that one can eliminate <math>1/2</math> by symmetry if you swap the ineq sign.<br />
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Another note by "andliu766": A shorter way to find the number of vertices and edges is to use the fact that the MAA logo is an icosahedron. :)<br />
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==Solution 6 (Case Work)==<br />
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[[File:2023AMC12AP21.png|center|200px]]<br />
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WLOG, let R be at the top-most vertex of the icosahedron. There are <math>2</math> cases where <math>d(Q, R) > d(R, S)</math>.<br />
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Case 1: <math>Q</math> is at the bottom-most vertex<br />
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If <math>Q</math> is at the bottom-most vertex, no matter where <math>S</math> is, <math>d(Q, R) > d(R, S)</math>. The probability that <math>Q</math> is at the bottom-most vertex is <math>\frac{1}{11}</math><br />
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Case 2: <math>Q</math> is at the second layer<br />
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If <math>Q</math> is at the second layer, <math>S</math> must be at the first layer, for <math>d(Q, R) > d(R, S)</math> to be true. The probability that <math>Q</math> is at the second layer, and <math>S</math> is at the first layer is <math>\frac{5}{11} \cdot \frac{5}{10} = \frac{5}{22}</math><br />
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<cmath>\frac{1}{11} + \frac{5}{22} = \boxed{\textbf{(A) }\frac{7}{22}}</cmath><br />
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
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==Solution 7 (efficient)==<br />
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Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and <br />
S basically the same, we can first count the probability that <math>d(Q,R) = d(R,S)</math>.<br />
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<math>\mathfrak{Case} \ \mathfrak{1}: d(Q,R) = d(R,S) = 1</math><br />
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There are 5 points <math>P</math> such that <math>d(Q,P) = 1</math>. There is <math>5 \times 4 = \boxed{20}</math> ways to choose Q and S in this case.<br />
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<math>\mathfrak{Case} \ \mathfrak{2}: d(Q,R) = d(R,S) = 2</math><br />
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There are 5 points <math>P</math> such that <math>d(Q,P) = 2</math>. There is <math>5 \times 4 = \boxed{20}</math> ways to choose Q and S in this case.<br />
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<math>\mathfrak{Case} \ \mathfrak{3}: d(Q,R) = d(R,S) = 3</math><br />
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There is 1 point <math>P</math> such that <math>d(Q,P) = 3</math>. There is <math>1 \times 0 = \boxed{0}</math> ways to choose Q and S in this case.<br />
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<math>\mathfrak{Final} \ \mathfrak{solution}</math><br />
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There are 11 points <math>P</math> that are distinct from R. There is <math>11 \times 10 = \boxed{110}</math> ways to choose Q and S. There is <math>20 + 20 + 0 = \boxed{40}</math> ways to choose Q and S such that <math>d(Q,R) = d(R,S)</math>. There is <math>\frac{110-40}{2} = 35</math> ways to choose Q and S such that <math>d(Q, R) > d(R, S)</math>. The probability that <math>d(Q, R) > d(R, S)</math> is therefore <math>\frac{35}{110} = \frac{7}{22}</math> which corresponds to answer choice <math>\boxed{A}</math><br />
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~~[https://artofproblemsolving.com/wiki/index.php/User:Afly afly]<br />
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==Video Solution by epicbird08==<br />
https://youtu.be/s_6q0C0z6Ug<br />
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~EpicBird08<br />
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==Video Solution==<br />
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https://youtu.be/bS3tle-jP4g<br />
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
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==See also==<br />
{{AMC10 box|year=2023|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2023|ab=A|num-b=20|num-a=22}}<br />
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{{MAA Notice}}</div>Kagrathttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_6&diff=1445252019 AMC 10A Problems/Problem 62021-02-02T01:58:02Z<p>Kagrat: /* Solution 4 */</p>
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<div>==Problem==<br />
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For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?<br />
*a square<br />
*a rectangle that is not a square<br />
*a rhombus that is not a square<br />
*a parallelogram that is not a rectangle or a rhombus<br />
*an isosceles trapezoid that is not a parallelogram<br />
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<math>\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5</math><br />
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==Solution 1==<br />
This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is <math>\boxed{\textbf{(C) } 3}</math>.<br />
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==Solution 2==<br />
We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is <math>\boxed{\textbf{(C) } 3}</math>.<br />
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==Solution 3==<br />
The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is <math>\boxed{\textbf{(C) } 3}</math>.<br />
==Video Solution==<br />
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https://youtu.be/_BaA86QR5ws<br />
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Education, The Study of Everything<br />
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==Video Solution==<br />
https://youtu.be/8LldR2lCmV8<br />
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~savannahsolver<br />
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==See Also==<br />
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{{AMC10 box|year=2019|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Kagrathttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_4&diff=1445242019 AMC 10A Problems/Problem 42021-02-02T01:54:57Z<p>Kagrat: /* Solution 3 */</p>
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<div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #4]] and [[2019 AMC 12A Problems|2019 AMC 12A #3]]}}<br />
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==Problem==<br />
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A box contains <math>28</math> red balls, <math>20</math> green balls, <math>19</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least <math>15</math> balls of a single color will be drawn<math>?</math><br />
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<math>\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91</math><br />
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==Solution==<br />
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By choosing the maximum number of balls while getting <math><15</math> of each color, we could have chosen <math>14</math> red balls, <math>14</math> green balls, <math>14</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls, for a total of <math>75</math> balls. Picking one more ball guarantees that we will get <math>15</math> balls of a color -- either red, green, or yellow. Thus the answer is <math>75 + 1 = \boxed{\textbf{(B) } 76}</math>.<br />
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==Video Solution 1==<br />
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https://youtu.be/givTTqH8Cqo<br />
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Education, The Study of Everything<br />
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== Video Solution 2 ==<br />
https://youtu.be/8WrdYLw9_ns?t=23<br />
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~ pi_is_3.14<br />
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==Video Solution 3==<br />
https://youtu.be/2HmS3n1b4SI<br />
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~savannahsolver<br />
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==Solution 3==<br />
Pigeon Hole Principle. You want to assume that you have very bad luck. The worst the choosing balls could go would be choosing 14 red balls, 14 green, 14 yellow, 13 blue, 11 white, and 9 black. These add up to 75, now no matter what color you choose it makes 15. (There are no more blue, white, and black socks left) = <math>75+1=\boxed{\textbf{(B) } 76}</math><br />
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==See Also==<br />
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{{AMC10 box|year=2019|ab=A|num-b=3|num-a=5}}<br />
{{AMC12 box|year=2019|ab=A|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Kagrat