https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Amritvignesh0719062.0&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-21T12:02:02Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_4&diff=144734 1997 AIME Problems/Problem 4 2021-02-03T02:33:17Z <p>Amritvignesh0719062.0: </p> <hr /> <div>== Problem ==<br /> [[Circle]]s of [[radii]] &lt;math&gt;5, 5, 8,&lt;/math&gt; and &lt;math&gt;\frac mn&lt;/math&gt; are mutually externally tangent, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n.&lt;/math&gt;<br /> <br /> == Solution ==<br /> [[Image:1997_AIME-4.png]]<br /> <br /> If (in the diagram above) we draw the line going through the centers of the circles with radii &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;\frac mn = r&lt;/math&gt;, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii &lt;math&gt;5&lt;/math&gt;. Then we form two [[right triangles]], of lengths &lt;math&gt;5, x, 5+r&lt;/math&gt; and &lt;math&gt;5, 8+r+x, 13&lt;/math&gt;, wher &lt;math&gt;x&lt;/math&gt; is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii &lt;math&gt;5&lt;/math&gt;. By the [[Pythagorean Theorem]], we now have two equations with two unknowns:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}<br /> 5^2 + x^2 &amp;=&amp; (5+r)^2 \\<br /> x &amp;=&amp; \sqrt{10r + r^2} \\<br /> &amp;&amp; \\<br /> (8 + r + \sqrt{10r+r^2})^2 + 5^2 &amp;=&amp; 13^2\\<br /> 8 + r + \sqrt{10r+r^2} &amp;=&amp; 12\\<br /> \sqrt{10r+r^2}&amp;=&amp; 4-r\\<br /> 10r+r^2 &amp;=&amp; 16 - 8r + r^2\\<br /> r &amp;=&amp; \frac{8}{9}<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;m+n = \boxed{17}&lt;/math&gt;.<br /> <br /> <br /> NOTE: It can be seen that there is no apparent need to use the variable x as a 5,12,13 right triangle has been formed.<br /> == Solution 2 ==<br /> We may also use Descartes' theorem, &lt;math&gt;k_4=k_1+k_2+k_3\pm 2\sqrt{k_1k_2+k_2k_3+k_3k_1}&lt;/math&gt; where each of &lt;math&gt;k_i&lt;/math&gt; is the curvature of a circle with radius &lt;math&gt;r_i&lt;/math&gt;, and the curvature is defined as &lt;math&gt;k_i=\frac{1}{r_i}&lt;/math&gt;. The larger solution for &lt;math&gt;k_4&lt;/math&gt; will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get &lt;math&gt;k_4=\frac15+\frac15+\frac18+2\sqrt{\frac{1}{40}+\frac{1}{40}+\frac{1}{25}}=\frac{21}{40}+2\sqrt{\frac{45}{500}}=\frac{45}{40}&lt;/math&gt;. Thus, &lt;math&gt;r_4=\frac{1}{k_4}=\frac{40}{45}=\frac89&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{017}&lt;/math&gt;<br /> <br /> ==Video Solution 1==<br /> https://www.youtube.com/watch?v=KUs44jwBLD8&amp;t=118s<br /> -amritvignesh0719062.0<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Amritvignesh0719062.0 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_24&diff=144266 2020 AMC 10A Problems/Problem 24 2021-02-01T01:10:53Z <p>Amritvignesh0719062.0: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer greater than &lt;math&gt;1000&lt;/math&gt; for which<br /> <br /> &lt;cmath&gt;\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.&lt;/cmath&gt;<br /> <br /> What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24&lt;/math&gt;<br /> <br /> == Solution 1 == <br /> We know that &lt;math&gt;(n+57,63)=21, (n-57, 120)= 60&lt;/math&gt; by the Euclidean Algorithm. Hence, let &lt;math&gt;n+57=21\alpha, n-57=60 \gamma, (\alpha,3)=1, (\gamma,2)=1&lt;/math&gt;. Subtracting the &lt;math&gt;2&lt;/math&gt; equations, &lt;math&gt;38=7\alpha-20\gamma&lt;/math&gt;. Letting &lt;math&gt;\gamma = 2s+1&lt;/math&gt;, &lt;math&gt;58=7\alpha-40s&lt;/math&gt;. Taking &lt;math&gt;\mod{40}&lt;/math&gt;, we have &lt;math&gt;\alpha \equiv{14} \pmod{40}&lt;/math&gt;. We are given &lt;math&gt;n=21\alpha -57 &gt;1000 \implies \alpha \geq 51&lt;/math&gt;. Notice that if &lt;math&gt;\alpha =54&lt;/math&gt; then the condition &lt;math&gt;(\alpha,3)=1&lt;/math&gt; is violated. The next possible value of &lt;math&gt;\alpha = 94&lt;/math&gt; satisfies the given condition, giving us the answer &lt;math&gt;\boxed{1917}&lt;/math&gt;. Alternatively, we could have said &lt;math&gt;\alpha = 40k+14 \equiv{0} \pmod{3}&lt;/math&gt; for &lt;math&gt;k \equiv{1} \pmod{3}&lt;/math&gt; only, so &lt;math&gt;k \equiv{0,2} \pmod{3}&lt;/math&gt;, giving us our answer.<br /> <br /> <br /> ~Prabh1512<br /> <br /> == Solution 2==<br /> <br /> We know that &lt;math&gt;gcd(63, n+120)=21&lt;/math&gt;, so we can write &lt;math&gt;n+120\equiv0\pmod {21}&lt;/math&gt;. Simplifying, we get &lt;math&gt;n\equiv6\pmod {21}&lt;/math&gt;. Similarly, we can write &lt;math&gt;n+63\equiv0\pmod {60}&lt;/math&gt;, or &lt;math&gt;n\equiv-3\pmod {60}&lt;/math&gt;. Solving these two modular congruences, &lt;math&gt;n\equiv237\pmod {420}&lt;/math&gt; which we know is the only solution by CRT (Chinese Remainder Theorem used to so,be a system of MODULAR CONGURENCES). Now, since the problem is asking for the least positive integer greater than &lt;math&gt;1000&lt;/math&gt;, we find the least solution is &lt;math&gt;n=1077&lt;/math&gt;. However, we are have not considered cases where &lt;math&gt;gcd(63, n+120) =63&lt;/math&gt; or &lt;math&gt;gcd(n+63, 120) =120&lt;/math&gt;. &lt;math&gt;{1077+120}\equiv0\pmod {63}&lt;/math&gt; so we try &lt;math&gt;n=1077+420=1497&lt;/math&gt;. &lt;math&gt;{1497+63}\equiv0\pmod {120}&lt;/math&gt; so again we add &lt;math&gt;420&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. It turns out that &lt;math&gt;n=1497+420=1917&lt;/math&gt; does indeed satisfy the original conditions, so our answer is &lt;math&gt;1+9+1+7=\boxed{\textbf{(C) }18}&lt;/math&gt;.<br /> <br /> ==Solution 3 (bashing)==<br /> <br /> We are given that &lt;math&gt;\gcd(63, n+120)=21&lt;/math&gt; and &lt;math&gt;\gcd(n+63,120) = 60&lt;/math&gt;. This tells us that &lt;math&gt;n+120&lt;/math&gt; is divisible by &lt;math&gt;21&lt;/math&gt; but not &lt;math&gt;63&lt;/math&gt;. It also tells us that &lt;math&gt;n+63&lt;/math&gt; is divisible by 60 but not 120. Starting, we find the least value of &lt;math&gt;n+120&lt;/math&gt; which is divisible by &lt;math&gt;21&lt;/math&gt; which satisfies the conditions for &lt;math&gt;n&lt;/math&gt;, which is &lt;math&gt;1134&lt;/math&gt;, making &lt;math&gt;n=1014&lt;/math&gt;. We then now keep on adding &lt;math&gt;21&lt;/math&gt; until we get a number which satisfies the second equation. This number turns out to be &lt;math&gt;1917&lt;/math&gt;, whose digits add up to &lt;math&gt;\boxed{\textbf{(C) } 18}&lt;/math&gt;.<br /> <br /> -Midnight<br /> <br /> ==Solution 4 (bashing but worse)==<br /> <br /> Assume that &lt;math&gt;n&lt;/math&gt; has 4 digits. Then &lt;math&gt;n = abcd&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; represent digits of the number (not to get confused with &lt;math&gt;a * b * c * d&lt;/math&gt;). As given the problem, &lt;math&gt;gcd(63, n + 120) = 21&lt;/math&gt; and &lt;math&gt;gcd(n + 63, 120) = 60&lt;/math&gt;. So we know that &lt;math&gt;d = 7&lt;/math&gt; (last digit of &lt;math&gt;n&lt;/math&gt;). That means that &lt;math&gt;12 + abc \equiv0\pmod {7}&lt;/math&gt; and &lt;math&gt;7 + abc\equiv0\pmod {6}&lt;/math&gt;. We can bash this after this. We just want to find all pairs of numbers &lt;math&gt;(x, y)&lt;/math&gt; such that &lt;math&gt;x&lt;/math&gt; is a multiple of 7 that is &lt;math&gt;5&lt;/math&gt; greater than a multiple of &lt;math&gt;6&lt;/math&gt;. Our equation for &lt;math&gt;12 + abc&lt;/math&gt; would be &lt;math&gt;42*j + 35 = x&lt;/math&gt; and our equation for &lt;math&gt;7 + abc&lt;/math&gt; would be &lt;math&gt; 42*j + 30 = y&lt;/math&gt;, where &lt;math&gt;j&lt;/math&gt; is any integer. We plug this value in until we get a value of &lt;math&gt;abc&lt;/math&gt; that makes &lt;math&gt;n = abc7&lt;/math&gt; satisfy the original problem statement (remember, &lt;math&gt;abc &gt; 100&lt;/math&gt;). After bashing for hopefully a couple minutes, we find that &lt;math&gt;abc = 191&lt;/math&gt; works. So &lt;math&gt;n = 1917&lt;/math&gt; which means that the sum of its digits is &lt;math&gt;\boxed{\textbf{(C) } 18}&lt;/math&gt;.<br /> <br /> ~ Baolan<br /> <br /> ==Solution 5==<br /> The conditions of the problem reduce to the following. &lt;math&gt;n+120 = 21k&lt;/math&gt; where &lt;math&gt;gcd(k,3) = 1&lt;/math&gt; and &lt;math&gt;n+63 = 60l&lt;/math&gt; where &lt;math&gt;gcd(l,2) = 1&lt;/math&gt;. From these equations, we see that &lt;math&gt;21k - 60l = 57&lt;/math&gt;. Solving this diophantine equation gives us that &lt;math&gt;k = 20a + 17&lt;/math&gt;, &lt;math&gt;l = 7a + 5&lt;/math&gt; form. Since, &lt;math&gt;n&lt;/math&gt; is greater than &lt;math&gt;1000&lt;/math&gt;, we can do some bounding and get that &lt;math&gt;k &gt; 53&lt;/math&gt; and &lt;math&gt;l &gt; 17&lt;/math&gt;. Now we start the bash by plugging in numbers that satisfy these conditions. We get &lt;math&gt;l = 33&lt;/math&gt;, &lt;math&gt;k = 97&lt;/math&gt;. So the answer is &lt;math&gt;1917 \implies 1+9+1+7=<br /> \boxed{\textbf{(C) } 18}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> You can first find that n must be congruent to &lt;math&gt;6\equiv0\pmod {21}&lt;/math&gt; and &lt;math&gt;57\equiv0\pmod {60}&lt;/math&gt;. The we can find that &lt;math&gt;n=21x+6&lt;/math&gt; and &lt;math&gt;n=60y+57&lt;/math&gt;, where x and y are integers. Then we can find that y must be odd, since if it was even the gcd will be 120, not 60. Also, the unit digit of n has to be 7, since the unit digit of 60y is always 0 and the unit digit of 57 is 7. Therefore, you can find that x must end in 1 to satisfy n having a unit digit of 7. Also, you can find that x must not be a multiple of three or else the gcd will be 63. Therefore, you can test values for x and you can find that x=91 satisfies all these conditions.Therefore, n is 1917 and &lt;math&gt;1+9+1+7 = \boxed{\textbf{(C) } 18}&lt;/math&gt;.-happykeeper<br /> <br /> ==Solution 7 (Reverse Euclidean Algorithm)==<br /> We are given that &lt;math&gt;\gcd(63, n+120) =21&lt;/math&gt; and &lt;math&gt;\gcd(n+63, 120)=60.&lt;/math&gt; By applying the Euclidean algorithm, but in reverse, we have &lt;cmath&gt;\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21&lt;/cmath&gt; and &lt;cmath&gt;\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60.&lt;/cmath&gt;<br /> <br /> We now know that &lt;math&gt;n+183&lt;/math&gt; must be divisible by &lt;math&gt;21&lt;/math&gt; and &lt;math&gt;60,&lt;/math&gt; so it is divisible by &lt;math&gt;\text{lcm}(21, 60) = 420.&lt;/math&gt; Therefore, &lt;math&gt;n+183 = 420k&lt;/math&gt; for some integer &lt;math&gt;k.&lt;/math&gt; We know that &lt;math&gt;3 \nmid k,&lt;/math&gt; or else the first condition won't hold (&lt;math&gt;\gcd&lt;/math&gt; will be &lt;math&gt;63&lt;/math&gt;) and &lt;math&gt;2 \nmid k,&lt;/math&gt; or else the second condition won't hold (&lt;math&gt;\gcd&lt;/math&gt; will be &lt;math&gt;120&lt;/math&gt;). Since &lt;math&gt;k = 1&lt;/math&gt; gives us too small of an answer, then &lt;math&gt;k=5 \implies n = 1917,&lt;/math&gt; so the answer is &lt;math&gt;1+9+1+7 = \boxed{\textbf{(C) } 18}.&lt;/math&gt;<br /> <br /> ==Solution 8==<br /> &lt;math&gt;\gcd(n+63,120)=60&lt;/math&gt; tells us &lt;math&gt;n+63\equiv60\pmod {120}&lt;/math&gt;. The smallest &lt;math&gt;n+63&lt;/math&gt; that satisfies the previous condition and &lt;math&gt;n&gt;1000&lt;/math&gt; is &lt;math&gt;1140&lt;/math&gt;, so we start from there. If &lt;math&gt;n+63=1140&lt;/math&gt;, then &lt;math&gt;n+120=1197&lt;/math&gt;. Because &lt;math&gt;\gcd(n+120,63)=21&lt;/math&gt;, &lt;math&gt;n+120\equiv21\pmod {63}&lt;/math&gt; or &lt;math&gt;n+120\equiv42\pmod {63}&lt;/math&gt;. We see that &lt;math&gt;1197\equiv0\pmod {63}&lt;/math&gt;, which does not fulfill the requirement for &lt;math&gt;n+120&lt;/math&gt;, so we continue by keep on adding &lt;math&gt;120&lt;/math&gt; to &lt;math&gt;1197&lt;/math&gt;, in order to also fulfill the requirement for &lt;math&gt;n+63&lt;/math&gt;. Soon, we see that &lt;math&gt;n+120\pmod {63}&lt;/math&gt; decreases by &lt;math&gt;6&lt;/math&gt; every time we add &lt;math&gt;120&lt;/math&gt;, so we can quickly see that &lt;math&gt;n=1917&lt;/math&gt; because at that point &lt;math&gt;n+120\equiv21\pmod {63}&lt;/math&gt;. Adding up all the digits in &lt;math&gt;1917&lt;/math&gt;, we have &lt;math&gt;\boxed{\textbf{(C) } 18}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> ==Solution 9==<br /> We are able to set-up the following system-of-congruences: <br /> &lt;cmath&gt;n \equiv 6 \pmod {21},&lt;/cmath&gt; <br /> &lt;cmath&gt;n \equiv 57 \pmod {60}.&lt;/cmath&gt; <br /> Therefore, by definition, we are able to set-up the following system of equations: <br /> &lt;cmath&gt;n = 21a + 6,&lt;/cmath&gt; <br /> &lt;cmath&gt;n = 60b + 57.&lt;/cmath&gt; <br /> Thus, <br /> &lt;cmath&gt;21a + 6 = 60b + 57&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies 7a + 2 = 20b + 19.&lt;/cmath&gt;<br /> We know &lt;math&gt;7a \equiv 0 \pmod {7},&lt;/math&gt; and since &lt;math&gt;7a = 20b + 17,&lt;/math&gt; therefore &lt;math&gt;20b + 17 \equiv 0 \pmod{7}.&lt;/math&gt; Simplifying this congruence further, we have <br /> &lt;cmath&gt;5b \equiv 1 \pmod{7}&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies b \equiv 3 \pmod {7}.&lt;/cmath&gt;<br /> Thus, by definition, &lt;math&gt;b = 7x + 3.&lt;/math&gt; Substituting this back into our original equation, <br /> &lt;cmath&gt;n = 60(7x + 3) + 57&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies n = 420x + 180 + 57&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies n = 420x + 237.&lt;/cmath&gt;<br /> By definition, we are able to set-up the following congruence:<br /> &lt;cmath&gt;n \equiv 237 \pmod{420}.&lt;/cmath&gt;<br /> Thus, &lt;math&gt;n = 1917&lt;/math&gt;, so our answer is simply &lt;math&gt;\boxed{18}&lt;/math&gt;.<br /> <br /> (Remarks. &lt;math&gt;n \equiv 6 \pmod{21}&lt;/math&gt; since &lt;math&gt;n \equiv -120 \pmod{21},&lt;/math&gt; by definition &amp; &lt;math&gt;n \equiv 57 \pmod{60}&lt;/math&gt; since &lt;math&gt;n \equiv -63 \pmod{60},&lt;/math&gt; by definition. <br /> <br /> Remember, &lt;math&gt;5b \equiv 1 \pmod{7} \implies 5b \equiv 15 \pmod{7} \implies b \equiv 3 \pmod{7}.&lt;/math&gt;<br /> <br /> Lastly, the reason why &lt;math&gt;n \neq 1077&lt;/math&gt; is &lt;math&gt;n + 120&lt;/math&gt; would be divisible by &lt;math&gt;63&lt;/math&gt;, which is not possible due to the certain condition.) <br /> <br /> ~ nikenissan<br /> <br /> == Solution 10==<br /> <br /> First, we find &lt;math&gt;n&lt;/math&gt;. We know that it is greater than &lt;math&gt;1000&lt;/math&gt;, so we first input &lt;math&gt;n = 1000&lt;/math&gt;. From the first equation, &lt;math&gt;gcd(63, n + 120) = 21&lt;/math&gt;, we know that if &lt;math&gt;n&lt;/math&gt; is correct, after we add &lt;math&gt;120&lt;/math&gt; to it, it should be divisible by &lt;math&gt;21&lt;/math&gt;, but not &lt;math&gt;63&lt;/math&gt;. <br /> &lt;cmath&gt;\frac{n + 120}{21}, &lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1120}{21}, &lt;/cmath&gt;<br /> &lt;cmath&gt;53 r 7. &lt;/cmath&gt;<br /> Uh oh. To get to the nearest number divisible by &lt;math&gt;21&lt;/math&gt;, we have to add &lt;math&gt;14&lt;/math&gt; to cancel out the remainder. (Note that we don't subtract &lt;math&gt;7&lt;/math&gt; to get to &lt;math&gt;53&lt;/math&gt;; &lt;math&gt;n&lt;/math&gt; is already at its lowest possible value!)<br /> Adding &lt;math&gt;14&lt;/math&gt; to &lt;math&gt;1000&lt;/math&gt; gives us &lt;math&gt;n = 1014&lt;/math&gt;. (Note: &lt;math&gt;n&lt;/math&gt; is currently divisible by 63, but that's fine since we'll be changing it in the next step.)<br /> <br /> Now using, the second equation, &lt;math&gt;gcd(n + 63, 120) = 60&lt;/math&gt;, we know that if &lt;math&gt;n&lt;/math&gt; is correct, after we add &lt;math&gt;63&lt;/math&gt; to it, it should be divisible by &lt;math&gt;60&lt;/math&gt;, but not &lt;math&gt;120&lt;/math&gt;.<br /> &lt;cmath&gt;\frac{n + 63}{60}, &lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1077}{60}, &lt;/cmath&gt;<br /> &lt;cmath&gt;17r57. &lt;/cmath&gt;<br /> Uh oh (again). This requires some guessing and checking. We can add &lt;math&gt;21&lt;/math&gt; over and over again until &lt;math&gt;n&lt;/math&gt; is valid. This changes &lt;math&gt;n&lt;/math&gt; while also maintaining that &lt;math&gt;\frac{n + 120}{21}&lt;/math&gt; has no remainders. <br /> After adding &lt;math&gt;21&lt;/math&gt; once, we get &lt;math&gt;18 r 18&lt;/math&gt;. By pure luck, adding &lt;math&gt;21&lt;/math&gt; two more times gives us &lt;math&gt;19&lt;/math&gt; with no remainders. <br /> We now have &lt;math&gt;1077 + 21 + 21 + 21 = 1140&lt;/math&gt;. However, this number is divisible by &lt;math&gt;120&lt;/math&gt;. To get the next possible number, we add the LCM of &lt;math&gt;21&lt;/math&gt; and &lt;math&gt;60&lt;/math&gt; (once again, to maintain divisibility), which is &lt;math&gt;420&lt;/math&gt;. Unfortunately, &lt;math&gt;1140 + 420 = 1560&lt;/math&gt; is still divisible by &lt;math&gt;120&lt;/math&gt;. Adding &lt;math&gt;420&lt;/math&gt; again gives us &lt;math&gt;1980&lt;/math&gt;, which is valid. However, remember that this is equal to &lt;math&gt;n + 63&lt;/math&gt;, so subtracting &lt;math&gt;63&lt;/math&gt; from &lt;math&gt;1980&lt;/math&gt; gives us &lt;math&gt;1917&lt;/math&gt;, which is &lt;math&gt;n&lt;/math&gt;. <br /> <br /> The sum of its digits are &lt;math&gt;1 + 9 + 1 + 7 = 18&lt;/math&gt;.<br /> <br /> So, our answer is &lt;math&gt;\boxed{\textbf{(C) }18}&lt;/math&gt;. ~ primegn <br /> <br /> == Video Solution 1 ==<br /> <br /> https://youtu.be/tk3yOGG2K-s ~ Richard Rusczyk<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/8mNMKH0T9W0 - Happytwin<br /> <br /> ==Video Solution 3==<br /> Education The Study of Everything<br /> <br /> https://youtu.be/e5BJKMEIPEM<br /> <br /> == Video Solution 4 ==<br /> https://www.youtube.com/watch?v=gdGmSyzR908&amp;list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&amp;index=5 ~ MathEx<br /> <br /> == Video Solution 5 ==<br /> https://youtu.be/R220vbM_my8?t=899<br /> ~ amritvignesh0719062.0<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Amritvignesh0719062.0 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=144264 2020 AMC 10A Problems/Problem 20 2021-02-01T01:08:39Z <p>Amritvignesh0719062.0: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #18]] and [[2020 AMC 10A Problems|2020 AMC 10A #20]]}}<br /> <br /> == Problem ==<br /> Quadrilateral ABCD satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution 1 (Just Drop An Altitude)==<br /> <br /> &lt;asy&gt;<br /> size(15cm,0);<br /> import olympiad;<br /> draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);<br /> label(&quot;A&quot;, (0,2), NW);<br /> label(&quot;B&quot;, (0,0), SW);<br /> label(&quot;C&quot;, (4,0), SE);<br /> label(&quot;D&quot;, (6,4), NE);<br /> label(&quot;E&quot;, (1.714,1.143), N);<br /> label(&quot;F&quot;, (1,1.5), N);<br /> draw((0,2)--(4,0), dashed);<br /> draw((0,0)--(6,4), dashed);<br /> draw((0,0)--(1,1.5), dashed);<br /> label(&quot;20&quot;, (0,2)--(4,0), SW);<br /> label(&quot;30&quot;, (4,0)--(6,4), SE);<br /> label(&quot;$x$&quot;, (1,1.5)--(1.714,1.143), NE);<br /> draw(rightanglemark((0,2),(0,0),(4,0)));<br /> draw(rightanglemark((0,2),(4,0),(6,4)));<br /> draw(rightanglemark((0,0),(1,1.5),(0,2)));<br /> &lt;/asy&gt;<br /> <br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt; because of the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}&lt;/math&gt;<br /> <br /> <br /> <br /> ~ Solution by Ultraman<br /> <br /> ~ Diagram by ciceronii<br /> <br /> ==Solution 2 (Coordinates)==<br /> &lt;asy&gt;<br /> size(10cm,0);<br /> draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br /> draw((-20,0)--(20,0));<br /> draw((0,-15)--(0,35));<br /> draw((10,30)--(-8,-6));<br /> draw(circle((0,0),10));<br /> label(&quot;E&quot;,(-4.05,-.25),S);<br /> label(&quot;D&quot;,(10,30),NE);<br /> label(&quot;C&quot;,(10,0),NE);<br /> label(&quot;B&quot;,(-8,-6),SW);<br /> label(&quot;A&quot;,(-10,0),NW);<br /> label(&quot;5&quot;,(-10,0)--(-5,0), NE);<br /> label(&quot;15&quot;,(-5,0)--(10,0), N);<br /> label(&quot;30&quot;,(10,0)--(10,30), E);<br /> dot((-5,0));<br /> dot((-10,0));<br /> dot((-8,-6));<br /> dot((10,0));<br /> dot((10,30));<br /> &lt;/asy&gt;<br /> Let the points be &lt;math&gt;A(-10,0)&lt;/math&gt;, &lt;math&gt;\:B(x,y)&lt;/math&gt;, &lt;math&gt;\:C(10,0)&lt;/math&gt;, &lt;math&gt;\:D(10,30)&lt;/math&gt;,and &lt;math&gt;\:E(-5,0)&lt;/math&gt;, respectively. Since &lt;math&gt;B&lt;/math&gt; lies on line &lt;math&gt;DE&lt;/math&gt;, we know that &lt;math&gt;y=2x+10&lt;/math&gt;. Furthermore, since &lt;math&gt;\angle{ABC}=90^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; lies on the circle with diameter &lt;math&gt;AC&lt;/math&gt;, so &lt;math&gt;x^2+y^2=100&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with these equations, we get the solutions &lt;math&gt;(0,10)&lt;/math&gt; and &lt;math&gt;(-8,-6)&lt;/math&gt;. We immediately discard the &lt;math&gt;(0,10)&lt;/math&gt; solution as &lt;math&gt;y&lt;/math&gt; should be negative. Thus, we conclude that &lt;math&gt;[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Trigonometry)==<br /> Let &lt;math&gt;\angle C = \angle{ACB}&lt;/math&gt; and &lt;math&gt;\angle{B} = \angle{CBE}.&lt;/math&gt; Using Law of Sines on &lt;math&gt;\triangle{BCE}&lt;/math&gt; we get &lt;cmath&gt;\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}&lt;/cmath&gt; and LoS on &lt;math&gt;\triangle{ABE}&lt;/math&gt; yields &lt;cmath&gt;\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.&lt;/cmath&gt; Divide the two to get &lt;math&gt;\tan{B} = 3 \tan{C}.&lt;/math&gt; Now, &lt;cmath&gt;\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}&lt;/cmath&gt; and solve the quadratic, taking the positive solution (C is acute) to get &lt;math&gt;\tan{C} = \frac{1}{3}.&lt;/math&gt; So if &lt;math&gt;AB = a,&lt;/math&gt; then &lt;math&gt;BC = 3a&lt;/math&gt; and &lt;math&gt;[ABC] = \frac{3a^2}{2}.&lt;/math&gt; By Pythagorean Theorem, &lt;math&gt;10a^2 = 400 \iff \frac{3a^2}{2} = 60&lt;/math&gt; and the answer is &lt;math&gt;300 + 60 \iff \boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> (This solution is incomplete, can someone complete it please-Lingjun) ok<br /> Latex edited by kc5170<br /> <br /> We could use the famous m-n rule in trigonometry in &lt;math&gt;\triangle ABC&lt;/math&gt; with Point &lt;math&gt;E&lt;/math&gt; <br /> [Unable to write it here.Could anybody write the expression]<br /> . We will find that &lt;math&gt;\overrightarrow{BD}&lt;/math&gt; is an angle bisector of &lt;math&gt;\triangle ABC&lt;/math&gt; (because we will get &lt;math&gt;\tan(x) = 1&lt;/math&gt;). <br /> Therefore by converse of angle bisector theorem &lt;math&gt;AB:BC = 1:3&lt;/math&gt;. By using Pythagorean theorem, we have values of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;. <br /> Computing &lt;math&gt;AB \cdot AC = 120&lt;/math&gt;. Adding the areas of &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;ACD&lt;/math&gt;, hence the answer is &lt;math&gt;\boxed{\textbf{(D)}\:360}&lt;/math&gt;.<br /> <br /> By: Math-Amaze<br /> Latex: Catoptrics.<br /> <br /> ==Solution 4 (Answer Choices)==<br /> We know that the big triangle has area 300. Using the answer choices would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. We guess that the legs are equal to &lt;math&gt;\sqrt{20a}&lt;/math&gt; and &lt;math&gt;\sqrt{20b}&lt;/math&gt;, and because the hypotenuse is 20 we get &lt;math&gt;a+b=20&lt;/math&gt;. Testing small numbers, we get that when &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=18&lt;/math&gt;, &lt;math&gt;ab&lt;/math&gt; is indeed a square. The area of the triangle is thus &lt;math&gt;60&lt;/math&gt;, so the answer is &lt;math&gt;\boxed {\textbf{(D) }360}&lt;/math&gt;.<br /> <br /> ~tigershark22<br /> <br /> ==Solution 5 (LoC)==<br /> <br /> &lt;asy&gt;<br /> import olympiad;<br /> pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798);<br /> dot(&quot;$A$&quot;, A, W); dot(&quot;$B$&quot;, B, S); dot(&quot;$C$&quot;, C, E); dot(&quot;$D$&quot;, D, N);dot(&quot;$E$&quot;,(140, 140), N);<br /> draw(A--B--C--D--A);<br /> draw(A--C, dotted); draw(B--D, dotted);<br /> &lt;/asy&gt;<br /> <br /> Denote &lt;math&gt;EB&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;. By the Law of Cosine:<br /> &lt;cmath&gt;AB^2 = 25 + x^2 - 10x\cos(\angle DEC)&lt;/cmath&gt;<br /> &lt;cmath&gt;BC^2 = 225 + x^2 + 30x\cos(\angle DEC)&lt;/cmath&gt;<br /> <br /> Adding these up yields:<br /> &lt;cmath&gt;400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0&lt;/cmath&gt;<br /> By the quadratic formula, &lt;math&gt;x = 3\sqrt5&lt;/math&gt;.<br /> <br /> Observe:<br /> &lt;cmath&gt;[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60&lt;/cmath&gt;.<br /> <br /> Thus the desired area is &lt;math&gt;\frac{1}{2}(30)(20) + 60 = \boxed{\textbf{(D) } 360}&lt;/math&gt;<br /> <br /> ~qwertysri987<br /> <br /> ==Solution 6 (basic vectors/coordinates)==<br /> <br /> Let &lt;math&gt;C = (0, 0)&lt;/math&gt; and &lt;math&gt;D = (0, 30)&lt;/math&gt;. Then &lt;math&gt;E = (-15, 0), A = (-20, 0),&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; lies on the line &lt;math&gt;y=2x+30.&lt;/math&gt; So the coordinates of &lt;math&gt;B&lt;/math&gt; are &lt;cmath&gt;(x, 2x+30).&lt;/cmath&gt;<br /> <br /> We can make this a vector problem.<br /> &lt;math&gt;\overrightarrow{\mathbf{B}} = \begin{pmatrix}<br /> x \\<br /> 2x+30<br /> \end{pmatrix}.&lt;/math&gt; We notice that point &lt;math&gt;B&lt;/math&gt; forms a right angle, meaning vectors &lt;math&gt;\overrightarrow{\mathbf{BC}}&lt;/math&gt; and &lt;math&gt;\overrightarrow{\mathbf{BA}}&lt;/math&gt; are orthogonal, and their dot-product is &lt;math&gt;0&lt;/math&gt;.<br /> <br /> We determine &lt;math&gt;\overrightarrow{\mathbf{BC}}&lt;/math&gt; and &lt;math&gt;\overrightarrow{\mathbf{BA}}&lt;/math&gt; to be &lt;math&gt;\begin{pmatrix}<br /> -x \\<br /> -2x-30<br /> \end{pmatrix}&lt;/math&gt; and &lt;math&gt;\begin{pmatrix}<br /> -20-x \\<br /> -2x-30<br /> \end{pmatrix}&lt;/math&gt; , respectively. (To get this, we use the fact that &lt;math&gt;\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}&lt;/math&gt; and similarly, &lt;math&gt;\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.&lt;/math&gt; )<br /> <br /> Equating the cross-product to &lt;math&gt;0&lt;/math&gt; gets us the quadratic &lt;math&gt;-x(-20-x)+(-2x-30)(-2x-30)=0.&lt;/math&gt; The solutions are &lt;math&gt;x=-18, -10.&lt;/math&gt; Since &lt;math&gt;B&lt;/math&gt; clearly has a more negative x-coordinate than &lt;math&gt;E&lt;/math&gt;, we take &lt;math&gt;x=-18&lt;/math&gt;. So &lt;math&gt;B = (-18, -6).&lt;/math&gt;<br /> <br /> From here, there are multiple ways to get the area of &lt;math&gt;\Delta{ABC}&lt;/math&gt; to be &lt;math&gt;60&lt;/math&gt;, and since the area of &lt;math&gt;\Delta{ACD}&lt;/math&gt; is &lt;math&gt;300&lt;/math&gt;, we get our final answer to be &lt;cmath&gt;60 + 300 = \boxed{\text{(D) } 360}.&lt;/cmath&gt;<br /> <br /> -PureSwag<br /> <br /> ==Video Solution 1==<br /> On The Spot STEM<br /> <br /> https://www.youtube.com/watch?v=hIdNde2Vln4<br /> <br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=sHrjx968ZaM&amp;list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&amp;index=2 ~ MathEx<br /> <br /> ==Video Solution 3==<br /> Education, The Study of Everything<br /> <br /> https://youtu.be/5lb8kk1qbaA<br /> <br /> ==Video Solution 4==<br /> The Beauty Of Math<br /> https://www.youtube.com/watch?v=RKlG6oZq9so&amp;ab_channel=TheBeautyofMath<br /> <br /> ==Video Solution 5==<br /> https://youtu.be/R220vbM_my8?t=658<br /> <br /> (amritvignesh0719062.0)<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Amritvignesh0719062.0 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_17&diff=144263 2020 AMC 10A Problems/Problem 17 2021-02-01T01:06:32Z <p>Amritvignesh0719062.0: </p> <hr /> <div>== Problem ==<br /> <br /> Define &lt;cmath&gt;P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).&lt;/cmath&gt; How many integers &lt;math&gt;n&lt;/math&gt; are there such that &lt;math&gt;P(n)\leq 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100&lt;/math&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> Notice that &lt;math&gt;P(x)&lt;/math&gt; is a product of many integers. We either need one factor to be 0 or an odd number of negative factors.<br /> <br /> Case 1: There are 100 integers &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;P(x)=0&lt;/math&gt;<br /> <br /> Case 2: For there to be an odd number of negative factors, &lt;math&gt;n&lt;/math&gt; must be between an odd number squared and an even number squared. This means that there are &lt;math&gt;2+6+10+\dots+194+198&lt;/math&gt; total possible values of &lt;math&gt;n&lt;/math&gt;. Simplifying, there are &lt;math&gt;5000&lt;/math&gt; possible numbers.<br /> <br /> Summing, there are &lt;math&gt;\boxed{\textbf{(E) } 5100}&lt;/math&gt; total possible values of &lt;math&gt;n&lt;/math&gt;. ~PCChess<br /> <br /> === Solution 2 ===<br /> Notice that &lt;math&gt;P(x)&lt;/math&gt; is nonpositive when &lt;math&gt;x&lt;/math&gt; is between &lt;math&gt;100^2&lt;/math&gt; and &lt;math&gt;99^2&lt;/math&gt;, &lt;math&gt;98^2&lt;/math&gt; and &lt;math&gt;97^2 \ldots&lt;/math&gt; , &lt;math&gt;2^2&lt;/math&gt; and &lt;math&gt;1^2&lt;/math&gt; (inclusive), which means that the number of values equals &lt;math&gt;((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \ldots + ((2+1)(2-1)+1)&lt;/math&gt;.<br /> <br /> This reduces to &lt;math&gt;200 + 196 + 192 + \ldots + 4 = 4(1+2+\ldots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}&lt;/math&gt;<br /> <br /> ~Zeric<br /> <br /> === Solution 3 (end behavior) ===<br /> <br /> We know that &lt;math&gt;P(x)&lt;/math&gt; is a &lt;math&gt;100&lt;/math&gt;-degree function with a positive leading coefficient. That is, &lt;math&gt;P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}&lt;/math&gt;.<br /> <br /> Since the degree of &lt;math&gt;P(x)&lt;/math&gt; is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach &lt;math&gt;\infty&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt; goes in either direction.<br /> <br /> &lt;cmath&gt;\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty&lt;/cmath&gt;<br /> <br /> So the first time &lt;math&gt;P(x)&lt;/math&gt; is going to be negative is when it intersects the &lt;math&gt;x&lt;/math&gt;-axis at an &lt;math&gt;x&lt;/math&gt;-intercept and it's going to dip below. This happens at &lt;math&gt;1^2&lt;/math&gt;, which is the smallest intercept.<br /> <br /> However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at &lt;math&gt;2^2&lt;/math&gt;. And when it hits &lt;math&gt;3^2&lt;/math&gt;, it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until &lt;math&gt;100^2&lt;/math&gt;.<br /> <br /> To get the amount of integers below and/or on the &lt;math&gt;x&lt;/math&gt;-axis, we simply need to count the integers. For example, the amount of integers in between the &lt;math&gt;[1^2,2^2]&lt;/math&gt; interval we got earlier, we subtract and add one. &lt;math&gt;(2^2-1^2+1)=4&lt;/math&gt; integers, so there are four integers in this interval that produce a negative result. <br /> <br /> Doing this with all of the other intervals, we have<br /> <br /> &lt;math&gt;(2^2-1^2+1)+(4^2-3^2+1)+...+(100^2-99^2+1)&lt;/math&gt;. Proceed with Solution 2. ~quacker88<br /> <br /> === Video Solution ===<br /> https://youtu.be/3dfbWzOfJAI?t=4026<br /> <br /> ~ pi_is_3.14<br /> <br /> https://youtu.be/zl5rtHnk0rY<br /> <br /> ~Education, The Study of Everything<br /> <br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> https://www.youtube.com/watch?v=YDMMhSguq0w&amp;list=PLeFyQ1uCoINM4D5Lgi5Y3KkfvQuYuIbj<br /> <br /> -Walt S.<br /> <br /> https://youtu.be/chDmeTQBxq8<br /> <br /> ~savannahsolver<br /> <br /> https://youtu.be/R220vbM_my8?t=463<br /> <br /> ~ amritvignesh0719062.0<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2020|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Amritvignesh0719062.0 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=144260 2020 AMC 10A Problems/Problem 16 2021-02-01T01:04:58Z <p>Amritvignesh0719062.0: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #16]] and [[2020 AMC 10A Problems|2020 AMC 10A #16]]}}<br /> <br /> == Problem ==<br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> == Solutions ==<br /> ==== Diagram ====<br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray);<br /> draw(arc((1,0), 0.3989, 90, 180));<br /> filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray);<br /> draw(arc((1,1), 0.3989, 180, 270));<br /> filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray);<br /> draw(arc((0,1), 0.3989, 270, 360));<br /> filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> Diagram by [[User:Mathandski|MathandSki]] Using Asymptote<br /> <br /> Note: The diagram represents each unit square of the given &lt;math&gt;2020 \times 2020&lt;/math&gt; square.<br /> <br /> ==== Solution 1 ====<br /> We consider an individual one-by-one block.<br /> <br /> If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius &lt;math&gt;d&lt;/math&gt;, the area covered by the circles should be &lt;math&gt;0.5&lt;/math&gt;. Because of this, and the fact that there are four circles, we write<br /> <br /> &lt;cmath&gt;4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;d&lt;/math&gt;, we obtain &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;, where with &lt;math&gt;\pi \approx 3&lt;/math&gt;, we get &lt;math&gt;d = \frac{1}{\sqrt{6}}&lt;/math&gt;, and from here, we simplify and see that &lt;math&gt;d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}&lt;/math&gt; ~Crypthes<br /> <br /> &lt;math&gt;\textbf{Note:}&lt;/math&gt; To be more rigorous, note that &lt;math&gt;d&lt;0.5&lt;/math&gt; since if &lt;math&gt;d\geq0.5&lt;/math&gt; then clearly the probability is greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This would make sure the above solution works, as if &lt;math&gt;d\geq0.5&lt;/math&gt; there is overlap with the quartercircles. &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> As in the previous solution, we obtain the equation &lt;math&gt;4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}&lt;/math&gt;, which simplifies to &lt;math&gt;\pi d^2 = \frac{1}{2} = 0.5&lt;/math&gt;. Since &lt;math&gt;\pi&lt;/math&gt; is slightly more than &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;d^2&lt;/math&gt; is slightly less than &lt;math&gt;\frac{0.5}{3} = 0.1\bar{6}&lt;/math&gt;. We notice that &lt;math&gt;0.1\bar{6}&lt;/math&gt; is slightly more than &lt;math&gt;0.4^2 = 0.16&lt;/math&gt;, so &lt;math&gt;d&lt;/math&gt; is roughly &lt;math&gt;\boxed{\textbf{(B) } 0.4}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> === Solution 3 (Estimating) ===<br /> As above, we find that we need to estimate &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;. <br /> <br /> Note that we can approximate &lt;math&gt;2\pi \approx 6.28318 \approx 6.25&lt;/math&gt; and so &lt;math&gt;\frac{1}{\sqrt{2\pi}}&lt;/math&gt; &lt;math&gt;\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4&lt;/math&gt;.<br /> <br /> And so our answer is &lt;math&gt;\boxed{\textbf{(B) } 0.4}&lt;/math&gt;.<br /> <br /> ~Silverdragon<br /> <br /> === Solution 4 (Estimating but a bit different) ===<br /> We only need to figure out the probability for a unit square, as it will scale up to the &lt;math&gt;2020\times 2020&lt;/math&gt; square. Since we want to find the probability that a point inside a unit square that is &lt;math&gt;d&lt;/math&gt; units away from a lattice point (a corner of the square) is &lt;math&gt;\frac{1}{2}&lt;/math&gt;, we can find which answer will come the closest to covering &lt;math&gt;\frac{1}{2}&lt;/math&gt; of the area. <br /> <br /> Since the closest is &lt;math&gt;0.4&lt;/math&gt; which turns out to be &lt;math&gt;(0.4)^2\times \pi = 0.16 \times \pi&lt;/math&gt; which is about &lt;math&gt;0.502&lt;/math&gt;, we find that the answer rounded to the nearest tenth is &lt;math&gt;0.4&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt;.<br /> <br /> ~RuiyangWu<br /> <br /> === Solution 5 (Estimating but differently again) ===<br /> As per the above diagram, realize that &lt;math&gt;\pi d^2 = \frac{1}{2}&lt;/math&gt;, so &lt;math&gt;d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sqrt{2} \approx 1.4 = \frac{7}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sqrt{\pi}&lt;/math&gt; is between &lt;math&gt;1.7&lt;/math&gt; and &lt;math&gt;1.8&lt;/math&gt; &lt;math&gt;((1.7)^2 = 2.89&lt;/math&gt; and &lt;math&gt;(1.8)^2 = 3.24)&lt;/math&gt;, so we can say &lt;math&gt;\sqrt{\pi} \approx 1.75 = \frac{7}{4}&lt;/math&gt;.<br /> <br /> So &lt;math&gt;d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}&lt;/math&gt;. This is slightly above &lt;math&gt;\boxed{\textbf{(B) } 0.4}&lt;/math&gt;, since &lt;math&gt;\frac{20}{49} \approx \frac{2}{5}&lt;/math&gt;.<br /> <br /> -Solution by Joeya<br /> <br /> == Video Solution ==<br /> === Video Solution 1 ===<br /> Education, The Study of Everything<br /> <br /> https://youtu.be/napCkujyrac<br /> <br /> === Video Solution 2 ===<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> === Video Solution 3 ===<br /> https://youtu.be/R220vbM_my8?t=238<br /> <br /> ~ amritvignesh0719062.0<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> {{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Amritvignesh0719062.0 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=144259 2020 AMC 10A Problems/Problem 13 2021-02-01T01:03:40Z <p>Amritvignesh0719062.0: </p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is &lt;math&gt;\frac{1}{4} \cdot 1 = \frac{1}{4}&lt;/math&gt;. If the frog goes to the right, it will be in the center of the square at &lt;math&gt;(2,2)&lt;/math&gt;, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. The probability of this happening is &lt;math&gt;\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}&lt;/math&gt;.<br /> <br /> <br /> If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Because there's a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of the frog going up and down, the total probability for this case is &lt;math&gt;\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}&lt;/math&gt; and summing up all the cases, &lt;math&gt;\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have &quot;succeeded&quot;, while B means that we have a half chance, we compute &lt;math&gt;1 \cdot C + \frac{1}{2} \cdot B&lt;/math&gt;. <br /> <br /> <br /> &lt;cmath&gt;1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{4} + \frac{3}{8}&lt;/cmath&gt;<br /> We get &lt;math&gt;\frac{5}{8}&lt;/math&gt;, or &lt;math&gt;B&lt;/math&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt; <br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{C A B O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O B O O O}&lt;/cmath&gt;<br /> &lt;cmath&gt;\text{O O O O O}&lt;/cmath&gt;<br /> -yeskay<br /> <br /> ==Solution 3==<br /> If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Since it starts on &lt;math&gt;(1,2)&lt;/math&gt;, there is a &lt;math&gt;\frac{3}{4}&lt;/math&gt; chance (up, down, or right) it will reach a diagonal on the first jump and &lt;math&gt;\frac{1}{4}&lt;/math&gt; chance (left) it will reach the vertical side. The probablity of landing on a vertical is &lt;math&gt;\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}&lt;/math&gt;.<br /> - Lingjun<br /> <br /> ==Solution 4 (Complete States)==<br /> Let &lt;math&gt;P_{(x,y)}&lt;/math&gt; denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at &lt;math&gt;(x,y)&lt;/math&gt;. Note that &lt;math&gt;P_{(1,2)}=P_{(3,2)}&lt;/math&gt; by reflective symmetry over the line &lt;math&gt;x=2&lt;/math&gt;. Similarly, &lt;math&gt;P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}&lt;/math&gt;, and &lt;math&gt;P_{(2,1)}=P_{(2,3)}&lt;/math&gt;. <br /> Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}&lt;/cmath&gt;<br /> We have a system of &lt;math&gt;4&lt;/math&gt; equations in &lt;math&gt;4&lt;/math&gt; variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> Plugging in the third equation into this gives <br /> &lt;cmath&gt;P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}&lt;/cmath&gt;<br /> Next, plugging in the second and third equation into the first equation yields <br /> &lt;cmath&gt;P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}&lt;/cmath&gt;<br /> Now plugging in (*) into this, we get <br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}&lt;/cmath&gt;<br /> -mathisawesome2169<br /> <br /> ==Solution 5 (Very fast)==<br /> <br /> We can immediately note that the probability of landing on any lattice point is equal to the probability of landing on another. With this in mind, we see that each vertical &quot;boundary&quot; has 5 lattice points. (Remember, the &quot;corners&quot; count!) There are two boundary lines, so there are &lt;math&gt;2 \times 5 = 10&lt;/math&gt; lattice points on our desired vertical boundary lines. The total amount of lattice points on the &lt;math&gt;4 \times 4&lt;/math&gt; boundary is &lt;math&gt;16&lt;/math&gt;. Using &lt;math&gt;\frac{P(desired)}{P(total)}&lt;/math&gt;, we get &lt;math&gt;\frac{10}{16} = \boxed{\textbf{(B) }\frac{5}{8}}&lt;/math&gt; &lt;cmath&gt;\phantom{}&lt;/cmath&gt;<br /> -hansenhe<br /> <br /> ==Video Solution 1==<br /> [https://www.youtube.com/watch?v=ZGwAasE32Y4&amp;t=280s IceMatrix's Solution (Starts at 4:40)]<br /> <br /> ==Video Solution 2==<br /> <br /> https://youtu.be/qNaN0BlIsw0<br /> <br /> ==Video Solution 3==<br /> On The Spot STEM<br /> <br /> https://youtu.be/xGs7BjQbGYU<br /> <br /> ==Video Solution 4==<br /> https://youtu.be/0m4lbXSUV1I<br /> <br /> ~savannahsolver<br /> <br /> == Video Solution 5 ==<br /> https://youtu.be/IRyWOZQMTV8?t=5173<br /> <br /> ~ pi_is_3.14<br /> <br /> == Video Solution 6 ==<br /> https://www.youtube.com/watch?v=R220vbM_my8<br /> <br /> ~ amritvignesh0719062.0<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Probability Problems]]<br /> {{MAA Notice}}</div> Amritvignesh0719062.0