https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hi+im+bob&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-25T03:27:54Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_14&diff=160567 2013 AIME II Problems/Problem 14 2021-08-19T03:52:26Z <p>Hi im bob: /* The Pattern */</p> <hr /> <div>==Problem 14==<br /> For positive integers &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, let &lt;math&gt;f(n, k)&lt;/math&gt; be the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k&lt;/math&gt;, and for &lt;math&gt;n &gt; 1&lt;/math&gt; let &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;. Find the remainder when &lt;math&gt;\sum\limits_{n=20}^{100} F(n)&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> ===The Pattern===<br /> <br /> We can find that <br /> <br /> &lt;math&gt;20\equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;21\equiv 5 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;22\equiv 6 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;23\equiv 7 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;24\equiv 6 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;25\equiv 7 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;26\equiv 8 \pmod{9}&lt;/math&gt;<br /> <br /> Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get<br /> <br /> &lt;math&gt;99\equiv 31 \pmod{34}&lt;/math&gt;<br /> <br /> &lt;math&gt;100\equiv 32 \pmod{34}&lt;/math&gt;<br /> <br /> So the sum is &lt;math&gt;5+3\times(6+...+31)+32+32=1512&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;.<br /> By: Kris17<br /> <br /> ===The Intuition===<br /> First, let's see what happens if we remove a restriction. Let's define &lt;math&gt;G(x)&lt;/math&gt; as<br /> <br /> &lt;math&gt;G(x):=\max_{\substack{1\le k}} f(n, k)&lt;/math&gt;<br /> <br /> Now, if you set &lt;math&gt;k&lt;/math&gt; as any number greater than &lt;math&gt;n&lt;/math&gt;, you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence &lt;math&gt;H(x)&lt;/math&gt; is defined as<br /> <br /> &lt;math&gt;H(x):=\max_{\substack{1\le k\le n}} f(n, k)&lt;/math&gt;<br /> <br /> Now, after some thought, we find that if we set &lt;math&gt;k=\lfloor \frac{n}{2} \rfloor+1&lt;/math&gt; we get a remainder of &lt;math&gt;\lfloor \frac{n-1}{2} \rfloor&lt;/math&gt;, the max possible. Once we have gotten this far, it is easy to see that the original equation, <br /> <br /> &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;<br /> <br /> has a solution with &lt;math&gt;k=\lfloor \frac{n}{3} \rfloor+1&lt;/math&gt;.<br /> <br /> &lt;math&gt;W^5&lt;/math&gt;~Rowechen<br /> <br /> ===The Proof===<br /> The solution presented above does not prove why &lt;math&gt;F(x)&lt;/math&gt; is found by dividing &lt;math&gt;x&lt;/math&gt; by &lt;math&gt;3&lt;/math&gt;. Indeed, that is the case, as rigorously shown below.<br /> <br /> Consider the case where &lt;math&gt;x = 3k&lt;/math&gt;. We shall prove that &lt;math&gt;F(x) = f(x, k+1)&lt;/math&gt;.<br /> For all &lt;math&gt;x/2\geq n &gt; k+1, x = 2n + q&lt;/math&gt;, where &lt;math&gt;0\leq q\leq n&lt;/math&gt;. This is because &lt;math&gt;x &gt; 3k + 3 = 3n&lt;/math&gt; and &lt;math&gt;x &lt; n&lt;/math&gt;. Also, as n increases, &lt;math&gt;q&lt;/math&gt; decreases. Thus, &lt;math&gt;q = f(x, n) &lt; f(x, k+1) = k - 2&lt;/math&gt; for all &lt;math&gt;n &gt; k+1&lt;/math&gt;.<br /> Consider all &lt;math&gt;n &lt; k+1. f(x, k) = 0&lt;/math&gt; and &lt;math&gt;f(x, k-1) = 3&lt;/math&gt;. Also, &lt;math&gt;0 &lt; f(x, k-2) &lt; k-2&lt;/math&gt;. Thus, for &lt;math&gt;k &gt; 5, f(x, k+1) &gt; f(x, n)&lt;/math&gt; for &lt;math&gt;n &lt; k+1&lt;/math&gt;.<br /> <br /> Similar proofs apply for &lt;math&gt;x = 3k + 1&lt;/math&gt; and &lt;math&gt;x = 3k + 2&lt;/math&gt;. The reader should feel free to derive these proofs themself.<br /> <br /> ===Generalized Solution===<br /> <br /> &lt;math&gt;Lemma:&lt;/math&gt; Highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n/2&lt;/math&gt; is obtained for &lt;math&gt;k_0 = (n + (3 - n \pmod{3}))/3&lt;/math&gt; and the remainder thus obtained is &lt;math&gt;(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]&lt;/math&gt;. <br /> <br /> &lt;math&gt;Note:&lt;/math&gt; This is the second highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n&lt;/math&gt; and the highest remainder occurs when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+1)/2&lt;/math&gt; for odd &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+2)/2&lt;/math&gt; for even &lt;math&gt;n&lt;/math&gt;.<br /> <br /> Using the lemma above:<br /> <br /> &lt;math&gt;\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}] &lt;/math&gt;<br /> &lt;math&gt;= [(120*81/2)/3 - 2*81 + (2/3)*81]<br /> = 1512&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;<br /> <br /> Proof of Lemma: It is similar to &lt;math&gt;The Proof&lt;/math&gt; stated above. <br /> <br /> Kris17<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_14&diff=160566 2013 AIME II Problems/Problem 14 2021-08-19T03:50:33Z <p>Hi im bob: /* The Pattern */</p> <hr /> <div>==Problem 14==<br /> For positive integers &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt;, let &lt;math&gt;f(n, k)&lt;/math&gt; be the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k&lt;/math&gt;, and for &lt;math&gt;n &gt; 1&lt;/math&gt; let &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;. Find the remainder when &lt;math&gt;\sum\limits_{n=20}^{100} F(n)&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> ===The Pattern===<br /> <br /> We can find that <br /> <br /> &lt;math&gt;20\equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;21\equiv 5 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;22\equiv 6 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;23\equiv 7 \pmod{8}&lt;/math&gt;<br /> <br /> &lt;math&gt;24\equiv 6 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;25\equiv 7 \pmod{9}&lt;/math&gt;<br /> <br /> &lt;math&gt;26\equiv 8 \pmod{9}&lt;/math&gt;<br /> <br /> Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get<br /> <br /> &lt;math&gt;99\equiv 31 \pmod{34}&lt;/math&gt;<br /> <br /> &lt;math&gt;100\equiv 32 \pmod{34}&lt;/math&gt;<br /> <br /> So the sum is &lt;math&gt;5+3\times(6+...+31)+31+32=1512&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;.<br /> By: Kris17<br /> <br /> ===The Intuition===<br /> First, let's see what happens if we remove a restriction. Let's define &lt;math&gt;G(x)&lt;/math&gt; as<br /> <br /> &lt;math&gt;G(x):=\max_{\substack{1\le k}} f(n, k)&lt;/math&gt;<br /> <br /> Now, if you set &lt;math&gt;k&lt;/math&gt; as any number greater than &lt;math&gt;n&lt;/math&gt;, you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence &lt;math&gt;H(x)&lt;/math&gt; is defined as<br /> <br /> &lt;math&gt;H(x):=\max_{\substack{1\le k\le n}} f(n, k)&lt;/math&gt;<br /> <br /> Now, after some thought, we find that if we set &lt;math&gt;k=\lfloor \frac{n}{2} \rfloor+1&lt;/math&gt; we get a remainder of &lt;math&gt;\lfloor \frac{n-1}{2} \rfloor&lt;/math&gt;, the max possible. Once we have gotten this far, it is easy to see that the original equation, <br /> <br /> &lt;math&gt;F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)&lt;/math&gt;<br /> <br /> has a solution with &lt;math&gt;k=\lfloor \frac{n}{3} \rfloor+1&lt;/math&gt;.<br /> <br /> &lt;math&gt;W^5&lt;/math&gt;~Rowechen<br /> <br /> ===The Proof===<br /> The solution presented above does not prove why &lt;math&gt;F(x)&lt;/math&gt; is found by dividing &lt;math&gt;x&lt;/math&gt; by &lt;math&gt;3&lt;/math&gt;. Indeed, that is the case, as rigorously shown below.<br /> <br /> Consider the case where &lt;math&gt;x = 3k&lt;/math&gt;. We shall prove that &lt;math&gt;F(x) = f(x, k+1)&lt;/math&gt;.<br /> For all &lt;math&gt;x/2\geq n &gt; k+1, x = 2n + q&lt;/math&gt;, where &lt;math&gt;0\leq q\leq n&lt;/math&gt;. This is because &lt;math&gt;x &gt; 3k + 3 = 3n&lt;/math&gt; and &lt;math&gt;x &lt; n&lt;/math&gt;. Also, as n increases, &lt;math&gt;q&lt;/math&gt; decreases. Thus, &lt;math&gt;q = f(x, n) &lt; f(x, k+1) = k - 2&lt;/math&gt; for all &lt;math&gt;n &gt; k+1&lt;/math&gt;.<br /> Consider all &lt;math&gt;n &lt; k+1. f(x, k) = 0&lt;/math&gt; and &lt;math&gt;f(x, k-1) = 3&lt;/math&gt;. Also, &lt;math&gt;0 &lt; f(x, k-2) &lt; k-2&lt;/math&gt;. Thus, for &lt;math&gt;k &gt; 5, f(x, k+1) &gt; f(x, n)&lt;/math&gt; for &lt;math&gt;n &lt; k+1&lt;/math&gt;.<br /> <br /> Similar proofs apply for &lt;math&gt;x = 3k + 1&lt;/math&gt; and &lt;math&gt;x = 3k + 2&lt;/math&gt;. The reader should feel free to derive these proofs themself.<br /> <br /> ===Generalized Solution===<br /> <br /> &lt;math&gt;Lemma:&lt;/math&gt; Highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n/2&lt;/math&gt; is obtained for &lt;math&gt;k_0 = (n + (3 - n \pmod{3}))/3&lt;/math&gt; and the remainder thus obtained is &lt;math&gt;(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]&lt;/math&gt;. <br /> <br /> &lt;math&gt;Note:&lt;/math&gt; This is the second highest remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1\leq k\leq n&lt;/math&gt; and the highest remainder occurs when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+1)/2&lt;/math&gt; for odd &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;k_M&lt;/math&gt; = &lt;math&gt;(n+2)/2&lt;/math&gt; for even &lt;math&gt;n&lt;/math&gt;.<br /> <br /> Using the lemma above:<br /> <br /> &lt;math&gt;\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}] &lt;/math&gt;<br /> &lt;math&gt;= [(120*81/2)/3 - 2*81 + (2/3)*81]<br /> = 1512&lt;/math&gt;<br /> <br /> So the answer is &lt;math&gt;\boxed{512}&lt;/math&gt;<br /> <br /> Proof of Lemma: It is similar to &lt;math&gt;The Proof&lt;/math&gt; stated above. <br /> <br /> Kris17<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_11&diff=160558 2013 AIME II Problems/Problem 11 2021-08-18T23:49:07Z <p>Hi im bob: /* Solution */</p> <hr /> <div>==Problem 11==<br /> Let &lt;math&gt;A = \{1, 2, 3, 4, 5, 6, 7\}&lt;/math&gt;, and let &lt;math&gt;N&lt;/math&gt; be the number of functions &lt;math&gt;f&lt;/math&gt; from set &lt;math&gt;A&lt;/math&gt; to set &lt;math&gt;A&lt;/math&gt; such that &lt;math&gt;f(f(x))&lt;/math&gt; is a constant function. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> ==Solution==<br /> Any such function can be constructed by distributing the elements of &lt;math&gt;A&lt;/math&gt; on three tiers.<br /> <br /> The bottom tier contains the constant value, &lt;math&gt;c=f(f(x))&lt;/math&gt; for any &lt;math&gt;x&lt;/math&gt;. (Obviously &lt;math&gt;f(c)=c&lt;/math&gt;.)<br /> <br /> The middle tier contains &lt;math&gt;k&lt;/math&gt; elements &lt;math&gt;x\ne c&lt;/math&gt; such that &lt;math&gt;f(x)=c&lt;/math&gt;, where &lt;math&gt;1\le k\le 6&lt;/math&gt;.<br /> <br /> The top tier contains &lt;math&gt;6-k&lt;/math&gt; elements such that &lt;math&gt;f(x)&lt;/math&gt; equals an element on the middle tier.<br /> <br /> There are &lt;math&gt;7&lt;/math&gt; choices for &lt;math&gt;c&lt;/math&gt;. Then for a given &lt;math&gt;k&lt;/math&gt;, there are &lt;math&gt;\tbinom6k&lt;/math&gt; ways to choose the elements on the middle tier, and then &lt;math&gt;k^{6-k}&lt;/math&gt; ways to draw arrows down from elements on the top tier to elements on the middle tier.<br /> <br /> Thus &lt;math&gt;N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399&lt;/math&gt;, giving the answer &lt;math&gt;\boxed{399}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159616 2001 AIME I Problems/Problem 12 2021-08-05T01:55:25Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+C(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=C/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7-12/7)&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation &lt;math&gt;-(11r/7-12/7)=r&lt;/math&gt; to solve &lt;math&gt;r={2/3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{005}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159531 2001 AIME I Problems/Problem 12 2021-08-04T02:02:16Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+C(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=C/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7-12/7)&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation &lt;math&gt;-(11r/7-12/7)=r&lt;/math&gt; to solve &lt;math&gt;r={2/3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{005}&lt;/math&gt;.<br /> <br /> -hi_im_bob<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159530 2001 AIME I Problems/Problem 12 2021-08-04T02:02:01Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+C(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=C/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7-12/7)&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation &lt;math&gt;-(11r/7-12/7)=r&lt;/math&gt; to solve &lt;math&gt;r={2/3}&lt;/math&gt;, so the answer is &lt;cmath&gt;\boxed{005}&lt;/cmath&gt;.<br /> <br /> -hi_im_bob<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159529 2001 AIME I Problems/Problem 12 2021-08-04T01:56:43Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+C(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=C/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7-12/7)&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation &lt;math&gt;-(11r/7-12/7)=r&lt;/math&gt; to solve r=&lt;math&gt;\boxed{2/3}&lt;/math&gt;.<br /> <br /> -hi_im_bob<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159528 2001 AIME I Problems/Problem 12 2021-08-04T01:55:26Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+C(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=C/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7)-12/7&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation -&lt;math&gt;(11r/7)-12/7=r&lt;/math&gt; to solve r=&lt;math&gt;\boxed{2/3}&lt;/math&gt;.<br /> <br /> -hi_im_bob<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159527 2001 AIME I Problems/Problem 12 2021-08-04T01:54:45Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+C(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=C/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(sqrt(4/49+9/49+36/49))&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7)-12/7&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation -&lt;math&gt;(11r/7)-12/7=r&lt;/math&gt; to solve r=&lt;math&gt;\boxed{2/3}&lt;/math&gt;.<br /> <br /> -hi_im_bob<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159526 2001 AIME I Problems/Problem 12 2021-08-04T01:53:14Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+c(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=C/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(sqrt(4/49+9/49+36/49))&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7)-12/7&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation -&lt;math&gt;(11r/7)-12/7=r&lt;/math&gt; to solve r=&lt;math&gt;\boxed{2/3}&lt;/math&gt;.<br /> <br /> -hi_im_bob<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159525 2001 AIME I Problems/Problem 12 2021-08-04T01:52:28Z <p>Hi im bob: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through &lt;math&gt;(6,0,0), (0,0,2), (0,4,0)&lt;/math&gt;. The &quot;point-slope form&quot; is &lt;math&gt;A(6-x1)+B(0-y1)+c(0-z1)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,0,2)&lt;/math&gt; gives &lt;math&gt;A(6)+B(0)+C(-2)=0.&lt;/math&gt; Plugging in &lt;math&gt;(0,4,0)&lt;/math&gt; gives &lt;math&gt;A(6)+B(-4)+C(0)=0.&lt;/math&gt; We can then use Cramer's rule/cross multiplication to get &lt;math&gt;A/(0-8)=-B/(0+12)=c/(-24)=k.&lt;/math&gt; Solve for A, B, C to get &lt;math&gt;2k, 3k, 6k&lt;/math&gt; respectively. We can then get &lt;math&gt;2k(x-x1)+3k(y-y1)+6k(z-z1)=0.&lt;/math&gt; Cancel out k on both sides. Next, let us substitute &lt;math&gt;(0,0,2)&lt;/math&gt;. We can then get &lt;math&gt;2x+3y+6z=12 &lt;/math&gt;as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get &lt;math&gt;2x/7+2y/7+6z/7=12/7&lt;/math&gt; to be the normal form. Note that the point is going to be at &lt;math&gt;(r,r,r).&lt;/math&gt; We find the distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane as &lt;math&gt;2/7r+3/7r+6/7r-12/7/(sqrt(4/49+9/49+36/49))&lt;/math&gt;, which is &lt;math&gt;+/-(11r/7)-12/7&lt;/math&gt;. We take the negative value of this because if we plug in &lt;math&gt;(0,0,0)&lt;/math&gt; to the equation of the plane we get a negative value. We equate that value to r and we get the equation -&lt;math&gt;(11r/7)-12/7=r&lt;/math&gt; to solve r=&lt;math&gt;\boxed{2/3}&lt;/math&gt;.<br /> <br /> -hi_im_bob<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_12&diff=159524 2001 AIME I Problems/Problem 12 2021-08-04T01:50:16Z <p>Hi im bob: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are &lt;math&gt;A = (6,0,0), B = (0,4,0), C = (0,0,2),&lt;/math&gt; and &lt;math&gt;D = (0,0,0).&lt;/math&gt; The [[radius]] of the sphere is &lt;math&gt;m/n,&lt;/math&gt; 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 /> &lt;asy&gt;<br /> import three; <br /> currentprojection = perspective(-2,9,4);<br /> triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);<br /> triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);<br /> triple I = (2/3,2/3,2/3);<br /> triple J = (6/7,20/21,26/21);<br /> draw(C--A--D--C--B--D--B--A--C);<br /> draw(L--F--N--E--M--G--L--I--M--I--N--I--J);<br /> label(&quot;$I$&quot;,I,W);<br /> label(&quot;$A$&quot;,A,S);<br /> label(&quot;$B$&quot;,B,S);<br /> label(&quot;$C$&quot;,C,W*-1);<br /> label(&quot;$D$&quot;,D,W*-1);<br /> &lt;/asy&gt;<br /> <br /> The center &lt;math&gt;I&lt;/math&gt; of the insphere must be located at &lt;math&gt;(r,r,r)&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the sphere's radius.<br /> &lt;math&gt;I&lt;/math&gt; must also be a distance &lt;math&gt;r&lt;/math&gt; from the plane &lt;math&gt;ABC&lt;/math&gt;<br /> <br /> The signed distance between a plane and a point &lt;math&gt;I&lt;/math&gt; can be calculated as &lt;math&gt;\frac{(I-G) \cdot P}{|P|}&lt;/math&gt;, where G is any point on the plane, and P is a vector perpendicular to ABC.<br /> <br /> A vector &lt;math&gt;P&lt;/math&gt; perpendicular to plane &lt;math&gt;ABC&lt;/math&gt; can be found as &lt;math&gt;V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{(I-C) \cdot P}{|P|}=-r&lt;/math&gt; where the negative comes from the fact that we want &lt;math&gt;I&lt;/math&gt; to be in the opposite direction of &lt;math&gt;P&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\begin{align*}\frac{(I-C) \cdot P}{|P|}&amp;=-r\\<br /> \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&amp;=-r\\<br /> \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&amp;=-r\\<br /> \frac{44r -48}{28}&amp;=-r\\<br /> 44r-48&amp;=-28r\\<br /> 72r&amp;=48\\<br /> r&amp;=\frac{2}{3}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> <br /> Finally &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that we can split the tetrahedron into &lt;math&gt;4&lt;/math&gt; smaller tetrahedrons such that the height of each tetrahedron is &lt;math&gt;r&lt;/math&gt; and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be &lt;math&gt;V&lt;/math&gt; and surface area be &lt;math&gt;F&lt;/math&gt;, using the volume formula for each pyramid(base times height divided by 3) we have &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;. The surface area of the pyramid is &lt;math&gt;\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]&lt;/math&gt;. We know triangle ABC's side lengths, &lt;math&gt;\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},&lt;/math&gt; and &lt;math&gt;\sqrt{4^{2}+6^{2}}&lt;/math&gt;, so using the expanded form of heron's formula, &lt;math&gt;[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14&lt;/math&gt;. Therefore, the surface area is &lt;math&gt;14+22=36&lt;/math&gt;, and the volume is &lt;math&gt;\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8&lt;/math&gt;, and using the formula above that &lt;math&gt;\dfrac{rF}{3}=V&lt;/math&gt;, we have &lt;math&gt;12r=8&lt;/math&gt; and thus &lt;math&gt;r=\dfrac{2}{3}&lt;/math&gt;, so the desired answer is &lt;math&gt;2+3=\boxed{005}&lt;/math&gt;.<br /> <br /> (Solution by Shaddoll)<br /> <br /> ==Solution 3==<br /> The intercept form equation of the plane &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.&lt;/math&gt; Its normal form is &lt;math&gt;\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0&lt;/math&gt; (square sum of the coefficients equals 1). The distance from &lt;math&gt;(r,r,r)&lt;/math&gt; to the plane is &lt;math&gt;\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |&lt;/math&gt;. Since &lt;math&gt;(r,r,r)&lt;/math&gt; and &lt;math&gt;(0,0,0)&lt;/math&gt; are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in &lt;math&gt;(0,0,0)&lt;/math&gt;). Therefore we have <br /> &lt;math&gt;-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.&lt;/math&gt; So &lt;math&gt;r=\dfrac{2}{3},&lt;/math&gt; which solves the problem.<br /> <br /> Additionally, if &lt;math&gt;(r,r,r)&lt;/math&gt; is on the other side of &lt;math&gt;ABC&lt;/math&gt;, we have &lt;math&gt;\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r&lt;/math&gt;, which yields &lt;math&gt;r=\dfrac{12}{5},&lt;/math&gt; corresponding an &quot;ex-sphere&quot; that is tangent to face &lt;math&gt;ABC&lt;/math&gt; as well as the extensions of the other 3 faces.<br /> <br /> -JZ<br /> <br /> ==Solution 4==<br /> First let us find the equation of the plane passing through (6,0,0), (0,0,2), (0,4,0). The &quot;point-slope form&quot; is A(6-x1)+B(0-y1)+c(0-z1)=0. Plugging in (0,0,2) gives A(6)+B(0)+C(-2)=0. Plugging in (0,4,0) gives A(6)+B(-4)+C(0)=0. We can then use Cramer's rule/cross multiplication to get A/(0-8)=-B/(0+12)=c/(-24)=k. Solve for A, B, C to get 2k, 3k, 6k respectively. We can then get 2k(x-x1)+3k(y-y1)+6k(z-z1)=0. Cancel out k on both sides. Next, let us substitute (0,0,2). We can then get 2x+3y+6z=12 as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get 2x/7+2y/7+6z/7=12/7 to be the normal form. Note that the point is going to be at (r,r,r). We find the distance from (r,r,r) to the plane as 2/7r+3/7r+6/7r-12/7/(sqrt(4/49+9/49+36/49)), which is +/-(11r/7)-12/7. We take the negative value of this because if we plug in 0,0,0 to the equation of the plane we get a negative value. We equate that value to r and we get the equation -(11r/7)-12/7=r to solve r=2/3.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_8&diff=159478 2001 AIME I Problems/Problem 8 2021-08-03T04:24:20Z <p>Hi im bob: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Call a positive integer &lt;math&gt;N&lt;/math&gt; a ''7-10 double'' if the digits of the base-&lt;math&gt;7&lt;/math&gt; representation of &lt;math&gt;N&lt;/math&gt; form a base-&lt;math&gt;10&lt;/math&gt; number that is twice &lt;math&gt;N&lt;/math&gt;. For example, &lt;math&gt;51&lt;/math&gt; is a 7-10 double because its base-&lt;math&gt;7&lt;/math&gt; representation is &lt;math&gt;102&lt;/math&gt;. What is the largest 7-10 double?<br /> <br /> == Solution ==<br /> We let &lt;math&gt;N_7 = \overline{a_na_{n-1}\cdots a_0}_7&lt;/math&gt;; we are given that<br /> <br /> &lt;cmath&gt;2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}&lt;/cmath&gt; (This is because the digits in &lt;math&gt;N&lt;/math&gt; ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)<br /> <br /> Expanding, we find that<br /> <br /> &lt;cmath&gt;2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0&lt;/cmath&gt;<br /> <br /> or re-arranging,<br /> <br /> &lt;cmath&gt;a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n&lt;/cmath&gt;<br /> <br /> Since the &lt;math&gt;a_i&lt;/math&gt;s are base-&lt;math&gt;7&lt;/math&gt; digits, it follows that &lt;math&gt;a_i &lt; 7&lt;/math&gt;, and the LHS is less than or equal to &lt;math&gt;30&lt;/math&gt;. Hence our number can have at most &lt;math&gt;3&lt;/math&gt; digits in base-&lt;math&gt;7&lt;/math&gt;. Letting &lt;math&gt;a_2 = 6&lt;/math&gt;, we find that &lt;math&gt;630_7 = \boxed{315}_{10}&lt;/math&gt; is our largest 7-10 double.<br /> <br /> ==Solution 2 (Guess and Check)==<br /> Let &lt;math&gt;A&lt;/math&gt; be the base &lt;math&gt;10&lt;/math&gt; representation of our number, and let &lt;math&gt;B&lt;/math&gt; be its base &lt;math&gt;7&lt;/math&gt; representation.<br /> <br /> Given this is an AIME problem, &lt;math&gt;A&lt;1000&lt;/math&gt;. If we look at &lt;math&gt;B&lt;/math&gt; in base &lt;math&gt;10&lt;/math&gt;, it must be equal to &lt;math&gt;2A&lt;/math&gt;, so &lt;math&gt;B&lt;2000&lt;/math&gt; when &lt;math&gt;B&lt;/math&gt; is looked at in base &lt;math&gt;10.&lt;/math&gt;<br /> <br /> If &lt;math&gt;B&lt;/math&gt; in base &lt;math&gt;10&lt;/math&gt; is less than &lt;math&gt;2000&lt;/math&gt;, then &lt;math&gt;B&lt;/math&gt; as a number in base &lt;math&gt;7&lt;/math&gt; must be less than &lt;math&gt;2*7^3=686&lt;/math&gt;.<br /> <br /> &lt;math&gt;686&lt;/math&gt; is non-existent in base &lt;math&gt;7&lt;/math&gt;, so we're gonna have to bump that down to &lt;math&gt;666_7&lt;/math&gt;. <br /> <br /> This suggests that &lt;math&gt;A&lt;/math&gt; is less than &lt;math&gt;\frac{666}{2}=333&lt;/math&gt;. <br /> <br /> Guess and check shows that &lt;math&gt;A&lt;320&lt;/math&gt;, and checking values in that range produces &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> <br /> <br /> <br /> ==Solution 3== <br /> Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be &lt;cmath&gt;abc&lt;/cmath&gt; in base 7. Then the number in expanded form is &lt;cmath&gt;49a+7b+c&lt;/cmath&gt; in base 7 and &lt;cmath&gt;100a+10b+c&lt;/cmath&gt; in base 10. Since the number in base 7 is half the number in base 10, we get the following equation.<br /> &lt;cmath&gt;98b+14b+2c=100a+10b+c&lt;/cmath&gt; which simplifies to &lt;cmath&gt;2a=4b+c.&lt;/cmath&gt; <br /> The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, &lt;math&gt;b&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt;. Therefore, the largest 7-10 double is 630 in base 7, or &lt;math&gt;\boxed{315}&lt;/math&gt; in base 10.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_8&diff=159477 2001 AIME I Problems/Problem 8 2021-08-03T04:23:43Z <p>Hi im bob: /* Solution 2 (Guess and Check) */</p> <hr /> <div>== Problem ==<br /> Call a positive integer &lt;math&gt;N&lt;/math&gt; a ''7-10 double'' if the digits of the base-&lt;math&gt;7&lt;/math&gt; representation of &lt;math&gt;N&lt;/math&gt; form a base-&lt;math&gt;10&lt;/math&gt; number that is twice &lt;math&gt;N&lt;/math&gt;. For example, &lt;math&gt;51&lt;/math&gt; is a 7-10 double because its base-&lt;math&gt;7&lt;/math&gt; representation is &lt;math&gt;102&lt;/math&gt;. What is the largest 7-10 double?<br /> <br /> == Solution ==<br /> We let &lt;math&gt;N_7 = \overline{a_na_{n-1}\cdots a_0}_7&lt;/math&gt;; we are given that<br /> <br /> &lt;cmath&gt;2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}&lt;/cmath&gt; (This is because the digits in &lt;math&gt;N&lt;/math&gt; ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)<br /> <br /> Expanding, we find that<br /> <br /> &lt;cmath&gt;2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0&lt;/cmath&gt;<br /> <br /> or re-arranging,<br /> <br /> &lt;cmath&gt;a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n&lt;/cmath&gt;<br /> <br /> Since the &lt;math&gt;a_i&lt;/math&gt;s are base-&lt;math&gt;7&lt;/math&gt; digits, it follows that &lt;math&gt;a_i &lt; 7&lt;/math&gt;, and the LHS is less than or equal to &lt;math&gt;30&lt;/math&gt;. Hence our number can have at most &lt;math&gt;3&lt;/math&gt; digits in base-&lt;math&gt;7&lt;/math&gt;. Letting &lt;math&gt;a_2 = 6&lt;/math&gt;, we find that &lt;math&gt;630_7 = \boxed{315}_{10}&lt;/math&gt; is our largest 7-10 double.<br /> <br /> ==Solution 2 (Guess and Check)==<br /> Let &lt;math&gt;A&lt;/math&gt; be the base &lt;math&gt;10&lt;/math&gt; representation of our number, and let &lt;math&gt;B&lt;/math&gt; be its base &lt;math&gt;7&lt;/math&gt; representation.<br /> <br /> Given this is an AIME problem, &lt;math&gt;A&lt;1000&lt;/math&gt;. If we look at &lt;math&gt;B&lt;/math&gt; in base &lt;math&gt;10&lt;/math&gt;, it must be equal to &lt;math&gt;2A&lt;/math&gt;, so &lt;math&gt;B&lt;2000&lt;/math&gt; when &lt;math&gt;B&lt;/math&gt; is looked at in base &lt;math&gt;10.&lt;/math&gt;<br /> <br /> If &lt;math&gt;B&lt;/math&gt; in base &lt;math&gt;10&lt;/math&gt; is less than &lt;math&gt;2000&lt;/math&gt;, then &lt;math&gt;B&lt;/math&gt; as a number in base &lt;math&gt;7&lt;/math&gt; must be less than &lt;math&gt;2*7^3=686&lt;/math&gt;.<br /> <br /> &lt;math&gt;686&lt;/math&gt; is non-existent in base &lt;math&gt;7&lt;/math&gt;, so we're gonna have to bump that down to &lt;math&gt;666_7&lt;/math&gt;. <br /> <br /> This suggests that &lt;math&gt;A&lt;/math&gt; is less than &lt;math&gt;\frac{666}{2}=333&lt;/math&gt;. <br /> <br /> Guess and check shows that &lt;math&gt;A&lt;320&lt;/math&gt;, and checking values in that range produces &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> <br /> <br /> <br /> ==Solution 3== <br /> Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be &lt;cmath&gt;abc&lt;/cmath&gt; in base 7. Then the number in expanded form is &lt;cmath&gt;49a+7b+c&lt;/cmath&gt; in base 7 and &lt;cmath&gt;100a+10b+c&lt;/cmath&gt; in base 10. Since the number in base 7 is half the number in base 10, we get the following equation.<br /> &lt;cmath&gt;98b+14b+2c=100a+10b+c&lt;/cmath&gt;, which simplifies to &lt;cmath&gt;2a=4b+c&lt;/cmath&gt;. <br /> The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, &lt;math&gt;b&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt;. Therefore, the largest 7-10 double is 630 in base 7, or &lt;math&gt;\boxed{315}&lt;/math&gt; in base 10.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_8&diff=159476 2001 AIME I Problems/Problem 8 2021-08-03T04:23:03Z <p>Hi im bob: /* Solution 2 (Guess and Check) */</p> <hr /> <div>== Problem ==<br /> Call a positive integer &lt;math&gt;N&lt;/math&gt; a ''7-10 double'' if the digits of the base-&lt;math&gt;7&lt;/math&gt; representation of &lt;math&gt;N&lt;/math&gt; form a base-&lt;math&gt;10&lt;/math&gt; number that is twice &lt;math&gt;N&lt;/math&gt;. For example, &lt;math&gt;51&lt;/math&gt; is a 7-10 double because its base-&lt;math&gt;7&lt;/math&gt; representation is &lt;math&gt;102&lt;/math&gt;. What is the largest 7-10 double?<br /> <br /> == Solution ==<br /> We let &lt;math&gt;N_7 = \overline{a_na_{n-1}\cdots a_0}_7&lt;/math&gt;; we are given that<br /> <br /> &lt;cmath&gt;2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}&lt;/cmath&gt; (This is because the digits in &lt;math&gt;N&lt;/math&gt; ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)<br /> <br /> Expanding, we find that<br /> <br /> &lt;cmath&gt;2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0&lt;/cmath&gt;<br /> <br /> or re-arranging,<br /> <br /> &lt;cmath&gt;a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n&lt;/cmath&gt;<br /> <br /> Since the &lt;math&gt;a_i&lt;/math&gt;s are base-&lt;math&gt;7&lt;/math&gt; digits, it follows that &lt;math&gt;a_i &lt; 7&lt;/math&gt;, and the LHS is less than or equal to &lt;math&gt;30&lt;/math&gt;. Hence our number can have at most &lt;math&gt;3&lt;/math&gt; digits in base-&lt;math&gt;7&lt;/math&gt;. Letting &lt;math&gt;a_2 = 6&lt;/math&gt;, we find that &lt;math&gt;630_7 = \boxed{315}_{10}&lt;/math&gt; is our largest 7-10 double.<br /> <br /> ==Solution 2 (Guess and Check)==<br /> Let &lt;math&gt;A&lt;/math&gt; be the base &lt;math&gt;10&lt;/math&gt; representation of our number, and let &lt;math&gt;B&lt;/math&gt; be its base &lt;math&gt;7&lt;/math&gt; representation.<br /> <br /> Given this is an AIME problem, &lt;math&gt;A&lt;1000&lt;/math&gt;. If we look at &lt;math&gt;B&lt;/math&gt; in base &lt;math&gt;10&lt;/math&gt;, it must be equal to &lt;math&gt;2A&lt;/math&gt;, so &lt;math&gt;B&lt;2000&lt;/math&gt; when &lt;math&gt;B&lt;/math&gt; is looked at in base &lt;math&gt;10.&lt;/math&gt;<br /> <br /> If &lt;math&gt;B&lt;/math&gt; in base &lt;math&gt;10&lt;/math&gt; is less than &lt;math&gt;2000&lt;/math&gt;, then &lt;math&gt;B&lt;/math&gt; as a number in base &lt;math&gt;7&lt;/math&gt; must be less than &lt;math&gt;2*7^3=686&lt;/math&gt;.<br /> <br /> &lt;math&gt;686&lt;/math&gt; is non-existent in base &lt;math&gt;7&lt;/math&gt;, so we're gonna have to bump that down to &lt;math&gt;666_7&lt;/math&gt;. <br /> <br /> This suggests that &lt;math&gt;A&lt;/math&gt; is less than &lt;math&gt;\frac{666}{2}=333&lt;/math&gt;. <br /> <br /> Guess and check shows that &lt;math&gt;A&lt;320&lt;/math&gt;, and checking values in that range produces &lt;math&gt;\boxed{315}&lt;/math&gt;.<br /> <br /> <br /> <br /> ==Solution 3==7, or <br /> Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be &lt;cmath&gt;abc&lt;/cmath&gt; in base 7. Then the number in expanded form is &lt;cmath&gt;49a+7b+c&lt;/cmath&gt; in base 7 and &lt;cmath&gt;100a+10b+c&lt;/cmath&gt; in base 10. Since the number in base 7 is half the number in base 10, we get the following equation.<br /> &lt;cmath&gt;98b+14b+2c=100a+10b+c&lt;/cmath&gt;, which simplifies to &lt;math&gt;&lt;/math&gt;2a=4b+c&lt;math&gt;. <br /> The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, &lt;/math&gt;b&lt;math&gt; is &lt;/math&gt;3&lt;math&gt; and &lt;/math&gt;c&lt;math&gt; is &lt;/math&gt;0&lt;math&gt;. Therefore, the largest 7-10 double is 630 in base 7, or &lt;/math&gt;\boxed{315}in base 10.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157073 2017 AIME I Problems/Problem 15 2021-06-30T02:46:35Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> <br /> <br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> <br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2.&lt;/cmath&gt;<br /> <br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2).&lt;/cmath&gt;<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1.&lt;/cmath&gt; We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}.&lt;/cmath&gt; We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}.&lt;/cmath&gt; This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being &lt;cmath&gt;(a, b), (11, 7\sqrt{3}).&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67.&lt;/cmath&gt; From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt; which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> <br /> -hi_im_bob<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157071 2017 AIME I Problems/Problem 15 2021-06-30T02:46:03Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> <br /> <br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> <br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2.&lt;/cmath&gt;<br /> <br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2).&lt;/cmath&gt;<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1.&lt;/cmath&gt; We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}.&lt;/cmath&gt; We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}.&lt;/cmath&gt; This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being &lt;cmath&gt;(a, b), (11, 7\sqrt{3}).&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67.&lt;/cmath&gt; From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt; which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157069 2017 AIME I Problems/Problem 15 2021-06-30T02:45:38Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> <br /> <br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> <br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2.&lt;/cmath&gt;<br /> <br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2).&lt;/cmath&gt;<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1.&lt;/cmath&gt; We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}.&lt;/cmath&gt; We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}.&lt;/cmath&gt; This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being &lt;cmath&gt;{a, b}, {11, 7\sqrt{3}}.&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67.&lt;/cmath&gt; From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt; which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157067 2017 AIME I Problems/Problem 15 2021-06-30T02:45:03Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> <br /> <br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> <br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2.&lt;/cmath&gt;<br /> <br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2).&lt;/cmath&gt;<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1.&lt;/cmath&gt; We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}.&lt;/cmath&gt; We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}.&lt;/cmath&gt; This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being {a, b} and &lt;cmath&gt;{11, 7\sqrt{3}}.&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67.&lt;/cmath&gt; From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt; which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157065 2017 AIME I Problems/Problem 15 2021-06-30T02:44:45Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> <br /> <br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> <br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2.&lt;/cmath&gt;<br /> <br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2).&lt;/cmath&gt;<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1.&lt;/cmath&gt; We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}.&lt;/cmath&gt; We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}&lt;/cmath&gt;. This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being {a, b} and &lt;cmath&gt;{11, 7\sqrt{3}}.&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67.&lt;/cmath&gt; From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt; which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157064 2017 AIME I Problems/Problem 15 2021-06-30T02:43:50Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> <br /> <br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> <br /> <br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2&lt;/cmath&gt;.<br /> <br /> <br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2)&lt;/cmath&gt;.<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1&lt;/cmath&gt;. We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}&lt;/cmath&gt;. We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}&lt;/cmath&gt;. This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being {a, b} and &lt;cmath&gt;{11, 7\sqrt{3}}.&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67&lt;/cmath&gt;. From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt;, which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157062 2017 AIME I Problems/Problem 15 2021-06-30T02:43:33Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> <br /> <br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2&lt;/cmath&gt;.<br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2)&lt;/cmath&gt;.<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1&lt;/cmath&gt;. We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}&lt;/cmath&gt;. We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}&lt;/cmath&gt;. This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being {a, b} and &lt;cmath&gt;{11, 7\sqrt{3}}.&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67&lt;/cmath&gt;. From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt;, which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157061 2017 AIME I Problems/Problem 15 2021-06-30T02:42:59Z <p>Hi im bob: /* Solution 8 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be &lt;cmath&gt;x\sqrt{3}/2&lt;/cmath&gt;.<br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2)&lt;/cmath&gt;.<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1&lt;/cmath&gt;. We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}&lt;/cmath&gt;. We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}&lt;/cmath&gt;. This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being {a, b} and &lt;cmath&gt;{11, 7\sqrt{3}}.&lt;/cmath&gt; Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67&lt;/cmath&gt;. From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt;, which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_15&diff=157060 2017 AIME I Problems/Problem 15 2021-06-30T02:42:14Z <p>Hi im bob: /* Solution 7 */</p> <hr /> <div>==Problem 15==<br /> <br /> The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37},&lt;/math&gt; as shown, is &lt;math&gt;\frac{m\sqrt{p}}{n},&lt;/math&gt; where &lt;math&gt;m,~n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are positive integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m+n+p.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0);<br /> real t = .385, s = 3.5*t-1;<br /> pair R = A*t+B*(1-t), P=B*s;<br /> pair Q = dir(-60) * (R-P) + P;<br /> fill(P--Q--R--cycle,gray);<br /> draw(A--B--C--A^^P--Q--R--P);<br /> dot(A--B--C--P--Q--R);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> Let's start by proving a lemma: If &lt;math&gt;x,y&lt;/math&gt; satisfy &lt;math&gt;px+qy=1&lt;/math&gt;, then the minimal value of &lt;math&gt;\sqrt{x^2+y^2}&lt;/math&gt; is &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> Proof: Recall that the distance between the point &lt;math&gt;(x_0,y_0)&lt;/math&gt; and the line &lt;math&gt;px+qy+r = 0&lt;/math&gt; is given by &lt;math&gt;\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}&lt;/math&gt;. In particular, the distance between the origin and any point &lt;math&gt;(x,y)&lt;/math&gt; on the line &lt;math&gt;px+qy=1&lt;/math&gt; is at least &lt;math&gt;\frac{1}{\sqrt{p^2+q^2}}&lt;/math&gt;.<br /> <br /> ---<br /> <br /> Let the vertices of the right triangle be &lt;math&gt;(0,0),(5,0),(0,2\sqrt{3}),&lt;/math&gt; and let &lt;math&gt;(a,0),(0,b)&lt;/math&gt; be the two vertices of the equilateral triangle on the legs of the right triangle. Then, the third vertex of the equilateral triangle is &lt;math&gt;\left(\frac{a+b\sqrt{3}}{2},\frac{a\sqrt{3}+b}{2}\right)&lt;/math&gt;. This point must lie on the hypotenuse &lt;math&gt;\frac{x}{5} + \frac{y}{2\sqrt{3}} = 1&lt;/math&gt;, i.e. &lt;math&gt;a,b&lt;/math&gt; must satisfy<br /> &lt;cmath&gt; \frac{a+b\sqrt{3}}{10}+\frac{a\sqrt{3}+b}{4\sqrt{3}} = 1,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;\frac{7}{20}a + \frac{11\sqrt{3}}{60}b = 1.&lt;/cmath&gt;<br /> <br /> By the lemma, the minimal value of &lt;math&gt;\sqrt{a^2+b^2}&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{1}{\sqrt{\left(\frac{7}{20}\right)^2 + \left(\frac{11\sqrt{3}}{60}\right)^2}} = \frac{10\sqrt{3}}{\sqrt{67}},&lt;/cmath&gt;<br /> so the minimal area of the equilateral triangle is<br /> &lt;cmath&gt; \frac{\sqrt{3}}{4} \cdot \left(\frac{10\sqrt{3}}{\sqrt{67}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},&lt;/cmath&gt;<br /> and hence the answer is &lt;math&gt;75+3+67=\boxed{145}&lt;/math&gt;.<br /> <br /> ==Solution 2 (No Coordinates)==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the triangle with side lengths &lt;math&gt;2\sqrt{3},~5,&lt;/math&gt; and &lt;math&gt;\sqrt{37}&lt;/math&gt;.<br /> <br /> We will think about this problem backwards, by constructing a triangle as large as possible (We will call it &lt;math&gt;T&lt;/math&gt;, for convenience) which is similar to &lt;math&gt;S&lt;/math&gt; with vertices outside of a unit equilateral triangle &lt;math&gt;\triangle ABC&lt;/math&gt;, such that each vertex of the equilateral triangle lies on a side of &lt;math&gt;T&lt;/math&gt;. After we find the side lengths of &lt;math&gt;T&lt;/math&gt;, we will use ratios to trace back towards the original problem.<br /> <br /> First of all, let &lt;math&gt;\theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\alpha = \arctan\left(\frac{2\sqrt{3}}{5}\right)&lt;/math&gt;, and &lt;math&gt;\beta = \arctan\left(\frac{5}{2\sqrt{3}}\right)&lt;/math&gt; (These three angles are simply the angles of triangle &lt;math&gt;S&lt;/math&gt;; out of these three angles, &lt;math&gt;\alpha&lt;/math&gt; is the smallest angle, and &lt;math&gt;\theta&lt;/math&gt; is the largest angle). Then let us consider a point &lt;math&gt;P&lt;/math&gt; inside &lt;math&gt;\triangle ABC&lt;/math&gt; such that &lt;math&gt;\angle APB = 180^{\circ} - \theta&lt;/math&gt;, &lt;math&gt;\angle BPC = 180^{\circ} - \alpha&lt;/math&gt;, and &lt;math&gt;\angle APC = 180^{\circ} - \beta&lt;/math&gt;. Construct the circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; of triangles &lt;math&gt;APB, ~BPC,&lt;/math&gt; and &lt;math&gt;APC&lt;/math&gt; respectively. <br /> <br /> From here, we will prove the lemma that if we choose points &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on circumcircles &lt;math&gt;\omega_{AB}, ~\omega_{BC},&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt; respectively such that &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;Y&lt;/math&gt; are collinear and &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are collinear, then &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; must be collinear. First of all, if we let &lt;math&gt;\angle PAX = m&lt;/math&gt;, then &lt;math&gt;\angle PBX = 180^{\circ} - m&lt;/math&gt; (by the properties of cyclic quadrilaterals), &lt;math&gt;\angle PBY = m&lt;/math&gt; (by adjacent angles), &lt;math&gt;\angle PCY = 180^{\circ} - m&lt;/math&gt; (by cyclic quadrilaterals), &lt;math&gt;\angle PCZ = m&lt;/math&gt; (adjacent angles), and &lt;math&gt;\angle PAZ = 180^{\circ} - m&lt;/math&gt; (cyclic quadrilaterals). Since &lt;math&gt;\angle PAX&lt;/math&gt; and &lt;math&gt;\angle PAZ&lt;/math&gt; are supplementary, &lt;math&gt;Z&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear as desired. Hence, &lt;math&gt;\triangle XYZ&lt;/math&gt; has an inscribed equilateral triangle &lt;math&gt;ABC&lt;/math&gt;.<br /> <br /> In addition, now we know that all triangles &lt;math&gt;XYZ&lt;/math&gt; (as described above) must be similar to triangle &lt;math&gt;S&lt;/math&gt;, as &lt;math&gt;\angle AXB = \theta&lt;/math&gt; and &lt;math&gt;\angle BYC = \alpha&lt;/math&gt;, so we have developed &lt;math&gt;AA&lt;/math&gt; similarity between the two triangles. Thus, &lt;math&gt;\triangle XYZ&lt;/math&gt; is the triangle similar to &lt;math&gt;S&lt;/math&gt; which we were desiring. Our goal now is to maximize the length of &lt;math&gt;XY&lt;/math&gt;, in order to maximize the area of &lt;math&gt;XYZ&lt;/math&gt;, to achieve our original goal.<br /> <br /> Note that, all triangles &lt;math&gt;PYX&lt;/math&gt; are similar to each other if &lt;math&gt;Y&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; are collinear. This is because &lt;math&gt;\angle PYB&lt;/math&gt; is constant, and &lt;math&gt;\angle PXB&lt;/math&gt; is also a constant value. Then we have &lt;math&gt;AA&lt;/math&gt; similarity between this set of triangles. To maximize &lt;math&gt;XY&lt;/math&gt;, we can instead maximize &lt;math&gt;PY&lt;/math&gt;, which is simply the diameter of &lt;math&gt;\omega_{BC}&lt;/math&gt;. From there, we can determine that &lt;math&gt;\angle PBY = 90^{\circ}&lt;/math&gt;, and with similar logic, &lt;math&gt;PA&lt;/math&gt;, &lt;math&gt;PB&lt;/math&gt;, and &lt;math&gt;PC&lt;/math&gt; are perpendicular to &lt;math&gt;ZX&lt;/math&gt;, &lt;math&gt;XY&lt;/math&gt;, and &lt;math&gt;YZ&lt;/math&gt; respectively We have found our desired largest possible triangle &lt;math&gt;T&lt;/math&gt;.<br /> <br /> All we have to do now is to calculate &lt;math&gt;YZ&lt;/math&gt;, and use ratios from similar triangles to determine the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt;. First of all, we will prove that &lt;math&gt;\angle ZPY = \angle ACB + \angle AXB&lt;/math&gt;. By the properties of cyclic quadrilaterals, &lt;math&gt;\angle AXB = \angle PAB + \angle PBA&lt;/math&gt;, which means that &lt;math&gt;\angle ACB + \angle AXB = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Now we will show that &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt;. Note that, by cyclic quadrilaterals, &lt;math&gt;\angle YZP = \angle PAC&lt;/math&gt; and &lt;math&gt;\angle ZYP = \angle PBC&lt;/math&gt;. Hence, &lt;math&gt;\angle ZPY = 180^{\circ} - \angle PAC - \angle PBC&lt;/math&gt; (since &lt;math&gt;\angle ZPY + \angle YZP + \angle ZYP = 180^{\circ}&lt;/math&gt;), proving the aforementioned claim. Then, since &lt;math&gt;\angle ACB = 60^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AXB = \theta = 90^{\circ}&lt;/math&gt;, &lt;math&gt;\angle ZPY = 150^{\circ}&lt;/math&gt;.<br /> <br /> Now we calculate &lt;math&gt;PY&lt;/math&gt; and &lt;math&gt;PZ&lt;/math&gt;, which are simply the diameters of circumcircles &lt;math&gt;\omega_{BC}&lt;/math&gt; and &lt;math&gt;\omega_{AC}&lt;/math&gt;, respectively. By the extended law of sines, &lt;math&gt;PY = \frac{BC}{\sin{BPC}} = \frac{\sqrt{37}}{2\sqrt{3}}&lt;/math&gt; and &lt;math&gt;PZ = \frac{CA}{\sin{CPA}} = \frac{\sqrt{37}}{5}&lt;/math&gt;.<br /> <br /> We can now solve for &lt;math&gt;ZY&lt;/math&gt; with the law of cosines:<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} - \left(\frac{37}{5\sqrt{3}}\right)\left(-\frac{\sqrt{3}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37}{25} + \frac{37}{12} + \frac{37}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(ZY)^2 = \frac{37 \cdot 67}{300}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;ZY = \sqrt{37} \cdot \frac{\sqrt{67}}{10\sqrt{3}}&lt;/cmath&gt;<br /> <br /> Now we will apply this discovery towards our original triangle &lt;math&gt;S&lt;/math&gt;. Since the ratio between &lt;math&gt;ZY&lt;/math&gt; and the hypotenuse of &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{67}}{10\sqrt{3}}&lt;/math&gt;, the side length of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;\frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt; (as &lt;math&gt;S&lt;/math&gt; is simply as scaled version of &lt;math&gt;XYZ&lt;/math&gt;, and thus their corresponding inscribed equilateral triangles must be scaled by the same factor). Then the area of the equilateral triangle inscribed within &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;\frac{75\sqrt{3}}{67}&lt;/math&gt;, implying that the answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> == Solution 3 ==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be the right triangle with sides &lt;math&gt;AB = x&lt;/math&gt;, &lt;math&gt;AC = y&lt;/math&gt;, and &lt;math&gt;BC = z&lt;/math&gt; and right angle at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Let an equilateral triangle touch &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt; respectively, having side lengths of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> Now, call &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;. Thus, &lt;math&gt;DB = x-a&lt;/math&gt; and &lt;math&gt;EC = y-b&lt;/math&gt;.<br /> <br /> By Law of Sines on triangles &lt;math&gt;\triangle DBF&lt;/math&gt; and &lt;math&gt;ECF&lt;/math&gt;,<br /> <br /> &lt;math&gt;BF = \frac{z(a\sqrt{3}+b)} {2y}&lt;/math&gt; and &lt;math&gt;CF = \frac{z(a+b\sqrt{3})} {2x}&lt;/math&gt;.<br /> <br /> Summing, <br /> <br /> &lt;math&gt;BF+CF = \frac{z(a\sqrt{3}+b)} {2y} + \frac{z(a+b\sqrt{3})} {2x} = BC = z&lt;/math&gt;.<br /> <br /> Now substituting &lt;math&gt;AB = x = 2\sqrt{3}&lt;/math&gt;, &lt;math&gt;AC = y = 5&lt;/math&gt;, and &lt;math&gt;BC = \sqrt{37}&lt;/math&gt; and solving,<br /> &lt;math&gt;\frac{7a}{20} + \frac{11b\sqrt{3}}{60} = 1&lt;/math&gt;.<br /> <br /> We seek to minimize &lt;math&gt;[DEF] = c^2 \frac{\sqrt{3}}{4} = (a^2 + b^2) \frac{\sqrt{3}}{4}&lt;/math&gt;.<br /> <br /> This is equivalent to minimizing &lt;math&gt;a^2+b^2&lt;/math&gt;.<br /> <br /> Using the lemma from solution 1, we conclude that &lt;math&gt;\sqrt{a^2+b^2} = \frac{10\sqrt{3}}{\sqrt{67}}&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;[DEF] = \frac{75\sqrt{3}}{67}&lt;/math&gt; and our final answer is &lt;math&gt;\boxed{145}&lt;/math&gt;<br /> <br /> - Awsomness2000<br /> <br /> == Solution 4 ==<br /> We will use complex numbers. Set the vertex at the right angle to be the origin, and set the axes so the other two vertices are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;2\sqrt{3}i&lt;/math&gt;, respectively. Now let the vertex of the equilateral triangle on the real axis be &lt;math&gt;a&lt;/math&gt; and let the vertex of the equilateral triangle on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, the third vertex of the equilateral triangle is given by:<br /> &lt;cmath&gt;(bi-a)e^{-\frac{\pi}{3}i}+a=(bi-a)(\frac{1}{2}-\frac{\sqrt{3}}{2}i)+a=(\frac{a}{2}+\frac{b\sqrt{3}}{2})+(\frac{a\sqrt{3}}{2}+\frac{1}{2})i&lt;/cmath&gt;.<br /> <br /> For this to be on the hypotenuse of the right triangle, we also have the following:<br /> &lt;cmath&gt;\frac{\frac{a\sqrt{3}}{2}+\frac{1}{2}}{\frac{a}{2}+\frac{b\sqrt{3}}{2}-5}=-\frac{2\sqrt{3}}{5}\iff 7\sqrt{3}a+11b=20\sqrt{3}&lt;/cmath&gt;<br /> <br /> Note that the area of the equilateral triangle is given by &lt;math&gt;\frac{\sqrt{3}(a^2+b^2)}{4}&lt;/math&gt;, so we seek to minimize &lt;math&gt;a^2+b^2&lt;/math&gt;. This can be done by using the Cauchy Schwarz Inequality on the relation we derived above:<br /> &lt;cmath&gt;1200=(7\sqrt{3}a+11b)^2\leq ((7\sqrt{3})^2+11^2)(a^2+b^2)\implies a^2+b^2\geq \frac{1200}{268}&lt;/cmath&gt;<br /> <br /> Thus, the minimum we seek is simply &lt;math&gt;\frac{\sqrt{3}}{4}\cdot\frac{1200}{268}=\frac{75\sqrt{3}}{67}&lt;/math&gt;, so the desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> == Solution 5 (Alcumus)==<br /> In the complex plane, let the vertices of the triangle be &lt;math&gt;a = 5,&lt;/math&gt; &lt;math&gt;b = 2i \sqrt{3},&lt;/math&gt; and &lt;math&gt;c = 0.&lt;/math&gt; Let &lt;math&gt;e&lt;/math&gt; be one of the vertices, where &lt;math&gt;e&lt;/math&gt; is real. A point on the line passing through &lt;math&gt;a = 5&lt;/math&gt; and &lt;math&gt;b = 2i \sqrt{3}&lt;/math&gt; can be expressed in the form<br /> &lt;cmath&gt;f = (1 - t) a + tb = 5(1 - t) + 2ti \sqrt{3}.&lt;/cmath&gt;We want the third vertex &lt;math&gt;d&lt;/math&gt; to lie on the line through &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; which is the imaginary axis, so its real part is 0.<br /> Since the small triangle is equilateral, &lt;math&gt;d - e = \operatorname{cis} 60^\circ \cdot (f - e),&lt;/math&gt; or<br /> &lt;cmath&gt;d - e = \frac{1 + i \sqrt{3}}{2} \cdot (5(1 - t) - e + 2ti \sqrt{3}).&lt;/cmath&gt;Then the real part of &lt;math&gt;d&lt;/math&gt; is<br /> &lt;cmath&gt;\frac{5(1 - t) - e}{2} - 3t + e = 0.&lt;/cmath&gt;Solving for &lt;math&gt;t&lt;/math&gt; in terms of &lt;math&gt;e,&lt;/math&gt; we find<br /> &lt;cmath&gt;t = \frac{e + 5}{11}.&lt;/cmath&gt;Then<br /> &lt;cmath&gt;f = \frac{5(6 - e)}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;f - e = \frac{30 - 16e}{11} + \frac{2(e + 5) \sqrt{3}}{11} i,&lt;/cmath&gt;so<br /> &lt;cmath&gt;\begin{align*}<br /> |f - e|^2 &amp;= \left( \frac{30 - 16e}{11} \right)^2 + \left( \frac{2(e + 5) \sqrt{3}}{11} \right)^2 \\<br /> &amp;= \frac{268e^2 - 840e + 1200}{121}.<br /> \end{align*}&lt;/cmath&gt;This quadratic is minimized when &lt;math&gt;e = \frac{840}{2 \cdot 268} = \frac{105}{67},&lt;/math&gt; and the minimum is &lt;math&gt;\frac{300}{67},&lt;/math&gt; so the smallest area of the equilateral triangle is<br /> &lt;cmath&gt;\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \boxed{\frac{75 \sqrt{3}}{67}}.&lt;/cmath&gt;<br /> <br /> ==Solution 6==<br /> Employ the same complex bash as in Solution 4, but instead note that minimizing &lt;math&gt;x^2+y^2&lt;/math&gt; is the same as minimizing the distance from <br /> 0,0 to x,y, since they are the same quantity. We use point to plane instead, which gives you the required distance.<br /> <br /> ==Solution 7==<br /> We can use complex numbers. Set the origin at the right angle. Let the point on the real axis be &lt;math&gt;a&lt;/math&gt; and the point on the imaginary axis be &lt;math&gt;bi&lt;/math&gt;. Then, we see that &lt;math&gt;(a-bi)\left(\text{cis}\frac{\pi}{3}\right)+bi=(a-bi)\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)+bi=\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b\right)+i\left(\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right).&lt;/math&gt; Now we switch back to Cartesian coordinates. The equation of the hypotenuse is &lt;math&gt;y=-\frac{2\sqrt{3}}{5}x+2\sqrt{3}.&lt;/math&gt; This means that the point &lt;math&gt;\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}b,\frac{\sqrt{3}}{2}a+\frac{1}{2}b\right)&lt;/math&gt; is on the line. Plugging the numbers in, we have &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{1}{2}b=-\frac{\sqrt{3}}{5}a-\frac{3}{5}b+2\sqrt{3} \implies 7\sqrt{3}a+11b=20\sqrt{3}.&lt;/math&gt; Now, we note that the side length of the equilateral triangle is &lt;math&gt;a^2+b^2&lt;/math&gt; so it suffices to minimize that. By Cauchy-Schwarz, we have &lt;math&gt;(a^2+b^2)(147+121)\geq(7\sqrt{3}a+11b)^2 \implies (a^2+b^2)\geq\frac{300}{67}.&lt;/math&gt; Thus, the area of the smallest triangle is &lt;math&gt;\frac{300}{67}\cdot\frac{\sqrt{3}}{4}=\frac{75\sqrt{3}}{67}&lt;/math&gt; so our desired answer is &lt;math&gt;\boxed{145}&lt;/math&gt;.<br /> <br /> (Solution by Pleaseletmetwin, but not added to the Wiki by Pleaseletmetwin)<br /> <br /> <br /> <br /> <br /> ==Solution 8==<br /> We will use coordinates. Let the right triangle's lower left point be at the origin. Notice that 2 points will determine a unique equilateral triangle. Let 2 points be on the x-axis (B) and y-axis (A) and label them (b, 0) and (0, a) respectively. The third point (C) will then be located on the hypotenuse. We proceed to find the third point's coordinates in terms of a and b. (There are many ways to do it the following is how I did it).<br /> 1. Find the slope of AB and take the negative reciprocal of it to find the slope of the line containing C. Notice the line contains the midpoint of AB so we can then have an equation of the line.<br /> 2. Let AB=x. For ABC to be an equilateral triangle, the altitude from C to AB must be x\sqrt{3}/2.<br /> 3. We then have two equations and two unknowns and can solve for C's coordinates. <br /> <br /> We can find C is &lt;cmath&gt;((a+b\sqrt{3})/2, (b+a\sqrt{3})/2)&lt;/cmath&gt;.<br /> Also, note that C must be on the hypotenuse of the triangle, whose equation is &lt;cmath&gt;x/5+y/(2\sqrt{3})=1&lt;/cmath&gt;. We can plug in x and y as the coordinates of C. We can then simplify the equation to &lt;cmath&gt;11b+7\sqrt{3}a=20\sqrt{3}&lt;/cmath&gt;. We aim to minimize the side length of the triangle, which is &lt;cmath&gt;\sqrt{a^2+b^2}&lt;/cmath&gt;. This situation is perfect for Cauchy Schwartz Inequality. We can have the 2 sets being {a, b} and {&lt;cmath&gt;11, 7\sqrt{3}&lt;/cmath&gt;}. Applying Cauchy Schwartz gives us the minimum value of the side is &lt;cmath&gt;300/67&lt;/cmath&gt;. From here, we can find the minimum area is &lt;cmath&gt;\frac{75\sqrt{3}}{67}&lt;/cmath&gt;, which gives us the final answer of &lt;cmath&gt;\boxed{145}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AIME box|year=2017|n=I|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_18&diff=146839 2021 AMC 12A Problems/Problem 18 2021-02-15T17:51:15Z <p>Hi im bob: /* Solution 4 (Most Comprehensive, Similar to Solution 3) */</p> <hr /> <div>{{duplicate|[[2021 AMC 10A Problems#Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems#Problem 18|2021 AMC 12A #18]]}}<br /> <br /> ==Problem==<br /> Let &lt;math&gt;f&lt;/math&gt; be a function defined on the set of positive rational numbers with the property that &lt;math&gt;f(a\cdot b) = f(a)+f(b)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Furthermore, suppose that &lt;math&gt;f&lt;/math&gt; also has the property that &lt;math&gt;f(p)=p&lt;/math&gt; for every prime number &lt;math&gt;p&lt;/math&gt;. For which of the following numbers &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;f(x) &lt; 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Looking through the solutions we can see that &lt;math&gt;f(\frac{25}{11})&lt;/math&gt; can be expressed as &lt;math&gt;f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})&lt;/math&gt; so using the prime numbers to piece together what we have we can get &lt;math&gt;10=11+f(\frac{25}{11})&lt;/math&gt;, so &lt;math&gt;f(\frac{25}{11})=-1&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> -Lemonie<br /> <br /> &lt;math&gt;f(\frac{25}{11} \cdot 11) = f(25) = f(5) + f(5) = 10&lt;/math&gt;<br /> <br /> - awesomediabrine<br /> <br /> ==Solution 2==<br /> We know that &lt;math&gt;f(p) = f(p \cdot 1) = f(p) + f(1)&lt;/math&gt;. By transitive, we have &lt;cmath&gt;f(p) = f(p) + f(1).&lt;/cmath&gt;<br /> Subtracting &lt;math&gt;f(p)&lt;/math&gt; from both sides gives &lt;math&gt;0 = f(1).&lt;/math&gt;<br /> Also<br /> &lt;cmath&gt;f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2&lt;/cmath&gt;<br /> &lt;cmath&gt;f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3&lt;/cmath&gt;<br /> &lt;cmath&gt;f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11&lt;/cmath&gt;<br /> In &lt;math&gt;\textbf{(A)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(B)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(C)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(D)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(E)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{(E)} \frac{25}{11}}&lt;/math&gt;<br /> <br /> ~JHawk0224 ~awesomediabrine<br /> <br /> ==Solution 3 (Deeper)==<br /> Consider the rational &lt;math&gt;\frac{a}{b}&lt;/math&gt;, for &lt;math&gt;a,b&lt;/math&gt; integers. We have &lt;math&gt;f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)&lt;/math&gt;. So &lt;math&gt;f\left(\frac{a}{b}\right)=f(a)-f(b)&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; be a prime. Notice that &lt;math&gt;f(p^k)=kf(p)&lt;/math&gt;. And &lt;math&gt;f(p)=p&lt;/math&gt;. So if &lt;math&gt;a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}&lt;/math&gt;, &lt;math&gt;f(a)=a_1p_1+a_2p_2+....+a_kp_k&lt;/math&gt;. We simply need this to be greater than what we have for &lt;math&gt;f(b)&lt;/math&gt;. Notice that for answer choices &lt;math&gt;A,B,C, &lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, the numerator &lt;math&gt;(a)&lt;/math&gt; has less prime factors than the denominator, and so they are less likely to work. We check &lt;math&gt;E&lt;/math&gt; first, and it works, therefore the answer is &lt;math&gt;\boxed{\textbf{(E)}}&lt;/math&gt;. <br /> <br /> ~yofro<br /> <br /> ==Solution 4 (Most Comprehensive, Similar to Solution 3)==<br /> We have the following important results:<br /> <br /> &lt;math&gt;(1) \ f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)&lt;/math&gt; for all positive integers &lt;math&gt;k&lt;/math&gt;<br /> <br /> &lt;math&gt;(2) \ f\left(a^n\right)=nf(a)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt;<br /> <br /> &lt;math&gt;(3) \ f(1)=0&lt;/math&gt;<br /> <br /> &lt;math&gt;(4) \ f\left({\frac 1a}\right)=-f(a)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Proofs&lt;/u&gt;&lt;/b&gt;<br /> <br /> Result &lt;math&gt;(1)&lt;/math&gt; can be shown by induction.<br /> <br /> Result &lt;math&gt;(2):&lt;/math&gt; Since powers are just repeated multiplication, we will use result &lt;math&gt;(1)&lt;/math&gt; to prove result &lt;math&gt;(2):&lt;/math&gt;<br /> &lt;cmath&gt;f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).&lt;/cmath&gt;<br /> <br /> Result &lt;math&gt;(3):&lt;/math&gt; For all positive rational numbers &lt;math&gt;a,&lt;/math&gt; we have &lt;cmath&gt;f(a)=f(a\cdot1)=f(a)+f(1).&lt;/cmath&gt; Therefore, we get &lt;math&gt;f(1)=0.&lt;/math&gt; So, result &lt;math&gt;(3)&lt;/math&gt; is true.<br /> <br /> Result &lt;math&gt;(4):&lt;/math&gt; For all positive rational numbers &lt;math&gt;a,&lt;/math&gt; we have &lt;cmath&gt;f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.&lt;/cmath&gt; It follows that &lt;math&gt;f\left({\frac 1a}\right)=-f(a),&lt;/math&gt; and result &lt;math&gt;(4)&lt;/math&gt; is true.<br /> <br /> For all positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y,&lt;/math&gt; suppose &lt;math&gt;\prod_{k=1}^{m}{p_k}^{e_k}&lt;/math&gt; and &lt;math&gt;\prod_{k=1}^{n}{q_k}^{d_k}&lt;/math&gt; are their prime factorizations, respectively, we have <br /> &lt;cmath&gt;\begin{align*}<br /> f\left(\frac xy\right)&amp;=f(x)+f\left(\frac 1y\right) \\<br /> &amp;=f(x)-f(y) \\<br /> &amp;=f\left(\prod_{k=1}^{m}{p_k}^{e_k}\right)-f\left(\prod_{k=1}^{n}{q_k}^{d_k}\right) \\<br /> &amp;=\left[\sum_{k=1}^{m}f\left({p_k}^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left({q_k}^{d_k}\right)\right] \\<br /> &amp;=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] \\<br /> &amp;=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right].<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> We apply function &lt;math&gt;f&lt;/math&gt; on each fraction in the choices:<br /> <br /> &lt;cmath&gt;\begin{array}{cccccccc}<br /> \textbf{(A) } &amp; f\left(\frac{17}{32}\right) &amp; = &amp; f\left(\frac{17^1}{2^5}\right) &amp; = &amp; [1(17)]-[5(2)] &amp; = &amp; 7 \\ [2ex]<br /> \textbf{(B) } &amp; f\left(\frac{11}{16}\right) &amp; = &amp; f\left(\frac{11^1}{2^4}\right) &amp; = &amp; [1(11)]-[4(2)] &amp; = &amp; 3 \\ [2ex]<br /> \textbf{(C) } &amp; f\left(\frac{7}{9}\right) &amp; = &amp; f\left(\frac{7^1}{3^2}\right) &amp; = &amp; [1(7)]-[2(3)] &amp; = &amp; 1 \\ [2ex]<br /> \textbf{(D) } &amp; f\left(\frac{7}{6}\right) &amp; = &amp; f\left(\frac{7^1}{2^1\cdot3^1}\right) &amp; = &amp; [1(7)]-[1(2)+1(3)] &amp; = &amp; 2 \\ [2ex]<br /> \textbf{(E) } &amp; f\left(\frac{25}{11}\right) &amp; = &amp; f\left(\frac{5^2}{11^1}\right) &amp; = &amp; [2(5)]-[1(11)] &amp; = &amp; -1.<br /> \end{array}&lt;/cmath&gt;<br /> Therefore, the answer is &lt;math&gt;\boxed{\textbf{(E) }\frac{25}{11}}.&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 5==<br /> The problem gives us that f(p)=p. If we let a=p and b=1, we get f(p)=f(p)+f(1), which implies f(1)=0. Notice that the answer choices are all fractions, which means we will have to multiply an integer by a fraction to be able to solve it. Therefore, let's try plugging in fractions and try to solve them. Note that if we plug in a=p and b=1/p, we get f(1)=f(p)+f(1/p). We can solve for f(1/p) as -f(p)! This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of E.<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=dvlTA8Ncp58<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtu.be/8gGcj95rlWY<br /> <br /> == Video Solution by OmegaLearn (Using Functions and manipulations) ==<br /> https://youtu.be/aGv99CLzguE<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_18&diff=146824 2021 AMC 12A Problems/Problem 18 2021-02-15T17:41:47Z <p>Hi im bob: /* Solution 1 (but where do you get 10=11+f(\frac{25}{11}) */</p> <hr /> <div>{{duplicate|[[2021 AMC 10A Problems#Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems#Problem 18|2021 AMC 12A #18]]}}<br /> <br /> ==Problem==<br /> Let &lt;math&gt;f&lt;/math&gt; be a function defined on the set of positive rational numbers with the property that &lt;math&gt;f(a\cdot b) = f(a)+f(b)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Furthermore, suppose that &lt;math&gt;f&lt;/math&gt; also has the property that &lt;math&gt;f(p)=p&lt;/math&gt; for every prime number &lt;math&gt;p&lt;/math&gt;. For which of the following numbers &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;f(x) &lt; 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Looking through the solutions we can see that &lt;math&gt;f(\frac{25}{11})&lt;/math&gt; can be expressed as &lt;math&gt;f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})&lt;/math&gt; so using the prime numbers to piece together what we have we can get &lt;math&gt;10=11+f(\frac{25}{11})&lt;/math&gt;, so &lt;math&gt;f(\frac{25}{11})=-1&lt;/math&gt; or &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> -Lemonie<br /> <br /> &lt;math&gt;f(\frac{25}{11} \cdot 11) = f(25) = f(5) + f(5) = 10&lt;/math&gt;<br /> <br /> - awesomediabrine<br /> <br /> ==Solution 2==<br /> We know that &lt;math&gt;f(p) = f(p \cdot 1) = f(p) + f(1)&lt;/math&gt;. By transitive, we have &lt;cmath&gt;f(p) = f(p) + f(1).&lt;/cmath&gt;<br /> Subtracting &lt;math&gt;f(p)&lt;/math&gt; from both sides gives &lt;math&gt;0 = f(1).&lt;/math&gt;<br /> Also<br /> &lt;cmath&gt;f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2&lt;/cmath&gt;<br /> &lt;cmath&gt;f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3&lt;/cmath&gt;<br /> &lt;cmath&gt;f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11&lt;/cmath&gt;<br /> In &lt;math&gt;\textbf{(A)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(B)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(C)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(D)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\textbf{(E)}&lt;/math&gt; we have &lt;math&gt;f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{(E)} \frac{25}{11}}&lt;/math&gt;<br /> <br /> ~JHawk0224 ~awesomediabrine<br /> <br /> ==Solution 3 (Deeper)==<br /> Consider the rational &lt;math&gt;\frac{a}{b}&lt;/math&gt;, for &lt;math&gt;a,b&lt;/math&gt; integers. We have &lt;math&gt;f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)&lt;/math&gt;. So &lt;math&gt;f\left(\frac{a}{b}\right)=f(a)-f(b)&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; be a prime. Notice that &lt;math&gt;f(p^k)=kf(p)&lt;/math&gt;. And &lt;math&gt;f(p)=p&lt;/math&gt;. So if &lt;math&gt;a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}&lt;/math&gt;, &lt;math&gt;f(a)=a_1p_1+a_2p_2+....+a_kp_k&lt;/math&gt;. We simply need this to be greater than what we have for &lt;math&gt;f(b)&lt;/math&gt;. Notice that for answer choices &lt;math&gt;A,B,C, &lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, the numerator &lt;math&gt;(a)&lt;/math&gt; has less prime factors than the denominator, and so they are less likely to work. We check &lt;math&gt;E&lt;/math&gt; first, and it works, therefore the answer is &lt;math&gt;\boxed{\textbf{(E)}}&lt;/math&gt;. <br /> <br /> ~yofro<br /> <br /> ==Solution 4 (Most Comprehensive, Similar to Solution 3)==<br /> We have the following important results:<br /> <br /> &lt;math&gt;(1) \ f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)&lt;/math&gt; for all positive integers &lt;math&gt;k&lt;/math&gt;<br /> <br /> &lt;math&gt;(2) \ f\left(a^n\right)=nf(a)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt;<br /> <br /> &lt;math&gt;(3) \ f(1)=0&lt;/math&gt;<br /> <br /> &lt;math&gt;(4) \ f\left({\frac 1a}\right)=-f(a)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt;<br /> <br /> &lt;b&gt;&lt;u&gt;Proofs&lt;/u&gt;&lt;/b&gt;<br /> <br /> Result &lt;math&gt;(1)&lt;/math&gt; can be shown by induction.<br /> <br /> Result &lt;math&gt;(2):&lt;/math&gt; Since powers are just repeated multiplication, we will use result &lt;math&gt;(1)&lt;/math&gt; to prove result &lt;math&gt;(2):&lt;/math&gt;<br /> &lt;cmath&gt;f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).&lt;/cmath&gt;<br /> <br /> Result &lt;math&gt;(3):&lt;/math&gt; For all positive rational numbers &lt;math&gt;a,&lt;/math&gt; we have &lt;cmath&gt;f(a)=f(a\cdot1)=f(a)+f(1).&lt;/cmath&gt; Therefore, we get &lt;math&gt;f(1)=0.&lt;/math&gt; So, result &lt;math&gt;(3)&lt;/math&gt; is true.<br /> <br /> Result &lt;math&gt;(4):&lt;/math&gt; For all positive rational numbers &lt;math&gt;a,&lt;/math&gt; we have &lt;cmath&gt;f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.&lt;/cmath&gt; It follows that &lt;math&gt;f\left({\frac 1a}\right)=-f(a),&lt;/math&gt; and result &lt;math&gt;(4)&lt;/math&gt; is true.<br /> <br /> For all positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y,&lt;/math&gt; suppose &lt;math&gt;\prod_{k=1}^{m}{p_k}^{e_k}&lt;/math&gt; and &lt;math&gt;\prod_{k=1}^{n}{q_k}^{d_k}&lt;/math&gt; are their prime factorizations, respectively, we have <br /> &lt;cmath&gt;\begin{align*}<br /> f\left(\frac xy\right)&amp;=f(x)+f\left(\frac 1y\right) \\<br /> &amp;=f(x)-f(y) \\<br /> &amp;=f\left(\prod_{k=1}^{m}{p_k}^{e_k}\right)-f\left(\prod_{k=1}^{n}{q_k}^{d_k}\right) \\<br /> &amp;=\left[\sum_{k=1}^{m}f\left({p_k}^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left({q_k}^{d_k}\right)\right] \\<br /> &amp;=\left[\sum_{k=1}^{m}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] \\<br /> &amp;=\left[\sum_{k=1}^{m}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right].<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> We apply function &lt;math&gt;f&lt;/math&gt; on each fraction in the choices:<br /> <br /> &lt;cmath&gt;\begin{array}{cccccccc}<br /> \textbf{(A) } &amp; f\left(\frac{17}{32}\right) &amp; = &amp; f\left(\frac{17^1}{2^5}\right) &amp; = &amp; [1(17)]-[5(2)] &amp; = &amp; 7 \\ [2ex]<br /> \textbf{(B) } &amp; f\left(\frac{11}{16}\right) &amp; = &amp; f\left(\frac{11^1}{2^4}\right) &amp; = &amp; [1(11)]-[4(2)] &amp; = &amp; 3 \\ [2ex]<br /> \textbf{(C) } &amp; f\left(\frac{7}{9}\right) &amp; = &amp; f\left(\frac{7^1}{3^2}\right) &amp; = &amp; [1(7)]-[2(3)] &amp; = &amp; 1 \\ [2ex]<br /> \textbf{(D) } &amp; f\left(\frac{7}{6}\right) &amp; = &amp; f\left(\frac{7^1}{2^1\cdot3^1}\right) &amp; = &amp; [1(7)]-[1(2)+1(3)] &amp; = &amp; 2 \\ [2ex]<br /> \textbf{(E) } &amp; f\left(\frac{25}{11}\right) &amp; = &amp; f\left(\frac{5^2}{11^1}\right) &amp; = &amp; [2(5)]-[1(11)] &amp; = &amp; -1.<br /> \end{array}&lt;/cmath&gt;<br /> Therefore, the answer is &lt;math&gt;\boxed{\textbf{(E) }\frac{25}{11}}.&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=dvlTA8Ncp58<br /> <br /> ==Video Solution by Punxsutawney Phil==<br /> https://youtu.be/8gGcj95rlWY<br /> <br /> == Video Solution by OmegaLearn (Using Functions and manipulations) ==<br /> https://youtu.be/aGv99CLzguE<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=136411 2019 AMC 8 Problems/Problem 24 2020-11-02T16:28:10Z <p>Hi im bob: /* Solution 18 */</p> <hr /> <div>==Problem 24==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, point &lt;math&gt;D&lt;/math&gt; divides side &lt;math&gt;\overline{AC}&lt;/math&gt; so that &lt;math&gt;AD:DC=1:2&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt; and let &lt;math&gt;F&lt;/math&gt; be the point of intersection of line &lt;math&gt;BC&lt;/math&gt; and line &lt;math&gt;AE&lt;/math&gt;. Given that the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, what is the area of &lt;math&gt;\triangle EBF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{AF}&lt;/math&gt; such that &lt;math&gt;XD&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;. That makes triangles &lt;math&gt;BEF&lt;/math&gt; and &lt;math&gt;EXD&lt;/math&gt; congruent since &lt;math&gt;BE = ED&lt;/math&gt;. &lt;math&gt;FC=3XD&lt;/math&gt; so &lt;math&gt;BC=4BF&lt;/math&gt;. Since &lt;math&gt;AF=3EF&lt;/math&gt; (&lt;math&gt;XE=EF&lt;/math&gt; and &lt;math&gt;AX=\frac13 AF&lt;/math&gt;, so &lt;math&gt;XE=EF=\frac13 AF&lt;/math&gt;), the altitude of triangle &lt;math&gt;BEF&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{3}&lt;/math&gt; of the altitude of &lt;math&gt;ABC&lt;/math&gt;. The area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, so the area of &lt;math&gt;BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}&lt;/math&gt; ~[[User:heeeeeeeheeeee|heeeeeeeheeeee]]<br /> <br /> ==Solution 2 (Mass Points)==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(7cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br /> draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br /> draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br /> draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br /> draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br /> draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((0.28,2.39),dotstyle); <br /> label(&quot;$A$&quot;, (0.36,2.59), NE * labelscalefactor); <br /> dot((-2.8,-1.17),dotstyle); <br /> label(&quot;$B$&quot;, (-2.72,-0.97), NE * labelscalefactor); <br /> dot((3.78,-1.05),dotstyle); <br /> label(&quot;$C$&quot;, (3.86,-0.85), NE * labelscalefactor); <br /> dot((1.2887445398528459,1.3985482236874887),dotstyle); <br /> label(&quot;$D$&quot;, (1.36,1.59), NE * labelscalefactor); <br /> dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br /> label(&quot;$F$&quot;, (-0.64,-0.93), NE * labelscalefactor); <br /> dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-0.2,0.57), NE * labelscalefactor); <br /> label(&quot;2&quot;, (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br /> label(&quot;1&quot;, (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (1.9,1.45), NE * labelscalefactor,wrwrwr); <br /> label(&quot;3&quot;, (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br /> label(&quot;6&quot;, (0.08,0.03), NE * labelscalefactor,wrwrwr); <br /> label(&quot;4&quot;, (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br /> <br /> First, we assign a mass of &lt;math&gt;2&lt;/math&gt; to point &lt;math&gt;A&lt;/math&gt;. We figure out that &lt;math&gt;C&lt;/math&gt; has a mass of &lt;math&gt;1&lt;/math&gt; since &lt;math&gt;2\times1 = 1\times2&lt;/math&gt;. Then, by adding &lt;math&gt;1+2 = 3&lt;/math&gt;, we get that point &lt;math&gt;D&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt;. By equality, point &lt;math&gt;B&lt;/math&gt; has a mass of &lt;math&gt;3&lt;/math&gt; also. <br /> <br /> Now, we add &lt;math&gt;3+3 = 6&lt;/math&gt; for point &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;3+1 = 4&lt;/math&gt; for point &lt;math&gt;F&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;BF&lt;/math&gt; is a common base for triangles &lt;math&gt;ABF&lt;/math&gt; and &lt;math&gt;EBF&lt;/math&gt;, so we figure out that the ratios of the areas is the ratios of the heights which is &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;. So, &lt;math&gt;EBF&lt;/math&gt;'s area is one third the area of &lt;math&gt;ABF&lt;/math&gt;, and we know the area of &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; the area of &lt;math&gt;ABC&lt;/math&gt; since they have the same heights but different bases.<br /> <br /> So we get the area of &lt;math&gt;EBF&lt;/math&gt; as &lt;math&gt;\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of &lt;math&gt;EBF&lt;/math&gt; over the product of the mass points of &lt;math&gt;ABC&lt;/math&gt; which is &lt;math&gt;\frac{2\times3\times1}{3\times6\times4}\times360&lt;/math&gt; which also yields &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;<br /> -Brudder<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}&lt;/math&gt;. The area of triangle &lt;math&gt;ABE&lt;/math&gt; is equal to &lt;math&gt;60&lt;/math&gt; because it is equal to on half of the area of triangle &lt;math&gt;ABD&lt;/math&gt;, which is equal to one third of the area of triangle &lt;math&gt;ABC&lt;/math&gt;, which is &lt;math&gt;360&lt;/math&gt;. The area of triangle &lt;math&gt;ACE&lt;/math&gt; is the sum of the areas of triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;CED&lt;/math&gt;, which is respectively &lt;math&gt;60&lt;/math&gt; and &lt;math&gt;120&lt;/math&gt;. So, &lt;math&gt;\frac{BF}{FC}&lt;/math&gt; is equal to &lt;math&gt;\frac{60}{180}&lt;/math&gt;=&lt;math&gt;\frac{1}{3}&lt;/math&gt;, so the area of triangle &lt;math&gt;ABF&lt;/math&gt; is &lt;math&gt;90&lt;/math&gt;. That minus the area of triangle &lt;math&gt;ABE&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. ~~SmileKat32<br /> <br /> ==Solution 4 (Similar Triangles)==<br /> Extend &lt;math&gt;\overline{BD}&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that &lt;math&gt;\overline{AG} \parallel \overline{BC}&lt;/math&gt; as shown:<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> &lt;/asy&gt;<br /> Then &lt;math&gt;\triangle ADG \sim \triangle CDB&lt;/math&gt; and &lt;math&gt;\triangle AEG \sim \triangle FEB&lt;/math&gt;. Since &lt;math&gt;CD = 2AD&lt;/math&gt;, triangle &lt;math&gt;CDB&lt;/math&gt; has four times the area of triangle &lt;math&gt;ADG&lt;/math&gt;. Since &lt;math&gt;[CDB] = 240&lt;/math&gt;, we get &lt;math&gt;[ADG] = 60&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;[AED]&lt;/math&gt; is also &lt;math&gt;60&lt;/math&gt;, we have &lt;math&gt;ED = DG&lt;/math&gt; because triangles &lt;math&gt;AED&lt;/math&gt; and &lt;math&gt;ADG&lt;/math&gt; have the same height and same areas and so their bases must be the congruent. Thus triangle &lt;math&gt;AEG&lt;/math&gt; has twice the side lengths and therefore four times the area of triangle &lt;math&gt;BEF&lt;/math&gt;, giving &lt;math&gt;[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F, G;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> G = (4.5, 3);<br /> <br /> draw(A--B--C--A--G--B);<br /> draw(A--F);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$G$&quot;, G, ENE);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$60$&quot;, (A+G+D)/3);<br /> label(&quot;$30$&quot;, (B+E+F)/3);<br /> &lt;/asy&gt;<br /> (Credit to MP8148 for the idea)<br /> <br /> ==Solution 5 (Area Ratios)==<br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A, B, C, D, E, F;<br /> B = (0,0);<br /> A = (2, 3);<br /> C = (5, 0);<br /> D = (3, 2);<br /> E = (1.5, 1);<br /> F = (1.25, 0);<br /> <br /> draw(A--B--C--A--D--B);<br /> draw(A--F);<br /> draw(E--C);<br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, WSW);<br /> label(&quot;$C$&quot;, C, ESE);<br /> label(&quot;$D$&quot;, D, dir(0)*1.5);<br /> label(&quot;$E$&quot;, E, SSE);<br /> label(&quot;$F$&quot;, F, S);<br /> label(&quot;$60$&quot;, (A+E+D)/3);<br /> label(&quot;$60$&quot;, (A+E+B)/3);<br /> label(&quot;$120$&quot;, (D+E+C)/3);<br /> label(&quot;$x$&quot;, (B+E+F)/3);<br /> label(&quot;$120-x$&quot;, (F+E+C)/3);<br /> &lt;/asy&gt;<br /> As before we figure out the areas labeled in the diagram. Then we note that &lt;cmath&gt;\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.&lt;/cmath&gt;Solving gives &lt;math&gt;x = \boxed{\textbf{(B) }30}&lt;/math&gt;. <br /> (Credit to scrabbler94 for the idea)<br /> <br /> ==Solution 6 (Coordinate Bashing)==<br /> Let &lt;math&gt;ADB&lt;/math&gt; be a right triangle, and &lt;math&gt;BD=CD&lt;/math&gt;<br /> <br /> Let &lt;math&gt;A=(-2\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;B=(0, 4\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;C=(4\sqrt{30}, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;D=(0, 0)&lt;/math&gt;<br /> <br /> &lt;math&gt;E=(0, 2\sqrt{30})&lt;/math&gt;<br /> <br /> &lt;math&gt;F=(\sqrt{30}, 3\sqrt{30})&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{AE}&lt;/math&gt; can be described with the equation &lt;math&gt;y=x-2\sqrt{30}&lt;/math&gt;<br /> <br /> The line &lt;math&gt;\overleftrightarrow{BC}&lt;/math&gt; can be described with &lt;math&gt;x+y=4\sqrt{30}&lt;/math&gt;<br /> <br /> Solving, we get &lt;math&gt;x=3\sqrt{30}&lt;/math&gt; and &lt;math&gt;y=\sqrt{30}&lt;/math&gt;<br /> <br /> Now we can find &lt;math&gt;EF=BF=2\sqrt{15}&lt;/math&gt;<br /> <br /> &lt;math&gt;[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare&lt;/math&gt;<br /> <br /> -Trex4days<br /> <br /> == Solution 7 ==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw(circle((0,0), 5), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> draw((-4,-3)--(0,5), linewidth(2)); <br /> draw((0,5)--(4,3), linewidth(2)); <br /> draw((12,-1)--(-4,-3), linewidth(2)); <br /> draw((0,5)--(0,-5), linewidth(2)); <br /> draw((-4,-3)--(0,-5), linewidth(2)); <br /> draw((4,3)--(0,2.48), linewidth(2)); <br /> draw((4,3)--(12,-1), linewidth(2)); <br /> draw((-4,-3)--(4,3), linewidth(2)); <br /> /* dots and labels */<br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((-5,0),dotstyle); <br /> dot((-4,-3),dotstyle); <br /> label(&quot;B&quot;, (-4.45,-3.38), NE * labelscalefactor); <br /> dot((4,3),dotstyle); <br /> label(&quot;$D$&quot;, (4.15,3.2), NE * labelscalefactor); <br /> dot((0,5),dotstyle); <br /> label(&quot;A&quot;, (-0.09,5.26), NE * labelscalefactor); <br /> dot((12,-1),dotstyle); <br /> label(&quot;C&quot;, (12.23,-1.24), NE * labelscalefactor); <br /> dot((0,-5),dotstyle); <br /> label(&quot;$G$&quot;, (0.19,-4.82), NE * labelscalefactor); <br /> dot((0,2.48),dotstyle); <br /> label(&quot;I&quot;, (-0.33,2.2), NE * labelscalefactor); <br /> dot((0,0),dotstyle); <br /> label(&quot;E&quot;, (0.27,-0.24), NE * labelscalefactor); <br /> dot((0,-2.5),dotstyle); <br /> label(&quot;F&quot;, (0.23,-2.2), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;A[\Delta XYZ]&lt;/math&gt; = &lt;math&gt;Area&lt;/math&gt; &lt;math&gt;of&lt;/math&gt; &lt;math&gt;Triangle&lt;/math&gt; &lt;math&gt;XYZ&lt;/math&gt; <br /> <br /> <br /> &lt;math&gt;A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;A[\Delta ABE] = A[\Delta AED] = 60&lt;/math&gt; (the median divides the area of the triangle into two equal parts)<br /> <br /> <br /> Construction: Draw a circumcircle around &lt;math&gt;\Delta ABD&lt;/math&gt; with &lt;math&gt;BD&lt;/math&gt; as is diameter. Extend &lt;math&gt;AF&lt;/math&gt; to &lt;math&gt;G&lt;/math&gt; such that it meets the circle at &lt;math&gt;G&lt;/math&gt;. Draw line &lt;math&gt;BG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;A[\Delta ABD] = A[\Delta ABG] = 120&lt;/math&gt; (Since &lt;math&gt;\square ABGD&lt;/math&gt; is cyclic)<br /> <br /> <br /> But &lt;math&gt;A[\Delta ABE]&lt;/math&gt; is common in both with an area of 60. So, &lt;math&gt;A[\Delta AED] = A[\Delta BEG]&lt;/math&gt;.<br /> <br /> \therefore &lt;math&gt;A[\Delta AED] \cong A[\Delta BEG]&lt;/math&gt; (SAS Congruency Theorem).<br /> <br /> In &lt;math&gt;\Delta AED&lt;/math&gt;, let &lt;math&gt;DI&lt;/math&gt; be the median of &lt;math&gt;\Delta AED&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta AID] = 30 = A[\Delta EID]&lt;/math&gt;<br /> <br /> <br /> Rotate &lt;math&gt;\Delta DEA&lt;/math&gt; to meet &lt;math&gt;D&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; at &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;DE&lt;/math&gt; will fit exactly in &lt;math&gt;BE&lt;/math&gt; (both are radii of the circle). From the above solutions, &lt;math&gt;\frac{AE}{EF} = 2:1&lt;/math&gt;.<br /> <br /> &lt;math&gt;AE&lt;/math&gt; is a radius and &lt;math&gt;EF&lt;/math&gt; is half of it implies &lt;math&gt;EF&lt;/math&gt; = &lt;math&gt;\frac{radius}{2}&lt;/math&gt;.<br /> <br /> Which means &lt;math&gt;A[\Delta BEF] \cong A[\Delta DEI]&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;A[\Delta BEF] = \boxed{\textbf{(B) }30}&lt;/math&gt;<br /> <br /> <br /> ~phoenixfire &amp; flamewavelight<br /> <br /> == Solution 8 ==<br /> &lt;asy&gt;<br /> import geometry;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF, M;<br /> B = (0,0); C = (3,0); M = (1.45,0);<br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> draw(EE--M,StickIntervalMarker(1,1));<br /> label(&quot;$M$&quot;,M,S);<br /> draw(A--DD,invisible,StickIntervalMarker(1,1));<br /> dot((DD+C)/2);<br /> draw(DD--C,invisible,StickIntervalMarker(2,1));<br /> &lt;/asy&gt;<br /> Using the ratio of &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, we find the area of &lt;math&gt;\triangle ADB&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt; and the area of &lt;math&gt;\triangle BDC&lt;/math&gt; is &lt;math&gt;240&lt;/math&gt;. Also using the fact that &lt;math&gt;E&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;, we know &lt;math&gt;\triangle ADE = \triangle ABE = 60&lt;/math&gt;.<br /> Let &lt;math&gt;M&lt;/math&gt; be a point such &lt;math&gt;\overline{EM}&lt;/math&gt; is parellel to &lt;math&gt;\overline{CD}&lt;/math&gt;. We immediatley know that &lt;math&gt;\triangle BEM \sim BDC&lt;/math&gt; by &lt;math&gt;2&lt;/math&gt;. Using that we can conclude &lt;math&gt;EM&lt;/math&gt; has ratio &lt;math&gt;1&lt;/math&gt;. Using &lt;math&gt;\triangle EFM \sim \triangle AFC&lt;/math&gt;, we get &lt;math&gt;EF:AE = 1:2&lt;/math&gt;. Therefore using the fact that &lt;math&gt;\triangle EBF&lt;/math&gt; is in &lt;math&gt;\triangle ABF&lt;/math&gt;, the area has ratio &lt;math&gt;\triangle BEF : \triangle ABE=1:2&lt;/math&gt; and we know &lt;math&gt;\triangle ABE&lt;/math&gt; has area &lt;math&gt;60&lt;/math&gt; so &lt;math&gt;\triangle BEF&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(B) }30}&lt;/math&gt;. - fath2012<br /> <br /> ==Solution 9 (Menelaus's Theorem)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> By Menelaus's Theorem on triangle &lt;math&gt;BCD&lt;/math&gt;, we have &lt;cmath&gt;\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.&lt;/cmath&gt; Therefore, &lt;cmath&gt;[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.&lt;/cmath&gt;<br /> <br /> ==Solution 10 (Graph Paper)==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,D,E,F,a,b,c,d,e,f;<br /> A = (2,3);<br /> B = (0,2); <br /> C = (2,0);<br /> D = (2/3)*A+(1/3)*C;<br /> E = (B+D)/2;<br /> F = intersectionpoint(B--C,A--A+2*(E-A));<br /> a = (0,0);<br /> b = (1,0);<br /> c = (2,1);<br /> d = (1,3);<br /> e = (0,3);<br /> f = (0,1);<br /> draw(a--C,dashed);<br /> draw(f--c,dashed);<br /> draw(e--A,dashed);<br /> draw(a--e,dashed);<br /> draw(b--d,dashed);<br /> draw(A--B--C--cycle);<br /> draw(A--F); <br /> draw(B--D);<br /> dot(A); <br /> label(&quot;$A$&quot;,A,NE);<br /> dot(B); <br /> label(&quot;$B$&quot;,B,dir(180));<br /> dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(D); <br /> label(&quot;$D$&quot;,D,dir(0));<br /> dot(E); <br /> label(&quot;$E$&quot;,E,SE);<br /> dot(F); <br /> label(&quot;$F$&quot;,F,SW);<br /> &lt;/asy&gt;<br /> &lt;b&gt;Note:&lt;/b&gt; If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.&lt;br&gt;<br /> &lt;br&gt;<br /> As triangle &lt;math&gt;ABC&lt;/math&gt; is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.&lt;br&gt;<br /> &lt;br&gt;<br /> As point &lt;math&gt;D&lt;/math&gt; splits line segment &lt;math&gt;\overline{AC}&lt;/math&gt; in a &lt;math&gt;1:2&lt;/math&gt; ratio, we draw &lt;math&gt;\overline{AC}&lt;/math&gt; as a vertical line segment &lt;math&gt;3&lt;/math&gt; units long. Point &lt;math&gt;D&lt;/math&gt; is thus &lt;math&gt;1&lt;/math&gt; unit below point &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; units above point &lt;math&gt;C&lt;/math&gt;. By definition, Point &lt;math&gt;E&lt;/math&gt; splits line segment &lt;math&gt;\overline{BD}&lt;/math&gt; in a &lt;math&gt;1:1&lt;/math&gt; ratio, so we draw &lt;math&gt;\overline{BD}&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; units long directly left of &lt;math&gt;D&lt;/math&gt; and draw &lt;math&gt;E&lt;/math&gt; directly between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt; unit away from both.&lt;br&gt;<br /> &lt;br&gt;<br /> We then draw line segments &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. We can easily tell that triangle &lt;math&gt;ABC&lt;/math&gt; occupies &lt;math&gt;3&lt;/math&gt; square units of space. Constructing line &lt;math&gt;AE&lt;/math&gt; and drawing &lt;math&gt;F&lt;/math&gt; at the intersection of &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, we can easily see that triangle &lt;math&gt;EBF&lt;/math&gt; forms a right triangle occupying &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a square unit of space.&lt;br&gt;<br /> &lt;br&gt;<br /> The ratio of the areas of triangle &lt;math&gt;EBF&lt;/math&gt; and triangle &lt;math&gt;ABC&lt;/math&gt; is thus &lt;math&gt;\frac{1}{4}\div3=\frac{1}{12}&lt;/math&gt;, and since the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;360&lt;/math&gt;, this means that the area of triangle &lt;math&gt;EBF&lt;/math&gt; is &lt;math&gt;\frac{1}{12}\times360=\boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]&lt;br&gt;<br /> &lt;br&gt;<br /> &lt;b&gt;Additional note:&lt;/b&gt; There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 11==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF,G;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> G = (1.5,0);<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(B--DD); <br /> draw(G--DD);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,<br /> B,SW); <br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,DD,NE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> label(&quot;$F$&quot;,FF,S);<br /> label(&quot;$G$&quot;,G,S);<br /> &lt;/asy&gt;<br /> We know that &lt;math&gt;AD = \dfrac{1}{3} AC&lt;/math&gt;, so &lt;math&gt;[ABD] = \dfrac{1}{3} [ABC] = 120&lt;/math&gt;. Using the same method, since &lt;math&gt;BE = \dfrac{1}{2} BD&lt;/math&gt;, &lt;math&gt;[ABE] = \dfrac{1}{2} [ABD] = 60&lt;/math&gt;. Next, we draw &lt;math&gt;G&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{DG}&lt;/math&gt; is parallel to &lt;math&gt;\overline{AF}&lt;/math&gt; and create segment &lt;math&gt;DG&lt;/math&gt;. We then observe that &lt;math&gt;\triangle AFC \sim \triangle DGC&lt;/math&gt;, and since &lt;math&gt;AD:DC = 1:2&lt;/math&gt;, &lt;math&gt;FG:GC&lt;/math&gt; is also equal to &lt;math&gt;1:2&lt;/math&gt;. Similarly (no pun intended), &lt;math&gt;\triangle DBG \sim \triangle EBF&lt;/math&gt;, and since &lt;math&gt;BE:ED = 1:1&lt;/math&gt;, &lt;math&gt;BF:FG&lt;/math&gt; is also equal to &lt;math&gt;1:1&lt;/math&gt;. Combining the information in these two ratios, we find that &lt;math&gt;BF:FG:GC = 1:1:2&lt;/math&gt;, or equivalently, &lt;math&gt;BF = \dfrac{1}{4} BC&lt;/math&gt;. Thus, &lt;math&gt;[BFA] = \dfrac{1}{4} [BCA] = 90&lt;/math&gt;. We already know that &lt;math&gt;[ABE] = 60&lt;/math&gt;, so the area of &lt;math&gt;\triangle EBF&lt;/math&gt; is &lt;math&gt;[BFA] - [ABE] = \boxed{\textbf{(B) }30}&lt;/math&gt;. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br /> <br /> ==Solution 12 (Fastest Solution if you have no time)==<br /> The picture is misleading. Assume that the triangle ABC is right. <br /> <br /> Then find two factors of &lt;math&gt;720&lt;/math&gt; that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near &lt;math&gt;720&lt;/math&gt; to use difference of squares, we find &lt;math&gt;24&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt; as our numbers. Then the coordinates of D are &lt;math&gt;(10,16)&lt;/math&gt;(note, A=0,0). E is then &lt;math&gt;(5,8)&lt;/math&gt;. Then the equation of the line AE is &lt;math&gt;-16x/5+24=y&lt;/math&gt;. Plugging in &lt;math&gt;y=0&lt;/math&gt;, we have &lt;math&gt;x=\dfrac{15}{2}&lt;/math&gt;. Now notice that we have both the height and the base of EBF. <br /> <br /> Solving for the area, we have &lt;math&gt;(8)(15/2)(1/2)=30&lt;/math&gt;.<br /> <br /> == Solution 13 ==<br /> AD : DC = 1:2, so ADB has area 120 and CDB has area 240. BE=ED so area of ABE = area of ADE = 60.<br /> Draw &lt;math&gt;\overline{DG}&lt;/math&gt; parallel to &lt;math&gt;\overline{AF}&lt;/math&gt;.&lt;br&gt; <br /> Set area of BEF = &lt;math&gt;x&lt;/math&gt;. BEF is similar to BDG in ratio of 1:2&lt;br&gt; <br /> so area of BDG = &lt;math&gt;4x&lt;/math&gt;, area of EFDG=&lt;math&gt;3x&lt;/math&gt;, and area of CDG&lt;math&gt;=240-4x&lt;/math&gt;.&lt;br&gt;<br /> CDG is similar to CAF in ratio of 2:3 so area CDG = &lt;math&gt;4/9&lt;/math&gt; area CAF, and area AFDG=&lt;math&gt;5/4&lt;/math&gt; area CDG.&lt;br&gt; <br /> Thus &lt;math&gt;60+3x=5/4(240-4x)&lt;/math&gt; and &lt;math&gt;x=30&lt;/math&gt;.<br /> ~EFrame<br /> <br /> <br /> == Solution 14 - Geometry &amp; Algebra==<br /> &lt;asy&gt;<br /> unitsize(2cm);<br /> pair A,B,C,DD,EE,FF;<br /> B = (0,0); C = (3,0); <br /> A = (1.2,1.7);<br /> DD = (2/3)*A+(1/3)*C;<br /> EE = (B+DD)/2;<br /> FF = intersectionpoint(B--C,A--A+2*(EE-A));<br /> draw(A--B--C--cycle);<br /> draw(A--FF); <br /> draw(DD--FF,blue);<br /> draw(B--DD);dot(A); <br /> label(&quot;$A$&quot;,A,N);<br /> dot(B); <br /> label(&quot;$B$&quot;,<br /> B,SW);dot(C); <br /> label(&quot;$C$&quot;,C,SE);<br /> dot(DD); <br /> label(&quot;$D$&quot;,DD,NE);<br /> dot(EE); <br /> label(&quot;$E$&quot;,EE,NW);<br /> dot(FF); <br /> label(&quot;$F\$&quot;,FF,S);<br /> &lt;/asy&gt;<br /> <br /> We draw line &lt;math&gt;FD&lt;/math&gt; so that we can define a variable &lt;math&gt;x&lt;/math&gt; for the area of &lt;math&gt; \triangle BEF = \triangle DEF&lt;/math&gt;. Knowing that &lt;math&gt; \triangle ABE&lt;/math&gt; and &lt;math&gt; \triangle ADE&lt;/math&gt; share both their height and base, we get that &lt;math&gt;ABE = ADE = 60&lt;/math&gt;.<br /> <br /> Since we have a rule where 2 triangles, (&lt;math&gt;\triangle A&lt;/math&gt; which has base &lt;math&gt;a&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;), and (&lt;math&gt;\triangle B&lt;/math&gt; which has Base &lt;math&gt;b&lt;/math&gt; and vertex &lt;math&gt;c&lt;/math&gt;)who share the same vertex (which is vertex &lt;math&gt;c&lt;/math&gt; in this case), and share a common height, their relationship is : Area of &lt;math&gt;A : B = a : b&lt;/math&gt; (the length of the two bases), we can list the equation where &lt;math&gt;\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}&lt;/math&gt;. Substituting &lt;math&gt;x&lt;/math&gt; into the equation we get: <br /> <br /> &lt;cmath&gt;\frac{x+60}{300-x} = \frac{2x}{240-2x}&lt;/cmath&gt;. &lt;cmath&gt;(2x)(300-x) = (60+x)(240-x).&lt;/cmath&gt; &lt;cmath&gt;600-2x^2 = 14400 - 120x + 240x - 2x^2.&lt;/cmath&gt; &lt;cmath&gt;480x = 14400.&lt;/cmath&gt; and we now have that &lt;math&gt; \triangle BEF=30.&lt;/math&gt; <br /> ~&lt;math&gt;\bold{\color{blue}{onionheadjr}}&lt;/math&gt;<br /> <br /> ==Solution 15==<br /> https://youtu.be/Ns34Jiq9ofc<br /> —DSA_Catachu<br /> <br /> ==Solution 16==<br /> https://www.youtube.com/watch?v=nm-Vj_fsXt4<br /> - Happytwin (Another video solution)<br /> <br /> ==Solution 17==<br /> https://www.youtube.com/watch?v=nyevg9w-CCI&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=6 ~ MathEx<br /> <br /> <br /> ==Solution 18==<br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==Solution 19==<br /> https://www.youtube.com/watch?v=m04K0Q2SNXY&amp;t=1s<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=23|num-a=25}}<br /> <br /> [[Category: Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=136409 2019 AMC 8 Problems/Problem 25 2020-11-02T16:26:32Z <p>Hi im bob: /* Videos explaining solution */</p> <hr /> <div>==Problem 25==<br /> Alice has &lt;math&gt;24&lt;/math&gt; apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?<br /> &lt;math&gt;\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We use [[stars and bars]]. Let Alice get &lt;math&gt;k&lt;/math&gt; apples, let Becky get &lt;math&gt;r&lt;/math&gt; apples, let Chris get &lt;math&gt;y&lt;/math&gt; apples.<br /> &lt;cmath&gt;\implies k + r + y = 24&lt;/cmath&gt;We can manipulate this into an equation which can be solved using stars and bars.<br /> <br /> All of them get at least &lt;math&gt;2&lt;/math&gt; apples, so we can subtract &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;r&lt;/math&gt;, and &lt;math&gt;2&lt;/math&gt; from &lt;math&gt;y&lt;/math&gt;.<br /> &lt;cmath&gt;\implies (k - 2) + (r - 2) + (y - 2) = 18&lt;/cmath&gt;Let &lt;math&gt;k' = k - 2&lt;/math&gt;, let &lt;math&gt;r' = r - 2&lt;/math&gt;, let &lt;math&gt;y' = y - 2&lt;/math&gt;.<br /> &lt;cmath&gt;\implies k' + r' + y' = 18&lt;/cmath&gt;We can allow either of them to equal to &lt;math&gt;0&lt;/math&gt;, hence this can be solved by stars and bars.<br /> <br /> <br /> By Stars and Bars, our answer is just &lt;math&gt;\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Without loss of generality]], let's assume that Alice has &lt;math&gt;2&lt;/math&gt; apples. There are &lt;math&gt;19&lt;/math&gt; ways to split the rest of the apples with Becky and Chris. If Alice has &lt;math&gt;3&lt;/math&gt; apples, there are &lt;math&gt;18&lt;/math&gt; ways to split the rest of the apples with Becky and Chris. If Alice has &lt;math&gt;4&lt;/math&gt; apples, there are &lt;math&gt;17&lt;/math&gt; ways to split the rest. So the total number of ways to split &lt;math&gt;24&lt;/math&gt; apples between the three friends is equal to &lt;math&gt;19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Let's assume that the three of them have &lt;math&gt;x, y, z&lt;/math&gt; apples. Since each of them has to have at least &lt;math&gt;2&lt;/math&gt; apples, we say that &lt;math&gt;a+2=x, b+2=y&lt;/math&gt; and &lt;math&gt;c+2=z&lt;/math&gt;. Thus, &lt;math&gt;a+b+c+6=24 \implies a+b+c=18&lt;/math&gt;, and so by stars and bars, the number of solutions for this is &lt;math&gt;{n+k-1 \choose k} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2} = \boxed{\textbf{(C)}\ 190}&lt;/math&gt; - aops5234 <br /> <br /> ==Solution 4==<br /> <br /> Since we have to give each of the &lt;math&gt;3&lt;/math&gt; friends at least &lt;math&gt;2&lt;/math&gt; apples, we need to spend a total of &lt;math&gt;2+2+2=6&lt;/math&gt; apples to solve the restriction. Now we have &lt;math&gt;24-6=18&lt;/math&gt; apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have &lt;math&gt;18&lt;/math&gt; stones and &lt;math&gt;2&lt;/math&gt; sticks, which have a total of &lt;math&gt;\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}&lt;/math&gt; ways to arrange. <br /> <br /> ~by sakshamsethi<br /> <br /> ==Videos explaining solution==<br /> <br /> https://www.youtube.com/watch?v=wJ7uvypbB28<br /> <br /> https://www.youtube.com/watch?v=2dBUklyUaNI<br /> <br /> https://www.youtube.com/watch?v=EJzSOPXULBc<br /> <br /> https://youtu.be/ZsCRGK4VgBE ~DSA_Catachu<br /> <br /> https://www.youtube.com/watch?v=3qp0wTq-LI0&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=7 ~ MathEx<br /> <br /> https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=24|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=122366 User:Piphi 2020-05-14T00:10:33Z <p>Hi im bob: /* User Count */</p> <hr /> <div>&lt;center&gt;[[File:Piphi-Avatar.png]]&lt;/center&gt;<br /> <br /> &lt;div style=&quot;border:2px solid black; background:#eeeeee;&quot;&gt;<br /> ::::&lt;font style=&quot;font-family: Verdana, sans-serif&quot;&gt;[[User:Piphi|Userpage]] | [[User talk:Piphi|Talk]] | [[Special:Contributions/Piphi|Contributions]]&lt;/font&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#dddddd; align:center&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;User Count&lt;/font&gt;==<br /> &lt;font color=&quot;black&quot;&gt;If this is your first time visting this page, edit it by incrementing the user count below by one.<br /> <br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;99999999999999999999999999o41&lt;/font&gt;&lt;/center&gt;<br /> &lt;/font&gt; <br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;About Me&lt;/font&gt;==<br /> &lt;font color=&quot;black&quot;&gt;PM me if you want to find out about some cool things you can do with the AoPS wiki.<br /> <br /> My main project on the AoPS wiki is [[AoPS_Administrators#Current_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&amp;action=edit&amp;section=1 here]. I also added most of the info in the [[Reaper Archives]].&lt;/font&gt; <br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;Asymptote&lt;/font&gt;==<br /> <br /> Here is a list of the different drawings I have made using Asymptote.<br /> <br /> * [[User:Piphi/Firefox | Firefox Logo]] (Started April 25th, 2020, Finished April 28th, 2020)<br /> * [[User:Piphi/Eclipse | Eclipse Logo]] (Started April 29th, 2020, Finished April 29th, 2020)<br /> * [[User:Piphi/Screencast | Screencast Logo]] (Started April 29th, 2020, Finished April 29th, 2020)<br /> * [[User:Piphi/Whatsapp | Whatsapp Logo]] (Started May 2nd, 2020, Finished May 2nd, 2020)<br /> * [[User:Piphi/Office 365 | Office 365 Logo]] (Started May 4th, 2020, Finished May 4th, 2020)<br /> * [[User:Piphi/LG | LG Logo]] (Started May 5th, 2020, Finished May 6th, 2020)<br /> <br /> Work In Progress<br /> <br /> * Wolfram Logo (Not Started Yet)</div> Hi im bob https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=122205 User:Piphi 2020-05-09T17:51:12Z <p>Hi im bob: /* User Count */</p> <hr /> <div>&lt;center&gt;[[File:Piphi-Avatar.png]]&lt;/center&gt;<br /> <br /> &lt;div style=&quot;border:2px solid black; background:#eeeeee;&quot;&gt;<br /> ::::&lt;font style=&quot;font-family: Verdana, sans-serif&quot;&gt;[[User:Piphi|Userpage]] | [[User talk:Piphi|Talk]] | [[Special:Contributions/Piphi|Contributions]]&lt;/font&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#dddddd; align:center&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;User Count&lt;/font&gt;==<br /> &lt;font color=&quot;black&quot;&gt;If this is your first time visting this page, edit it by incrementing the user count below by one.<br /> <br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;4234126489327419467294826347124623894164389422&lt;/font&gt;&lt;/center&gt;<br /> &lt;/font&gt; <br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;About Me&lt;/font&gt;==<br /> &lt;font color=&quot;black&quot;&gt;PM me if you want to find out about some cool things you can do with the AoPS wiki.<br /> <br /> My main project on the AoPS wiki is [[AoPS_Administrators#Current_Admins | a list of all the AoPS admins]], everyone is welcome to add more admins to the list by clicking [https://artofproblemsolving.com/wiki/index.php?title=AoPS_Administrators&amp;action=edit&amp;section=1 here]. I also added most of the info in the [[Reaper Archives]].&lt;/font&gt; <br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: MV Boli, Verdana&quot;&gt;Asymptote&lt;/font&gt;==<br /> <br /> Here is a list of the different drawings I have made using Asymptote.<br /> <br /> * [[User:Piphi/Firefox | Firefox Logo]] (Started April 25th, 2020, Finished April 28th, 2020)<br /> * [[User:Piphi/Eclipse | Eclipse Logo]] (Started April 29th, 2020, Finished April 29th, 2020)<br /> * [[User:Piphi/Screencast | Screencast Logo]] (Started April 29th, 2020, Finished April 29th, 2020)<br /> * [[User:Piphi/Whatsapp | Whatsapp Logo]] (Started May 2nd, 2020, Finished May 2nd, 2020)<br /> * [[User:Piphi/Office 365 | Office 365 Logo]] (Started May 4th, 2020, Finished May 4th, 2020)<br /> <br /> Work In Progress<br /> <br /> * Wolfram Logo (Not Started Yet)</div> Hi im bob