https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Zhaom&feedformat=atom AoPS Wiki - User contributions [en] 2022-08-16T09:20:45Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_15&diff=176672 2018 AIME I Problems/Problem 15 2022-08-04T14:59:38Z <p>Zhaom: /* Solution 4 */</p> <hr /> <div>==Problem 15==<br /> David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, &lt;math&gt;A,\text{ }B,\text{ }C&lt;/math&gt;, which can each be inscribed in a circle with radius &lt;math&gt;1&lt;/math&gt;. Let &lt;math&gt;\varphi_A&lt;/math&gt; denote the measure of the acute angle made by the diagonals of quadrilateral &lt;math&gt;A&lt;/math&gt;, and define &lt;math&gt;\varphi_B&lt;/math&gt; and &lt;math&gt;\varphi_C&lt;/math&gt; similarly. Suppose that &lt;math&gt;\sin\varphi_A=\frac{2}{3}&lt;/math&gt;, &lt;math&gt;\sin\varphi_B=\frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\sin\varphi_C=\frac{6}{7}&lt;/math&gt;. All three quadrilaterals have the same area &lt;math&gt;K&lt;/math&gt;, which can be written in the form &lt;math&gt;\dfrac{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 1==<br /> <br /> Suppose our four sides lengths cut out arc lengths of &lt;math&gt;2a&lt;/math&gt;, &lt;math&gt;2b&lt;/math&gt;, &lt;math&gt;2c&lt;/math&gt;, and &lt;math&gt;2d&lt;/math&gt;, where &lt;math&gt;a+b+c+d=180^\circ&lt;/math&gt;. Then, we only have to consider which arc is opposite &lt;math&gt;2a&lt;/math&gt;. These are our three cases, so<br /> &lt;cmath&gt;\varphi_A=a+c&lt;/cmath&gt;<br /> &lt;cmath&gt;\varphi_B=a+b&lt;/cmath&gt;<br /> &lt;cmath&gt;\varphi_C=a+d&lt;/cmath&gt;<br /> Our first case involves quadrilateral &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;\overarc{AB}=2a&lt;/math&gt;, &lt;math&gt;\overarc{BC}=2b&lt;/math&gt;, &lt;math&gt;\overarc{CD}=2c&lt;/math&gt;, and &lt;math&gt;\overarc{DA}=2d&lt;/math&gt;.<br /> <br /> Then, by Law of Sines, &lt;math&gt;AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)&lt;/math&gt; and &lt;math&gt;BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)&lt;/math&gt;. Therefore,<br /> <br /> &lt;cmath&gt;K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},&lt;/cmath&gt;<br /> so our answer is &lt;math&gt;24+35=\boxed{059}&lt;/math&gt;.<br /> <br /> Note that the conditions of the problem are satisfied when the lengths of the four sticks are about &lt;math&gt;0.32, 0.91, 1.06, 1.82&lt;/math&gt;.<br /> <br /> By S.B.<br /> <br /> ==Solution 2==<br /> <br /> Let the four stick lengths be &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt;. WLOG, let’s say that quadrilateral &lt;math&gt;A&lt;/math&gt; has sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; opposite each other, quadrilateral &lt;math&gt;B&lt;/math&gt; has sides &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; opposite each other, and quadrilateral &lt;math&gt;C&lt;/math&gt; has sides &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; opposite each other. The area of a convex quadrilateral can be written as &lt;math&gt;\frac{1}{2} d_1 d_2 \sin{\theta}&lt;/math&gt;, where &lt;math&gt;d_1&lt;/math&gt; and &lt;math&gt;d_2&lt;/math&gt; are the lengths of the diagonals of the quadrilateral and &lt;math&gt;\theta&lt;/math&gt; is the angle formed by the intersection of &lt;math&gt;d_1&lt;/math&gt; and &lt;math&gt;d_2&lt;/math&gt;. By Ptolemy's theorem &lt;math&gt;d_1 d_2 = ad+bc&lt;/math&gt; for quadrilateral &lt;math&gt;A&lt;/math&gt;, so, defining &lt;math&gt;K_A&lt;/math&gt; as the area of &lt;math&gt;A&lt;/math&gt;,<br /> &lt;cmath&gt;K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}&lt;/cmath&gt;<br /> Similarly, for quadrilaterals &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;,<br /> &lt;cmath&gt;K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}&lt;/cmath&gt;<br /> Multiplying the three equations and rearranging, we see that<br /> &lt;cmath&gt;K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}&lt;/cmath&gt;<br /> &lt;cmath&gt;K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)&lt;/cmath&gt;<br /> The circumradius &lt;math&gt;R&lt;/math&gt; of a cyclic quadrilateral with side lengths &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; and area &lt;math&gt;K&lt;/math&gt; can be computed as &lt;math&gt;R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}&lt;/math&gt;.<br /> Inserting what we know,<br /> &lt;cmath&gt;1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}&lt;/cmath&gt;<br /> &lt;cmath&gt;4K = \sqrt{\frac{70}{3}K^3}&lt;/cmath&gt;<br /> &lt;cmath&gt;16K^2 = \frac{70}{3}K^3&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{24}{35} = K&lt;/cmath&gt;<br /> So our answer is &lt;math&gt;24 + 35 = \boxed{059}&lt;/math&gt;.<br /> <br /> ~Solution by divij04<br /> <br /> ==Solution 3 (No words)==<br /> [[File:2018 AIME I 15.png|900px]]<br /> <br /> '''Shelomovskii, vvsss, www.deoma-cmd.ru'''<br /> <br /> ==Solution 4==<br /> Let the sides of the quadrilaterals be &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; in some order such that &lt;math&gt;A&lt;/math&gt; has &lt;math&gt;a&lt;/math&gt; opposite of &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; has &lt;math&gt;a&lt;/math&gt; opposite of &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; has &lt;math&gt;a&lt;/math&gt; opposite of &lt;math&gt;d&lt;/math&gt;. Then, let the diagonals of &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;e&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt;. Similarly to solution &lt;math&gt;2&lt;/math&gt;, we get that &lt;math&gt;\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K&lt;/math&gt;, but this is also equal to &lt;math&gt;2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}&lt;/math&gt; using the area formula for a triangle using the circumradius and the sides, so &lt;math&gt;\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)&lt;/math&gt; and &lt;math&gt;\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)&lt;/math&gt;. Solving for &lt;math&gt;e&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt;, we get that &lt;math&gt;e=\tfrac{6}{5}&lt;/math&gt; and &lt;math&gt;f=\tfrac{12}{7}&lt;/math&gt;, but &lt;math&gt;K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef&lt;/math&gt;, similarly to solution &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;K=\tfrac{24}{35}&lt;/math&gt; and the answer is &lt;math&gt;24+35=\boxed{059}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=14|after=Last question}}<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_15&diff=176671 2018 AIME I Problems/Problem 15 2022-08-04T14:59:15Z <p>Zhaom: </p> <hr /> <div>==Problem 15==<br /> David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, &lt;math&gt;A,\text{ }B,\text{ }C&lt;/math&gt;, which can each be inscribed in a circle with radius &lt;math&gt;1&lt;/math&gt;. Let &lt;math&gt;\varphi_A&lt;/math&gt; denote the measure of the acute angle made by the diagonals of quadrilateral &lt;math&gt;A&lt;/math&gt;, and define &lt;math&gt;\varphi_B&lt;/math&gt; and &lt;math&gt;\varphi_C&lt;/math&gt; similarly. Suppose that &lt;math&gt;\sin\varphi_A=\frac{2}{3}&lt;/math&gt;, &lt;math&gt;\sin\varphi_B=\frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\sin\varphi_C=\frac{6}{7}&lt;/math&gt;. All three quadrilaterals have the same area &lt;math&gt;K&lt;/math&gt;, which can be written in the form &lt;math&gt;\dfrac{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 1==<br /> <br /> Suppose our four sides lengths cut out arc lengths of &lt;math&gt;2a&lt;/math&gt;, &lt;math&gt;2b&lt;/math&gt;, &lt;math&gt;2c&lt;/math&gt;, and &lt;math&gt;2d&lt;/math&gt;, where &lt;math&gt;a+b+c+d=180^\circ&lt;/math&gt;. Then, we only have to consider which arc is opposite &lt;math&gt;2a&lt;/math&gt;. These are our three cases, so<br /> &lt;cmath&gt;\varphi_A=a+c&lt;/cmath&gt;<br /> &lt;cmath&gt;\varphi_B=a+b&lt;/cmath&gt;<br /> &lt;cmath&gt;\varphi_C=a+d&lt;/cmath&gt;<br /> Our first case involves quadrilateral &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;\overarc{AB}=2a&lt;/math&gt;, &lt;math&gt;\overarc{BC}=2b&lt;/math&gt;, &lt;math&gt;\overarc{CD}=2c&lt;/math&gt;, and &lt;math&gt;\overarc{DA}=2d&lt;/math&gt;.<br /> <br /> Then, by Law of Sines, &lt;math&gt;AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)&lt;/math&gt; and &lt;math&gt;BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)&lt;/math&gt;. Therefore,<br /> <br /> &lt;cmath&gt;K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},&lt;/cmath&gt;<br /> so our answer is &lt;math&gt;24+35=\boxed{059}&lt;/math&gt;.<br /> <br /> Note that the conditions of the problem are satisfied when the lengths of the four sticks are about &lt;math&gt;0.32, 0.91, 1.06, 1.82&lt;/math&gt;.<br /> <br /> By S.B.<br /> <br /> ==Solution 2==<br /> <br /> Let the four stick lengths be &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt;. WLOG, let’s say that quadrilateral &lt;math&gt;A&lt;/math&gt; has sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; opposite each other, quadrilateral &lt;math&gt;B&lt;/math&gt; has sides &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; opposite each other, and quadrilateral &lt;math&gt;C&lt;/math&gt; has sides &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; opposite each other. The area of a convex quadrilateral can be written as &lt;math&gt;\frac{1}{2} d_1 d_2 \sin{\theta}&lt;/math&gt;, where &lt;math&gt;d_1&lt;/math&gt; and &lt;math&gt;d_2&lt;/math&gt; are the lengths of the diagonals of the quadrilateral and &lt;math&gt;\theta&lt;/math&gt; is the angle formed by the intersection of &lt;math&gt;d_1&lt;/math&gt; and &lt;math&gt;d_2&lt;/math&gt;. By Ptolemy's theorem &lt;math&gt;d_1 d_2 = ad+bc&lt;/math&gt; for quadrilateral &lt;math&gt;A&lt;/math&gt;, so, defining &lt;math&gt;K_A&lt;/math&gt; as the area of &lt;math&gt;A&lt;/math&gt;,<br /> &lt;cmath&gt;K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}&lt;/cmath&gt;<br /> Similarly, for quadrilaterals &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;,<br /> &lt;cmath&gt;K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}&lt;/cmath&gt;<br /> Multiplying the three equations and rearranging, we see that<br /> &lt;cmath&gt;K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}&lt;/cmath&gt;<br /> &lt;cmath&gt;K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)&lt;/cmath&gt;<br /> The circumradius &lt;math&gt;R&lt;/math&gt; of a cyclic quadrilateral with side lengths &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; and area &lt;math&gt;K&lt;/math&gt; can be computed as &lt;math&gt;R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}&lt;/math&gt;.<br /> Inserting what we know,<br /> &lt;cmath&gt;1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}&lt;/cmath&gt;<br /> &lt;cmath&gt;4K = \sqrt{\frac{70}{3}K^3}&lt;/cmath&gt;<br /> &lt;cmath&gt;16K^2 = \frac{70}{3}K^3&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{24}{35} = K&lt;/cmath&gt;<br /> So our answer is &lt;math&gt;24 + 35 = \boxed{059}&lt;/math&gt;.<br /> <br /> ~Solution by divij04<br /> <br /> ==Solution 3 (No words)==<br /> [[File:2018 AIME I 15.png|900px]]<br /> <br /> '''Shelomovskii, vvsss, www.deoma-cmd.ru'''<br /> <br /> ==Solution 4==<br /> Let the sides of the quadrilaterals be &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; in some order such that &lt;math&gt;A&lt;/math&gt; has &lt;math&gt;a&lt;/math&gt; opposite of &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; has &lt;math&gt;a&lt;/math&gt; opposite of &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; has &lt;math&gt;a&lt;/math&gt; opposite of &lt;math&gt;d&lt;/math&gt;. Then let the diagonals of &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;e&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt;. Similarly to solution &lt;math&gt;2&lt;/math&gt;, we get that &lt;math&gt;\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K&lt;/math&gt;, but this is also equal to &lt;math&gt;2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}&lt;/math&gt; using the area formula for a triangle using the circumradius and the sides, so &lt;math&gt;\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)&lt;/math&gt; and &lt;math&gt;\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)&lt;/math&gt;. Solving for &lt;math&gt;e&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt;, we get that &lt;math&gt;e=\tfrac{6}{5}&lt;/math&gt; and &lt;math&gt;f=\tfrac{12}{7}&lt;/math&gt;, but &lt;math&gt;K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef&lt;/math&gt;, similarly to solution &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;K=\tfrac{24}{35}&lt;/math&gt; and the answer is &lt;math&gt;24+35=\boxed{059}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2018|n=I|num-b=14|after=Last question}}<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_6&diff=176558 2016 AIME I Problems/Problem 6 2022-08-02T14:08:26Z <p>Zhaom: /* Solution */</p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; let &lt;math&gt;I&lt;/math&gt; be the center of the inscribed circle, and let the bisector of &lt;math&gt;\angle ACB&lt;/math&gt; intersect &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;L&lt;/math&gt;. The line through &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; intersects the circumscribed circle of &lt;math&gt;\triangle ABC&lt;/math&gt; at the two points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. If &lt;math&gt;LI=2&lt;/math&gt; and &lt;math&gt;LD=3&lt;/math&gt;, then &lt;math&gt;IC=\tfrac{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 /> ==Solution 1==<br /> Suppose we label the angles as shown below.<br /> &lt;asy&gt;<br /> size(150);<br /> import olympiad;<br /> real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;<br /> pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));<br /> pair C=intersectionpoints(circle(A,b),circle(B,a));<br /> pair I=incenter(A,B,C);<br /> pair L=extension(C,D,A,B);<br /> dot(I^^A^^B^^C^^D);<br /> draw(C--D);<br /> path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}<br /> draw(A--B--D--cycle);<br /> draw(circumcircle(A,B,D));<br /> draw(A--C--B);<br /> draw(A--I--B^^C--I);<br /> draw(incircle(A,B,C));<br /> label(&quot;$A$&quot;,A,SW,fontsize(8));<br /> label(&quot;$B$&quot;,B,SE,fontsize(8));<br /> label(&quot;$C$&quot;,C,N,fontsize(8));<br /> label(&quot;$D$&quot;,D,S,fontsize(8));<br /> label(&quot;$I$&quot;,I,NE,fontsize(8));<br /> label(&quot;$L$&quot;,L,SW,fontsize(8));<br /> label(&quot;$\alpha$&quot;,A,5*dir(midangle(C,A,I)),fontsize(8));<br /> label(&quot;$\alpha$&quot;,A,5*dir(midangle(I,A,B)),fontsize(8));<br /> label(&quot;$\beta$&quot;,B,12*dir(midangle(A,B,I)),fontsize(8));<br /> label(&quot;$\beta$&quot;,B,12*dir(midangle(I,B,C)),fontsize(8));<br /> label(&quot;$\gamma$&quot;,C,5*dir(midangle(A,C,I)),fontsize(8));<br /> label(&quot;$\gamma$&quot;,C,5*dir(midangle(I,C,B)),fontsize(8));<br /> &lt;/asy&gt;<br /> As &lt;math&gt;\angle BCD&lt;/math&gt; and &lt;math&gt;\angle BAD&lt;/math&gt; intercept the same arc, we know that &lt;math&gt;\angle BAD=\gamma&lt;/math&gt;. Similarly, &lt;math&gt;\angle ABD=\gamma&lt;/math&gt;. Also, using &lt;math&gt;\triangle ICA&lt;/math&gt;, we find &lt;math&gt;\angle CIA=180-\alpha-\gamma&lt;/math&gt;. Therefore, &lt;math&gt;\angle AID=\alpha+\gamma&lt;/math&gt;. Therefore, &lt;math&gt;\angle DAI=\angle AID=\alpha+\gamma&lt;/math&gt;, so &lt;math&gt;\triangle AID&lt;/math&gt; must be isosceles with &lt;math&gt;AD=ID=5&lt;/math&gt;. Similarly, &lt;math&gt;BD=ID=5&lt;/math&gt;. Then &lt;math&gt;\triangle DLB \sim \triangle ALC&lt;/math&gt;, hence &lt;math&gt;\frac{AL}{AC} = \frac{3}{5}&lt;/math&gt;. Also, &lt;math&gt;AI&lt;/math&gt; bisects &lt;math&gt;\angle LAC&lt;/math&gt;, so by the Angle Bisector Theorem &lt;math&gt;\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}&lt;/math&gt;. Thus &lt;math&gt;CI = \frac{10}{3}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> WLOG assume &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles. Then, &lt;math&gt;L&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt;, and &lt;math&gt;\angle CLB=\angle CLA=90^\circ&lt;/math&gt;. Draw the perpendicular from &lt;math&gt;I&lt;/math&gt; to &lt;math&gt;CB&lt;/math&gt;, and let it meet &lt;math&gt;CB&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. Since &lt;math&gt;IL=2&lt;/math&gt;, &lt;math&gt;IE&lt;/math&gt; is also &lt;math&gt;2&lt;/math&gt; (they are both inradii). Set &lt;math&gt;BD&lt;/math&gt; as &lt;math&gt;x&lt;/math&gt;. Then, triangles &lt;math&gt;BLD&lt;/math&gt; and &lt;math&gt;CEI&lt;/math&gt; are similar, and &lt;math&gt;\tfrac{2}{3}=\tfrac{CI}{x}&lt;/math&gt;. Thus, &lt;math&gt;CI=\tfrac{2x}{3}&lt;/math&gt;. &lt;math&gt;\triangle CBD \sim \triangle CEI&lt;/math&gt;, so &lt;math&gt;\tfrac{IE}{DB}=\tfrac{CI}{CD}&lt;/math&gt;. Thus &lt;math&gt;\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt;, we have:<br /> &lt;math&gt;x^2-2x-15=0&lt;/math&gt;, or &lt;math&gt;x=5, -3&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is positive, so &lt;math&gt;x=5&lt;/math&gt;. As a result, &lt;math&gt;CI=\tfrac{2x}{3}=\tfrac{10}{3}&lt;/math&gt; and the answer is &lt;math&gt;\boxed{013}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> WLOG assume &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles (with vertex &lt;math&gt;C&lt;/math&gt;). Let &lt;math&gt;O&lt;/math&gt; be the center of the circumcircle, &lt;math&gt;R&lt;/math&gt; the circumradius, and &lt;math&gt;r&lt;/math&gt; the inradius. A simple sketch will reveal that &lt;math&gt;\triangle ABC&lt;/math&gt; must be obtuse (as an acute triangle will result in &lt;math&gt;LI&lt;/math&gt; being greater than &lt;math&gt;DL&lt;/math&gt;) and that &lt;math&gt;O&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; are collinear. Next, if &lt;math&gt;OI=d&lt;/math&gt;, &lt;math&gt;DO+OI=R+d&lt;/math&gt; and &lt;math&gt;R+d=DL+LI=5&lt;/math&gt;. Euler gives us that &lt;math&gt;d^{2}=R(R-2r)&lt;/math&gt;, and in this case, &lt;math&gt;r=LI=2&lt;/math&gt;. Thus, &lt;math&gt;d=\sqrt{R^{2}-4R}&lt;/math&gt;. Solving for &lt;math&gt;d&lt;/math&gt;, we have &lt;math&gt;R+\sqrt{R^{2}-4R}=5&lt;/math&gt;, then &lt;math&gt;R^{2}-4R=25-10R+R^{2}&lt;/math&gt;, yielding &lt;math&gt;R=\frac{25}{6}&lt;/math&gt;. Next, &lt;math&gt;R+d=5&lt;/math&gt; so &lt;math&gt;d=\frac{5}{6}&lt;/math&gt;. Finally, &lt;math&gt;OC=OI+IC&lt;/math&gt; gives us &lt;math&gt;R=d+IC&lt;/math&gt;, and &lt;math&gt;IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}&lt;/math&gt;. Our answer is then &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Since &lt;math&gt;\angle{LAD} = \angle{BDC}&lt;/math&gt; and &lt;math&gt;\angle{DLA}=\angle{DBC}&lt;/math&gt;, &lt;math&gt;\triangle{DLA}\sim\triangle{DBC}&lt;/math&gt;. Also, &lt;math&gt;\angle{DAC}=\angle{BLC}&lt;/math&gt; and &lt;math&gt;\angle{ACD}=\angle{LCB}&lt;/math&gt; so &lt;math&gt;\triangle{DAC}\sim\triangle{BLC}&lt;/math&gt;. Now we can call &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. By angle bisector theorem, &lt;math&gt;\frac{AD}{DB}=\frac{AC}{BC}&lt;/math&gt;. So let &lt;math&gt;AD=bk&lt;/math&gt; and &lt;math&gt;DB=ak&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt;. Now call &lt;math&gt;IC=x&lt;/math&gt;. By the similar triangles we found earlier, &lt;math&gt;\frac{3}{ak}=\frac{bk}{x+2}&lt;/math&gt; and &lt;math&gt;\frac{b}{x+5}=\frac{x+2}{a}&lt;/math&gt;. We can simplify this to &lt;math&gt;abk^2=3x+6&lt;/math&gt; and &lt;math&gt;ab=(x+5)(x+2)&lt;/math&gt;. So we can plug the &lt;math&gt;ab&lt;/math&gt; into the first equation and get &lt;math&gt;(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3&lt;/math&gt;. We can now draw a line through &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;I&lt;/math&gt; that intersects &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. By mass points, we can assign a mass of &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;a+b&lt;/math&gt; to &lt;math&gt;D&lt;/math&gt;. We can also assign a mass of &lt;math&gt;(a+b)k&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; by angle bisector theorem. So the ratio of &lt;math&gt;\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}&lt;/math&gt;. So since &lt;math&gt;k=\frac{2}{x}&lt;/math&gt;, we can plug this back into the original equation to get &lt;math&gt;\left(\frac{2}{x}\right)^2(x+5)=3&lt;/math&gt;. This means that &lt;math&gt;\frac{3x^2}{4}-x-5=0&lt;/math&gt; which has roots -2 and &lt;math&gt;\frac{10}{3}&lt;/math&gt; which means our &lt;math&gt;CI=\frac{10}{3}&lt;/math&gt; and our answer is &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Since &lt;math&gt;\angle BCD&lt;/math&gt; and &lt;math&gt;\angle BAD&lt;/math&gt; both intercept arc &lt;math&gt;BD&lt;/math&gt;, it follows that &lt;math&gt;\angle BAD=\gamma&lt;/math&gt;. Note that &lt;math&gt;\angle AID=\alpha+\gamma&lt;/math&gt; by the external angle theorem. It follows that &lt;math&gt;\angle DAI=\angle AID=\alpha+\gamma&lt;/math&gt;, so we must have that &lt;math&gt;\triangle AID&lt;/math&gt; is isosceles, yielding &lt;math&gt;AD=ID=5&lt;/math&gt;. Note that &lt;math&gt;\triangle DLA \sim \triangle DAC&lt;/math&gt;, so &lt;math&gt;\frac{DA}{DL} = \frac{DC}{DA}&lt;/math&gt;. This yields &lt;math&gt;DC = \frac{25}{3}&lt;/math&gt;. It follows that &lt;math&gt;CI = DC - DI = \frac{10}{3}&lt;/math&gt;, giving a final answer of &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> Let &lt;math&gt;I_C&lt;/math&gt; be the excenter opposite to &lt;math&gt;C&lt;/math&gt; in &lt;math&gt;ABC&lt;/math&gt;. By the incenter-excenter lemma &lt;math&gt;DI=DC \therefore&lt;/math&gt; &lt;math&gt;LI_C=8,LI=2,II_C=10&lt;/math&gt;. Its well known that &lt;math&gt;(I_C,I,L,C)=-1 \implies \dfrac{LI_C}{LI}.\dfrac{CI}{CI_C}=-1 \implies \dfrac{CI}{CI+10}=\dfrac{1}{4} \implies \boxed{CI=\dfrac{10}{3}}&lt;/math&gt;.&lt;math&gt;\blacksquare&lt;/math&gt;<br /> ~Pluto1708<br /> <br /> Alternate solution: &quot;We can use the angle bisector theorem on &lt;math&gt;\triangle CBL&lt;/math&gt; and bisector &lt;math&gt;BI&lt;/math&gt; to get that &lt;math&gt;\tfrac{CI}{IL}=\tfrac{CI}{2}=\tfrac{BC}{BL}&lt;/math&gt;. Since &lt;math&gt;\triangle CBL \sim \triangle ADL&lt;/math&gt;, we get &lt;math&gt;\tfrac{BC}{BL}=\tfrac{AD}{DL}=\tfrac{5}{3}&lt;/math&gt;. Thus, &lt;math&gt;CI=\tfrac{10}{3}&lt;/math&gt; and &lt;math&gt;p+q=\boxed{13}&lt;/math&gt;.&quot;<br /> (https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)<br /> <br /> ==Solution 7==<br /> We can just say that quadrilateral &lt;math&gt;ADBC&lt;/math&gt; is a right kite with right angles at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Let us construct another similar right kite with the points of tangency on &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; called &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; respectively, point &lt;math&gt;I&lt;/math&gt;, and point &lt;math&gt;C&lt;/math&gt;. Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call &lt;math&gt;CI&lt;/math&gt; &lt;math&gt;x&lt;/math&gt; for simplicity's sake. Based on the fact that &lt;math&gt;\triangle BCD&lt;/math&gt; is similar to &lt;math&gt;\triangle FCI&lt;/math&gt; we can use triangle proportionality to say that &lt;math&gt;BD&lt;/math&gt; is &lt;math&gt;2\frac{x+5}{x}&lt;/math&gt;. Using geometric mean theorem we can show that &lt;math&gt;BL&lt;/math&gt; must be &lt;math&gt;\sqrt{3x+6}&lt;/math&gt;. With Pythagorean Theorem we can say that &lt;math&gt;3x+6+9=4{(\frac{x+5}{x})}^2&lt;/math&gt;. Multiplying both sides by &lt;math&gt;x^2&lt;/math&gt; and moving everything to LHS will give you &lt;math&gt;3{x}^3+11{x}^2-40x-100=0&lt;/math&gt; Since &lt;math&gt;x&lt;/math&gt; must be in the form &lt;math&gt;\frac{p}{q}&lt;/math&gt; we can assume that &lt;math&gt;x&lt;/math&gt; is most likely a positive fraction in the form &lt;math&gt;\frac{p}{3}&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; is a factor of &lt;math&gt;100&lt;/math&gt;. Testing the factors in synthetic division would lead &lt;math&gt;x = \frac{10}{3}&lt;/math&gt;, giving us our desired answer &lt;math&gt;\boxed{013}&lt;/math&gt;. ~Lopkiloinm<br /> <br /> ==Solution 8 (Cyclic Quadrilaterals)==<br /> <br /> &lt;asy&gt;<br /> size(150);<br /> import olympiad;<br /> real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;<br /> pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2));<br /> pair C=intersectionpoints(circle(A,b),circle(B,a));<br /> pair I=incenter(A,B,C);<br /> pair L=extension(C,D,A,B);<br /> dot(I^^A^^B^^C^^D);<br /> draw(C--D);<br /> path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}<br /> draw(A--B--D--cycle);<br /> draw(circumcircle(A,B,D));<br /> draw(A--C--B);<br /> draw(A--I--B^^C--I);<br /> draw(incircle(A,B,C));<br /> label(&quot;$A$&quot;,A,SW,fontsize(8));<br /> label(&quot;$B$&quot;,B,SE,fontsize(8));<br /> label(&quot;$C$&quot;,C,N,fontsize(8));<br /> label(&quot;$D$&quot;,D,S,fontsize(8));<br /> label(&quot;$I$&quot;,I,NE,fontsize(8));<br /> label(&quot;$L$&quot;,L,SW,fontsize(8));<br /> label(&quot;$\alpha$&quot;,A,5*dir(midangle(C,A,I)),fontsize(8));<br /> label(&quot;$\alpha$&quot;,A,5*dir(midangle(I,A,B)),fontsize(8));<br /> label(&quot;$\beta$&quot;,B,12*dir(midangle(A,B,I)),fontsize(8));<br /> label(&quot;$\beta$&quot;,B,12*dir(midangle(I,B,C)),fontsize(8));<br /> label(&quot;$\gamma$&quot;,C,5*dir(midangle(A,C,I)),fontsize(8));<br /> label(&quot;$\gamma$&quot;,C,5*dir(midangle(I,C,B)),fontsize(8));<br /> &lt;/asy&gt;<br /> First, using the diagram above, extend the bisector of &lt;math&gt;\angle ACB&lt;/math&gt; to pass through points &lt;math&gt;L&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; as stated in the problem. Then, connect &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; to form quadrilateral &lt;math&gt;ACBD&lt;/math&gt;. Note, that since points &lt;math&gt;D,A,C,B&lt;/math&gt; are concyclic, quadrilateral &lt;math&gt;ACBD&lt;/math&gt; is cyclic. <br /> <br /> As a result, apply Ptolemy's Theorem on the quadrilateral, denoting the length of &lt;math&gt;BD&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; as &lt;math&gt;z&lt;/math&gt; (they both must be equal, as &lt;math&gt;\angle ABD&lt;/math&gt; and &lt;math&gt;\angle DAB&lt;/math&gt; are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at &lt;math&gt;D&lt;/math&gt;), and the lengths of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;AB&lt;/math&gt;, and &lt;math&gt;CI&lt;/math&gt; as &lt;math&gt;a,b,c, x&lt;/math&gt;, respectively.<br /> <br /> After applying Ptolemy's, one will get that:<br /> <br /> &lt;cmath&gt;z(a+b)=c(x+5)&lt;/cmath&gt;<br /> <br /> Next, since &lt;math&gt;ACBD&lt;/math&gt; is cyclic, triangles &lt;math&gt;ALD&lt;/math&gt; and &lt;math&gt;CLB&lt;/math&gt; are similar, yielding the following equation once simplifications are made to the equation &lt;math&gt;\frac{AD}{CB}=\frac{AL}{BL}&lt;/math&gt;, with the length of &lt;math&gt;BL&lt;/math&gt; written in terms of &lt;math&gt;a,b,c&lt;/math&gt; using the angle bisector theorem on triangle &lt;math&gt;ABC&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;zc=3(a+b)&lt;/cmath&gt;<br /> <br /> Next, drawing in the bisector of &lt;math&gt;\angle BAC&lt;/math&gt; to the incenter &lt;math&gt;I&lt;/math&gt;, and applying the angle bisector theorem, we have that:<br /> <br /> &lt;cmath&gt;cx=2(a+b)&lt;/cmath&gt;<br /> <br /> Now, solving for &lt;math&gt;z&lt;/math&gt; in the second equation, and &lt;math&gt;x&lt;/math&gt; in the third equation and plugging them both back into the first equation, and making the substitution &lt;math&gt;w=\frac{a+b}{c}&lt;/math&gt;, we get the quadratic equation:<br /> <br /> &lt;cmath&gt;3w^2-2w-5=0&lt;/cmath&gt;<br /> <br /> Solving, we get &lt;math&gt;w=5/3&lt;/math&gt;, which gives &lt;math&gt;z=5&lt;/math&gt; and &lt;math&gt;x=10/3&lt;/math&gt;, when we rewrite the above equations in terms of &lt;math&gt;w&lt;/math&gt;. Thus, our answer is &lt;math&gt;\boxed{013}&lt;/math&gt; and we're done.<br /> <br /> -mathislife52<br /> <br /> ==Solution 9(Visual)==<br /> [[File:2016 AIME I 6b.png|500px]]<br /> '''Shelomovskii, vvsss, www.deoma-cmd.ru'''<br /> <br /> ==Solution 10==<br /> Let &lt;math&gt;AB=c,BC=a,CA=b&lt;/math&gt;, and &lt;math&gt;x=\tfrac{a+b}{c}&lt;/math&gt;. Then, notice that &lt;math&gt;\tfrac{CI}{IL}=\tfrac{a+b}{c}=x&lt;/math&gt;, so &lt;math&gt;CI=IL\cdot{}x=2x&lt;/math&gt;. Also, by the incenter-excenter lemma, &lt;math&gt;AD=BD=ID=IL+LD=5&lt;/math&gt;. Therefore, by Ptolemy's Theorem on cyclic quadrilateral &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;5a+5b=c(2x+5)&lt;/math&gt;, so &lt;math&gt;5\left(\tfrac{a+b}{c}\right)=2x+5&lt;/math&gt;, so &lt;math&gt;5x=2x+5&lt;/math&gt;. Solving, we get that &lt;math&gt;x=\tfrac{5}{3}&lt;/math&gt;, so &lt;math&gt;CI=\tfrac{10}{3}&lt;/math&gt; and the answer is &lt;math&gt;10+3=\boxed{013}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2016|n=I|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_12&diff=176525 2015 AMC 10A Problems/Problem 12 2022-08-01T19:25:47Z <p>Zhaom: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Points &lt;math&gt;( \sqrt{\pi} , a)&lt;/math&gt; and &lt;math&gt;( \sqrt{\pi} , b)&lt;/math&gt; are distinct points on the graph of &lt;math&gt;y^2 + x^4 = 2x^2 y + 1&lt;/math&gt;. What is &lt;math&gt;|a-b|&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since points on the graph make the equation true, substitute &lt;math&gt;\sqrt{\pi}&lt;/math&gt; in to the equation and then solve to find &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> &lt;math&gt;y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y^2 + \pi^2 = 2\pi y + 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y^2 - 2\pi y + \pi^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(y-\pi)^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y-\pi = \pm 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y = \pi + 1&lt;/math&gt; <br /> <br /> &lt;math&gt;y = \pi - 1&lt;/math&gt;<br /> <br /> There are only two solutions to the equation &lt;math&gt;(y-\pi)^2 = 1&lt;/math&gt;, so one of them is the value of &lt;math&gt;a&lt;/math&gt; and the other is &lt;math&gt;b&lt;/math&gt;. The order does not matter because of the absolute value sign.<br /> <br /> &lt;math&gt;| (\pi + 1) - (\pi - 1) | = 2&lt;/math&gt;<br /> <br /> The answer is &lt;math&gt;\boxed{\textbf{(C) }2}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.<br /> <br /> &lt;math&gt;y^2 + x^4 = 2x^2 y + 1&lt;/math&gt; can be written as &lt;math&gt;x^4-2x^2y+y^2=1&lt;/math&gt;. Recognizing that this is a binomial square, simplify this to &lt;math&gt;(x^2-y)^2=1&lt;/math&gt;. This gives us two equations:<br /> <br /> &lt;math&gt;x^2-y=1&lt;/math&gt; and &lt;math&gt;x^2-y=-1&lt;/math&gt;.<br /> <br /> One of these &lt;math&gt;y&lt;/math&gt;'s is &lt;math&gt;a&lt;/math&gt; and one is &lt;math&gt;b&lt;/math&gt;. Substituting &lt;math&gt;\sqrt{\pi}&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;a=\pi+1&lt;/math&gt; and &lt;math&gt;b=\pi-1&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;|a-b|=|(\pi+1)-(\pi-1)|=2&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;\boxed{\textbf{(C) }2}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> This solution is similar to Solution #1 but uses a different way to find &lt;math&gt;y&lt;/math&gt; at the end.<br /> <br /> Just like Solution #1, we arrive at the conclusion that &lt;math&gt;y^2 - 2\pi y + \pi^2 = 1&lt;/math&gt;.<br /> <br /> Simplifying we get:<br /> <br /> &lt;math&gt;y^2 - 2\pi y + \pi^2 -1 = 0&lt;/math&gt;<br /> <br /> We now can factor this quadratic. We must find two terms that multiply to &lt;math&gt;\pi^2 -1&lt;/math&gt; and add to &lt;math&gt;2\pi&lt;/math&gt;. <br /> <br /> These terms are &lt;math&gt;\pi+1&lt;/math&gt; and &lt;math&gt;\pi-1&lt;/math&gt;. <br /> <br /> Subtracting one from the other, we get &lt;math&gt;2&lt;/math&gt;.<br /> <br /> Thus, the answer is &lt;math&gt;\boxed{\textbf{(C) }2}&lt;/math&gt;<br /> <br /> -DuckDuckGooseGoose<br /> <br /> ==Video Solution==<br /> https://youtu.be/gKzliDi3zgk<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_12&diff=176524 2015 AMC 10A Problems/Problem 12 2022-08-01T19:25:26Z <p>Zhaom: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Points &lt;math&gt;( \sqrt{\pi} , a)&lt;/math&gt; and &lt;math&gt;( \sqrt{\pi} , b)&lt;/math&gt; are distinct points on the graph of &lt;math&gt;y^2 + x^4 = 2x^2 y + 1&lt;/math&gt;. What is &lt;math&gt;|a-b|&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since points on the graph make the equation true, substitute &lt;math&gt;\sqrt{\pi}&lt;/math&gt; in to the equation and then solve to find &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;.<br /> <br /> &lt;math&gt;y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y^2 + \pi^2 = 2\pi y + 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y^2 - 2\pi y + \pi^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(y-\pi)^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y-\pi = \pm 1&lt;/math&gt;<br /> <br /> &lt;math&gt;y = \pi + 1&lt;/math&gt; <br /> <br /> &lt;math&gt;y = \pi - 1&lt;/math&gt;<br /> <br /> There are only two solutions to the equation &lt;math&gt;(y-\pi)^2 = 1&lt;/math&gt;, so one of them is the value of &lt;math&gt;a&lt;/math&gt; and the other is &lt;math&gt;b&lt;/math&gt;. The order does not matter because of the absolute value sign.<br /> <br /> &lt;math&gt;| (\pi + 1) - (\pi - 1) | = 2&lt;/math&gt;<br /> <br /> The answer is &lt;math&gt;\boxed{\textbf{(C) }2}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.<br /> <br /> &lt;math&gt;y^2 + x^4 = 2x^2 y + 1&lt;/math&gt; can be written as &lt;math&gt;x^4-2x^2y+y^2=1&lt;/math&gt;. Recognizing that this is a binomial square, simplify this to &lt;math&gt;(x^2-y)^2=1&lt;/math&gt;. This gives us two equations:<br /> <br /> &lt;math&gt;x^2-y=1&lt;/math&gt; and &lt;math&gt;x^2-y=-1&lt;/math&gt;.<br /> <br /> One of these &lt;math&gt;y&lt;/math&gt;'s is &lt;math&gt;a&lt;/math&gt; and one is &lt;math&gt;b&lt;/math&gt;. Substituting &lt;math&gt;\sqrt{\pi}&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;a=\pi+1&lt;/math&gt; and &lt;math&gt;b=\pi-1&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;|a-b|=|(\pi+1)-(\pi-1)|=2&lt;/math&gt;.<br /> <br /> The answer is &lt;math&gt;\boxed{\textbf{(C) }2}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> This solution is similar to Solution #1 but uses a different way to find &lt;math&gt;y&lt;/math&gt; at the end.<br /> <br /> Just like Solution #1, we arrive at the conclusion that &lt;math&gt;y^2 - 2\pi y + \pi^2 = 1&lt;/math&gt;.<br /> <br /> Simplifying we get:<br /> <br /> &lt;math&gt;y^2 - 2\pi y + \pi^2 -1 = 0&lt;/math&gt;<br /> <br /> We now can factor this quadratic. We must find two terms that multiply to &lt;math&gt;\pi^2 -1&lt;/math&gt; and add to &lt;math&gt;-2\pi&lt;/math&gt;. <br /> <br /> These terms are &lt;math&gt;\pi+1&lt;/math&gt; and &lt;math&gt;\pi-1&lt;/math&gt;. <br /> <br /> Subtracting one from the other, we get &lt;math&gt;2&lt;/math&gt;.<br /> <br /> Thus, the answer is &lt;math&gt;\boxed{\textbf{(C) }2}&lt;/math&gt;<br /> <br /> -DuckDuckGooseGoose<br /> <br /> ==Video Solution==<br /> https://youtu.be/gKzliDi3zgk<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=11|num-a=13}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_11&diff=176523 2019 AIME II Problems/Problem 11 2022-08-01T19:22:24Z <p>Zhaom: /* Solution 6 (Inversion simplified) */</p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has side lengths &lt;math&gt;AB=7, BC=8, &lt;/math&gt; and &lt;math&gt;CA=9.&lt;/math&gt; Circle &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;B&lt;/math&gt; and is tangent to line &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Circle &lt;math&gt;\omega_2&lt;/math&gt; passes through &lt;math&gt;C&lt;/math&gt; and is tangent to line &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;A.&lt;/math&gt; Let &lt;math&gt;K&lt;/math&gt; be the intersection of circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; not equal to &lt;math&gt;A.&lt;/math&gt; Then &lt;math&gt;AK=\tfrac mn,&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 1==<br /> &lt;asy&gt;<br /> unitsize(20);<br /> pair B = (0,0);<br /> pair A = (2,sqrt(45));<br /> pair C = (8,0);<br /> draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));<br /> draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));<br /> draw(B--A--C--cycle);<br /> label(&quot;$A$&quot;,A,dir(105));<br /> label(&quot;$B$&quot;,B,dir(-135));<br /> label(&quot;$C$&quot;,C,dir(-75));<br /> dot((2.68,2.25));<br /> label(&quot;$K$&quot;,(2.68,2.25),dir(-150));<br /> label(&quot;$\omega_1$&quot;,(-6,1));<br /> label(&quot;$\omega_2$&quot;,(14,6));<br /> label(&quot;$7$&quot;,(A+B)/2,dir(140));<br /> label(&quot;$8$&quot;,(B+C)/2,dir(-90));<br /> label(&quot;$9$&quot;,(A+C)/2,dir(60));<br /> &lt;/asy&gt;<br /> -Diagram by Brendanb4321<br /> <br /> <br /> Note that from the tangency condition that the supplement of &lt;math&gt;\angle CAB&lt;/math&gt; with respects to lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are equal to &lt;math&gt;\angle AKB&lt;/math&gt; and &lt;math&gt;\angle AKC&lt;/math&gt;, respectively, so from tangent-chord, &lt;cmath&gt;\angle AKC=\angle AKB=180^{\circ}-\angle BAC&lt;/cmath&gt; Also note that &lt;math&gt;\angle ABK=\angle KAC&lt;/math&gt;, so &lt;math&gt;\triangle AKB\sim \triangle CKA&lt;/math&gt;. Using similarity ratios, we can easily find &lt;cmath&gt;AK^2=BK*KC&lt;/cmath&gt; However, since &lt;math&gt;AB=7&lt;/math&gt; and &lt;math&gt;CA=9&lt;/math&gt;, we can use similarity ratios to get &lt;cmath&gt;BK=\frac{7}{9}AK, CK=\frac{9}{7}AK&lt;/cmath&gt; <br /> <br /> *Now we use Law of Cosines on &lt;math&gt;\triangle AKB&lt;/math&gt;: From reverse Law of Cosines, &lt;math&gt;\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}&lt;/math&gt;. This gives us &lt;cmath&gt;AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;\implies \frac{196}{81}AK^2=49&lt;/cmath&gt; &lt;cmath&gt;AK=\frac{9}{2}&lt;/cmath&gt; so our answer is &lt;math&gt;9+2=\boxed{011}&lt;/math&gt;.<br /> -franchester<br /> *The motivation for using the Law of Cosines (&quot;LoC&quot;) is after finding the similar triangles it's hard to figure out what to do with &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CK&lt;/math&gt; yet we know &lt;math&gt;BC&lt;/math&gt; which somehow has to help us solve the problem--a common theme in solving geometry problems is figuring out how to use what you haven't used yet. We know all three sides of some triangle though, and we're dealing with angles (that's how we found similarity), so why not try the Law of Cosines? This is to help with motivation--the solution is franchester's and I learned about using LoC from reading his solution (I only solved half the problem and got stuck). To anyone in the future reading this, math is beautiful.<br /> -First<br /> <br /> 11111111:L)xiexie<br /> <br /> ==Solution 2 (Inversion)==<br /> Consider an inversion with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;r=AK&lt;/math&gt;. Then, we have &lt;math&gt;AB\cdot AB^*=AK^2&lt;/math&gt;, or &lt;math&gt;AB^*=\frac{AK^2}{7}&lt;/math&gt;. Similarly, &lt;math&gt;AC^*=\frac{AK^2}{9}&lt;/math&gt;. Notice that &lt;math&gt;AB^*KC^*&lt;/math&gt; is a parallelogram, since &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are tangent to &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Thus, &lt;math&gt;AC^*=B^*K&lt;/math&gt;. Now, we get that<br /> &lt;cmath&gt;\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}&lt;/cmath&gt;<br /> so by Law of Cosines on &lt;math&gt;\triangle AB^*K&lt;/math&gt; we have <br /> &lt;cmath&gt;(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{22AK^2}{63\cdot21}&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow AK=\frac{9}{2}&lt;/cmath&gt;<br /> Then, our answer is &lt;math&gt;9+2=\boxed{11}&lt;/math&gt;. <br /> -brianzjk<br /> <br /> == Solution 3 (Death By Trig Bash) ==<br /> <br /> 14. Let the centers of the circles be &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; where the &lt;math&gt;O_{1}&lt;/math&gt; has the side length &lt;math&gt;7&lt;/math&gt; contained in the circle. Now let &lt;math&gt;\angle BAC =x.&lt;/math&gt; This implies &lt;cmath&gt;\angle AO_{1}B = \angle AO_{2}C = 2x&lt;/cmath&gt; by the angle by by tangent. Then we also know that &lt;cmath&gt;\angle O_{1}AB = \angle O_{1}BA = \angle O_{2}AC = \angle O_{2}CA = 90^{\circ}-x&lt;/cmath&gt; Now we first find &lt;math&gt;\cos x.&lt;/math&gt; We use law of cosines on &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; to obtain &lt;cmath&gt;64 = 81 + 48 - 2 \cdot 9 \cdot 7 \cdot \cos{x}&lt;/cmath&gt; &lt;cmath&gt;\implies \cos{x} =\frac{11}{21}&lt;/cmath&gt; &lt;cmath&gt;\implies \sin{x} =\frac{8\sqrt{5}}{21}&lt;/cmath&gt; Then applying law of sines on &lt;math&gt;\bigtriangleup AO_{1}B&lt;/math&gt; we obtain &lt;cmath&gt;\frac{7}{\sin{2x}} =\frac{OB_{1}}{\sin{90^{\circ}-x}}&lt;/cmath&gt; &lt;cmath&gt;\implies\frac{7}{2\sin{x}\cos{x}} =\frac{OB_{1}}{\cos{x}}&lt;/cmath&gt; &lt;cmath&gt;\implies OB_{1} = O_{1}A=\frac{147}{16\sqrt{5}}&lt;/cmath&gt; Using similar logic we obtain &lt;math&gt;OA_{1} =\frac{189}{16\sqrt{5}}.&lt;/math&gt; <br /> <br /> Now we know that &lt;math&gt;\angle O_{1}AO_{2}=180^{\circ}-x.&lt;/math&gt; Thus using law of cosines on &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; yields &lt;cmath&gt;O_{1}O_{2} =\sqrt{\left(\frac{147}{16\sqrt{5}}\right)^2+\left(\frac{189}{16\sqrt{5}}\right)^2-2\:\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot -\frac{11}{21}}&lt;/cmath&gt; While this does look daunting we can write the above expression as &lt;cmath&gt;\sqrt{\left(\frac{189+147}{16\sqrt{5}}\right)^2 - 2\cdot \left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}&lt;/cmath&gt; Then factoring yields &lt;cmath&gt;\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{5}}&lt;/cmath&gt; The area &lt;cmath&gt;[O_{1}AO_{2}] =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot \sin(180^{\circ}-x) =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}&lt;/cmath&gt; Now &lt;math&gt;AK&lt;/math&gt; is twice the length of the altitude of &lt;math&gt;\bigtriangleup O_{1}AO_{2}&lt;/math&gt; so we let the altitude be &lt;math&gt;h&lt;/math&gt; and we have &lt;cmath&gt;\frac{1}{2} \cdot h \cdot\frac{147\sqrt{5}}{8\sqrt{5}} =\frac{1}{2} \cdot\frac{147}{16\sqrt{5}} \cdot\frac{189}{16\sqrt{5}} \cdot\frac{8\sqrt{5}}{21}&lt;/cmath&gt; &lt;cmath&gt;\implies h =\frac{9}{4}&lt;/cmath&gt; Thus our desired length is &lt;math&gt;\frac{9}{2} \implies n+n = \boxed{11}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Video)==<br /> <br /> Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI<br /> <br /> ==Solution 5 (Olympiad Geometry)==<br /> <br /> By the definition of &lt;math&gt;K&lt;/math&gt;, it is the spiral center mapping &lt;math&gt;BA\to AC&lt;/math&gt;, which means that it is the midpoint of the &lt;math&gt;A&lt;/math&gt;-symmedian chord. In particular, if &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;M'&lt;/math&gt; is the reflection of &lt;math&gt;A&lt;/math&gt; across &lt;math&gt;K&lt;/math&gt;, we have &lt;math&gt;\triangle ABM'\sim\triangle AMC&lt;/math&gt;. By Stewart's Theorem, it then follows that<br /> &lt;cmath&gt;AK = \frac{AM'}{2} = \frac{AC\cdot AB}{2AM} = \frac{7\cdot 9}{2\sqrt{\frac{9^2\cdot 4 + 7^2\cdot 4 - 4^2\cdot 8}{8}}} = \frac{7\cdot 9}{2\sqrt{49}} = \frac{9}{2}\implies m + n = \boxed{11}.&lt;/cmath&gt;<br /> <br /> ==Solution 6 (Inversion simplified)==<br /> [[File:2019 AIME II 11.png|500px|right]]<br /> The median of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;AM = \sqrt{\frac {AB^2 + AC^2 }{2} – \frac{BC^2}{4}} = 7.&lt;/math&gt;<br /> <br /> Consider an inversion with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AK&lt;/math&gt; (inversion with respect the red circle). <br /> Let &lt;math&gt;K, B',&lt;/math&gt; and &lt;math&gt;C'&lt;/math&gt; be inverse points for &lt;math&gt;K, B,&lt;/math&gt; and &lt;math&gt;C,&lt;/math&gt; respectively.<br /> <br /> Image of line &lt;math&gt;AB&lt;/math&gt; is line &lt;math&gt;AB, B'&lt;/math&gt; lies on this line. <br /> <br /> Image of &lt;math&gt;\omega_2&lt;/math&gt; is line &lt;math&gt;KC'||AB&lt;/math&gt; (circle &lt;math&gt;\omega_2&lt;/math&gt; passes through K, C and is tangent to the line &lt;math&gt;AB&lt;/math&gt; at point &lt;math&gt;A.&lt;/math&gt; Diagram shows circle and its image using same color).<br /> <br /> Similarly, &lt;math&gt;AC||B'K (B'K&lt;/math&gt; is the image of the circle &lt;math&gt;\omega_1&lt;/math&gt;).<br /> <br /> Therefore &lt;math&gt;AB'KC'&lt;/math&gt; is a parallelogram, &lt;math&gt;AF&lt;/math&gt; is median of &lt;math&gt;\triangle AB'C'&lt;/math&gt; and &lt;math&gt;AK = 2 AF.&lt;/math&gt;<br /> Then, we have &lt;math&gt;AB'=\frac{AK^2}{7}&lt;/math&gt;. &lt;math&gt;\triangle ABC \sim \triangle AC'B'&lt;/math&gt; with coefficient &lt;math&gt;k =\frac {AB'}{AC} = \frac{AK^2}{7\cdot 9}.&lt;/math&gt;<br /> <br /> So median &lt;cmath&gt;AF = k AM \implies \frac {AK}{2} = AM \cdot k = 7\cdot \frac{AK^2}{7\cdot 9} \implies AK = \frac{9}{2}.&lt;/cmath&gt;<br /> '''Shelomovskii, vvsss, www.deoma-cmd.ru'''<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=10|num-a=12}}<br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_4&diff=176520 2020 AIME I Problems/Problem 4 2022-08-01T18:02:01Z <p>Zhaom: /* Solution 3 */</p> <hr /> <div><br /> == Problem ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;N&lt;/math&gt; with the property that the last four digits of &lt;math&gt;N&lt;/math&gt; are &lt;math&gt;2020,&lt;/math&gt; and when the last four digits are removed, the result is a divisor of &lt;math&gt;N.&lt;/math&gt; For example, &lt;math&gt;42,020&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; because &lt;math&gt;4&lt;/math&gt; is a divisor of &lt;math&gt;42,020.&lt;/math&gt; Find the sum of all the digits of all the numbers in &lt;math&gt;S.&lt;/math&gt; For example, the number &lt;math&gt;42,020&lt;/math&gt; contributes &lt;math&gt;4+2+0+2+0=8&lt;/math&gt; to this total.<br /> <br /> == Solution 1 ==<br /> <br /> We note that any number in &lt;math&gt;S&lt;/math&gt; can be expressed as &lt;math&gt;a(10,000) + 2,020&lt;/math&gt; for some integer &lt;math&gt;a&lt;/math&gt;. The problem requires that &lt;math&gt;a&lt;/math&gt; divides this number, and since we know &lt;math&gt;a&lt;/math&gt; divides &lt;math&gt;a(10,000)&lt;/math&gt;, we need that &lt;math&gt;a&lt;/math&gt; divides 2020. Each number contributes the sum of the digits of &lt;math&gt;a&lt;/math&gt;, as well as &lt;math&gt;2 + 0 + 2 +0 = 4&lt;/math&gt;. Since &lt;math&gt;2020&lt;/math&gt; can be prime factorized as &lt;math&gt;2^2 \cdot 5 \cdot 101&lt;/math&gt;, it has &lt;math&gt;(2+1)(1+1)(1+1) = 12&lt;/math&gt; factors. So if we sum all the digits of all possible &lt;math&gt;a&lt;/math&gt; values, and add &lt;math&gt;4 \cdot 12 = 48&lt;/math&gt;, we obtain the answer.<br /> <br /> Now we list out all factors of &lt;math&gt;2,020&lt;/math&gt;, or all possible values of &lt;math&gt;a&lt;/math&gt;. &lt;math&gt;1,2,4,5,10,20,101,202,404,505,1010,2020&lt;/math&gt;. If we add up these digits, we get &lt;math&gt;45&lt;/math&gt;, for a final answer of &lt;math&gt;45+48=\boxed{093}&lt;/math&gt;.<br /> <br /> -molocyxu<br /> <br /> ==Solution 2 (Official MAA)==<br /> Suppose that &lt;math&gt;N&lt;/math&gt; has the required property. Then there are positive integers &lt;math&gt;k&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt; such that &lt;math&gt;N = 10^4m + 2020 = k\cdot m&lt;/math&gt;. Thus &lt;math&gt;(k - 10^4)m = 2020&lt;/math&gt;, which holds exactly when &lt;math&gt;m&lt;/math&gt; is a positive divisor of &lt;math&gt;2020.&lt;/math&gt; The number &lt;math&gt;2020 = 2^2\cdot 5\cdot 101&lt;/math&gt; has &lt;math&gt;12&lt;/math&gt; divisors: &lt;math&gt;1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010&lt;/math&gt;, and &lt;math&gt;2020.&lt;/math&gt; The requested sum is therefore the sum of the digits in these divisors plus &lt;math&gt;12&lt;/math&gt; times the sum of the digits in &lt;math&gt;2020,&lt;/math&gt; which is<br /> &lt;cmath&gt;(1+2+4+5+1+2+2+4+8+10+2+4)+12\cdot4 = 93.&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> Note that for all &lt;math&gt;N \in S&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt; can be written as &lt;math&gt;N=10000x+2020=20(500x+101)&lt;/math&gt; for some positive integer &lt;math&gt;x&lt;/math&gt;. Because &lt;math&gt;N&lt;/math&gt; must be divisible by &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;\frac{20(500x+101)}{x}&lt;/math&gt; is an integer. We now let &lt;math&gt;x=ab&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; is a divisor of &lt;math&gt;20&lt;/math&gt;. Then &lt;math&gt;\frac{20(500x+101)}{x}=(\frac{20}{a})( \frac{500x}{b}+\frac{101}{b})&lt;/math&gt;. We know &lt;math&gt;\frac{20}{a}&lt;/math&gt; and &lt;math&gt;\frac{500x}{b}&lt;/math&gt; are integers, so for &lt;math&gt;N&lt;/math&gt; to be an integer, &lt;math&gt;\frac{101}{b}&lt;/math&gt; must be an integer. For this to happen, &lt;math&gt;b&lt;/math&gt; must be a divisor of &lt;math&gt;101&lt;/math&gt;. &lt;math&gt;101&lt;/math&gt; is prime, so &lt;math&gt;b\in \left \{ 1, 101 \right \}&lt;/math&gt;. Because &lt;math&gt;a&lt;/math&gt; is a divisor of &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;a \in \left \{ 1,2,4,5,10,20\right\}&lt;/math&gt;. So &lt;math&gt;x \in \left\{1,2,4,5,10,20,101,202,404,505,1010,2020\right\}&lt;/math&gt;. Be know that all &lt;math&gt;N&lt;/math&gt; end in &lt;math&gt;2020&lt;/math&gt;, so the sum of the digits of each &lt;math&gt;N&lt;/math&gt; is the sum of the digits of each &lt;math&gt;x&lt;/math&gt; plus &lt;math&gt;2+0+2+0=4&lt;/math&gt;. Hence the sum of all of the digits of the numbers in &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;12 \cdot 4 +45=\boxed{093}&lt;/math&gt;.<br /> <br /> ==Video Solutions==<br /> *https://youtu.be/5b9Nw4bQt_k<br /> *https://youtu.be/djWzRC-jGYw<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_4&diff=176479 2022 AIME II Problems/Problem 4 2022-08-01T13:36:41Z <p>Zhaom: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> <br /> There is a positive real number &lt;math&gt;x&lt;/math&gt; not equal to either &lt;math&gt;\tfrac{1}{20}&lt;/math&gt; or &lt;math&gt;\tfrac{1}{2}&lt;/math&gt; such that&lt;cmath&gt;\log_{20x} (22x)=\log_{2x} (202x).&lt;/cmath&gt;The value &lt;math&gt;\log_{20x} (22x)&lt;/math&gt; can be written as &lt;math&gt;\log_{10} (\tfrac{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 /> ==Solution 1==<br /> Define &lt;math&gt;a&lt;/math&gt; to be &lt;math&gt;\log_{20x} (22x) = \log_{2x} (202x)&lt;/math&gt;, what we are looking for. Then, by the definition of the logarithm,<br /> &lt;cmath&gt;\begin{cases} <br /> (20x)^{a} &amp;= 22x \\<br /> (2x)^{a} &amp;= 202x. <br /> \end{cases}&lt;/cmath&gt;<br /> Dividing the first equation by the second equation gives us &lt;math&gt;10^a = \frac{11}{101}&lt;/math&gt;, so by the definition of logs, &lt;math&gt;a = \log_{10} \frac{11}{101}&lt;/math&gt;. This is what the problem asked for, so the fraction &lt;math&gt;\frac{11}{101}&lt;/math&gt; gives us &lt;math&gt;m+n = \boxed{112}&lt;/math&gt;.<br /> <br /> ~ihatemath123<br /> <br /> ==Solution 2==<br /> We could assume a variable &lt;math&gt;v&lt;/math&gt; which equals to both &lt;math&gt;\log_{20x} (22x)&lt;/math&gt; and &lt;math&gt;\log_{2x} (202x)&lt;/math&gt;.<br /> <br /> So that &lt;math&gt;(20x)^v=22x \textcircled{1}&lt;/math&gt;<br /> and &lt;math&gt;(2x)^v=202x \textcircled{2}&lt;/math&gt;<br /> <br /> Express &lt;math&gt;\textcircled{1}&lt;/math&gt; as: &lt;math&gt;(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot (10^v)=22x \textcircled{3}&lt;/math&gt;<br /> <br /> Substitute &lt;math&gt;\textcircled{2}&lt;/math&gt; to &lt;math&gt;\textcircled{3}&lt;/math&gt;: &lt;math&gt;202x \cdot (10^v)=22x&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;v=\log_{10} (\frac{22x}{202x})= \log_{10} (\frac{11}{101})&lt;/math&gt;, where &lt;math&gt;m=11&lt;/math&gt; and &lt;math&gt;n=101&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;m+n = \boxed{112}&lt;/math&gt;.<br /> <br /> ~DSAERF-CALMIT (https://binaryphi.site)<br /> <br /> ==Solution 3==<br /> <br /> We have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \log_{20x} (22x)<br /> &amp; = \frac{\log_k 22x}{\log_k 20x} \\<br /> &amp; = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> We have<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \log_{2x} (202x)<br /> &amp; = \frac{\log_k 202x}{\log_k 2x} \\<br /> &amp; = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;\log_{20x} (22x)=\log_{2x} (202x)&lt;/math&gt;, we get<br /> &lt;cmath&gt;<br /> $<br /> \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20}<br /> = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} .<br />$<br /> &lt;/cmath&gt;<br /> <br /> We denote this common value as &lt;math&gt;\lambda&lt;/math&gt;.<br /> <br /> By solving the equality &lt;math&gt;\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \lambda&lt;/math&gt;, we get &lt;math&gt;\log_k x = \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}&lt;/math&gt;.<br /> <br /> By solving the equality &lt;math&gt;\frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} = \lambda&lt;/math&gt;, we get &lt;math&gt;\log_k x = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1}&lt;/math&gt;.<br /> <br /> By equating these two equations, we get<br /> &lt;cmath&gt;<br /> $<br /> \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}<br /> = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1} .<br />$<br /> &lt;/cmath&gt;<br /> <br /> Therefore,<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> \log_{20x} (22x)<br /> &amp; = \lambda \\<br /> &amp; = \frac{\log_k 22 - \log_k 202}{\log_k 20 - \log_k 2} \\<br /> &amp; = \frac{\log_k \frac{11}{101}}{\log_k 10} \\<br /> &amp; = \log_{10} \frac{11}{101} .<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Therefore, the answer is &lt;math&gt;11 + 101 = \boxed{\textbf{112}}&lt;/math&gt;.<br /> <br /> ~Steven Chen (www.professorchenedu.com)<br /> <br /> ==Solution 4==<br /> By the change of base rule, we have &lt;math&gt;\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}&lt;/math&gt;, or &lt;math&gt;\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k&lt;/math&gt;. We also know that if &lt;math&gt;a/b=c/d&lt;/math&gt;, then this also equals &lt;math&gt;\frac{a-b}{c-d}&lt;/math&gt;. We use this identity and find that &lt;math&gt;k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{22}=\log\frac{11}{101}&lt;/math&gt;. The requested sum is &lt;math&gt;11+101=\boxed{112}.&lt;/math&gt;<br /> <br /> ~MathIsFun286<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=4qJyvyZN630<br /> <br /> ==Video Solution by Power of Logic==<br /> https://youtu.be/8eb0ycrVWIM<br /> <br /> ~Hayabusa1<br /> <br /> ==See Also==<br /> {{AIME box|year=2022|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_5&diff=176359 2011 AIME II Problems/Problem 5 2022-07-29T17:41:37Z <p>Zhaom: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of the first &lt;math&gt;2011&lt;/math&gt; terms of a [[geometric sequence]] is &lt;math&gt;200&lt;/math&gt;. The sum of the first &lt;math&gt;4022&lt;/math&gt; terms is &lt;math&gt;380&lt;/math&gt;. Find the sum of the first &lt;math&gt;6033&lt;/math&gt; terms.<br /> <br /> ==Solution==<br /> Since the sum of the first &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;200&lt;/math&gt;, and the sum of the fist &lt;math&gt;4022&lt;/math&gt; terms is &lt;math&gt;380&lt;/math&gt;, the sum of the second &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;180&lt;/math&gt;.<br /> This is decreasing from the first 2011, so the common ratio is less than one.<br /> <br /> Because it is a geometric sequence and the sum of the first 2011 terms is &lt;math&gt;200&lt;/math&gt;, second &lt;math&gt;2011&lt;/math&gt; is &lt;math&gt;180&lt;/math&gt;, the ratio of the second &lt;math&gt;2011&lt;/math&gt; terms to the first &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;\frac{9}{10}&lt;/math&gt;. Following the same pattern, the sum of the third &lt;math&gt;2011&lt;/math&gt; terms is &lt;math&gt;\frac{9}{10}*180 = 162&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;200+180+162=542&lt;/math&gt;, so the sum of the first &lt;math&gt;6033&lt;/math&gt; terms is &lt;math&gt;\boxed{542}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Solution by e_power_pi_times_i<br /> <br /> The sum of the first &lt;math&gt;2011&lt;/math&gt; terms can be written as &lt;math&gt;\dfrac{a_1(1-k^{2011})}{1-k}&lt;/math&gt;, and the first &lt;math&gt;4022&lt;/math&gt; terms can be written as &lt;math&gt;\dfrac{a_1(1-k^{4022})}{1-k}&lt;/math&gt;. Dividing these equations, we get &lt;math&gt;\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}&lt;/math&gt;. Noticing that &lt;math&gt;k^{4022}&lt;/math&gt; is just the square of &lt;math&gt;k^{2011}&lt;/math&gt;, we substitute &lt;math&gt;x = k^{2011}&lt;/math&gt;, so &lt;math&gt;\dfrac{1}{x+1} = \dfrac{10}{19}&lt;/math&gt;. That means that &lt;math&gt;k^{2011} = \dfrac{9}{10}&lt;/math&gt;. Since the sum of the first &lt;math&gt;6033&lt;/math&gt; terms can be written as &lt;math&gt;\dfrac{a_1(1-k^{6033})}{1-k}&lt;/math&gt;, dividing gives &lt;math&gt;\dfrac{1-k^{2011}}{1-k^{6033}}&lt;/math&gt;. Since &lt;math&gt;k^{6033} = \dfrac{729}{1000}&lt;/math&gt;, plugging all the values in gives &lt;math&gt;\boxed{542}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> The sum of the first 2011 terms of the sequence is expressible as &lt;math&gt;a_1 + a_1r + a_1r^2 + a_1r^3&lt;/math&gt; .... until &lt;math&gt;a_1r^{2010}&lt;/math&gt;. The sum of the 2011 terms following the first 2011 is expressible as &lt;math&gt;a_1r^{2011} + a_1r^{2012} + a_1r^{2013}&lt;/math&gt; .... until &lt;math&gt;a_1r^{4021}&lt;/math&gt;. Notice that the latter sum of terms can be expressed as &lt;math&gt;(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})&lt;/math&gt;. We also know that the latter sum of terms can be obtained by subtracting 200 from 380, which then means that &lt;math&gt;r^{2011} = 9/10&lt;/math&gt;. The terms from 4023 to 6033 can be expressed as &lt;math&gt;(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})&lt;/math&gt;, which is equivalent to &lt;math&gt;((9/10)^2)(200) = 162&lt;/math&gt;. Adding 380 and 162 gives the answer of &lt;math&gt;\boxed{542}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=rpYphKOIKRs&amp;t=186s<br /> ~anellipticcurveoverq<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_5&diff=176356 2019 AIME II Problems/Problem 5 2022-07-29T15:40:05Z <p>Zhaom: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Four ambassadors and one advisor for each of them are to be seated at a round table with &lt;math&gt;12&lt;/math&gt; chairs numbered in order &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt;. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are &lt;math&gt;N&lt;/math&gt; ways for the &lt;math&gt;8&lt;/math&gt; people to be seated at the table under these conditions. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> ==Solution 1==<br /> Let us first consider the &lt;math&gt;4&lt;/math&gt; ambassadors and the &lt;math&gt;6&lt;/math&gt; even-numbered chairs. If we consider only their relative positions, they can sit in one of &lt;math&gt;3&lt;/math&gt; distinct ways: Such that the &lt;math&gt;2&lt;/math&gt; empty even-numbered chairs are adjacent, such that they are separated by one occupied even-numbered chair, and such that they are opposite each other. For each way, there are &lt;math&gt;4!=24&lt;/math&gt; ways to assign the &lt;math&gt;4&lt;/math&gt; ambassadors to the &lt;math&gt;4&lt;/math&gt; selected seats.<br /> <br /> In the first way, there are &lt;math&gt;6&lt;/math&gt; distinct orientations. The &lt;math&gt;4&lt;/math&gt; advisors can be placed in any of the &lt;math&gt;5&lt;/math&gt; odd-numbered chairs adjacent to the ambassadors, and for each placement, there is exactly one way to assign them to the ambassadors. This means that there are &lt;math&gt;24\cdot6\cdot\binom{5}{4}=720&lt;/math&gt; total seating arrangements for this way.<br /> <br /> In the second way, there are &lt;math&gt;6&lt;/math&gt; distinct orientations. &lt;math&gt;3&lt;/math&gt; advisors can be placed in any of the &lt;math&gt;4&lt;/math&gt; chairs adjacent to the &quot;chunk&quot; of &lt;math&gt;3&lt;/math&gt; ambassadors, and &lt;math&gt;1&lt;/math&gt; advisor can be placed in either of the &lt;math&gt;2&lt;/math&gt; chairs adjacent to the &quot;lonely&quot; ambassador. Once again, for each placement, there is exactly one way to assign the advisors to the ambassadors. This means that there are &lt;math&gt;24\cdot6\cdot\binom{4}{3}\cdot\binom{2}{1}=1152&lt;/math&gt; total seating arrangements for this way.<br /> <br /> In the third way, there are &lt;math&gt;3&lt;/math&gt; distinct orientations. For both &quot;chunks&quot; of &lt;math&gt;2&lt;/math&gt; ambassadors, &lt;math&gt;2&lt;/math&gt; advisors can be placed in any of the &lt;math&gt;3&lt;/math&gt; chairs adjacent to them, and once again, there is exactly one way to assign them for each placement. This means that there are &lt;math&gt;24\cdot3\cdot\binom{3}{2}\cdot\binom{3}{2}=648&lt;/math&gt; total seating arrangements for this way.<br /> <br /> Totalling up the arrangements, there are &lt;math&gt;720+1152+648=2520&lt;/math&gt; total ways to seat the people, and the remainder when &lt;math&gt;2520&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;\boxed{520}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 2==<br /> [[File:Ambassador_Table.png|200px]]<br /> <br /> In the diagram, the seats are numbered 1...12. Rather than picking seats for each person, however, each ambassador/assistant team picks a gap between the seats (A...L) and the ambassador sits in the even seat while the assistant sits in the odd seat. For example, if team 1 picks gap C then Ambassador 1 will sit in seat 2 while assistant 1 will sit in seat 3. No two teams can pick adjacent gaps. For example, if team 1 chooses gap C then team 2 cannot pick gaps B or D. In the diagram, the teams have picked gaps C, F, H and J. Note that the gap-gaps - distances between the chosen gaps - (in the diagram, 2, 1, 1, 4) must sum to 8. So, to get the number of seatings, we:<br /> #Choose a gap for team 1 (&lt;math&gt;12&lt;/math&gt; options)<br /> #Choose 3 other gaps around the table with positive gap-gaps. The number of ways to do this is the number of ways to partition 8 with 4 positive integers. This is the same as partitioning 4 with 4 non-negative integers, and using stars-and-bars, this is &lt;math&gt;\dbinom{4+4-1}{4-1}=\dbinom{7}{3}=35&lt;/math&gt;<br /> #Place the other 3 teams in the chosen gaps (&lt;math&gt;6&lt;/math&gt; permutations)<br /> So the total is &lt;math&gt;12\cdot35\cdot6=2520&lt;/math&gt;<br /> And the remainder is &lt;math&gt;\boxed{520}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> There are &lt;math&gt;360\cdot16&lt;/math&gt; total ways to let everyone sit. However this may lead to advisors sitting in the same chair, leading to awkward situations. So we find how many ways this happens. There are 6 ways to choose which advisors end up sitting together, times 12 ways to find neighboring even seats and sitting down, 12 ways for the rest of the ambassador to sit, and 4 ways for their advisors to sit to get 3456 ways for this to happen. However we overcounted the case when two pairs of advisors run out of room to sit, where there are &lt;math&gt;6\cdot9\cdot4=216&lt;/math&gt; ways to happen. So the answer is &lt;math&gt;5760 - 3456 + 216 = 2520&lt;/math&gt;, which has a remainder of &lt;math&gt;\boxed{520}&lt;/math&gt; when divided by 1000.<br /> <br /> <br /> ~minor &lt;math&gt;LaTeX&lt;/math&gt; fixes by virjoy2001<br /> <br /> ==Solution 4==<br /> We see that for every 2 adjacent spots on the table, there is exactly one way for an ambassador and his or her partner to sit. There are 2 cases:<br /> #There is an ambassador at the 12th chair and his or her partner at the 1st chair<br /> #There is no pair that has their chairs numbered as 12 and 1<br /> For the first case, there are &lt;math&gt;\frac{7!}{3!4!} = 35&lt;/math&gt; ways to arrange the pairs.<br /> <br /> For the second case, there are &lt;math&gt;\frac{8!}{4!4!} = 70&lt;/math&gt; ways to arrange the pairs.<br /> <br /> Finally, we have to assign ambassadors and their advisors to the pairs, which has &lt;math&gt;4!=24&lt;/math&gt; ways, so &lt;math&gt;N = (35+70)\cdot 24 = 2520&lt;/math&gt;, or &lt;math&gt;\boxed{520}&lt;/math&gt;<br /> ~PenguinMoosey<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=2017_JBMO_Problems/Problem_2&diff=176354 2017 JBMO Problems/Problem 2 2022-07-29T14:05:44Z <p>Zhaom: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;x,y,z&lt;/math&gt; be positive integers such that &lt;math&gt;x\neq y\neq z \neq x&lt;/math&gt; .Prove that &lt;cmath&gt;(x+y+z)(xy+yz+zx-2)\geq 9xyz.&lt;/cmath&gt;<br /> When does the equality hold?<br /> <br /> == Solution ==<br /> Since the equation is symmetric and &lt;math&gt;x,y,z&lt;/math&gt; are distinct integers WLOG we can assume that &lt;math&gt;x\geq y+1\geq z+2&lt;/math&gt;. <br /> &lt;cmath&gt;\begin{align*}<br /> x+y+z\geq 3(z+1)\\<br /> xy+yz+xz-2 = y(x+z)+xy-2 \geq (z+1)(2z+2)+z(z+2)-2 \\<br /> xy+yz+xz-2 \geq 3z(z+2)<br /> \end{align*}&lt;/cmath&gt;<br /> Hence &lt;cmath&gt;(x+y+z)(xy+yz+xz-2)\geq 9(z)(z+1)(z+2)&lt;/cmath&gt;<br /> <br /> == See also ==<br /> <br /> {{JBMO box|year=2017|num-b=1|num-a=3|five=}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Zhaom&diff=156688 User:Zhaom 2021-06-23T19:06:09Z <p>Zhaom: Blanked the page</p> <hr /> <div></div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Zhaom&diff=140692 User:Zhaom 2020-12-27T01:41:18Z <p>Zhaom: Blanked the page</p> <hr /> <div></div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=Chicken_McNugget_Theorem&diff=132948 Chicken McNugget Theorem 2020-09-01T18:56:42Z <p>Zhaom: /* Simple */</p> <hr /> <div>The '''Chicken McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s &lt;math&gt;m,n&lt;/math&gt;, the greatest integer that cannot be written in the form &lt;math&gt;am + bn&lt;/math&gt; for [[nonnegative]] integers &lt;math&gt;a, b&lt;/math&gt; is &lt;math&gt;mn-m-n&lt;/math&gt;.<br /> <br /> A consequence of the theorem is that there are exactly &lt;math&gt;\frac{(m - 1)(n - 1)}{2}&lt;/math&gt; positive integers which cannot be expressed in the form &lt;math&gt;am + bn&lt;/math&gt;. The proof is based on the fact that in each pair of the form &lt;math&gt;(k, (m - 1)(n - 1) - k+1)&lt;/math&gt;, exactly one element is expressible.<br /> <br /> == Origins ==<br /> There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins.<br /> <br /> <br /> <br /> <br /> <br /> ==Proof Without Words==<br /> &lt;math&gt;\begin{array}{ccccccc}<br /> 0\mod{m}&amp;1\mod{m}&amp;2\mod{m}&amp;...&amp;...&amp;...&amp;(m-1)\mod{m}\\<br /> \hline<br /> \cancel{0n}&amp;1&amp;2&amp;&amp;...&amp;&amp;m-1\\<br /> \cancel{0n+m}&amp;...&amp;&amp;\vdots&amp;&amp;...&amp;\\<br /> \cancel{0n+2m}&amp;...&amp;&amp;\cancel{1n}&amp;&amp;...&amp;\\<br /> \cancel{0n+3m}&amp;&amp;&amp;\cancel{1n+m}&amp;&amp;\vdots&amp;\\<br /> \cancel{0n+4m}&amp;&amp;&amp;\cancel{1n+2m}&amp;&amp;\cancel{2n}&amp;\\<br /> \cancel{0n+5m}&amp;&amp;&amp;\cancel{1n+3m}&amp;&amp;\cancel{2n+m}&amp;\\<br /> \vdots&amp;&amp;&amp;\vdots&amp;&amp;\vdots&amp;\\<br /> \cancel{\qquad}&amp;\cancel{\qquad}&amp;\cancel{ \qquad}&amp;\cancel{ \qquad}&amp;\mathbf{(m-1)n-m}&amp;\cancel{\qquad }&amp;\cancel{\qquad }\\<br /> \cancel{\qquad}&amp;\cancel{\qquad}&amp;\cancel{ \qquad}&amp;\cancel{ \qquad}&amp;\cancel{(m-1)n}&amp;\cancel{\qquad }&amp;\cancel{\qquad }<br /> \end{array}&lt;/math&gt;<br /> <br /> ==Proof 1==<br /> &lt;b&gt;Definition&lt;/b&gt;. An integer &lt;math&gt;N \in \mathbb{Z}&lt;/math&gt; will be called &lt;i&gt;purchasable&lt;/i&gt; if there exist nonnegative integers &lt;math&gt;a,b&lt;/math&gt; such that &lt;math&gt;am+bn = N&lt;/math&gt;.<br /> <br /> We would like to prove that &lt;math&gt;mn-m-n&lt;/math&gt; is the largest non-purchasable integer. We are required to show that (1) &lt;math&gt;mn-m-n&lt;/math&gt; is non-purchasable, and (2) every &lt;math&gt;N &gt; mn-m-n&lt;/math&gt; is purchasable. <br /> Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. Let &lt;math&gt;A_{N} \subset \mathbb{Z} \times \mathbb{Z}&lt;/math&gt; be the set of solutions &lt;math&gt;(x,y)&lt;/math&gt; to &lt;math&gt;xm+yn = N&lt;/math&gt;. Then &lt;math&gt;A_{N} = \{(x+kn,y-km) \;:\; k \in \mathbb{Z}\}&lt;/math&gt; for any &lt;math&gt;(x,y) \in A_{N}&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: By [[Bezout's Lemma]], there exist integers &lt;math&gt;x',y'&lt;/math&gt; such that &lt;math&gt;x'm+y'n = 1&lt;/math&gt;. Then &lt;math&gt;(Nx')m+(Ny')n = N&lt;/math&gt;. Hence &lt;math&gt;A_{N}&lt;/math&gt; is nonempty. It is easy to check that &lt;math&gt;(Nx'+kn,Ny'-km) \in A_{N}&lt;/math&gt; for all &lt;math&gt;k \in \mathbb{Z}&lt;/math&gt;. We now prove that there are no others. Suppose &lt;math&gt;(x_{1},y_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2})&lt;/math&gt; are solutions to &lt;math&gt;xm+yn=N&lt;/math&gt;. Then &lt;math&gt;x_{1}m+y_{1}n = x_{2}m+y_{2}n&lt;/math&gt; implies &lt;math&gt;m(x_{1}-x_{2}) = n(y_{2}-y_{1})&lt;/math&gt;. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime and &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;n(y_{2}-y_{1})&lt;/math&gt;, &lt;math&gt;m&lt;/math&gt; divides &lt;math&gt;y_{2}-y_{1}&lt;/math&gt; and &lt;math&gt;y_{2} \equiv y_{1} \pmod{m}&lt;/math&gt;. Similarly &lt;math&gt;x_{2} \equiv x_{1} \pmod{n}&lt;/math&gt;. Let &lt;math&gt;k_{1},k_{2}&lt;/math&gt; be integers such that &lt;math&gt;x_{2}-x_{1} = k_{1}n&lt;/math&gt; and &lt;math&gt;y_{2}-y_{1} = k_{2}m&lt;/math&gt;. Then &lt;math&gt;m(-k_{1}n) = n(k_{2}m)&lt;/math&gt; implies &lt;math&gt;k_{1} = -k_{2}.&lt;/math&gt; We have the desired result. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. For any integer &lt;math&gt;N&lt;/math&gt;, there exists unique &lt;math&gt;(a_{N},b_{N}) \in \mathbb{Z} \times \{0,1,\ldots,m-1\}&lt;/math&gt; such that &lt;math&gt;a_{N}m + b_{N}n = N&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: By the division algorithm, there exists one and only one &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;0 \le y-km \le m-1&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;. &lt;math&gt;N&lt;/math&gt; is purchasable if and only if &lt;math&gt;a_{N} \ge 0&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: If &lt;math&gt;a_{N} \ge 0&lt;/math&gt;, then we may simply pick &lt;math&gt;(a,b) = (a_{N},b_{N})&lt;/math&gt; so &lt;math&gt;N&lt;/math&gt; is purchasable. If &lt;math&gt;a_{N} &lt; 0&lt;/math&gt;, then &lt;math&gt;a_{N}+kn &lt; 0&lt;/math&gt; if &lt;math&gt;k \le 0&lt;/math&gt; and &lt;math&gt;b_{N}-km &lt; 0&lt;/math&gt; if &lt;math&gt;k &gt; 0&lt;/math&gt;, hence at least one coordinate of &lt;math&gt;(a_{N}+kn,b_{N}-km)&lt;/math&gt; is negative for all &lt;math&gt;k \in \mathbb{Z}&lt;/math&gt;. Thus &lt;math&gt;N&lt;/math&gt; is not purchasable. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Thus the set of non-purchasable integers is &lt;math&gt;\{xm+yn \;:\; x&lt;0,0 \le y \le m-1\}&lt;/math&gt;. We would like to find the maximum of this set. <br /> Since both &lt;math&gt;m,n&lt;/math&gt; are positive, the maximum is achieved when &lt;math&gt;x = -1&lt;/math&gt; and &lt;math&gt;y = m-1&lt;/math&gt; so that &lt;math&gt;xm+yn = (-1)m+(m-1)n = mn-m-n&lt;/math&gt;.<br /> <br /> ==Proof 2==<br /> We start with this statement taken from [[Fermat%27s_Little_Theorem#Proof_2_.28Inverses.29|Proof 2 of Fermat's Little Theorem]]:<br /> <br /> &quot;Let &lt;math&gt;S = \{1,2,3,\cdots, p-1\}&lt;/math&gt;. Then, we claim that the set &lt;math&gt;a \cdot S&lt;/math&gt;, consisting of the product of the elements of &lt;math&gt;S&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;, taken modulo &lt;math&gt;p&lt;/math&gt;, is simply a permutation of &lt;math&gt;S&lt;/math&gt;. In other words, <br /> <br /> &lt;center&gt;&lt;cmath&gt;S \equiv \{1a, 2a, \cdots, (p-1)a\} \pmod{p}.&lt;/cmath&gt;&lt;/center&gt;&lt;br&gt;<br /> <br /> Clearly none of the &lt;math&gt;ia&lt;/math&gt; for &lt;math&gt;1 \le i \le p-1&lt;/math&gt; are divisible by &lt;math&gt;p&lt;/math&gt;, so it suffices to show that all of the elements in &lt;math&gt;a \cdot S&lt;/math&gt; are distinct. Suppose that &lt;math&gt;ai \equiv aj \pmod{p}&lt;/math&gt; for &lt;math&gt;i \neq j&lt;/math&gt;. Since &lt;math&gt;\text{gcd}\, (a,p) = 1&lt;/math&gt;, by the cancellation rule, that reduces to &lt;math&gt;i \equiv j \pmod{p}&lt;/math&gt;, which is a contradiction.&quot;<br /> <br /> Because &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime, we know that multiplying the residues of &lt;math&gt;m&lt;/math&gt; by &lt;math&gt;n&lt;/math&gt; simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form &lt;math&gt;am+bn&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is the original residue. We now prove the following lemma.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt;: For any nonnegative integer &lt;math&gt;c &lt; m&lt;/math&gt;, &lt;math&gt;cn&lt;/math&gt; is the least purchasable number &lt;math&gt;\equiv cn \bmod m&lt;/math&gt;.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: Any number that is less than &lt;math&gt;cn&lt;/math&gt; and congruent to it &lt;math&gt;\bmod m&lt;/math&gt; can be represented in the form &lt;math&gt;cn-dm&lt;/math&gt;, where &lt;math&gt;d&lt;/math&gt; is a positive integer. If this is purchasable, we can say &lt;math&gt;cn-dm=am+bn&lt;/math&gt; for some nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. This can be rearranged into &lt;math&gt;(a+d)m=(c-b)n&lt;/math&gt;, which implies that &lt;math&gt;(a+d)&lt;/math&gt; is a multiple of &lt;math&gt;n&lt;/math&gt; (since &lt;math&gt;\gcd(m, n)=1&lt;/math&gt;). We can say that &lt;math&gt;(a+d)=gn&lt;/math&gt; for some positive integer &lt;math&gt;g&lt;/math&gt;, and substitute to get &lt;math&gt;gmn=(c-b)n&lt;/math&gt;. Because &lt;math&gt;c &lt; m&lt;/math&gt;, &lt;math&gt;(c-b)n &lt; mn&lt;/math&gt;, and &lt;math&gt;gmn &lt; mn&lt;/math&gt;. We divide by &lt;math&gt;mn&lt;/math&gt; to get &lt;math&gt;g&lt;1&lt;/math&gt;. However, we defined &lt;math&gt;g&lt;/math&gt; to be a positive integer, and all positive integers are greater than or equal to &lt;math&gt;1&lt;/math&gt;. Therefore, we have a contradiction, and &lt;math&gt;cn&lt;/math&gt; is the least purchasable number congruent to &lt;math&gt;cn \bmod m&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> This means that because &lt;math&gt;cn&lt;/math&gt; is purchasable, every number that is greater than &lt;math&gt;cn&lt;/math&gt; and congruent to it &lt;math&gt;\bmod m&lt;/math&gt; is also purchasable (because these numbers are in the form &lt;math&gt;am+bn&lt;/math&gt; where &lt;math&gt;b=c&lt;/math&gt;). Another result of this Lemma is that &lt;math&gt;cn-m&lt;/math&gt; is the greatest number &lt;math&gt;\equiv cn \bmod m&lt;/math&gt; that is not purchasable. &lt;math&gt;c \leq m-1&lt;/math&gt;, so &lt;math&gt;cn-m \leq (m-1)n-m=mn-m-n&lt;/math&gt;, which shows that &lt;math&gt;mn-m-n&lt;/math&gt; is the greatest number in the form &lt;math&gt;cn-m&lt;/math&gt;. Any number greater than this and congruent to some &lt;math&gt;cn \bmod m&lt;/math&gt; is purchasable, because that number is greater than &lt;math&gt;cn&lt;/math&gt;. All numbers are congruent to some &lt;math&gt;cn&lt;/math&gt;, and thus all numbers greater than &lt;math&gt;mn-m-n&lt;/math&gt; are purchasable.<br /> <br /> Putting it all together, we can say that for any coprime &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;mn-m-n&lt;/math&gt; is the greatest number not representable in the form &lt;math&gt;am + bn&lt;/math&gt; for nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Corollary==<br /> This corollary is based off of Proof 2, so it is necessary to read that proof before this corollary. We prove the following lemma.<br /> <br /> &lt;b&gt;Lemma&lt;/b&gt; For any integer &lt;math&gt;k&lt;/math&gt;, exactly one of the integers &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;mn-m-n-k&lt;/math&gt; is not purchasable.<br /> <br /> &lt;i&gt;Proof&lt;/i&gt;: Because every number is congruent to some residue of &lt;math&gt;m&lt;/math&gt; permuted by &lt;math&gt;n&lt;/math&gt;, we can set &lt;math&gt;k \equiv cn \bmod m&lt;/math&gt; for some &lt;math&gt;c&lt;/math&gt;. We can break this into two cases.<br /> <br /> &lt;i&gt;Case 1&lt;/i&gt;: &lt;math&gt;k \leq cn-m&lt;/math&gt;. This implies that &lt;math&gt;k&lt;/math&gt; is not purchasable, and that &lt;math&gt;mn-m-n-k \geq mn-m-n-(cn-m) = n(m-1-c)&lt;/math&gt;. &lt;math&gt;n(m-1-c)&lt;/math&gt; is a permuted residue, and a result of the lemma in Proof 2 was that a permuted residue is the least number congruent to itself &lt;math&gt;\bmod m&lt;/math&gt; that is purchasable. Therefore, &lt;math&gt;mn-m-n-k \equiv n(m-1-c) \bmod m&lt;/math&gt; and &lt;math&gt;mn-m-n-k \geq n(m-1-c)&lt;/math&gt;, so &lt;math&gt;mn-m-n-k&lt;/math&gt; is purchasable.<br /> <br /> &lt;i&gt;Case 2&lt;/i&gt;: &lt;math&gt;k &gt; cn-m&lt;/math&gt;. This implies that &lt;math&gt;k&lt;/math&gt; is purchasable, and that &lt;math&gt;mn-m-n-k &lt; mn-m-n-(cn-m) = n(m-1-c)&lt;/math&gt;. Again, because &lt;math&gt;n(m-1-c)&lt;/math&gt; is the least number congruent to itself &lt;math&gt;\bmod m&lt;/math&gt; that is purchasable, and because &lt;math&gt;mn-m-n-k \equiv n(m-1-c) \bmod m&lt;/math&gt; and &lt;math&gt;mn-m-n-k &lt; n(m-1-c)&lt;/math&gt;, &lt;math&gt;mn-m-n-k&lt;/math&gt; is not purchasable.<br /> <br /> We now limit the values of &lt;math&gt;k&lt;/math&gt; to all integers &lt;math&gt;0 \leq k \leq \frac{mn-m-n}{2}&lt;/math&gt;, which limits the values of &lt;math&gt;mn-m-n-k&lt;/math&gt; to &lt;math&gt;mn-m-n \geq mn-m-n-k \geq \frac{mn-m-n}{2}&lt;/math&gt;. Because &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are coprime, only one of them can be a multiple of &lt;math&gt;2&lt;/math&gt;. Therefore, &lt;math&gt;mn-m-n \equiv (0)(1)-0-1 \equiv -1 \equiv 1 \bmod 2&lt;/math&gt;, showing that &lt;math&gt;\frac{mn-m-n}{2}&lt;/math&gt; is not an integer and that &lt;math&gt;\frac{mn-m-n-1}{2}&lt;/math&gt; and &lt;math&gt;\frac{mn-m-n+1}{2}&lt;/math&gt; are integers. We can now set limits that are equivalent to the previous on the values of &lt;math&gt;k&lt;/math&gt; and &lt;math&gt;mn-m-n-k&lt;/math&gt; so that they cover all integers form &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt; without overlap: &lt;math&gt;0 \leq k \leq \frac{mn-m-n-1}{2}&lt;/math&gt; and &lt;math&gt;\frac{mn-m-n+1}{2} \leq mn-m-n-k \leq mn-m-n&lt;/math&gt;. There are &lt;math&gt;\frac{mn-m-n-1}{2}+1=\frac{(m-1)(n-1)}{2}&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt;, and each is paired with a value of &lt;math&gt;mn-m-n-k&lt;/math&gt;, so we can make &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; different ordered pairs of the form &lt;math&gt;(k, mn-m-n-k)&lt;/math&gt;. The coordinates of these ordered pairs cover all integers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt; inclusive, and each contains exactly one not-purchasable integer, so that means that there are &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; different not-purchasable integers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;mn-m-n&lt;/math&gt;. All integers greater than &lt;math&gt;mn-m-n&lt;/math&gt; are purchasable, so that means there are a total of &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; integers &lt;math&gt;\geq 0&lt;/math&gt; that are not purchasable.<br /> <br /> In other words, for every pair of coprime integers &lt;math&gt;m, n&lt;/math&gt;, there are exactly &lt;math&gt;\frac{(m-1)(n-1)}{2}&lt;/math&gt; nonnegative integers that cannot be represented in the form &lt;math&gt;am + bn&lt;/math&gt; for nonnegative integers &lt;math&gt;a, b&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Generalization==<br /> If &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are not relatively prime, then we can simply rearrange &lt;math&gt;am+bn&lt;/math&gt; into the form<br /> &lt;cmath&gt;\gcd(m,n) \left( a\frac{m}{\gcd(m,n)}+b\frac{n}{\gcd(m,n)} \right)&lt;/cmath&gt;<br /> &lt;math&gt;\frac{m}{\gcd(m,n)}&lt;/math&gt; and &lt;math&gt;\frac{n}{\gcd(m,n)}&lt;/math&gt; are relatively prime, so we apply Chicken McNugget to find a bound<br /> &lt;cmath&gt;\frac{mn}{\gcd(m,n)^{2}}-\frac{m}{\gcd(m,n)}-\frac{n}{\gcd(m,n)}&lt;/cmath&gt;<br /> We can simply multiply &lt;math&gt;\gcd(m,n)&lt;/math&gt; back into the bound to get<br /> &lt;cmath&gt;\frac{mn}{\gcd(m,n)}-m-n=\textrm{lcm}(m, n)-m-n&lt;/cmath&gt;<br /> Therefore, all multiples of &lt;math&gt;\gcd(m, n)&lt;/math&gt; greater than &lt;math&gt;\textrm{lcm}(m, n)-m-n&lt;/math&gt; are representable in the form &lt;math&gt;am+bn&lt;/math&gt; for some positive integers &lt;math&gt;a, b&lt;/math&gt;.<br /> <br /> =Problems=<br /> <br /> ===Simple===<br /> *Marcy buys paint jars in containers of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;. What's the largest number of paint jars that Marcy can't obtain? <br /> <br /> Answer: &lt;math&gt;5&lt;/math&gt; containers<br /> <br /> *Bay Area Rapid food sells chicken nuggets. You can buy packages of &lt;math&gt;11&lt;/math&gt; or &lt;math&gt;7&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; such that there is no way to buy exactly &lt;math&gt;n&lt;/math&gt; nuggets? Can you Generalize ?(ACOPS) <br /> <br /> Answer: &lt;math&gt;n=59&lt;/math&gt; <br /> <br /> *If a game of American Football has only scores of field goals (&lt;math&gt;3&lt;/math&gt; points) and touchdowns with the extra point (&lt;math&gt;7&lt;/math&gt; points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)?<br /> <br /> Answer: &lt;math&gt;11&lt;/math&gt; points<br /> <br /> *The town of Hamlet has &lt;math&gt;3&lt;/math&gt; people for each horse, &lt;math&gt;4&lt;/math&gt; sheep for each cow, and &lt;math&gt;3&lt;/math&gt; ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?<br /> <br /> &lt;math&gt;\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66&lt;/math&gt; [[2015 AMC 10B Problems/Problem 15|AMC 10B 2015 Problem 15]]<br /> <br /> Answer: &lt;math&gt;47\qquad\textbf{(B) }&lt;/math&gt;<br /> <br /> ===Intermediate===<br /> *Ninety-four bricks, each measuring &lt;math&gt;4''\times10''\times19'',&lt;/math&gt; are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes &lt;math&gt;4''\,&lt;/math&gt; or &lt;math&gt;10''\,&lt;/math&gt; or &lt;math&gt;19''\,&lt;/math&gt; to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? [[1994 AIME Problems/Problem 11|AIME]]<br /> <br /> *Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that, given an unlimited supply of stamps of denominations &lt;math&gt;5,n,&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; cents, &lt;math&gt;91&lt;/math&gt; cents is the greatest postage that cannot be formed. [[2019 AIME II Problems/Problem 14|AIME II 2019 Problem 14]]<br /> <br /> ===Olympiad===<br /> *On the real number line, paint red all points that correspond to integers of the form &lt;math&gt;81x+100y&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are positive integers. Paint the remaining integer points blue. Find a point &lt;math&gt;P&lt;/math&gt; on the line such that, for every integer point &lt;math&gt;T&lt;/math&gt;, the reflection of &lt;math&gt;T&lt;/math&gt; with respect to &lt;math&gt;P&lt;/math&gt; is an integer point of a different colour than &lt;math&gt;T&lt;/math&gt;. (India TST)<br /> <br /> *Let &lt;math&gt;S&lt;/math&gt; be a set of integers (not necessarily positive) such that<br /> <br /> (a) there exist &lt;math&gt;a,b \in S&lt;/math&gt; with &lt;math&gt;\gcd(a,b)=\gcd(a-2,b-2)=1&lt;/math&gt;;<br /> <br /> (b) if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are elements of &lt;math&gt;S&lt;/math&gt; (possibly equal), then &lt;math&gt;x^2-y&lt;/math&gt; also belongs to &lt;math&gt;S&lt;/math&gt;. <br /> <br /> Prove that &lt;math&gt;S&lt;/math&gt; is the set of all integers. (USAMO)<br /> <br /> ==See Also==<br /> *[[Theorem]]<br /> *[[Prime]]<br /> <br /> [[Category:Theorems]]<br /> [[Category:Number theory]]</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:OlympusHero&diff=132428 User:OlympusHero 2020-08-24T00:41:30Z <p>Zhaom: /* User Count */</p> <hr /> <div>OlympusHero's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;107px&quot;&gt; 101<br /> &lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;OlympusHero is currently borderline AIME.&lt;br&gt;<br /> <br /> OlympusHero is 10 years old.&lt;br&gt;<br /> <br /> OlympusHero scored 47/46 (got an extra point for finishing the test under 10 minutes) when mocking the 2019 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State.&lt;br&gt;<br /> <br /> OlympusHero is a pro at maths and chess<br /> <br /> OlympusHero for black mop 2021<br /> <br /> OlympusHero was #1 at the 2017 Chess World Cadets U8 (He was 7 years old)<br /> <br /> OlympusHero is pro believe it<br /> <br /> OlympusHero is much better than Rusczyk at maths yes<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;A User Count of 300<br /> <br /> Make AIME 2021 (Currently borderline)<br /> <br /> Get 5 or more on AIME I 2021 (Mocked a 3 on the 2020 AIME I a few months ago)<br /> <br /> Get to 40 or more on the MATHCOUNTS Trainer Nationals Level (25/40)<br /> <br /> Get a perfect score and First Place in IMO 2021.<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Rusczyk&diff=131379 User:Rusczyk 2020-08-10T19:07:58Z <p>Zhaom: /* User Count */</p> <hr /> <div>Rusczyk's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#4EC284&quot;&gt;<br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;&lt;font color=&quot;white&quot;&gt;<br /> &lt;center&gt;&lt;font size=&quot;101px&quot;&gt;57&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt;Rusczyk is currently borderline AIME.&lt;br&gt;<br /> <br /> Rusczyk just turned 13 years old.&lt;br&gt;<br /> <br /> Rusczyk scored 44/46 when mocking the 2018 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State held on AoPS.&lt;br&gt;<br /> <br /> Rusczyk is a pro at maths and physics<br /> <br /> Rusczyk has come world and country #1 in various international tournaments and competitions starting from 2017<br /> <br /> Rusczyk is better than me at maths<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;white&quot;&gt; A User Count of 500<br /> <br /> Make AIME 2021 (Currently borderline)<br /> <br /> Pass AP Calculus AB, BC and AP Physics exam<br /> <br /> Get in the Alcumus HoF in the next 6 months<br /> <br /> Get &lt;math&gt;\color{white}{2 \times}&lt;/math&gt; medals this year as compared to what they did last year. That is &lt;math&gt;\color{white}{2 \times 14 = \boxed{28}}&lt;/math&gt; which is nearly impossible.<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Zhaom&diff=130787 User:Zhaom 2020-08-06T15:52:06Z <p>Zhaom: </p> <hr /> <div>...<br /> nothing here<br /> ...&lt;br/&gt;<br /> why are you still here?<br /> ...&lt;br/&gt;<br /> really, why are you still here?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> um, do you have anything else to do?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> um..... still here, are you?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> why do you want info on me?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> um....<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> I'm between 0 and 100 years old. Now will you go?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> NOTHING HERE, AND THIS PAGE IS FORMATTED HORRIBLY</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Zhaom&diff=130786 User:Zhaom 2020-08-06T15:51:15Z <p>Zhaom: </p> <hr /> <div>...<br /> nothing here<br /> ...&lt;br/&gt;<br /> why are you still here?<br /> ...&lt;br/&gt;<br /> really, why are you still here?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> um, do you have anything else to do?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> um..... still here, are you?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> why do you want info on me?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> um....<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> I'm between 0 and 100 years old. Now will you go?<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> ...&lt;br/&gt;<br /> NOTHING HERE</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Zhaom&diff=130785 User:Zhaom 2020-08-06T15:49:03Z <p>Zhaom: </p> <hr /> <div>...<br /> nothing here<br /> ...<br /> why are you still here?<br /> ...<br /> really, why are you still here?<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> um, do you have anything else to do?<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> Really?<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> why?<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> I'm 10. Now will you go?<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> ...<br /> THERE'S NOTHING HERE</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Zhaom&diff=130784 User:Zhaom 2020-08-06T15:45:31Z <p>Zhaom: </p> <hr /> <div>...<br /> nothing here</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:Zhaom&diff=130783 User:Zhaom 2020-08-06T15:45:10Z <p>Zhaom: Created page with &quot;... nothhing here&quot;</p> <hr /> <div>...<br /> nothhing here</div> Zhaom https://artofproblemsolving.com/wiki/index.php?title=User:OlympusHero&diff=130782 User:OlympusHero 2020-08-06T15:43:25Z <p>Zhaom: /* User Count */</p> <hr /> <div>OlympusHero's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;101px&quot;&gt;14&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;OlympusHero is currently borderline AIME.&lt;br&gt;<br /> <br /> OlympusHero is 10 years old.&lt;br&gt;<br /> <br /> OlympusHero scored 38/46 when mocking the 2019 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State.&lt;br&gt;<br /> <br /> OlympusHero is a pro at maths and chess<br /> <br /> OlympusHero for black mop 2026<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;A User Count of 300<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Zhaom