https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ultraman&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-20T01:10:34Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=129037 User:Ultraman 2020-07-24T13:54:24Z <p>Ultraman: Blanked the page</p> <hr /> <div></div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2019_IMO_Problems/Problem_6&diff=126517 2019 IMO Problems/Problem 6 2020-06-25T21:04:49Z <p>Ultraman: /* Problem */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of acute triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB \neq AC&lt;/math&gt;. The incircle &lt;math&gt;\omega&lt;/math&gt; of &lt;math&gt;ABC&lt;/math&gt; is tangent to sides &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CA&lt;/math&gt;, and &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively. The line through &lt;math&gt;D&lt;/math&gt; perpendicular to &lt;math&gt;EF&lt;/math&gt; meets ω again at &lt;math&gt;R&lt;/math&gt;. Line &lt;math&gt;AR&lt;/math&gt; meets ω again at &lt;math&gt;P&lt;/math&gt;. The circumcircles of triangles &lt;math&gt;PCE&lt;/math&gt; and &lt;math&gt;PBF&lt;/math&gt; meet again at &lt;math&gt;Q&lt;/math&gt;.<br /> Prove that lines &lt;math&gt;DI&lt;/math&gt; and &lt;math&gt;PQ&lt;/math&gt; meet on the line through &lt;math&gt;A&lt;/math&gt; perpendicular to &lt;math&gt;AI&lt;/math&gt;.</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2019_IMO_Problems/Problem_6&diff=123053 2019 IMO Problems/Problem 6 2020-05-27T03:34:24Z <p>Ultraman: /* Problem */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of acute triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB \neq AC&lt;/math&gt;. The incircle ω of &lt;math&gt;ABC&lt;/math&gt; is tangent to sides &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CA&lt;/math&gt;, and &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively. The line through &lt;math&gt;D&lt;/math&gt; perpendicular to &lt;math&gt;EF&lt;/math&gt; meets ω again at &lt;math&gt;R&lt;/math&gt;. Line &lt;math&gt;AR&lt;/math&gt; meets ω again at &lt;math&gt;P&lt;/math&gt;. The circumcircles of triangles &lt;math&gt;PCE&lt;/math&gt; and &lt;math&gt;PBF&lt;/math&gt; meet again at &lt;math&gt;Q&lt;/math&gt;.<br /> Prove that lines &lt;math&gt;DI&lt;/math&gt; and &lt;math&gt;PQ&lt;/math&gt; meet on the line through &lt;math&gt;A&lt;/math&gt; perpendicular to &lt;math&gt;AI&lt;/math&gt;.</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2019_IMO_Problems/Problem_6&diff=123052 2019 IMO Problems/Problem 6 2020-05-27T03:31:14Z <p>Ultraman: /* Problem */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of acute triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB \neq AC&lt;/math&gt;. The incircle ω of &lt;math&gt;ABC&lt;/math&gt; is tangent to sides &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CA&lt;/math&gt;, and &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;F&lt;/math&gt;, respectively. The line through &lt;math&gt;D&lt;/math&gt; perpendicular to &lt;math&gt;EF&lt;/math&gt; meets ω again at &lt;math&gt;R&lt;/math&gt;. Line &lt;math&gt;AR&lt;/math&gt; meets ω again at P. The circumcircles of triangles PCE and PBF meet again at Q.<br /> Prove that lines &lt;math&gt;DI&lt;/math&gt; and &lt;math&gt;PQ&lt;/math&gt; meet on the line through &lt;math&gt;A&lt;/math&gt; perpendicular to &lt;math&gt;AI&lt;/math&gt;.</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2019_IMO_Problems/Problem_6&diff=123051 2019 IMO Problems/Problem 6 2020-05-27T03:27:38Z <p>Ultraman: Created page with &quot;==Problem== Let I be the incentre of acute triangle ABC with AB ̸= AC. The incircle ω of ABC is tangent to sides BC, CA, and AB at D, E, and F, respectively. The line throug...&quot;</p> <hr /> <div>==Problem==<br /> Let I be the incentre of acute triangle ABC with AB ̸= AC. The incircle ω of ABC is tangent to sides BC, CA, and AB at D, E, and F, respectively. The line through D perpendicular to EF meets ω again at R. Line AR meets ω again at P. The circumcircles of triangles PCE and PBF meet again at Q.<br /> Prove that lines DI and PQ meet on the line through A perpendicular to AI.</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=119835 User:Ultraman 2020-03-20T02:19:00Z <p>Ultraman: /* About Me */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, doing maths, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[2020 AMC 10A Problems/Problem 20]]<br /> <br /> [[2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1]]<br /> <br /> [[1966 AHSME Problems/Problem 24]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 4]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=116646 2020 AMC 10A Problems/Problem 20 2020-02-02T20:05:29Z <p>Ultraman: /* Solution 1 (Just Drop An Altitude) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #18]] and [[2020 AMC 10A Problems|2020 AMC 10A #20]]}}<br /> <br /> == Problem ==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution 1 (Just Drop An Altitude)==<br /> <br /> &lt;asy&gt;<br /> size(15cm,0);<br /> import olympiad;<br /> draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);<br /> label(&quot;A&quot;, (0,2), NW);<br /> label(&quot;B&quot;, (0,0), SW);<br /> label(&quot;C&quot;, (4,0), SE);<br /> label(&quot;D&quot;, (6,4), NE);<br /> label(&quot;E&quot;, (1.714,1.143), N);<br /> label(&quot;F&quot;, (1,1.5), N);<br /> draw((0,2)--(4,0), dashed);<br /> draw((0,0)--(6,4), dashed);<br /> draw((0,0)--(1,1.5), dashed);<br /> label(&quot;20&quot;, (0,2)--(4,0), SW);<br /> label(&quot;30&quot;, (4,0)--(6,4), SE);<br /> label(&quot;$x$&quot;, (1,1.5)--(1.714,1.143), NE);<br /> draw(rightanglemark((0,2),(0,0),(4,0)));<br /> draw(rightanglemark((0,2),(4,0),(6,4)));<br /> draw(rightanglemark((0,0),(1,1.5),(0,2)));<br /> &lt;/asy&gt;<br /> <br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt;. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}&lt;/math&gt;<br /> <br /> (I'm very sorry if you're a visual learner but now you have a diagram by ciceronii)<br /> <br /> ~ Solution by Ultraman<br /> <br /> ~ Diagram by ciceronii<br /> <br /> ==Solution 2 (Pro Guessing Strats)==<br /> We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to &lt;math&gt;\sqrt{20a}&lt;/math&gt; and &lt;math&gt;\sqrt{20b}&lt;/math&gt;, and because the hypotenuse is 20 we get &lt;math&gt;a+b=20&lt;/math&gt;. Testing small numbers, we get that when &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=18&lt;/math&gt;, &lt;math&gt;ab&lt;/math&gt; is indeed a square. The area of the triangle is thus 60, so the answer is &lt;math&gt;\boxed {\textbf{(D) }360}&lt;/math&gt;.<br /> <br /> ~tigershark22<br /> ~(edited by HappyHuman)<br /> <br /> ==Solution 3 (coordinates)==<br /> &lt;asy&gt;<br /> size(10cm,0);<br /> draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br /> draw((-20,0)--(20,0));<br /> draw((0,-15)--(0,35));<br /> draw((10,30)--(-8,-6));<br /> draw(circle((0,0),10));<br /> label(&quot;E&quot;,(-4.05,-.25),S);<br /> label(&quot;D&quot;,(10,30),NE);<br /> label(&quot;C&quot;,(10,0),NE);<br /> label(&quot;B&quot;,(-8,-6),SW);<br /> label(&quot;A&quot;,(-10,0),NW);<br /> label(&quot;5&quot;,(-10,0)--(-5,0), NE);<br /> label(&quot;15&quot;,(-5,0)--(10,0), N);<br /> label(&quot;30&quot;,(10,0)--(10,30), E);<br /> dot((-5,0));<br /> dot((-10,0));<br /> dot((-8,-6));<br /> dot((10,0));<br /> dot((10,30));<br /> &lt;/asy&gt;<br /> Let the points be &lt;math&gt;A(-10,0)&lt;/math&gt;, &lt;math&gt;\:B(x,y)&lt;/math&gt;, &lt;math&gt;\:C(10,0)&lt;/math&gt;, &lt;math&gt;\:D(10,30)&lt;/math&gt;,and &lt;math&gt;\:E(-5,0)&lt;/math&gt;, respectively. Since &lt;math&gt;B&lt;/math&gt; lies on line &lt;math&gt;DE&lt;/math&gt;, we know that &lt;math&gt;y=2x+10&lt;/math&gt;. Furthermore, since &lt;math&gt;\angle{ABC}=90^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; lies on the circle with diameter &lt;math&gt;AC&lt;/math&gt;, so &lt;math&gt;x^2+y^2=100&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with these equations, we get the solutions &lt;math&gt;(0,10)&lt;/math&gt; and &lt;math&gt;(-8,-6)&lt;/math&gt;. We immediately discard the &lt;math&gt;(0,10)&lt;/math&gt; solution as &lt;math&gt;y&lt;/math&gt; should be negative. Thus, we conclude that &lt;math&gt;[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Trigonometry)==<br /> Let &lt;math&gt;\angle C = \angle{ACB}&lt;/math&gt; and &lt;math&gt;\angle{B} = \angle{CBE}.&lt;/math&gt; Using Law of Sines on &lt;math&gt;\triangle{BCE}&lt;/math&gt; we get &lt;cmath&gt;\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}&lt;/cmath&gt; and LoS on &lt;math&gt;\triangle{ABE}&lt;/math&gt; yields &lt;cmath&gt;\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.&lt;/cmath&gt; Divide the two to get &lt;math&gt;\tan{B} = 3 \tan{C}.&lt;/math&gt; Now, &lt;cmath&gt;\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}&lt;/cmath&gt; and solve the quadratic, taking the positive solution (C is acute) to get &lt;math&gt;\tan{C} = \frac{1}{3}.&lt;/math&gt; So if &lt;math&gt;AB = a,&lt;/math&gt; then &lt;math&gt;BC = 3a&lt;/math&gt; and &lt;math&gt;[ABC] = \frac{3a^2}{2}.&lt;/math&gt; By Pythagorean Theorem, &lt;math&gt;10a^2 = 400 \iff \frac{3a^2}{2} = 60&lt;/math&gt; and the answer is &lt;math&gt;300 + 60 \iff \boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> (This solution is incomplete, can someone complete it please-Lingjun) ok<br /> Latex edited by kc5170<br /> <br /> ==Video Solution==<br /> On The Spot STEM<br /> https://www.youtube.com/watch?v=hIdNde2Vln4<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=116645 2020 AMC 10A Problems/Problem 20 2020-02-02T20:04:39Z <p>Ultraman: /* Solution 1 (Just Drop An Altitude) */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #18]] and [[2020 AMC 10A Problems|2020 AMC 10A #20]]}}<br /> <br /> == Problem ==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution 1 (Just Drop An Altitude)==<br /> <br /> &lt;asy&gt;<br /> size(15cm,0);<br /> import olympiad;<br /> draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);<br /> label(&quot;A&quot;, (0,2), NW);<br /> label(&quot;B&quot;, (0,0), SW);<br /> label(&quot;C&quot;, (4,0), SE);<br /> label(&quot;D&quot;, (6,4), NE);<br /> label(&quot;E&quot;, (1.714,1.143), N);<br /> label(&quot;F&quot;, (1,1.5), N);<br /> draw((0,2)--(4,0), dashed);<br /> draw((0,0)--(6,4), dashed);<br /> draw((0,0)--(1,1.5), dashed);<br /> label(&quot;20&quot;, (0,2)--(4,0), SW);<br /> label(&quot;30&quot;, (4,0)--(6,4), SE);<br /> label(&quot;$x$&quot;, (1,1.5)--(1.714,1.143), NE);<br /> draw(rightanglemark((0,2),(0,0),(4,0)));<br /> draw(rightanglemark((0,2),(4,0),(6,4)));<br /> draw(rightanglemark((0,0),(1,1.5),(0,2)));<br /> &lt;/asy&gt;<br /> <br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt;. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}&lt;/math&gt;<br /> <br /> (I'm very sorry if you're a visual learner)<br /> <br /> ~ Solution by Ultraman<br /> <br /> ~ Diagram by ciceronii<br /> <br /> ==Solution 2 (Pro Guessing Strats)==<br /> We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to &lt;math&gt;\sqrt{20a}&lt;/math&gt; and &lt;math&gt;\sqrt{20b}&lt;/math&gt;, and because the hypotenuse is 20 we get &lt;math&gt;a+b=20&lt;/math&gt;. Testing small numbers, we get that when &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=18&lt;/math&gt;, &lt;math&gt;ab&lt;/math&gt; is indeed a square. The area of the triangle is thus 60, so the answer is &lt;math&gt;\boxed {\textbf{(D) }360}&lt;/math&gt;.<br /> <br /> ~tigershark22<br /> ~(edited by HappyHuman)<br /> <br /> ==Solution 3 (coordinates)==<br /> &lt;asy&gt;<br /> size(10cm,0);<br /> draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br /> draw((-20,0)--(20,0));<br /> draw((0,-15)--(0,35));<br /> draw((10,30)--(-8,-6));<br /> draw(circle((0,0),10));<br /> label(&quot;E&quot;,(-4.05,-.25),S);<br /> label(&quot;D&quot;,(10,30),NE);<br /> label(&quot;C&quot;,(10,0),NE);<br /> label(&quot;B&quot;,(-8,-6),SW);<br /> label(&quot;A&quot;,(-10,0),NW);<br /> label(&quot;5&quot;,(-10,0)--(-5,0), NE);<br /> label(&quot;15&quot;,(-5,0)--(10,0), N);<br /> label(&quot;30&quot;,(10,0)--(10,30), E);<br /> dot((-5,0));<br /> dot((-10,0));<br /> dot((-8,-6));<br /> dot((10,0));<br /> dot((10,30));<br /> &lt;/asy&gt;<br /> Let the points be &lt;math&gt;A(-10,0)&lt;/math&gt;, &lt;math&gt;\:B(x,y)&lt;/math&gt;, &lt;math&gt;\:C(10,0)&lt;/math&gt;, &lt;math&gt;\:D(10,30)&lt;/math&gt;,and &lt;math&gt;\:E(-5,0)&lt;/math&gt;, respectively. Since &lt;math&gt;B&lt;/math&gt; lies on line &lt;math&gt;DE&lt;/math&gt;, we know that &lt;math&gt;y=2x+10&lt;/math&gt;. Furthermore, since &lt;math&gt;\angle{ABC}=90^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; lies on the circle with diameter &lt;math&gt;AC&lt;/math&gt;, so &lt;math&gt;x^2+y^2=100&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with these equations, we get the solutions &lt;math&gt;(0,10)&lt;/math&gt; and &lt;math&gt;(-8,-6)&lt;/math&gt;. We immediately discard the &lt;math&gt;(0,10)&lt;/math&gt; solution as &lt;math&gt;y&lt;/math&gt; should be negative. Thus, we conclude that &lt;math&gt;[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Trigonometry)==<br /> Let &lt;math&gt;\angle C = \angle{ACB}&lt;/math&gt; and &lt;math&gt;\angle{B} = \angle{CBE}.&lt;/math&gt; Using Law of Sines on &lt;math&gt;\triangle{BCE}&lt;/math&gt; we get &lt;cmath&gt;\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}&lt;/cmath&gt; and LoS on &lt;math&gt;\triangle{ABE}&lt;/math&gt; yields &lt;cmath&gt;\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.&lt;/cmath&gt; Divide the two to get &lt;math&gt;\tan{B} = 3 \tan{C}.&lt;/math&gt; Now, &lt;cmath&gt;\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}&lt;/cmath&gt; and solve the quadratic, taking the positive solution (C is acute) to get &lt;math&gt;\tan{C} = \frac{1}{3}.&lt;/math&gt; So if &lt;math&gt;AB = a,&lt;/math&gt; then &lt;math&gt;BC = 3a&lt;/math&gt; and &lt;math&gt;[ABC] = \frac{3a^2}{2}.&lt;/math&gt; By Pythagorean Theorem, &lt;math&gt;10a^2 = 400 \iff \frac{3a^2}{2} = 60&lt;/math&gt; and the answer is &lt;math&gt;300 + 60 \iff \boxed{\textbf{(D)}}.&lt;/math&gt;<br /> <br /> (This solution is incomplete, can someone complete it please-Lingjun) ok<br /> Latex edited by kc5170<br /> <br /> ==Video Solution==<br /> On The Spot STEM<br /> https://www.youtube.com/watch?v=hIdNde2Vln4<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=116303 2020 AMC 10A Problems/Problem 20 2020-02-01T19:29:02Z <p>Ultraman: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution 1 (Just Drop An Altitude)==<br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt;. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{D) }360}&lt;/math&gt;<br /> <br /> (I'm very sorry if you're a visual learner)<br /> <br /> ~Ultraman<br /> <br /> ==Solution 2 (Pro Guessing Strats)==<br /> We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to &lt;math&gt;\sqrt{20a}&lt;/math&gt; and &lt;math&gt;\sqrt{20b}&lt;/math&gt;, and because the hypotenuse is 20 we get &lt;math&gt;a+b=20&lt;/math&gt;. Testing small numbers, we get that when &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=18&lt;/math&gt;, &lt;math&gt;ab&lt;/math&gt; is indeed a square. The area of the triangle is thus 60, so the answer is &lt;math&gt;\boxed {\textbf{D) }360}&lt;/math&gt;.<br /> <br /> ~tigershark22<br /> ~(edited by HappyHuman)<br /> <br /> ==Solution 3 (coordinates)==<br /> &lt;asy&gt;<br /> draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));<br /> draw(circle((0,0),10));<br /> label(&quot;D&quot;,(10,30),N);<br /> label(&quot;C&quot;,(10,0),E);<br /> label(&quot;B&quot;,(-8,-6),S);<br /> label(&quot;A&quot;,(-10,0),W);<br /> &lt;/asy&gt;<br /> Let the points be &lt;math&gt;A(-10,0)&lt;/math&gt;, &lt;math&gt;\:B(x,y)&lt;/math&gt;, &lt;math&gt;\:C(10,0)&lt;/math&gt;, &lt;math&gt;\:D(10,30)&lt;/math&gt;,and &lt;math&gt;\:E(-5,0)&lt;/math&gt;, respectively. Since &lt;math&gt;B&lt;/math&gt; lies on line &lt;math&gt;DE&lt;/math&gt;, we know that &lt;math&gt;y=2x+10&lt;/math&gt;. Furthermore, since &lt;math&gt;\angle{ABC}=90^\circ&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; lies on the circle with diameter &lt;math&gt;AC&lt;/math&gt;, so &lt;math&gt;x^2+y^2=100&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with these equations, we get the solutions &lt;math&gt;(0,10)&lt;/math&gt; and &lt;math&gt;(-8,-6)&lt;/math&gt;. We immediately discard the &lt;math&gt;(0,10)&lt;/math&gt; solution as &lt;math&gt;y&lt;/math&gt; should be negative. Thus, we conclude that &lt;math&gt;[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}&lt;/math&gt;.<br /> <br /> ==Solution 4 (Trignometry)==<br /> Using the law of cosines, express &lt;math&gt;AB^2&lt;/math&gt; and &lt;math&gt;BC^2&lt;/math&gt; in terms of &lt;math&gt;\angle{AEB}&lt;/math&gt;. The sum of these two equations is &lt;math&gt;AC^2&lt;/math&gt; by the Pythagorean Theorem. Solving for &lt;math&gt;BE&lt;/math&gt;, we find &lt;math&gt;BE=3\sqrt5&lt;/math&gt;. Since &lt;math&gt;EC=15&lt;/math&gt; and &lt;math&gt;DC=30&lt;/math&gt;, &lt;math&gt;DE=15\sqrt3&lt;/math&gt;, which is five times &lt;math&gt;BE&lt;/math&gt;, so &lt;math&gt;[ABCD]=[ACD]+\frac{1/5}{[ACD]}=300+60=\boxed {\textbf{D) }360}&lt;/math&gt;<br /> <br /> (This solution is incomplete, can someone complete it please)<br /> Latex edited by kc5170<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2020|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=116162 2020 AMC 10A Problems/Problem 20 2020-02-01T05:00:17Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution ==<br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt;. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {D)360}&lt;/math&gt;<br /> <br /> (I'm very sorry if you're a visual learner)<br /> <br /> ~Ultraman<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=116160 2020 AMC 10A Problems/Problem 20 2020-02-01T04:59:33Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution ==<br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt;. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {D)360}&lt;/math&gt;<br /> <br /> (I'm very sorry if you're a visual learner - Ultraman)<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=116159 User:Ultraman 2020-02-01T04:57:51Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[2020 AMC 10A Problems/Problem 20]]<br /> <br /> [[2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1]]<br /> <br /> [[1966 AHSME Problems/Problem 24]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 4]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_20&diff=116157 2020 AMC 10A Problems/Problem 20 2020-02-01T04:57:15Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> == Solution ==<br /> It's crucial to draw a good diagram for this one. Since &lt;math&gt;AC=20&lt;/math&gt; and &lt;math&gt;CD=30&lt;/math&gt;, we get &lt;math&gt;[ACD]=300&lt;/math&gt;. Now we need to find &lt;math&gt;[ABC]&lt;/math&gt; to get the area of the whole quadrilateral. Drop an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and call the point of intersection &lt;math&gt;F&lt;/math&gt;. Let &lt;math&gt;FE=x&lt;/math&gt;. Since &lt;math&gt;AE=5&lt;/math&gt;, then &lt;math&gt;AF=5-x&lt;/math&gt;. By dropping this altitude, we can also see two similar triangles, &lt;math&gt;BFE&lt;/math&gt; and &lt;math&gt;DCE&lt;/math&gt;. Since &lt;math&gt;EC&lt;/math&gt; is &lt;math&gt;20-5=15&lt;/math&gt;, and &lt;math&gt;DC=30&lt;/math&gt;, we get that &lt;math&gt;BF=2x&lt;/math&gt;. Now, if we redraw another diagram just of &lt;math&gt;ABC&lt;/math&gt;, we get that &lt;math&gt;(2x)^2=(5-x)(15+x)&lt;/math&gt;. Now expanding, simplifying, and dividing by the GCF, we get &lt;math&gt;x^2+2x-15=0&lt;/math&gt;. This factors to &lt;math&gt;(x+5)(x-3)&lt;/math&gt;. Since lengths cannot be negative, &lt;math&gt;x=3&lt;/math&gt;. Since &lt;math&gt;x=3&lt;/math&gt;, &lt;math&gt;[ABC]=60&lt;/math&gt;. So &lt;math&gt;[ABCD]=[ACD]+[ABC]=300+60=\boxed {D)360}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2015_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=116132 2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2020-02-01T04:18:12Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In the sequence<br /> &lt;math&gt;1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, \cdots &lt;/math&gt;<br /> what number occupies position &lt;math&gt;2015&lt;/math&gt;?<br /> <br /> == Solution==<br /> Since there is &lt;math&gt;1&lt;/math&gt; one, &lt;math&gt;2&lt;/math&gt; twos, &lt;math&gt;4&lt;/math&gt; fours and so on, the number that occupies a position is &lt;math&gt;1+2+4+8+ \cdot \cdot \cdot&lt;/math&gt;. Since &lt;math&gt;2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1&lt;/math&gt;, we get the &lt;math&gt;1023&lt;/math&gt;rd number to be &lt;math&gt;512&lt;/math&gt;. Since the &lt;math&gt;2015&lt;/math&gt;th position is &lt;math&gt;992&lt;/math&gt; positions higher than the last &lt;math&gt;512&lt;/math&gt;, the answer is &lt;math&gt;\boxed{1024}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Beginning Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2015_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=116131 2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2020-02-01T04:17:56Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In the sequence<br /> &lt;math&gt;1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, \cdots &lt;/math&gt;<br /> what number occupies position &lt;math&gt;2015&lt;/math&gt;?<br /> <br /> == Solution==<br /> Since there is &lt;math&gt;1&lt;/math&gt; one, &lt;math&gt;2&lt;/math&gt; twos, &lt;math&gt;4&lt;/math&gt; fours and so on, the number that occupies a position is &lt;math&gt;1+2+4+8+ \cdot \cdot \cdot&lt;/math&gt;. Since &lt;math&gt;2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1&lt;/math&gt;, we get the &lt;math&gt;1023&lt;/math&gt;rd number to be &lt;math&gt;512&lt;/math&gt;. Since the &lt;math&gt;2015&lt;/math&gt;th position is &lt;math&gt;992&lt;/math&gt; positions higher than the last &lt;math&gt;512&lt;/math&gt;, the answer is &lt;math&gt;\boxed1024&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Beginning Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2015_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=116130 2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2020-02-01T04:17:43Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In the sequence<br /> &lt;math&gt;1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, \cdots &lt;/math&gt;<br /> what number occupies position &lt;math&gt;2015&lt;/math&gt;?<br /> <br /> == Solution==<br /> Since there is &lt;math&gt;1&lt;/math&gt; one, &lt;math&gt;2&lt;/math&gt; twos, &lt;math&gt;4&lt;/math&gt; fours and so on, the number that occupies a position is &lt;math&gt;1+2+4+8+ \cdot \cdot \cdot&lt;/math&gt;. Since &lt;math&gt;2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1&lt;/math&gt;, we get the &lt;math&gt;1023&lt;/math&gt;rd number to be &lt;math&gt;512&lt;/math&gt;. Since the &lt;math&gt;2015&lt;/math&gt;th position is &lt;math&gt;992&lt;/math&gt; positions higher than the last &lt;math&gt;512&lt;/math&gt;, the answer is \boxed&lt;math&gt;1024&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Beginning Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_1&diff=114304 2018 UNCO Math Contest II Problems/Problem 1 2020-01-04T22:59:09Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A printer used 1890 digits to number all the pages in the Seripian Puzzle Book. How many pages are in the book? (For example, to number the pages in a book with twelve pages, the printer would use fifteen digits.)<br /> <br /> == Solution ==<br /> We know that we use &lt;math&gt;1&lt;/math&gt; digit &lt;math&gt;9&lt;/math&gt; times, &lt;math&gt;2&lt;/math&gt; digits &lt;math&gt;90&lt;/math&gt; times, and &lt;math&gt;3&lt;/math&gt; digits &lt;math&gt;900&lt;/math&gt; times. So if we have &lt;math&gt;999&lt;/math&gt; pages, we have &lt;math&gt;1 \cdot 9 + 2 \cdot 90 + 3 \cdot 900 = 2889&lt;/math&gt; digits. Since we want to have &lt;math&gt;1890&lt;/math&gt; digits, we do &lt;math&gt;\frac{2889 - 1890}{3}=333&lt;/math&gt; pages less than &lt;math&gt;999&lt;/math&gt;. So, &lt;math&gt;999-333=\boxed {666}&lt;/math&gt; pages.<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_24&diff=114303 1966 AHSME Problems/Problem 24 2020-01-04T22:58:58Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;Log_M{N}=Log_N{M},M \ne N,MN&gt;0,M \ne 1, N \ne 1&lt;/math&gt;, then &lt;math&gt;MN&lt;/math&gt; equals:<br /> <br /> &lt;math&gt;\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}&lt;/math&gt;<br /> <br /> == Solution ==<br /> If we change the base of &lt;math&gt;Log_M{N}&lt;/math&gt; to base &lt;math&gt;N&lt;/math&gt;, we get &lt;math&gt;\frac{1}{log_N{M}}= Log_N{M}&lt;/math&gt;. Multiplying both sides by &lt;math&gt;Log_N{M}&lt;/math&gt;, we get &lt;math&gt;Log_N{M}^2=1&lt;/math&gt;. Since &lt;math&gt;N\not = M&lt;/math&gt;, &lt;math&gt;Log_N{M}=-1&lt;/math&gt;. So &lt;math&gt;N^{-1}=M \rightarrow \frac{1}{N} = M \rightarrow MN=1&lt;/math&gt;. So the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1966|num-b=23|num-a=25}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2015_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=114302 2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2020-01-04T22:58:45Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In the sequence<br /> &lt;math&gt;1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, \cdots &lt;/math&gt;<br /> what number occupies position &lt;math&gt;2015&lt;/math&gt;?<br /> <br /> == Solution==<br /> Since there is &lt;math&gt;1&lt;/math&gt; one, &lt;math&gt;2&lt;/math&gt; twos, &lt;math&gt;4&lt;/math&gt; fours and so on, the number that occupies a position is &lt;math&gt;1+2+4+8+ \cdot \cdot \cdot&lt;/math&gt;. Since &lt;math&gt;2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1&lt;/math&gt;, we get the &lt;math&gt;1023&lt;/math&gt;rd number to be &lt;math&gt;512&lt;/math&gt;. Since the &lt;math&gt;2015&lt;/math&gt;th position is &lt;math&gt;992&lt;/math&gt; positions higher than the last &lt;math&gt;512&lt;/math&gt;, the answer is &lt;math&gt;1024&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Beginning Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114121 User:Ultraman 2020-01-03T00:21:16Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1]]<br /> <br /> [[1966 AHSME Problems/Problem 24]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 4]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2015_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=114120 2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2020-01-03T00:20:36Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In the sequence<br /> &lt;math&gt;1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, \cdots &lt;/math&gt;<br /> what number occupies position &lt;math&gt;2015&lt;/math&gt;?<br /> <br /> == Solution==<br /> Since there is &lt;math&gt;1&lt;/math&gt; one, &lt;math&gt;2&lt;/math&gt; twos, &lt;math&gt;4&lt;/math&gt; fours and so on, the number that occupies a position is &lt;math&gt;1+2+4+8+ \cdot \cdot \cdot&lt;/math&gt;. Since &lt;math&gt;2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1&lt;/math&gt;, we get the &lt;math&gt;1023&lt;/math&gt;rd number to be &lt;math&gt;512&lt;/math&gt;. Since the &lt;math&gt;2015&lt;/math&gt;th position is &lt;math&gt;992&lt;/math&gt; positions higher than the last &lt;math&gt;512&lt;/math&gt;, the answer is &lt;math&gt;1024&lt;/math&gt;.<br /> <br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Beginning Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2015_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=114119 2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2020-01-03T00:20:22Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In the sequence<br /> &lt;math&gt;1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, \cdots &lt;/math&gt;<br /> what number occupies position &lt;math&gt;2015&lt;/math&gt;?<br /> <br /> == Solution==<br /> Since there is &lt;math&gt;1&lt;/math&gt; one, &lt;math&gt;2&lt;/math&gt; twos, &lt;math&gt;4&lt;/math&gt; fours and so on, the number that occupies a position is &lt;math&gt;1+2+4+8+ \cdot \cdot \cdot&lt;/math&gt;. Since &lt;math&gt;2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1&lt;/math&gt;, we get the &lt;math&gt;1023&lt;/math&gt;rd number to be &lt;math&gt;512&lt;/math&gt;. Since the &lt;math&gt;2015&lt;/math&gt;th position is &lt;math&gt;992&lt;/math&gt; positions higher than the last &lt;math&gt;512, the answer is &lt;/math&gt;1024\$.<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Beginning Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114104 User:Ultraman 2020-01-02T21:29:34Z <p>Ultraman: /* About Me */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[1966 AHSME Problems/Problem 24]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 4]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114103 User:Ultraman 2020-01-02T21:29:19Z <p>Ultraman: /* About Me */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using [[Unity]], math, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[1966 AHSME Problems/Problem 24]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 4]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=Badminton&diff=114102 Badminton 2020-01-02T21:28:59Z <p>Ultraman: </p> <hr /> <div>==Introduction==<br /> Badminton is the fastest racket sport in the world. All of it is math and physics. Like how fast you would have to be to run cross-court in 2 seconds, or whether it is possible or not to have a 500km/hr smash.<br /> <br /> ==Rules==</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=Badminton&diff=114101 Badminton 2020-01-02T21:27:41Z <p>Ultraman: Created page with &quot;Badminton is the fastest racket sport in the world. All of it is math and can be calculated by math. Like how fast you would have to be to run cross-court in 2 seconds, or whe...&quot;</p> <hr /> <div>Badminton is the fastest racket sport in the world. All of it is math and can be calculated by math. Like how fast you would have to be to run cross-court in 2 seconds, or whether it is possible or not to have a 500km/hr smash.</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_24&diff=114096 1966 AHSME Problems/Problem 24 2020-01-02T20:49:34Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;Log_M{N}=Log_N{M},M \ne N,MN&gt;0,M \ne 1, N \ne 1&lt;/math&gt;, then &lt;math&gt;MN&lt;/math&gt; equals:<br /> <br /> &lt;math&gt;\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}&lt;/math&gt;<br /> <br /> == Solution ==<br /> If we change the base of &lt;math&gt;Log_M{N}&lt;/math&gt; to base &lt;math&gt;N&lt;/math&gt;, we get &lt;math&gt;\frac{1}{log_N{M}}= Log_N{M}&lt;/math&gt;. Multiplying both sides by &lt;math&gt;Log_N{M}&lt;/math&gt;, we get &lt;math&gt;Log_N{M}^2=1&lt;/math&gt;. Since &lt;math&gt;N\not = M&lt;/math&gt;, &lt;math&gt;Log_N{M}=-1&lt;/math&gt;. So &lt;math&gt;N^{-1}=M \rightarrow \frac{1}{N} = M \rightarrow MN=1&lt;/math&gt;. So the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ~Ultraman<br /> <br /> == See also ==<br /> {{AHSME box|year=1966|num-b=23|num-a=25}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114094 User:Ultraman 2020-01-02T20:49:00Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[1966 AHSME Problems/Problem 24]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 4]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_24&diff=114093 1966 AHSME Problems/Problem 24 2020-01-02T20:48:28Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;Log_M{N}=Log_N{M},M \ne N,MN&gt;0,M \ne 1, N \ne 1&lt;/math&gt;, then &lt;math&gt;MN&lt;/math&gt; equals:<br /> <br /> &lt;math&gt;\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}&lt;/math&gt;<br /> <br /> == Solution ==<br /> If we change the base of &lt;math&gt;Log_M{N}&lt;/math&gt; to base &lt;math&gt;N&lt;/math&gt;, we get &lt;math&gt;\frac{1}{log_N{M}}= Log_N{M}&lt;/math&gt;. Multiplying both sides by &lt;math&gt;Log_N{M}&lt;/math&gt;, we get &lt;math&gt;Log_N{M}^2=1&lt;/math&gt;. Since &lt;math&gt;N\not = M&lt;/math&gt;, &lt;math&gt;Log_N{M}=-1&lt;/math&gt;. So &lt;math&gt;N^{-1}=M \rightarrow \frac{1}{N} = M \rightarrow MN=1&lt;/math&gt;. So the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1966|num-b=23|num-a=25}} <br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_12&diff=114068 2012 AMC 12B Problems/Problem 12 2020-01-02T19:13:01Z <p>Ultraman: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382&lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> There are &lt;math&gt;\binom{20}{2}&lt;/math&gt; selections; however, we count these twice, therefore<br /> <br /> &lt;math&gt;2\cdot\binom{20}{2} = 380&lt;/math&gt;. The wording of the question implies D not E.<br /> <br /> MAA decided to accept both D and E, however.<br /> <br /> ===Solution 2===<br /> <br /> Consider the 20 term sequence of &lt;math&gt;0&lt;/math&gt;'s and &lt;math&gt;1&lt;/math&gt;'s. Keeping all other terms 1, a sequence of &lt;math&gt;k&gt;0&lt;/math&gt; consecutive 0's can be placed in &lt;math&gt;21-k&lt;/math&gt; locations. That is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros. Hence there are &lt;math&gt;20+19+\cdots+1=\binom{21}{2}&lt;/math&gt; strings with consecutive zeros. The same argument shows there are &lt;math&gt;\binom{21}{2}&lt;/math&gt; strings with consecutive 1's. This yields &lt;math&gt;2\binom{21}{2}&lt;/math&gt; strings in all. However, we have counted twice those strings in which all the 1's and all the 0's are consecutive. These are the cases &lt;math&gt;01111...&lt;/math&gt;, &lt;math&gt;00111...&lt;/math&gt;, &lt;math&gt;000111...&lt;/math&gt;, ..., &lt;math&gt;000...0001&lt;/math&gt; (of which there are 19) as well as the cases &lt;math&gt;10000...&lt;/math&gt;, &lt;math&gt;11000...&lt;/math&gt;, &lt;math&gt;111000...&lt;/math&gt;, ..., &lt;math&gt;111...110&lt;/math&gt; (of which there are 19 as well). This yields &lt;math&gt;2\binom{21}{2}-2\cdot19=382&lt;/math&gt; so that the answer is &lt;math&gt;\framebox{E}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=University_of_Northern_Colorado_Math_Contest&diff=114066 University of Northern Colorado Math Contest 2020-01-02T19:09:16Z <p>Ultraman: /* Past Exams */</p> <hr /> <div>== Past Exams ==<br /> This is a list of all UNC Math Contests in the AoPSWiki. Users are encouraged to post additional solutions to the problems.<br /> <br /> * [[2006 UNCO Math Contest II]]<br /> * [[2007 UNCO Math Contest II]]<br /> * [[2008 UNCO Math Contest II]]<br /> * [[2009 UNCO Math Contest II]]<br /> * [[2010 UNCO Math Contest II]]<br /> * [[2011 UNCO Math Contest II]]<br /> * [[2012 UNCO Math Contest II]]<br /> * [[2013 UNCO Math Contest II]]<br /> * [[2014 UNCO Math Contest II]]<br /> * [[2015 UNCO Math Contest II]]<br /> * [[2016 UNCO Math Contest II]]<br /> * [[2017 UNCO Math Contest II]]<br /> * [[2018 UNCO Math Contest II]]<br /> <br /> ==Format &amp; Scoring==<br /> The contest consists of two rounds. Each round is a paper and pencil exam with about 9-12 problems. Students compete individually and no electronic devices are to be used in working the problems. The First Round of the contest is administered in early November in schools around the state and the answers are sent to UNC for grading. Students are allowed ninety minutes to work the First Round problems.<br /> Students who make high scores on the First Round of the contest are invited to the UNC campus in Greeley for the Final Round of the contest in late January. The Final Round is similar in spirit to the First Round, but the problems are more challenging. Students are allowed three hours to work the Final Round problems.<br /> Both rounds challenge students to exercise their creativity and ingenuity to solve problems in geometry, algebra, combinatorics, probability, and number theory. All the old contest exams are archived on the contest website.<br /> <br /> == Eligibility ==<br /> The contest is for all interested students in Colorado. It is designed for students in grades 7-12, but younger students are welcome to participate if they have an interest. All schools in the state, public, private, and home schools, are invited to administer the First Round of the contest in early November. <br /> <br /> == History ==<br /> The University of Northern Colorado Math Contest was founded in 1992 by Richard Grassl and has been directed by Ricardo Diaz since 2011-12.<br /> <br /> == Resources ==<br /> *[http://uncmathcontest.wordpress.com/ Here is the contest website]<br /> * [[American Mathematics Competitions]]<br /> * [[Colorado mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> [[Category: Math Contest Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=University_of_Northern_Colorado_Math_Contest&diff=114065 University of Northern Colorado Math Contest 2020-01-02T19:08:00Z <p>Ultraman: /* Past Exams */</p> <hr /> <div>== Past Exams ==<br /> This is a list of all UNC Math Contests in the AoPSWiki. Users are encouraged to post additional solutions to the problems.<br /> <br /> * [[2006 UNCO Math Contest II]]<br /> * [[2007 UNCO Math Contest II]]<br /> * [[2008 UNCO Math Contest II]]<br /> * [[2009 UNCO Math Contest II]]<br /> * [[2010 UNCO Math Contest II]]<br /> * [[2011 UNCO Math Contest II]]<br /> * [[2012 UNCO Math Contest II]]<br /> * [[2013 UNCO Math Contest II]]<br /> * [[2014 UNCO Math Contest II]]<br /> * [[2015 UNCO Math Contest II]]<br /> * [[2016 UNCO Math Contest II]]<br /> * [[2017 UNCO Math Contest II]]<br /> * [[2018 UNCO Math Contest II]]<br /> * [[2019 UNCO Math Contest II]]<br /> <br /> ==Format &amp; Scoring==<br /> The contest consists of two rounds. Each round is a paper and pencil exam with about 9-12 problems. Students compete individually and no electronic devices are to be used in working the problems. The First Round of the contest is administered in early November in schools around the state and the answers are sent to UNC for grading. Students are allowed ninety minutes to work the First Round problems.<br /> Students who make high scores on the First Round of the contest are invited to the UNC campus in Greeley for the Final Round of the contest in late January. The Final Round is similar in spirit to the First Round, but the problems are more challenging. Students are allowed three hours to work the Final Round problems.<br /> Both rounds challenge students to exercise their creativity and ingenuity to solve problems in geometry, algebra, combinatorics, probability, and number theory. All the old contest exams are archived on the contest website.<br /> <br /> == Eligibility ==<br /> The contest is for all interested students in Colorado. It is designed for students in grades 7-12, but younger students are welcome to participate if they have an interest. All schools in the state, public, private, and home schools, are invited to administer the First Round of the contest in early November. <br /> <br /> == History ==<br /> The University of Northern Colorado Math Contest was founded in 1992 by Richard Grassl and has been directed by Ricardo Diaz since 2011-12.<br /> <br /> == Resources ==<br /> *[http://uncmathcontest.wordpress.com/ Here is the contest website]<br /> * [[American Mathematics Competitions]]<br /> * [[Colorado mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> [[Category: Math Contest Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114063 User:Ultraman 2020-01-02T19:04:27Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 4]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_4&diff=114060 2018 UNCO Math Contest II Problems/Problem 4 2020-01-02T19:02:57Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> How many positive integer factors of &lt;math&gt;36,000,000&lt;/math&gt; are not perfect squares?<br /> <br /> == Solution ==<br /> We can use [[complementary counting]]. Taking the prime factorization of &lt;math&gt;36,000,000&lt;/math&gt;, we get &lt;math&gt;2^8\cdot3^2\cdot5^6&lt;/math&gt;.So the total number of factors of &lt;math&gt;36,000,000&lt;/math&gt; is &lt;math&gt;(8+1)(2+1)(6+1) = 189&lt;/math&gt; factors. Now we need to find the number of factors that are perfect squares. So back to the prime factorization, &lt;math&gt;2^8\cdot3^2\cdot5^6 = 4^4\cdot9^1\cdot5^3&lt;/math&gt;. Now we get &lt;math&gt;(4+1)(1+1)(3+1)=40&lt;/math&gt; factors that are perfect squares. So there are &lt;math&gt;189-40=\boxed{149}&lt;/math&gt; positive integer factors that are not perfect squares.<br /> <br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_4&diff=114059 2018 UNCO Math Contest II Problems/Problem 4 2020-01-02T19:02:23Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> How many positive integer factors of &lt;math&gt;36,000,000&lt;/math&gt; are not perfect squares?<br /> <br /> == Solution ==<br /> We can use [[complementary counting]]. Taking the prime factorization of &lt;math&gt;36,000,000&lt;/math&gt;, we get &lt;math&gt;2^8\cdot3^2\cdot5^6&lt;/math&gt;.So the total number of factors of &lt;math&gt;36,000,000&lt;/math&gt; is &lt;math&gt;(8+1)(2+1)(6+1) = 189&lt;/math&gt; factors. <br /> <br /> Now we need to find the number of factors that are perfect squares. So back to the prime factorization, &lt;math&gt;2^8\cdot3^2\cdot5^6 = 4^4\cdot9^1\cdot5^3&lt;/math&gt;. Now we get &lt;math&gt;(4+1)(1+1)(3+1)=40&lt;/math&gt; factors that are perfect squares.<br /> <br /> So there are &lt;math&gt;189-40=\boxed149&lt;/math&gt; positive integer factors that are not perfect squares.<br /> <br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114054 User:Ultraman 2020-01-02T18:54:50Z <p>Ultraman: /* About Me */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> Also, please improve my solutions...<br /> <br /> ==Solutions==<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114053 User:Ultraman 2020-01-02T18:54:29Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> ==Solutions==<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_3&diff=114052 2018 UNCO Math Contest II Problems/Problem 3 2020-01-02T18:53:45Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find all values of &lt;math&gt;B&lt;/math&gt; that have the property that if &lt;math&gt;(x, y)&lt;/math&gt; lies on the hyperbola &lt;math&gt;2y^2-x^2 = 1&lt;/math&gt;,<br /> then so does the point &lt;math&gt;(3x + 4y, 2x + By)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We can write a system of equations - <br /> &lt;cmath&gt;2y^2-x^2 = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;2(2x + By)^2 - (3x+4y)^2 = 1&lt;/cmath&gt;<br /> <br /> Expanding the second equation, we get &lt;math&gt;-x^2+8Bxy-24xy+2B^2y^2-16y^2=1&lt;/math&gt;.<br /> <br /> Since we want this to look like &lt;math&gt;2y^2-x^2=1&lt;/math&gt;, we plug in B's that would put it into that form. If we plug in &lt;math&gt;B=3&lt;/math&gt;, things cancel, and we get &lt;math&gt;-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1&lt;/math&gt;. So &lt;math&gt;\boxed{B=3}&lt;/math&gt;<br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114050 User:Ultraman 2020-01-02T18:53:14Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> ==Solutions==<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> (please improve them)<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114049 User:Ultraman 2020-01-02T18:52:57Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> ==Solutions==<br /> (please improve them)<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114048 User:Ultraman 2020-01-02T18:52:50Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> ==Solutions==<br /> (please improve them)<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 3]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_3&diff=114046 2018 UNCO Math Contest II Problems/Problem 3 2020-01-02T18:52:08Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find all values of &lt;math&gt;B&lt;/math&gt; that have the property that if &lt;math&gt;(x, y)&lt;/math&gt; lies on the hyperbola &lt;math&gt;2y^2-x^2 = 1&lt;/math&gt;,<br /> then so does the point &lt;math&gt;(3x + 4y, 2x + By)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We can write a system of equations - <br /> &lt;cmath&gt;2y^2-x^2 = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;2(2x + By)^2 - (3x+4y)^2 = 1&lt;/cmath&gt;<br /> <br /> Expanding the second equation, we get &lt;math&gt;-x^2+8Bxy-24xy+2B^2y^2-16y^2=1&lt;/math&gt;<br /> Since we want this to look like &lt;math&gt;2y^2-x^2=1&lt;/math&gt;, we plug in B's that would put it into that form. If we plug in &lt;math&gt;B=3&lt;/math&gt;, things cancel, and we get &lt;math&gt;-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1&lt;/math&gt;. So &lt;math&gt;\boxed{B=3}&lt;/math&gt;<br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_3&diff=114043 2018 UNCO Math Contest II Problems/Problem 3 2020-01-02T18:51:50Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> Find all values of &lt;math&gt;B&lt;/math&gt; that have the property that if &lt;math&gt;(x, y)&lt;/math&gt; lies on the hyperbola &lt;math&gt;2y^2-x^2 = 1&lt;/math&gt;,<br /> then so does the point &lt;math&gt;(3x + 4y, 2x + By)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We can write a system of equations - <br /> &lt;cmath&gt;2y^2-x^2 = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;2(2x + By)^2 - (3x+4y)^2 = 1&lt;/cmath&gt;<br /> <br /> Expanding the second equation, we get &lt;math&gt;-x^2+8Bxy-24xy+2B^2y^2-16y^2=1&lt;/math&gt;<br /> Since we want this to look like &lt;math&gt;2y^2-x^2=1&lt;/math&gt;, we plug in B's that would put it into that form. If we plug in &lt;math&gt;B=3&lt;/math&gt;, things cancel, and we get &lt;math&gt;-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1&lt;/math&gt; So &lt;math&gt;\boxed{B=3}&lt;/math&gt;<br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_3&diff=114042 2018 UNCO Math Contest II Problems/Problem 3 2020-01-02T18:05:50Z <p>Ultraman: /* Problem */</p> <hr /> <div>== Problem ==<br /> Find all values of &lt;math&gt;B&lt;/math&gt; that have the property that if &lt;math&gt;(x, y)&lt;/math&gt; lies on the hyperbola &lt;math&gt;2y^2-x^2 = 1&lt;/math&gt;,<br /> then so does the point &lt;math&gt;(3x + 4y, 2x + By)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> &lt;math&gt;B=3&lt;/math&gt;<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_2&diff=114041 2018 UNCO Math Contest II Problems/Problem 2 2020-01-02T18:04:55Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> &lt;asy&gt;<br /> pair D=(0,0),B=(0,12),C=(16,0),A=(-9,0);<br /> draw(A--B--C--A,dot);<br /> draw(D--B,dot);<br /> MP(&quot;D&quot;,D,S);MP(&quot;A&quot;,A,S);MP(&quot;C&quot;,C,S);MP(&quot;B&quot;,B,N);<br /> MP(&quot;16&quot;,(8,0),S);MP(&quot;15&quot;,(A+B)/2,NW);<br /> &lt;/asy&gt;<br /> <br /> Segment AB is perpendicular to segment BC and<br /> segment AC is perpendicular to segment BD. If segment AB has length 15 and segment DC has length 16,<br /> then what is the area of triangle ABC?<br /> <br /> == Solution ==<br /> Let the altitude &lt;math&gt;BD&lt;/math&gt; be called &lt;math&gt;h&lt;/math&gt;, &lt;math&gt;AD&lt;/math&gt; be called &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; called &lt;math&gt;y&lt;/math&gt;. Since &lt;math&gt;ABC&lt;/math&gt; is a right triangle, we can write a system of equations.<br /> &lt;cmath&gt;h^2=16x&lt;/cmath&gt;<br /> &lt;cmath&gt;15^2=x(x+16)&lt;/cmath&gt;<br /> &lt;cmath&gt;y^2=16(16+x)&lt;/cmath&gt;<br /> <br /> Solving the second equation, we get &lt;math&gt;x^2+16x-225=0 \rightarrow (x-9)(x+25)=0 \rightarrow x=9&lt;/math&gt; or &lt;math&gt;-25&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; has to be positive, &lt;math&gt;x=9&lt;/math&gt;. Plugging &lt;math&gt;9&lt;/math&gt; into the first equation, we get that &lt;math&gt;h^2=4^2\cdot3^2 \rightarrow h=12&lt;/math&gt; or &lt;math&gt;-12&lt;/math&gt;. Once again, side lengths must be positive, so &lt;math&gt;h=12&lt;/math&gt;. Now we have an altitude and a base, so the area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{(9+16)\cdot12}{2} = \boxed{150}&lt;/math&gt;<br /> <br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114040 User:Ultraman 2020-01-02T18:04:37Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> ==Solutions==<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2018 UNCO Math Contest II Problems/Problem 2]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_2&diff=114039 2018 UNCO Math Contest II Problems/Problem 2 2020-01-02T18:03:36Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> &lt;asy&gt;<br /> pair D=(0,0),B=(0,12),C=(16,0),A=(-9,0);<br /> draw(A--B--C--A,dot);<br /> draw(D--B,dot);<br /> MP(&quot;D&quot;,D,S);MP(&quot;A&quot;,A,S);MP(&quot;C&quot;,C,S);MP(&quot;B&quot;,B,N);<br /> MP(&quot;16&quot;,(8,0),S);MP(&quot;15&quot;,(A+B)/2,NW);<br /> &lt;/asy&gt;<br /> <br /> Segment AB is perpendicular to segment BC and<br /> segment AC is perpendicular to segment BD. If segment AB has length 15 and segment DC has length 16,<br /> then what is the area of triangle ABC?<br /> <br /> == Solution ==<br /> Let the altitude &lt;math&gt;BD&lt;/math&gt; be called &lt;math&gt;h&lt;/math&gt;, &lt;math&gt;AD&lt;/math&gt; be called &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;BC&lt;/math&gt; called &lt;math&gt;y&lt;/math&gt;. Since &lt;math&gt;ABC&lt;/math&gt; is a right triangle, we can write a system of equations.<br /> &lt;cmath&gt;h^2=16x&lt;/cmath&gt;<br /> &lt;cmath&gt;15^2=x(x+16)&lt;/cmath&gt;<br /> &lt;cmath&gt;y^2=16(16+x)&lt;/cmath&gt;<br /> <br /> Solving the second equation, we get &lt;math&gt;x^2+16x-225=0 \rightarrow (x-9)(x+25)=0 \rightarrow x=9&lt;/math&gt; or &lt;math&gt;-25&lt;/math&gt;. Since &lt;math&gt;x&lt;/math&gt; has to be positive, &lt;math&gt;x=9&lt;/math&gt;. Plugging &lt;math&gt;9&lt;/math&gt; into the first equation, we get that &lt;math&gt;h^2=4^2\cdot3^2 \rightarrow h=12&lt;/math&gt; or &lt;math&gt;-12&lt;/math&gt;. Once again, side lengths must be positive, so &lt;math&gt;h=12&lt;/math&gt;. Now we have an altitude and a base, so the area of &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;\frac{(9+16)\cdot12}{2} = \boxed{150}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114038 User:Ultraman 2020-01-02T17:35:10Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> ==Solutions==<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> <br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_1&diff=114037 2018 UNCO Math Contest II Problems/Problem 1 2020-01-02T17:32:42Z <p>Ultraman: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A printer used 1890 digits to number all the pages in the Seripian Puzzle Book. How many pages are in the book? (For example, to number the pages in a book with twelve pages, the printer would use fifteen digits.)<br /> <br /> == Solution ==<br /> We know that we use &lt;math&gt;1&lt;/math&gt; digit &lt;math&gt;9&lt;/math&gt; times, &lt;math&gt;2&lt;/math&gt; digits &lt;math&gt;90&lt;/math&gt; times, and &lt;math&gt;3&lt;/math&gt; digits &lt;math&gt;900&lt;/math&gt; times. So if we have &lt;math&gt;999&lt;/math&gt; pages, we have &lt;math&gt;1 \cdot 9 + 2 \cdot 90 + 3 \cdot 900 = 2889&lt;/math&gt; digits. Since we want to have &lt;math&gt;1890&lt;/math&gt; digits, we do &lt;math&gt;\frac{2889 - 1890}{3}=333&lt;/math&gt; pages less than &lt;math&gt;999&lt;/math&gt;. So, &lt;math&gt;999-333=\boxed {666}&lt;/math&gt; pages.<br /> ~Ultraman<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2018|n=II|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Ultraman https://artofproblemsolving.com/wiki/index.php?title=User:Ultraman&diff=114036 User:Ultraman 2020-01-02T17:32:28Z <p>Ultraman: /* Solutions */</p> <hr /> <div>==About Me==<br /> I currently live in Texas, and I love playing [[badminton]], violin, Greed Control, using Unity, math, and most importantly, sleep.<br /> <br /> ==Solutions==<br /> [[2018 UNCO Math Contest II Problems/Problem 1]]<br /> [[2012 UNCO Math Contest II Problems/Problem 7]]<br /> <br /> ==Edits==<br /> [[Power of a Point Theorem/Introductory Problem 1]]<br /> <br /> [[2005 CEMC Gauss (Grade 7) Problems/Problem 2a]]<br /> <br /> [[2019 AMC 12A Problems/Problem 14]]<br /> <br /> [[2002 AMC 12B Problems/Problem 7]]<br /> <br /> [[2005 PMWC Problems/Problem T8]]<br /> <br /> [[2017 AMC 10B Problems/Problem 14]]<br /> <br /> [[2004 AMC 10A Problems/Problem 16]]</div> Ultraman