https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hinna&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-17T12:53:49Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_9&diff=82693 2009 AMC 12B Problems/Problem 9 2017-02-04T23:51:42Z <p>Hinna: /* Solution */</p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has vertices &lt;math&gt;A = (3,0)&lt;/math&gt;, &lt;math&gt;B = (0,3)&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, where &lt;math&gt;C&lt;/math&gt; is on the line &lt;math&gt;x + y = 7&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad<br /> \mathrm{(B)}\ 8\qquad<br /> \mathrm{(C)}\ 10\qquad<br /> \mathrm{(D)}\ 12\qquad<br /> \mathrm{(E)}\ 14&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> Because the line &lt;math&gt;x + y = 7&lt;/math&gt; is parallel to &lt;math&gt;\overline {AB}&lt;/math&gt;, the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is independent of the location of &lt;math&gt;C&lt;/math&gt; on that line. Therefore it may be assumed that &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;(7,0)&lt;/math&gt;. In that case the triangle has base &lt;math&gt;AC = 4&lt;/math&gt; and altitude &lt;math&gt;3&lt;/math&gt;, so its area is &lt;math&gt;\frac 12 \cdot 4 \cdot 3 = \boxed {6}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> The base of the triangle is &lt;math&gt;AB = \sqrt{3^2 + 3^2} = 3\sqrt 2&lt;/math&gt;. Its altitude is the distance between the point &lt;math&gt;A&lt;/math&gt; and the parallel line &lt;math&gt;x + y = 7&lt;/math&gt;, which is &lt;math&gt;\frac 4{\sqrt 2} = 2\sqrt 2&lt;/math&gt;. Therefore its area is &lt;math&gt;\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(3,0), B=(0,3);<br /> draw ( (-1,0) -- (9,0), dashed );<br /> draw ( (0,-1) -- (0,9), dashed );<br /> dot(A); dot(B); draw(A--B);<br /> draw ( (-1,8) -- (8,-1) );<br /> label( &quot;$A$&quot;, A, S );<br /> label( &quot;$B$&quot;, B, W );<br /> label( &quot;$3$&quot;, A--(0,0), S );<br /> label( &quot;$3$&quot;, B--(0,0), W );<br /> label( &quot;$x+y=7$&quot;, (8,-1), SE );<br /> pair C = intersectionpoint(A--(10,7),(7,0)--(0,7));<br /> draw( A--C, dashed );<br /> draw(rightanglemark(A,C,(7,0)));<br /> draw(rightanglemark(C,A,B));<br /> label( &quot;$4$&quot;, A--(7,0), S );<br /> label( &quot;$3\sqrt 2$&quot;, 0.67*B+0.33*A, NE );<br /> label( &quot;$\frac 4{\sqrt 2}$&quot;, A--C, NW );<br /> label( &quot;$\frac 4{\sqrt 2}$&quot;, C--(7,0), NE );<br /> &lt;/asy&gt;<br /> <br /> === Solution 3 ===<br /> By Shoelace, our area is:<br /> &lt;cmath&gt;\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.&lt;/cmath&gt;<br /> We know &lt;math&gt;x+y=7&lt;/math&gt; so we get:<br /> &lt;cmath&gt;\frac {1}{2} \cdot |9-21|=\boxed 6&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=1998_AHSME_Problems/Problem_19&diff=82669 1998 AHSME Problems/Problem 19 2017-02-04T15:08:11Z <p>Hinna: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> How many triangles have area &lt;math&gt;10&lt;/math&gt; and vertices at &lt;math&gt;(-5,0),(5,0)&lt;/math&gt; and &lt;math&gt;(5\cos \theta, 5\sin \theta)&lt;/math&gt; for some angle &lt;math&gt;\theta&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The triangle can be seen as having the base on the &lt;math&gt;x&lt;/math&gt; axis and height &lt;math&gt;|5\sin\theta|&lt;/math&gt;. The length of the base is &lt;math&gt;10&lt;/math&gt;, thus the height must be &lt;math&gt;2&lt;/math&gt;. The equation &lt;math&gt;|\sin\theta| = \frac 25&lt;/math&gt; has &lt;math&gt;\boxed{4}&lt;/math&gt; solutions, one in each quadrant.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(250);<br /> defaultpen(0.8);<br /> <br /> pair A=(-5,0), B=(5,0);<br /> dot(A); dot(B); dot((0,0));<br /> <br /> pair ip1[] = intersectionpoints( circle((0,0),5), (-6,2) -- (6,2) );<br /> pair ip2[] = intersectionpoints( circle((0,0),5), (-6,-2) -- (6,-2) );<br /> <br /> draw ( (-6,0) -- (6,0) );<br /> draw ( A -- ip1 -- B, Dotted );<br /> draw ( circle((0,0),5), red );<br /> draw ( (-6,2) -- (6,2), blue );<br /> draw ( (-6,-2) -- (6,-2), blue );<br /> <br /> dot(ip1);<br /> dot(ip1);<br /> <br /> dot(ip2);<br /> dot(ip2);<br /> <br /> <br /> label(&quot;$$(-5,0)$$&quot;, A, SW );<br /> label(&quot;$$(5,0)$$&quot;, B, SE );<br /> label(&quot;$$(0,0)$$&quot;, (0,0), SE );<br /> label(&quot;$$y=2$$&quot;, (6,2), N, blue );<br /> label(&quot;$$y=-2$$&quot;, (6,-2), S, blue );<br /> label(&quot;$$y=-2$$&quot;, (6,-2), S, blue );<br /> label(&quot;$$(5\cos\theta,5\sin\theta)$$&quot;, 5*dir(-45), SE, red );<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> <br /> Visually, the set of points of the form &lt;math&gt;(5\cos \theta, 5\sin \theta)&lt;/math&gt; is a circle centered at &lt;math&gt;(0,0)&lt;/math&gt; with radius 5. The missing vertex of the triangle must lie on this circle. At the same time, its distance from the &lt;math&gt;x&lt;/math&gt; axis must be 2. The set of all such points are precisely the lines &lt;math&gt;y=2&lt;/math&gt; and &lt;math&gt;y=-2&lt;/math&gt;, and each of these lines intersects the circle in two points.<br /> <br /> == Solution 2==<br /> Alternatively, we use shoelace to get:<br /> &lt;cmath&gt;\frac {1}{2}|(-5*0+5*5\sin (\theta)+5 \cos (\theta) *0)-(0*5+0*5 \cos (\theta)-5*5\sin (\theta))|=10 \implies \frac {1}{2}|50\sin (\theta)|=10&lt;/cmath&gt;<br /> This means &lt;math&gt;\sin (\theta)=\pm \frac {2}{5}&lt;/math&gt;. We see that if it equals &lt;math&gt;\frac {2}{5}&lt;/math&gt;, then &lt;math&gt;\cos (\theta)=\pm \frac {\sqrt {21}}{5}&lt;/math&gt;. Likewise, we see that if &lt;math&gt;\sin (\theta)=-\frac {2}{5}&lt;/math&gt;, then &lt;math&gt;\cos (\theta)&lt;/math&gt; has &lt;math&gt;2&lt;/math&gt; solutions. Thus, there are &lt;math&gt;\boxed {4}&lt;/math&gt; unique points such that the triangle has an area of &lt;math&gt;10&lt;/math&gt;, or &lt;math&gt;C&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1998|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=1998_AHSME_Problems/Problem_19&diff=82668 1998 AHSME Problems/Problem 19 2017-02-04T15:04:04Z <p>Hinna: /* Solution */</p> <hr /> <div>== Problem ==<br /> How many triangles have area &lt;math&gt;10&lt;/math&gt; and vertices at &lt;math&gt;(-5,0),(5,0)&lt;/math&gt; and &lt;math&gt;(5\cos \theta, 5\sin \theta)&lt;/math&gt; for some angle &lt;math&gt;\theta&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The triangle can be seen as having the base on the &lt;math&gt;x&lt;/math&gt; axis and height &lt;math&gt;|5\sin\theta|&lt;/math&gt;. The length of the base is &lt;math&gt;10&lt;/math&gt;, thus the height must be &lt;math&gt;2&lt;/math&gt;. The equation &lt;math&gt;|\sin\theta| = \frac 25&lt;/math&gt; has &lt;math&gt;\boxed{4}&lt;/math&gt; solutions, one in each quadrant.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(250);<br /> defaultpen(0.8);<br /> <br /> pair A=(-5,0), B=(5,0);<br /> dot(A); dot(B); dot((0,0));<br /> <br /> pair ip1[] = intersectionpoints( circle((0,0),5), (-6,2) -- (6,2) );<br /> pair ip2[] = intersectionpoints( circle((0,0),5), (-6,-2) -- (6,-2) );<br /> <br /> draw ( (-6,0) -- (6,0) );<br /> draw ( A -- ip1 -- B, Dotted );<br /> draw ( circle((0,0),5), red );<br /> draw ( (-6,2) -- (6,2), blue );<br /> draw ( (-6,-2) -- (6,-2), blue );<br /> <br /> dot(ip1);<br /> dot(ip1);<br /> <br /> dot(ip2);<br /> dot(ip2);<br /> <br /> <br /> label(&quot;$$(-5,0)$$&quot;, A, SW );<br /> label(&quot;$$(5,0)$$&quot;, B, SE );<br /> label(&quot;$$(0,0)$$&quot;, (0,0), SE );<br /> label(&quot;$$y=2$$&quot;, (6,2), N, blue );<br /> label(&quot;$$y=-2$$&quot;, (6,-2), S, blue );<br /> label(&quot;$$y=-2$$&quot;, (6,-2), S, blue );<br /> label(&quot;$$(5\cos\theta,5\sin\theta)$$&quot;, 5*dir(-45), SE, red );<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> <br /> Visually, the set of points of the form &lt;math&gt;(5\cos \theta, 5\sin \theta)&lt;/math&gt; is a circle centered at &lt;math&gt;(0,0)&lt;/math&gt; with radius 5. The missing vertex of the triangle must lie on this circle. At the same time, its distance from the &lt;math&gt;x&lt;/math&gt; axis must be 2. The set of all such points are precisely the lines &lt;math&gt;y=2&lt;/math&gt; and &lt;math&gt;y=-2&lt;/math&gt;, and each of these lines intersects the circle in two points.<br /> <br /> == Solution 2==<br /> Alternatively, we use shoelace to get:<br /> &lt;cmath&gt;10=\frac {1}{2}|50\sin (\theta)|&lt;/cmath&gt;<br /> This means &lt;math&gt;\sin (\theta)=\pm \frac {2}{5}&lt;/math&gt;. We see that if it equals &lt;math&gt;\frac {2}{5}&lt;/math&gt;, then &lt;math&gt;\cos (\theta)=\pm \frac {\sqrt {21}}{5}&lt;/math&gt;. Likewise, we see that if &lt;math&gt;\sin (\theta)=-\frac {2}{5}&lt;/math&gt;, then &lt;math&gt;\cos (\theta)&lt;/math&gt; has &lt;math&gt;2&lt;/math&gt; solutions. Thus, there are &lt;math&gt;\boxed {4}&lt;/math&gt; unique points such that the triangle has an area of &lt;math&gt;10&lt;/math&gt;, or &lt;math&gt;C&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1998|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=Balkan_Mathematical_Olympiad&diff=82055 Balkan Mathematical Olympiad 2016-12-28T17:38:53Z <p>Hinna: /* Romania */</p> <hr /> <div>The '''Balkan Mathematical Olympiad''' (BMO) is an annual contest for students from one of the [[#Member countries|member countries]] (Balkan area). In recent years the hosts have also invited some non-member guest countries.<br /> <br /> == Member countries ==<br /> ===Albania===<br /> ===Bosnia &amp; Herzegovina===<br /> ===Bulgaria===<br /> ===Cyprus===<br /> ''Main article: [[Cyprus_mathematics_competitions#Junior high-school (Gymnasium) Competitions|Junior high-school (Gymnasium) Math Competitions in Cyprus]].<br /> <br /> In Cyprus '''Four provincial competitions''' and a '''National (Pancyprian) competition''' are held every year. During this procedure 30 students are selected and 4 '''Team Selection Tests''' are held to determine who will be the six member of national team for BMO.<br /> <br /> *In every competition or test there are four problem usually covering [[geometry]], [[number theory]], [[algebra]], and [[combinatorics]] (elementary level) and last four hours each.<br /> <br /> ===Hellas===<br /> * Θαλής ([[Thales|Thalis]]) - first round<br /> * Ευκλείδης ([[Euclides|Efklidis]]) - second round<br /> * Αρχιμήδης ([[Archimedes|Archimidis]]) - third round<br /> <br /> ===Former Yugoslav Republic of Macedonia===<br /> ===Republic of Moldova===<br /> ===Romania===<br /> <br /> ===Turkey===<br /> ==Past BMOs==<br /> <br /> * The [[1984_Balkan MO|1st BMO]] was held in Athens, Greece in 1984.<br /> * The [[1985_Balkan MO|2nd BMO]] was held in Sofia, Bulgaria in 1985.<br /> * The [[1986_Balkan MO|3rd BMO]] was held in Bucharest, Romania in 1986.<br /> * The [[1987_Balkan MO|4th BMO]] was held in Athens, Greece in 1987.<br /> * The [[1988_Balkan MO|5th BMO]] was held in Nicosia, Cyprus in 1988.<br /> * The [[1989_Balkan MO|6th BMO]] was held in Split, Yugoslavia in 1989.<br /> * The [[1990_Balkan MO|7th BMO]] was held in Sofia, Bulgaria in 1990.<br /> * The [[1991_Balkan MO|8th BMO]] was held in Constanţa, Romania in 1991.<br /> * The [[1992_Balkan MO|9th BMO]] was held in Athens, Greece in 1992.<br /> * The [[1993_Balkan MO|10th BMO]] was held in Nicosia, Cyprus in 1993.<br /> * The [[1994_Balkan MO|11th BMO]] was held in Novi Sad, Yugoslavia in 1994.<br /> * The [[1995_Balkan MO|12th BMO]] was held in Plovdiv, Bulgaria in 1995.<br /> * The [[1996_Balkan MO|13th BMO]] was held in Bacău, Romania in 1996.<br /> * The [[1997_Balkan MO|14th BMO]] was held in Kalambaka, Greece in 1997.<br /> * The [[1998_Balkan MO|15th BMO]] was held in Nicosia, Cyprus in 1998.<br /> * The [[1999_Balkan MO|16th BMO]] was held in Ohrid, Former Yugoslav Republic of Macedonia in 1999.<br /> * The [[2000_Balkan MO|17th BMO]] was held in Chişinău, Moldova in 2000.<br /> * The [[2001_Balkan MO|18th BMO]] was held in Belgrade, Yugoslavia in 2001.<br /> * The [[2002_Balkan MO|19th BMO]] was held in Antalya, Turkey in 2002.<br /> * The [[2003_Balkan MO|20th BMO]] was held in Tirana, Albania in 2003.<br /> * The [[2004_Balkan MO|21st BMO]] was held in Pleven, Bulgaria in 2004.<br /> * The [[2005_Balkan MO|22nd BMO]] was held in Iaşi, Romania on 4-10 May, 2005. [http://www.rms.unibuc.ro/balkan.html]<br /> * The [[2006_Balkan MO|23rd BMO]] was held in Agros, Cyprus on 27 April-3 May, 2006. [http://www.cms.org.cy/memannouncBMO06.htm]<br /> * The [[2007_Balkan MO|24th BMO]] will be held in Rhodes, Greece in 2007.<br /> <br /> ==See also==<br /> *[[Balkan MO Problems and Solutions, with authors]]<br /> *[[Junior Balkan Mathematical Olympiad]]<br /> *[[Mathematics competition resources]]<br /> *[[Math books]]<br /> *[[List of mathematics competitions]]</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=1972_USAMO_Problems/Problem_1&diff=81689 1972 USAMO Problems/Problem 1 2016-11-28T03:33:25Z <p>Hinna: /* Solution 3 (very long, but detailed) */</p> <hr /> <div>==Problem==<br /> The symbols &lt;math&gt;(a,b,\ldots,g)&lt;/math&gt; and &lt;math&gt;[a,b,\ldots, g]&lt;/math&gt; denote the greatest common divisor and least common multiple, respectively, of the positive integers &lt;math&gt;a,b,\ldots, g&lt;/math&gt;. For example, &lt;math&gt;(3,6,18)=3&lt;/math&gt; and &lt;math&gt;[6,15]=30&lt;/math&gt;. Prove that<br /> &lt;center&gt;&lt;math&gt;\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}&lt;/math&gt;.&lt;/center&gt;<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> As all of the values in the given equation are positive, we can invert it to get an equivalent equation:<br /> &lt;center&gt;&lt;math&gt;\frac{[a,b][b,c][c,a]}{[a,b,c]^2}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}&lt;/math&gt;.&lt;/center&gt;<br /> We will now prove that both sides are equal, and that they are integers.<br /> <br /> Consider an arbitrary prime &lt;math&gt;p&lt;/math&gt;. Let &lt;math&gt;p^\alpha&lt;/math&gt;, &lt;math&gt;p^\beta&lt;/math&gt;, and &lt;math&gt;p^\gamma&lt;/math&gt; be the greatest powers of &lt;math&gt;p&lt;/math&gt; that divide &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. <br /> WLOG let &lt;math&gt;\alpha \leq \beta \leq \gamma&lt;/math&gt;.<br /> <br /> We can now, for each of the expressions in our equation, easily determine the largest power of &lt;math&gt;p&lt;/math&gt; that divides it. In this way we will find that the largest power of &lt;math&gt;p&lt;/math&gt; that divides the left hand side is &lt;math&gt;\beta+\gamma+\gamma-2\gamma = \beta&lt;/math&gt;, and the largest power of &lt;math&gt;p&lt;/math&gt; that divides the right hand side is &lt;math&gt;\alpha + \beta + \alpha - 2\alpha = \beta&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> Let &lt;math&gt;p = (a, b, c)&lt;/math&gt;, &lt;math&gt;pq = (a, b)&lt;/math&gt;, &lt;math&gt;pr = (b, c)&lt;/math&gt;, and &lt;math&gt;ps = (c, a)&lt;/math&gt;. Then it follows that &lt;math&gt;q, r, s&lt;/math&gt; are pairwise coprime and &lt;math&gt;a = pqsa'&lt;/math&gt;, &lt;math&gt;b = pqrb'&lt;/math&gt;, and &lt;math&gt;c = prsc'&lt;/math&gt;, with &lt;math&gt;a', b', c'&lt;/math&gt; pairwise coprime as well. Then, we wish to show<br /> &lt;cmath&gt;\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)},&lt;/cmath&gt;<br /> which can be checked fairly easily.<br /> <br /> ===Solution 3 (very long, but detailed)===<br /> <br /> Note: This solution is more of a &quot;bashy&quot; solution, and is easier to think of than the first 2 solutions.<br /> <br /> Dividing both sides of the equation by &lt;math&gt;(a,b,c)^2&lt;/math&gt;, and then multiplying both sides by &lt;math&gt;[a,b][b,c][c,a]&lt;/math&gt; gives us &lt;cmath&gt;\left(\frac{[a,b,c]}{(a,b,c)}\right)^2 = \frac{[a,b][b,c][c,a]}{(a,b)(b,c)(c,a)}.&lt;/cmath&gt;<br /> Now, we look at the prime factorisations of &lt;math&gt;a,b,c&lt;/math&gt;. Let <br /> &lt;cmath&gt;a=2^{x_1}\cdot 3^{x_2}\cdot 5^{x_3}\cdots,&lt;/cmath&gt;<br /> &lt;cmath&gt;b=2^{y_1}\cdot 3^{y_2}\cdot 5^{y_3}\cdots,&lt;/cmath&gt;<br /> &lt;cmath&gt;c=2^{z_1}\cdot 3^{z_2}\cdot 5^{z_3}\cdots,&lt;/cmath&gt;<br /> where all of the &lt;math&gt;x_1, y_1, z_1&lt;/math&gt; are nonnegative integers (they could be 0).<br /> Thus, we have &lt;cmath&gt;[a,b,c]=2^{\max(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3)} \cdots&lt;/cmath&gt;<br /> and &lt;cmath&gt;(a,b,c)=2^{\min(x_1,y_1,z_1)} \cdot 3^{\min(x_2,y_2,z_2)} \cdot 5^{\min(x_3,y_3,z_3)} \cdots.&lt;/cmath&gt;<br /> Now, we see that &lt;math&gt;\left(\frac{[a,b,c]}{(a,b,c)}\right) = 2^{\max(x_1,y_1,z_1) - \min(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2) - \min(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3) - \min(x_3,y_3,z_3)} \cdots.&lt;/math&gt; Squaring the LHS will just double all the exponents on the RHS.<br /> <br /> <br /> Also, we have &lt;math&gt;[a,b]=2^{\max(x_1,y_1)} \cdot 3^{\max(x_2,y_2)} \cdot 5^{\max(x_3,y_3)} \cdots&lt;/math&gt;. We can also build similar equations for &lt;math&gt;[b,c],[c,a],(a,b),(b,c)(c,a)&lt;/math&gt;, using 2 of &lt;math&gt;x,y,z&lt;/math&gt; and either the minimum of the exponents or the maximum.<br /> We see that when we multiply &lt;math&gt;[a,b]&lt;/math&gt;, &lt;math&gt;[b,c]&lt;/math&gt;, and &lt;math&gt;[a,c]&lt;/math&gt; together, the exponents of the prime factorization will be &lt;math&gt;\max(x_i,y_i)+ \max(y_i,z_i)+ \max(z_i,x_1)&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt; is chosen for the &lt;math&gt;i&lt;/math&gt;'th prime. <br /> When we multiply &lt;math&gt;(a,b)&lt;/math&gt;, &lt;math&gt;(b,c)&lt;/math&gt;, and &lt;math&gt;(a,c)&lt;/math&gt; together, the exponents of the prime factorization will be &lt;math&gt;\min(x_i,y_i)+ \min(y_i,z_i)+ \min(z_i,x_1)&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt; is chosen for the &lt;math&gt;i&lt;/math&gt;'th prime.<br /> <br /> Thus, when we divide the former by the latter, we have &lt;math&gt;\max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))&lt;/math&gt;. We wish to prove that this is equal to the exponents we got from &lt;math&gt;([a,b,c]/(a,b,c))^2&lt;/math&gt;.<br /> <br /> In other words, this problem is now down to proving that &lt;cmath&gt;2(\max(x_i,y_i,z_i) - \min(x_i,y_i,z_i)) = \max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))&lt;/cmath&gt; for any nonnegative integers &lt;math&gt;x_i, y_i, z_i&lt;/math&gt;. WLOG, let &lt;math&gt;x_i \le y_i \le z_i&lt;/math&gt;. Thus, computing maximums and minimums, this equation turns into &lt;cmath&gt;2(z_i - x_i) = y_i + z_i + z_i - x_i - y_i - x_i.&lt;/cmath&gt;<br /> <br /> Finally, canceling out the &lt;math&gt;y_i&lt;/math&gt;'s and collecting like terms, we see the RHS is &lt;math&gt;2z_i - 2x_i&lt;/math&gt;, which is equal to the LHS. Thus, this equation is true, and we are done. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> {{alternate solutions}}<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1972|before=First Question|num-a=2}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=1972_USAMO_Problems/Problem_1&diff=81688 1972 USAMO Problems/Problem 1 2016-11-28T03:32:48Z <p>Hinna: /* Solution 3 (very long, but detailed) */</p> <hr /> <div>==Problem==<br /> The symbols &lt;math&gt;(a,b,\ldots,g)&lt;/math&gt; and &lt;math&gt;[a,b,\ldots, g]&lt;/math&gt; denote the greatest common divisor and least common multiple, respectively, of the positive integers &lt;math&gt;a,b,\ldots, g&lt;/math&gt;. For example, &lt;math&gt;(3,6,18)=3&lt;/math&gt; and &lt;math&gt;[6,15]=30&lt;/math&gt;. Prove that<br /> &lt;center&gt;&lt;math&gt;\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}&lt;/math&gt;.&lt;/center&gt;<br /> <br /> ==Solutions==<br /> <br /> ===Solution 1===<br /> <br /> As all of the values in the given equation are positive, we can invert it to get an equivalent equation:<br /> &lt;center&gt;&lt;math&gt;\frac{[a,b][b,c][c,a]}{[a,b,c]^2}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}&lt;/math&gt;.&lt;/center&gt;<br /> We will now prove that both sides are equal, and that they are integers.<br /> <br /> Consider an arbitrary prime &lt;math&gt;p&lt;/math&gt;. Let &lt;math&gt;p^\alpha&lt;/math&gt;, &lt;math&gt;p^\beta&lt;/math&gt;, and &lt;math&gt;p^\gamma&lt;/math&gt; be the greatest powers of &lt;math&gt;p&lt;/math&gt; that divide &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. <br /> WLOG let &lt;math&gt;\alpha \leq \beta \leq \gamma&lt;/math&gt;.<br /> <br /> We can now, for each of the expressions in our equation, easily determine the largest power of &lt;math&gt;p&lt;/math&gt; that divides it. In this way we will find that the largest power of &lt;math&gt;p&lt;/math&gt; that divides the left hand side is &lt;math&gt;\beta+\gamma+\gamma-2\gamma = \beta&lt;/math&gt;, and the largest power of &lt;math&gt;p&lt;/math&gt; that divides the right hand side is &lt;math&gt;\alpha + \beta + \alpha - 2\alpha = \beta&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> Let &lt;math&gt;p = (a, b, c)&lt;/math&gt;, &lt;math&gt;pq = (a, b)&lt;/math&gt;, &lt;math&gt;pr = (b, c)&lt;/math&gt;, and &lt;math&gt;ps = (c, a)&lt;/math&gt;. Then it follows that &lt;math&gt;q, r, s&lt;/math&gt; are pairwise coprime and &lt;math&gt;a = pqsa'&lt;/math&gt;, &lt;math&gt;b = pqrb'&lt;/math&gt;, and &lt;math&gt;c = prsc'&lt;/math&gt;, with &lt;math&gt;a', b', c'&lt;/math&gt; pairwise coprime as well. Then, we wish to show<br /> &lt;cmath&gt;\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \frac{p^2}{(pq)(pr)(ps)},&lt;/cmath&gt;<br /> which can be checked fairly easily.<br /> <br /> ===Solution 3 (very long, but detailed)===<br /> <br /> Note: This solution is more of a &quot;bashy&quot; solution, and is easier to think of than the first 2 solutions.<br /> <br /> Dividing both sides of the equation by &lt;math&gt;(a,b,c)^2&lt;/math&gt;, and then multiplying both sides by &lt;math&gt;[a,b][b,c][c,a]&lt;/math&gt; gives us &lt;cmath&gt;\left(\frac{[a,b,c]}{(a,b,c)}\right)^2 = \frac{[a,b][b,c][c,a]}{(a,b)(b,c)(c,a)}.&lt;/cmath&gt;<br /> Now, we look at the prime factorisations of &lt;math&gt;a,b,c&lt;/math&gt;. Let <br /> &lt;cmath&gt;a=2^{x_1}\cdot 3^{x_2}\cdot 5^{x_3}\cdots,&lt;/cmath&gt;<br /> &lt;cmath&gt;b=2^{y_1}\cdot 3^{y_2}\cdot 5^{y_3}\cdots,&lt;/cmath&gt;<br /> &lt;cmath&gt;c=2^{z_1}\cdot 3^{z_2}\cdot 5^{z_3}\cdots,&lt;/cmath&gt;<br /> where all of the &lt;math&gt;x_1, y_1, z_1&lt;/math&gt; are nonnegative integers (they could be 0).<br /> Thus, we have &lt;cmath&gt;[a,b,c]=2^{\max(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3)} \cdots&lt;/cmath&gt;<br /> and &lt;cmath&gt;(a,b,c)=2^{\min(x_1,y_1,z_1)} \cdot 3^{\min(x_2,y_2,z_2)} \cdot 5^{\min(x_3,y_3,z_3)} \cdots.&lt;/cmath&gt;<br /> Now, we see that &lt;math&gt;(\frac{[a,b,c]}{(a,b,c)}) = 2^{\max(x_1,y_1,z_1) - \min(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2) - \min(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3) - \min(x_3,y_3,z_3)} \cdots.&lt;/math&gt; Squaring the LHS will just double all the exponents on the RHS.<br /> <br /> <br /> Also, we have &lt;math&gt;[a,b]=2^{\max(x_1,y_1)} \cdot 3^{\max(x_2,y_2)} \cdot 5^{\max(x_3,y_3)} \cdots&lt;/math&gt;. We can also build similar equations for &lt;math&gt;[b,c],[c,a],(a,b),(b,c)(c,a)&lt;/math&gt;, using 2 of &lt;math&gt;x,y,z&lt;/math&gt; and either the minimum of the exponents or the maximum.<br /> We see that when we multiply &lt;math&gt;[a,b]&lt;/math&gt;, &lt;math&gt;[b,c]&lt;/math&gt;, and &lt;math&gt;[a,c]&lt;/math&gt; together, the exponents of the prime factorization will be &lt;math&gt;\max(x_i,y_i)+ \max(y_i,z_i)+ \max(z_i,x_1)&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt; is chosen for the &lt;math&gt;i&lt;/math&gt;'th prime. <br /> When we multiply &lt;math&gt;(a,b)&lt;/math&gt;, &lt;math&gt;(b,c)&lt;/math&gt;, and &lt;math&gt;(a,c)&lt;/math&gt; together, the exponents of the prime factorization will be &lt;math&gt;\min(x_i,y_i)+ \min(y_i,z_i)+ \min(z_i,x_1)&lt;/math&gt;, where &lt;math&gt;i&lt;/math&gt; is chosen for the &lt;math&gt;i&lt;/math&gt;'th prime.<br /> <br /> Thus, when we divide the former by the latter, we have &lt;math&gt;\max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))&lt;/math&gt;. We wish to prove that this is equal to the exponents we got from &lt;math&gt;([a,b,c]/(a,b,c))^2&lt;/math&gt;.<br /> <br /> In other words, this problem is now down to proving that &lt;cmath&gt;2(\max(x_i,y_i,z_i) - \min(x_i,y_i,z_i)) = \max(x_i,y_i)+ \max(y_i,z_i)+ \max(x_i,z_i) - (\min(x_i,y_i)+ \min(y_i,z_i)+ \min(x_i,z_i))&lt;/cmath&gt; for any nonnegative integers &lt;math&gt;x_i, y_i, z_i&lt;/math&gt;. WLOG, let &lt;math&gt;x_i \le y_i \le z_i&lt;/math&gt;. Thus, computing maximums and minimums, this equation turns into &lt;cmath&gt;2(z_i - x_i) = y_i + z_i + z_i - x_i - y_i - x_i.&lt;/cmath&gt;<br /> <br /> Finally, canceling out the &lt;math&gt;y_i&lt;/math&gt;'s and collecting like terms, we see the RHS is &lt;math&gt;2z_i - 2x_i&lt;/math&gt;, which is equal to the LHS. Thus, this equation is true, and we are done. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> {{alternate solutions}}<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1972|before=First Question|num-a=2}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_13&diff=81515 2016 AMC 8 Problems/Problem 13 2016-11-23T21:19:54Z <p>Hinna: /* Solution 2 */</p> <hr /> <div>Two different numbers are randomly selected from the set &lt;math&gt;{ - 2, -1, 0, 3, 4, 5}&lt;/math&gt; and multiplied together. What is the probability that the product is &lt;math&gt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The product can only be &lt;math&gt;0&lt;/math&gt; if one of the numbers is 0. Once we chose &lt;math&gt;0&lt;/math&gt;, there are &lt;math&gt;5&lt;/math&gt; ways we can chose the second number, or &lt;math&gt;6-1&lt;/math&gt;. There are &lt;math&gt;\dbinom{6}{2}&lt;/math&gt; ways we can chose &lt;math&gt;2&lt;/math&gt; numbers randomly, and that is &lt;math&gt;15&lt;/math&gt;. So, &lt;math&gt;\frac{5}{15}=\frac{1}{3}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(D)} \, \frac{1}{3}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> There are a total of &lt;math&gt;36&lt;/math&gt; possibilities. We want &lt;math&gt;0&lt;/math&gt; so one of the multiples is &lt;math&gt;0&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; possibilities where &lt;math&gt;0&lt;/math&gt; is chosen for the first number and there are &lt;math&gt;6&lt;/math&gt; ways for &lt;math&gt;0&lt;/math&gt; to be chosen as the second number. We seek &lt;math&gt;\frac {6+6}{36}=\frac {1}{3}&lt;/math&gt;<br /> <br /> {{AMC8 box|year=2016|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_13&diff=81514 2016 AMC 8 Problems/Problem 13 2016-11-23T21:19:43Z <p>Hinna: /* Solution */</p> <hr /> <div>Two different numbers are randomly selected from the set &lt;math&gt;{ - 2, -1, 0, 3, 4, 5}&lt;/math&gt; and multiplied together. What is the probability that the product is &lt;math&gt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The product can only be &lt;math&gt;0&lt;/math&gt; if one of the numbers is 0. Once we chose &lt;math&gt;0&lt;/math&gt;, there are &lt;math&gt;5&lt;/math&gt; ways we can chose the second number, or &lt;math&gt;6-1&lt;/math&gt;. There are &lt;math&gt;\dbinom{6}{2}&lt;/math&gt; ways we can chose &lt;math&gt;2&lt;/math&gt; numbers randomly, and that is &lt;math&gt;15&lt;/math&gt;. So, &lt;math&gt;\frac{5}{15}=\frac{1}{3}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(D)} \, \frac{1}{3}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> There are a total of &lt;math&gt;36&lt;/math&gt; possibilities. We want &lt;math&gt;0&lt;/math&gt; so one of the multiples is &lt;math&gt;0&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; possibilities where &lt;math&gt;0&lt;/math&gt; is chosen for the first number and there are &lt;math&gt;6&lt;/math&gt; ways for &lt;math&gt;0&lt;/math&gt; to be chosen as the second number. We seek &lt;math&gt;\frac {6+6}{36}=\frac {1}{3}&lt;/math&gt;<br /> <br /> {{AMC8 box|year=2016|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> ==Solution 2==<br /> There are a total of &lt;math&gt;36&lt;/math&gt; possibilities. We want &lt;math&gt;0&lt;/math&gt; so one of the multiples is &lt;math&gt;0&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; possibilities where &lt;math&gt;0&lt;/math&gt; is chosen for the first number and there are &lt;math&gt;6&lt;/math&gt; ways for &lt;math&gt;0&lt;/math&gt; to be chosen as the second number. We seek &lt;math&gt;\frac {6+6}{36}=\frac {1}{3}&lt;/math&gt;</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_13&diff=81513 2016 AMC 8 Problems/Problem 13 2016-11-23T21:18:43Z <p>Hinna: /* Solution */</p> <hr /> <div>Two different numbers are randomly selected from the set &lt;math&gt;{ - 2, -1, 0, 3, 4, 5}&lt;/math&gt; and multiplied together. What is the probability that the product is &lt;math&gt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The product can only be &lt;math&gt;0&lt;/math&gt; if one of the numbers is 0. Once we chose &lt;math&gt;0&lt;/math&gt;, there are &lt;math&gt;5&lt;/math&gt; ways we can chose the second number, or &lt;math&gt;6-1&lt;/math&gt;. There are &lt;math&gt;\dbinom{6}{2}&lt;/math&gt; ways we can chose &lt;math&gt;2&lt;/math&gt; numbers randomly, and that is &lt;math&gt;15&lt;/math&gt;. So, &lt;math&gt;\frac{5}{15}=\frac{1}{3}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(D)} \, \frac{1}{3}}&lt;/math&gt;<br /> {{AMC8 box|year=2016|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> ==Solution 2==<br /> There are a total of &lt;math&gt;36&lt;/math&gt; possibilities. We want &lt;math&gt;0&lt;/math&gt; so one of the multiples is &lt;math&gt;0&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; possibilities where &lt;math&gt;0&lt;/math&gt; is chosen for the first number and there are &lt;math&gt;6&lt;/math&gt; ways for &lt;math&gt;0&lt;/math&gt; to be chosen as the second number. We seek &lt;math&gt;\frac {6+6}{36}=\frac {1}{3}&lt;/math&gt;</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_5&diff=80357 1984 AIME Problems/Problem 5 2016-09-24T01:57:22Z <p>Hinna: /* Solution */</p> <hr /> <div>== Problem ==<br /> Determine the value of &lt;math&gt;ab&lt;/math&gt; if &lt;math&gt;\log_8a+\log_4b^2=5&lt;/math&gt; and &lt;math&gt;\log_8b+\log_4a^2=7&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> === Solution 1 ===<br /> Use the [[change of base formula]] to see that &lt;math&gt;\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5&lt;/math&gt;; combine [[denominator]]s to find that &lt;math&gt;\frac{\log ab^3}{3\log 2} = 5&lt;/math&gt;. Doing the same thing with the second equation yields that &lt;math&gt;\frac{\log a^3b}{3\log 2} = 7&lt;/math&gt;. This means that &lt;math&gt;\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}&lt;/math&gt; and that &lt;math&gt;\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}&lt;/math&gt;. If we multiply the two equations together, we get that &lt;math&gt;a^4b^4 = 2^{36}&lt;/math&gt;, so taking the fourth root of that, &lt;math&gt;ab = 2^9 = \boxed{512}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become &lt;math&gt;\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5&lt;/math&gt; and &lt;math&gt;\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7&lt;/math&gt;. Adding the equations and factoring, we get &lt;math&gt;(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12&lt;/math&gt;. Rearranging we see that &lt;math&gt;\ln ab = \frac{12}{\frac{1}{\ln 8}+\frac{2}{\ln 4}}&lt;/math&gt;. Again, we pull exponents out of our logarithms to get &lt;math&gt;\ln ab = \frac{12}{\frac{1}{3 \ln 2} + \frac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2&lt;/math&gt;. This means that &lt;math&gt;\frac{\ln ab}{\ln 2} = 9&lt;/math&gt;. The left-hand side can be interpreted as a base-2 logarithm, giving us &lt;math&gt;ab = 2^9 = \boxed{512}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> This solution is very similar to the above two, but it utilizes the well-known fact that &lt;math&gt; \log_{m^k}{n^k}= \log_m{n}. &lt;/math&gt; Thus, &lt;math&gt; \log_8a+\log_4b^2=5 \Rightarrow <br /> \log_{2^3}{(\sqrt{a})^3} + \log_{2^2}{b^2} = 5 \Rightarrow \log_2{\sqrt{a}} + \log_2{b} = 5 \Rightarrow \log_2{\sqrt{a}b} = 5. &lt;/math&gt; Similarly, &lt;math&gt; \log_8b+\log_4a^2=7 \Rightarrow \log_2{\sqrt{b}a} = 7. &lt;/math&gt; Adding these two equations, we have &lt;math&gt; \log_2{a^{\frac{4}{3}}b^{\frac{4}{3}}} = 12 \Rightarrow ab = 2^{12\times\frac{3}{4}} = 2^9 = \boxed{512} &lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> We can change everything to a common base, like so: &lt;math&gt;\log_8{a} + \log_8{b^3} = 5,&lt;/math&gt; &lt;math&gt;\log_8{b} + \log_8{a^3} = 7.&lt;/math&gt; We set the value of &lt;math&gt;\log_8{a}&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt;, and the value of &lt;math&gt;\log_8{b}&lt;/math&gt; to &lt;math&gt;y.&lt;/math&gt; Now we have a system of linear equations: &lt;cmath&gt;x + 3y = 5,&lt;/cmath&gt; &lt;cmath&gt;y + 3x = 7.&lt;/cmath&gt; We multiply the second equation by three and subtract the first equation from the second to get &lt;math&gt;8x = 16,&lt;/math&gt; which gives &lt;math&gt;x=2.&lt;/math&gt; We substitute our value into the first equation and find that &lt;math&gt;y = 1.&lt;/math&gt; Now we know that &lt;math&gt;\log_8{a} = 2,&lt;/math&gt; and &lt;math&gt;\log_8{b} = 1,&lt;/math&gt; so we find that &lt;math&gt;a = 64,&lt;/math&gt; and &lt;math&gt;b=8.&lt;/math&gt; Multiplying together gives us &lt;math&gt;ab = \boxed{512}.&lt;/math&gt;<br /> <br /> === Solution 5 ===<br /> Add the two equations to get &lt;math&gt;\log_8 {a}+ \log_8 {b}+ \log_{a^2}+\log_{b^2}=12&lt;/math&gt;. This can be simplified with the log property &lt;math&gt;\log_n {x}+\log_n {y}=log_n {xy}&lt;/math&gt;. Using this, we get &lt;math&gt;\log_8 {ab}+ \log_4 {a^2b^2}=12&lt;/math&gt;. Now let &lt;math&gt;\log_8 {ab}=c&lt;/math&gt; and &lt;math&gt;\log_4 {a^2b^2}=k&lt;/math&gt;. Converting to exponents, we get &lt;math&gt;8^c=ab&lt;/math&gt; and &lt;math&gt;4^k=(ab)^2&lt;/math&gt;. Sub in the &lt;math&gt;8^c&lt;/math&gt; to get &lt;math&gt;k=3c&lt;/math&gt;. So now we have that &lt;math&gt;k+c=12&lt;/math&gt; and &lt;math&gt;k=3c&lt;/math&gt; which gives &lt;math&gt;c=3&lt;/math&gt;, &lt;math&gt;k=9&lt;/math&gt;. This means &lt;math&gt;\log_4 {a^2b^2}=9&lt;/math&gt; so &lt;math&gt;4^9=(ab)^2 \implies ab=(2^2)^9 \implies 2^9 \implies \boxed {512}&lt;/math&gt;<br /> <br /> == See also ==<br /> *[[Logarithm]]<br /> <br /> {{AIME box|year=1984|num-b=4|num-a=6}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_17&diff=79906 2016 AMC 12B Problems/Problem 17 2016-08-08T23:02:05Z <p>Hinna: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; shown in the figure, &lt;math&gt;AB=7&lt;/math&gt;, &lt;math&gt;BC=8&lt;/math&gt;, &lt;math&gt;CA=9&lt;/math&gt;, and &lt;math&gt;\overline{AH}&lt;/math&gt; is an altitude. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively, so that &lt;math&gt;\overline{BD}&lt;/math&gt; and &lt;math&gt;\overline{CE}&lt;/math&gt; are angle bisectors, intersecting &lt;math&gt;\overline{AH}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, respectively. What is &lt;math&gt;PQ&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import graph; size(9cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -4.381056062031275, xmax = 15.020004395092375, ymin = -4.051697595316909, ymax = 10.663513514111651; /* image dimensions */<br /> <br /> <br /> draw((0.,0.)--(4.714285714285714,7.666518779999279)--(7.,0.)--cycle); <br /> /* draw figures */<br /> draw((0.,0.)--(4.714285714285714,7.666518779999279)); <br /> draw((4.714285714285714,7.666518779999279)--(7.,0.)); <br /> draw((7.,0.)--(0.,0.)); <br /> label(&quot;7&quot;,(3.2916797119724284,-0.07831656949355523),SE*labelscalefactor); <br /> label(&quot;9&quot;,(2.0037562070503783,4.196493361737088),SE*labelscalefactor); <br /> label(&quot;8&quot;,(6.114150371695219,3.785453945272603),SE*labelscalefactor); <br /> draw((0.,0.)--(6.428571428571427,1.9166296949998194)); <br /> draw((7.,0.)--(2.2,3.5777087639996634)); <br /> draw((4.714285714285714,7.666518779999279)--(3.7058823529411766,0.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-0.2432592696221352,-0.5715638692509372), NE * labelscalefactor); <br /> dot((7.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (7.0458397156813835,-0.48935598595804014), NE * labelscalefactor); <br /> dot((3.7058823529411766,0.),dotstyle); <br /> label(&quot;$E$&quot;, (3.8123296394941084,0.16830708038513573), NE * labelscalefactor); <br /> dot((4.714285714285714,7.666518779999279),dotstyle); <br /> label(&quot;$C$&quot;, (4.579603216894479,7.895848109917452), NE * labelscalefactor); <br /> dot((2.2,3.5777087639996634),linewidth(3.pt) + dotstyle); <br /> label(&quot;$D$&quot;, (2.1407693458718726,3.127790878929427), NE * labelscalefactor); <br /> dot((6.428571428571427,1.9166296949998194),linewidth(3.pt) + dotstyle); <br /> label(&quot;$H$&quot;, (6.004539860638023,1.9494778850645704), NE * labelscalefactor); <br /> dot((5.,1.49071198499986),linewidth(3.pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (4.935837377830365,1.7302568629501784), NE * labelscalefactor); <br /> dot((3.857142857142857,1.1499778169998918),linewidth(3.pt) + dotstyle); <br /> label(&quot;$P$&quot;, (3.538303361851119,1.2370095631927964), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad<br /> \textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad<br /> \textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad<br /> \textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad<br /> \textbf{(E)}\ \frac{6}{5}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Get the area of the triangle by heron's formula: <br /> &lt;cmath&gt;\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12)(3)(4)(5)} = 12\sqrt{5}&lt;/cmath&gt;<br /> Use the area to find the height AH with known base BC:<br /> &lt;cmath&gt;Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)&lt;/cmath&gt;<br /> &lt;cmath&gt;AH = 3\sqrt{5}&lt;/cmath&gt;<br /> &lt;cmath&gt;BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2&lt;/cmath&gt;<br /> &lt;cmath&gt;CH = BC - BH = 8 - 2 = 6&lt;/cmath&gt;<br /> Apply angle bisector theorem on triangle &lt;math&gt;ACH&lt;/math&gt; and triangle &lt;math&gt;ABH&lt;/math&gt;, we get &lt;math&gt;AP:PH = 9:6&lt;/math&gt; and &lt;math&gt;AQ:QH = 7:2&lt;/math&gt;, respectively.<br /> From now, you can simply use the answer choices because only choice &lt;math&gt;\textbf{D}&lt;/math&gt; has &lt;math&gt;\sqrt{5}&lt;/math&gt; in it and we know that &lt;math&gt;AH = 3\sqrt{5}&lt;/math&gt; the segments on it all have integral lengths so that &lt;math&gt;\sqrt{5}&lt;/math&gt; will remain there.<br /> However, by scaling up the length ratio:<br /> &lt;math&gt;AH:AP:PH = 45:27:18&lt;/math&gt; and &lt;math&gt;AQ:QH =45:35:10&lt;/math&gt;.<br /> we get &lt;math&gt;AH:PQ = 45:(18 - 10) = 45 : 8&lt;/math&gt;.<br /> &lt;cmath&gt;PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}&lt;/cmath&gt;<br /> <br /> Solution 2: (choices+faster)<br /> As in the above solution, let's find the area. By heron's, it is indeed &lt;math&gt;12\sqrt{5}&lt;/math&gt;. Because &lt;math&gt;\overline {AH}&lt;/math&gt; is an altitude and we are given a base, let's find &lt;math&gt;\overline {AH}&lt;/math&gt;. We set up the area is equal to &lt;math&gt;\frac {1}{2}<br /> \cdot b \cdot h \implies 12\sqrt5=\overline {AH}\cdot \frac {1}{2}\cdot 8 \implies \overline {AH}=3\sqrt5&lt;/math&gt; and since the answer should have &lt;math&gt;\sqrt 5&lt;/math&gt; in it, our answer is &lt;math&gt;D&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_17&diff=79905 2016 AMC 12B Problems/Problem 17 2016-08-08T23:01:02Z <p>Hinna: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; shown in the figure, &lt;math&gt;AB=7&lt;/math&gt;, &lt;math&gt;BC=8&lt;/math&gt;, &lt;math&gt;CA=9&lt;/math&gt;, and &lt;math&gt;\overline{AH}&lt;/math&gt; is an altitude. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively, so that &lt;math&gt;\overline{BD}&lt;/math&gt; and &lt;math&gt;\overline{CE}&lt;/math&gt; are angle bisectors, intersecting &lt;math&gt;\overline{AH}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, respectively. What is &lt;math&gt;PQ&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import graph; size(9cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -4.381056062031275, xmax = 15.020004395092375, ymin = -4.051697595316909, ymax = 10.663513514111651; /* image dimensions */<br /> <br /> <br /> draw((0.,0.)--(4.714285714285714,7.666518779999279)--(7.,0.)--cycle); <br /> /* draw figures */<br /> draw((0.,0.)--(4.714285714285714,7.666518779999279)); <br /> draw((4.714285714285714,7.666518779999279)--(7.,0.)); <br /> draw((7.,0.)--(0.,0.)); <br /> label(&quot;7&quot;,(3.2916797119724284,-0.07831656949355523),SE*labelscalefactor); <br /> label(&quot;9&quot;,(2.0037562070503783,4.196493361737088),SE*labelscalefactor); <br /> label(&quot;8&quot;,(6.114150371695219,3.785453945272603),SE*labelscalefactor); <br /> draw((0.,0.)--(6.428571428571427,1.9166296949998194)); <br /> draw((7.,0.)--(2.2,3.5777087639996634)); <br /> draw((4.714285714285714,7.666518779999279)--(3.7058823529411766,0.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-0.2432592696221352,-0.5715638692509372), NE * labelscalefactor); <br /> dot((7.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (7.0458397156813835,-0.48935598595804014), NE * labelscalefactor); <br /> dot((3.7058823529411766,0.),dotstyle); <br /> label(&quot;$E$&quot;, (3.8123296394941084,0.16830708038513573), NE * labelscalefactor); <br /> dot((4.714285714285714,7.666518779999279),dotstyle); <br /> label(&quot;$C$&quot;, (4.579603216894479,7.895848109917452), NE * labelscalefactor); <br /> dot((2.2,3.5777087639996634),linewidth(3.pt) + dotstyle); <br /> label(&quot;$D$&quot;, (2.1407693458718726,3.127790878929427), NE * labelscalefactor); <br /> dot((6.428571428571427,1.9166296949998194),linewidth(3.pt) + dotstyle); <br /> label(&quot;$H$&quot;, (6.004539860638023,1.9494778850645704), NE * labelscalefactor); <br /> dot((5.,1.49071198499986),linewidth(3.pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (4.935837377830365,1.7302568629501784), NE * labelscalefactor); <br /> dot((3.857142857142857,1.1499778169998918),linewidth(3.pt) + dotstyle); <br /> label(&quot;$P$&quot;, (3.538303361851119,1.2370095631927964), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad<br /> \textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad<br /> \textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad<br /> \textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad<br /> \textbf{(E)}\ \frac{6}{5}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Get the area of the triangle by heron's formula: <br /> &lt;cmath&gt;\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12)(3)(4)(5)} = 12\sqrt{5}&lt;/cmath&gt;<br /> Use the area to find the height AH with known base BC:<br /> &lt;cmath&gt;Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)&lt;/cmath&gt;<br /> &lt;cmath&gt;AH = 3\sqrt{5}&lt;/cmath&gt;<br /> &lt;cmath&gt;BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2&lt;/cmath&gt;<br /> &lt;cmath&gt;CH = BC - BH = 8 - 2 = 6&lt;/cmath&gt;<br /> Apply angle bisector theorem on triangle &lt;math&gt;ACH&lt;/math&gt; and triangle &lt;math&gt;ABH&lt;/math&gt;, we get &lt;math&gt;AP:PH = 9:6&lt;/math&gt; and &lt;math&gt;AQ:QH = 7:2&lt;/math&gt;, respectively.<br /> From now, you can simply use the answer choices because only choice &lt;math&gt;\textbf{D}&lt;/math&gt; has &lt;math&gt;\sqrt{5}&lt;/math&gt; in it and we know that &lt;math&gt;AH = 3\sqrt{5}&lt;/math&gt; the segments on it all have integral lengths so that &lt;math&gt;\sqrt{5}&lt;/math&gt; will remain there.<br /> However, by scaling up the length ratio:<br /> &lt;math&gt;AH:AP:PH = 45:27:18&lt;/math&gt; and &lt;math&gt;AQ:QH =45:35:10&lt;/math&gt;.<br /> we get &lt;math&gt;AH:PQ = 45:(18 - 10) = 45 : 8&lt;/math&gt;.<br /> &lt;cmath&gt;PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}&lt;/cmath&gt;<br /> <br /> Solution 2: (choices+faster)<br /> As in the above solution, let's find the area. By heron's, it is indeed &lt;math&gt;12\sqrt{5}&lt;/math&gt;. Because &lt;math&gt;\overline {AH}&lt;/math&gt; is an altitude and we are given a base, let's find &lt;math&gt;\overline {AH}&lt;/math&gt;. We set up the area is equal to &lt;math&gt;\frac {1}{2}<br /> cdot b \cdot h \implies 12/sqrt5=\overline {AH}\cdot \frac {1}{2}\cdot 8 \implies \overline {AH}=3\sqrt5&lt;/math&gt; and since the answer should have &lt;math&gt;\sqrt 5&lt;/math&gt; in it so our answer is &lt;math&gt;D&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Hinna https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_17&diff=79904 2016 AMC 12B Problems/Problem 17 2016-08-08T23:00:06Z <p>Hinna: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; shown in the figure, &lt;math&gt;AB=7&lt;/math&gt;, &lt;math&gt;BC=8&lt;/math&gt;, &lt;math&gt;CA=9&lt;/math&gt;, and &lt;math&gt;\overline{AH}&lt;/math&gt; is an altitude. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively, so that &lt;math&gt;\overline{BD}&lt;/math&gt; and &lt;math&gt;\overline{CE}&lt;/math&gt; are angle bisectors, intersecting &lt;math&gt;\overline{AH}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, respectively. What is &lt;math&gt;PQ&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import graph; size(9cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -4.381056062031275, xmax = 15.020004395092375, ymin = -4.051697595316909, ymax = 10.663513514111651; /* image dimensions */<br /> <br /> <br /> draw((0.,0.)--(4.714285714285714,7.666518779999279)--(7.,0.)--cycle); <br /> /* draw figures */<br /> draw((0.,0.)--(4.714285714285714,7.666518779999279)); <br /> draw((4.714285714285714,7.666518779999279)--(7.,0.)); <br /> draw((7.,0.)--(0.,0.)); <br /> label(&quot;7&quot;,(3.2916797119724284,-0.07831656949355523),SE*labelscalefactor); <br /> label(&quot;9&quot;,(2.0037562070503783,4.196493361737088),SE*labelscalefactor); <br /> label(&quot;8&quot;,(6.114150371695219,3.785453945272603),SE*labelscalefactor); <br /> draw((0.,0.)--(6.428571428571427,1.9166296949998194)); <br /> draw((7.,0.)--(2.2,3.5777087639996634)); <br /> draw((4.714285714285714,7.666518779999279)--(3.7058823529411766,0.)); <br /> /* dots and labels */<br /> dot((0.,0.),dotstyle); <br /> label(&quot;$A$&quot;, (-0.2432592696221352,-0.5715638692509372), NE * labelscalefactor); <br /> dot((7.,0.),dotstyle); <br /> label(&quot;$B$&quot;, (7.0458397156813835,-0.48935598595804014), NE * labelscalefactor); <br /> dot((3.7058823529411766,0.),dotstyle); <br /> label(&quot;$E$&quot;, (3.8123296394941084,0.16830708038513573), NE * labelscalefactor); <br /> dot((4.714285714285714,7.666518779999279),dotstyle); <br /> label(&quot;$C$&quot;, (4.579603216894479,7.895848109917452), NE * labelscalefactor); <br /> dot((2.2,3.5777087639996634),linewidth(3.pt) + dotstyle); <br /> label(&quot;$D$&quot;, (2.1407693458718726,3.127790878929427), NE * labelscalefactor); <br /> dot((6.428571428571427,1.9166296949998194),linewidth(3.pt) + dotstyle); <br /> label(&quot;$H$&quot;, (6.004539860638023,1.9494778850645704), NE * labelscalefactor); <br /> dot((5.,1.49071198499986),linewidth(3.pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (4.935837377830365,1.7302568629501784), NE * labelscalefactor); <br /> dot((3.857142857142857,1.1499778169998918),linewidth(3.pt) + dotstyle); <br /> label(&quot;$P$&quot;, (3.538303361851119,1.2370095631927964), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad<br /> \textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad<br /> \textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad<br /> \textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad<br /> \textbf{(E)}\ \frac{6}{5}&lt;/math&gt;<br /> <br /> ==Solution==<br /> Get the area of the triangle by heron's formula: <br /> &lt;cmath&gt;\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12)(3)(4)(5)} = 12\sqrt{5}&lt;/cmath&gt;<br /> Use the area to find the height AH with known base BC:<br /> &lt;cmath&gt;Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)&lt;/cmath&gt;<br /> &lt;cmath&gt;AH = 3\sqrt{5}&lt;/cmath&gt;<br /> &lt;cmath&gt;BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2&lt;/cmath&gt;<br /> &lt;cmath&gt;CH = BC - BH = 8 - 2 = 6&lt;/cmath&gt;<br /> Apply angle bisector theorem on triangle &lt;math&gt;ACH&lt;/math&gt; and triangle &lt;math&gt;ABH&lt;/math&gt;, we get &lt;math&gt;AP:PH = 9:6&lt;/math&gt; and &lt;math&gt;AQ:QH = 7:2&lt;/math&gt;, respectively.<br /> From now, you can simply use the answer choices because only choice &lt;math&gt;\textbf{D}&lt;/math&gt; has &lt;math&gt;\sqrt{5}&lt;/math&gt; in it and we know that &lt;math&gt;AH = 3\sqrt{5}&lt;/math&gt; the segments on it all have integral lengths so that &lt;math&gt;\sqrt{5}&lt;/math&gt; will remain there.<br /> However, by scaling up the length ratio:<br /> &lt;math&gt;AH:AP:PH = 45:27:18&lt;/math&gt; and &lt;math&gt;AQ:QH =45:35:10&lt;/math&gt;.<br /> we get &lt;math&gt;AH:PQ = 45:(18 - 10) = 45 : 8&lt;/math&gt;.<br /> &lt;cmath&gt;PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}&lt;/cmath&gt;<br /> <br /> Solution 2: (choices+faster)<br /> As in the above solution, let's find the area. By heron's, it is indeed &lt;math&gt;12\sqrt{5}&lt;/math&gt;. Because &lt;math&gt;\overline {AH}&lt;/math&gt; is an altitude and we are given a base, let's find &lt;math&gt;\overline {AH}&lt;/math&gt;. We set up the area is equal to &lt;math&gt;\frac {1}{2}<br /> cdot b \cdot h \implies &lt;/math&gt;12/sqrt5=\overline {AH}\cdot \frac {1}{2}\cdot 8 \implies \overline {AH}=3\sqrt5&lt;math&gt; and since the answer should have &lt;/math&gt;\sqrt 5&lt;math&gt; in it so our answer is &lt;/math&gt;D\$.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Hinna