https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pulusona&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-17T21:55:46Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_6&diff=98419 2011 AMC 12A Problems/Problem 6 2018-10-30T04:24:28Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was &lt;math&gt;61&lt;/math&gt; points. How many free throws did they make?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 13 \qquad<br /> \textbf{(B)}\ 14 \qquad<br /> \textbf{(C)}\ 15 \qquad<br /> \textbf{(D)}\ 16 \qquad<br /> \textbf{(E)}\ 17 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a &lt;math&gt;3:2&lt;/math&gt; ratio. Therefore, assume they made &lt;math&gt;3x&lt;/math&gt; and &lt;math&gt;2x&lt;/math&gt; two- and three- point shots, respectively, and thus &lt;math&gt;3x+1&lt;/math&gt; free throws. The total number of points is &lt;cmath&gt;2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1&lt;/cmath&gt;<br /> <br /> Set that equal to &lt;math&gt;61&lt;/math&gt;, we get &lt;math&gt;x = 4&lt;/math&gt;, and therefore the number of free throws they made &lt;math&gt;3 \times 4 + 1 = 13 \Rightarrow \boxed{A}&lt;/math&gt;<br /> <br /> == Solution 2 == <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of free throws. Then the number of points scored by two-pointers is &lt;math&gt;2(x-1)&lt;/math&gt; and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is &lt;math&gt;x+4(x-1) = 61 \Rightarrow x=13&lt;/math&gt;, giving us &lt;math&gt;\boxed{(A)}&lt;/math&gt; for an answer.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=A}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_6&diff=98418 2011 AMC 12A Problems/Problem 6 2018-10-30T04:24:03Z <p>Pulusona: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was &lt;math&gt;61&lt;/math&gt; points. How many free throws did they make?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 13 \qquad<br /> \textbf{(B)}\ 14 \qquad<br /> \textbf{(C)}\ 15 \qquad<br /> \textbf{(D)}\ 16 \qquad<br /> \textbf{(E)}\ 17 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a &lt;math&gt;3:2&lt;/math&gt; ratio. Therefore, assume they made &lt;math&gt;3x&lt;/math&gt; and &lt;math&gt;2x&lt;/math&gt; two- and three- point shots, respectively, and thus &lt;math&gt;3x+1&lt;/math&gt; free throws. The total number of points is &lt;cmath&gt;2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1&lt;/cmath&gt;<br /> <br /> Set that equal to &lt;math&gt;61&lt;/math&gt;, we get &lt;math&gt;x = 4&lt;/math&gt;, and therefore the number of free throws they made &lt;math&gt;3 \times 4 + 1 = 13 \Rightarrow \boxed{A}&lt;/math&gt;<br /> <br /> == Solution 2 == <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of free throws. Then the number of points scored by two-pointers is &lt;math&gt;2(x-1)&lt;/math&gt; and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is &lt;math&gt;x+4(x-1) = 61 \Rightarrow x=13&lt;/math&gt;, giving us &lt;math&gt;\boxed{(A)}&lt;/math&gt; for an answer.<br /> <br /> == Solution 2 == <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of free throws. Then the number of points scored by two-pointers is &lt;math&gt;2(x-1)&lt;/math&gt; and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is &lt;math&gt;x+4(x-1) = 61 \Rightarrow x=13&lt;/math&gt;, giving us &lt;math&gt;\boxed{(A)}&lt;/math&gt; for an answer.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=A}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_6&diff=98417 2011 AMC 12A Problems/Problem 6 2018-10-30T04:23:40Z <p>Pulusona: /* Solution */</p> <hr /> <div>== Problem ==<br /> The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was &lt;math&gt;61&lt;/math&gt; points. How many free throws did they make?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 13 \qquad<br /> \textbf{(B)}\ 14 \qquad<br /> \textbf{(C)}\ 15 \qquad<br /> \textbf{(D)}\ 16 \qquad<br /> \textbf{(E)}\ 17 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a &lt;math&gt;3:2&lt;/math&gt; ratio. Therefore, assume they made &lt;math&gt;3x&lt;/math&gt; and &lt;math&gt;2x&lt;/math&gt; two- and three- point shots, respectively, and thus &lt;math&gt;3x+1&lt;/math&gt; free throws. The total number of points is &lt;cmath&gt;2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1&lt;/cmath&gt;<br /> <br /> Set that equal to &lt;math&gt;61&lt;/math&gt;, we get &lt;math&gt;x = 4&lt;/math&gt;, and therefore the number of free throws they made &lt;math&gt;3 \times 4 + 1 = 13 \Rightarrow \boxed{A}&lt;/math&gt;<br /> <br /> == Solution 2 == <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of free throws. Then the number of points scored by two-pointers is &lt;math&gt;2(x-1)&lt;/math&gt; and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is &lt;math&gt;x+4(x-1) = 61 \Rightarrow x=13&lt;/math&gt;, giving us &lt;math&gt;\boxed{(A)}&lt;/math&gt; for an answer.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=5|num-a=7|ab=A}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_13&diff=97618 2018 AMC 10A Problems/Problem 13 2018-09-01T18:36:13Z <p>Pulusona: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point &lt;math&gt;A&lt;/math&gt; falls on point &lt;math&gt;B&lt;/math&gt;. What is the length in inches of the crease?<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,3), NE);<br /> label(&quot;$C$&quot;, (4,0), SE);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> First, we need to realize that the crease line is just the perpendicular bisector of side &lt;math&gt;AB&lt;/math&gt;, the hypotenuse of right triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;. Draw this line and call the intersection point with &lt;math&gt;AC&lt;/math&gt; as &lt;math&gt;E&lt;/math&gt;. Now, &lt;math&gt;\triangle ACB&lt;/math&gt; is similar to &lt;math&gt;\triangle ADE&lt;/math&gt; by &lt;math&gt;AA&lt;/math&gt; similarity. Setting up the ratios, we find that<br /> &lt;cmath&gt;\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> ~Nivek<br /> ===Note===<br /> In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of &lt;math&gt;AB&lt;/math&gt;, because &lt;math&gt;A&lt;/math&gt; must be reflected onto &lt;math&gt;B&lt;/math&gt;. (by pulusona)<br /> <br /> ==Solution 2==<br /> <br /> Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than &lt;math&gt;\frac{7}{4}&lt;/math&gt; units and somewhat less than &lt;math&gt;2&lt;/math&gt; units. The only answer choice in range is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> This is pretty much a cop-out, but it's allowed in the rules technically.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_13&diff=97617 2018 AMC 10A Problems/Problem 13 2018-09-01T18:35:29Z <p>Pulusona: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point &lt;math&gt;A&lt;/math&gt; falls on point &lt;math&gt;B&lt;/math&gt;. What is the length in inches of the crease?<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,3), NE);<br /> label(&quot;$C$&quot;, (4,0), SE);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> First, we need to realize that the crease line is just the perpendicular bisector of side &lt;math&gt;AB&lt;/math&gt;, the hypotenuse of right triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;. Draw this line and call the intersection point with &lt;math&gt;AC&lt;/math&gt; as &lt;math&gt;E&lt;/math&gt;. Now, &lt;math&gt;\triangle ACB&lt;/math&gt; is similar to &lt;math&gt;\triangle ADE&lt;/math&gt; by &lt;math&gt;AA&lt;/math&gt; similarity. Setting up the ratios, we find that<br /> &lt;cmath&gt;\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> ~Nivek<br /> In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of &lt;math&gt;AB&lt;/math&gt;, because &lt;math&gt;A&lt;/math&gt; must be reflected onto &lt;math&gt;B&lt;/math&gt;. (by pulusona)<br /> <br /> ==Solution 2==<br /> <br /> Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than &lt;math&gt;\frac{7}{4}&lt;/math&gt; units and somewhat less than &lt;math&gt;2&lt;/math&gt; units. The only answer choice in range is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> This is pretty much a cop-out, but it's allowed in the rules technically.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_6&diff=97616 2018 AMC 10A Problems/Problem 6 2018-09-01T18:27:20Z <p>Pulusona: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of 0, and the score increases by 1 for each like vote and decreases by 1 for each dislike vote. At one point Sangho saw that his video had a score of 90, and that &lt;math&gt;65\%&lt;/math&gt; of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?<br /> <br /> &lt;math&gt;\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600 &lt;/math&gt;<br /> <br /> ==Solution==<br /> If &lt;math&gt;65\%&lt;/math&gt; of the votes were likes, then &lt;math&gt;35\%&lt;/math&gt; of the votes were dislikes. &lt;math&gt;65\%-35\%=30\%&lt;/math&gt;, so &lt;math&gt;90&lt;/math&gt; votes is &lt;math&gt;30\%&lt;/math&gt; of the total number of votes. Doing quick (maths) arithmetic shows that the answer is &lt;math&gt;\boxed{\textbf{(B) } 300}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_19&diff=89923 2012 AMC 10A Problems/Problem 19 2018-01-22T02:03:45Z <p>Pulusona: /* Problem 19 */</p> <hr /> <div>{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #13]] and [[2012 AMC 10A Problems|2012 AMC 10A #19]]}}<br /> <br /> ==Problem 19==<br /> <br /> Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ &lt;/math&gt;<br /> <br /> ==Solution==<br /> Let Paula work at a rate of &lt;math&gt;p&lt;/math&gt;, the two helpers work at a combined rate of &lt;math&gt;h&lt;/math&gt;, and the time it takes to eat lunch be &lt;math&gt;L&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;h&lt;/math&gt; are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:<br /> <br /> &lt;cmath&gt;(8-L)(p+h)=50&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(6.2-L)h=24&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(11.2-L)p=26&lt;/cmath&gt;<br /> <br /> With three equations and three variables, we need to find the value of &lt;math&gt;L&lt;/math&gt;.<br /> Adding the second and third equations together gives us &lt;math&gt;6.2h+11.2p-L(p+h)=50&lt;/math&gt;. Subtracting the first equation from this new one gives us &lt;math&gt;-1.8h+3.2p=0&lt;/math&gt;, so we get &lt;math&gt;h=\frac{16}{9}p&lt;/math&gt;. <br /> Plugging into the second equation:<br /> <br /> &lt;cmath&gt;(6.2-L)\frac{16}{9}p=24&lt;/cmath&gt;<br /> &lt;cmath&gt;(6.2-L)p=\frac{27}{2}&lt;/cmath&gt;<br /> <br /> We can then subtract this from the third equation:<br /> <br /> &lt;cmath&gt;5p=26-\frac{27}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;p=\frac{5}{2}&lt;/cmath&gt;<br /> Plugging &lt;math&gt;p&lt;/math&gt; into our third equation gives: &lt;cmath&gt;L=\frac{4}{5}&lt;/cmath&gt;<br /> <br /> Converting &lt;math&gt;L&lt;/math&gt; from hours to minutes gives us &lt;math&gt;L=48&lt;/math&gt; minutes, which is &lt;math&gt;\boxed{\textbf{(D)}\ 48}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2012|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_15&diff=89722 2008 AMC 12A Problems/Problem 15 2018-01-13T20:40:27Z <p>Pulusona: /* Solution */</p> <hr /> <div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #15]] and [[2008 AMC 10A Problems/Problem 24|2008 AMC 10A #24]]}}<br /> ==Problem==<br /> Let &lt;math&gt;k={2008}^{2}+{2}^{2008}&lt;/math&gt;. What is the units digit of &lt;math&gt;k^2+2^k&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 /> &lt;math&gt;k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;k^2 \equiv 0 \pmod{10}&lt;/math&gt;. Since &lt;math&gt;k = 2008^2+2^{2008}&lt;/math&gt; is a multiple of four and the units digit of powers of two repeat in cycles of four, &lt;math&gt;2^k \equiv 2^4 \equiv 6 \pmod{10}&lt;/math&gt;. <br /> <br /> Therefore, &lt;math&gt;k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}&lt;/math&gt;. So the units digit is &lt;math&gt;6 \Rightarrow \boxed{D}&lt;/math&gt;.<br /> <br /> ==See Also== <br /> {{AMC12 box|year=2008|ab=A|num-b=14|num-a=16}}<br /> {{AMC10 box|year=2008|ab=A|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=89718 AoPS Wiki:FAQ 2018-01-13T05:39:43Z <p>Pulusona: /* Can I make more than one account? */</p> <hr /> <div>{{shortcut|[[A:FAQ]]}}<br /> <br /> This is a community created list of Frequently Asked Questions about Art of Problem Solving. 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Someone should get back to you within a couple days.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_22&diff=89717 2014 AMC 10A Problems/Problem 22 2018-01-13T05:12:26Z <p>Pulusona: /* Solution 4 (Measuring) */</p> <hr /> <div>==Problem==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;\overline{AB}=20&lt;/math&gt; and &lt;math&gt;\overline{BC}=10&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be a point on &lt;math&gt;\overline{CD}&lt;/math&gt; such that &lt;math&gt;\angle CBE=15^\circ&lt;/math&gt;. What is &lt;math&gt;\overline{AE}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution (Trigonometry)==<br /> Note that &lt;math&gt;\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3&lt;/math&gt;. (If you do not know the tangent half-angle formula, it is &lt;math&gt;\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}&lt;/math&gt;). Therefore, we have &lt;math&gt;DE=10\sqrt 3&lt;/math&gt;. Since &lt;math&gt;ADE&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, &lt;math&gt;AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}&lt;/math&gt;<br /> <br /> ==Solution 2 (Without Trigonometry)==<br /> <br /> Let &lt;math&gt;F&lt;/math&gt; be a point on line &lt;math&gt;\overline{CD}&lt;/math&gt; such that points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are distinct and that &lt;math&gt;\angle EBF = 15^\circ&lt;/math&gt;. By the angle bisector theorem, &lt;math&gt;\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}&lt;/math&gt;. Since &lt;math&gt;\triangle BFC&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle, &lt;math&gt;\overline{CF} = \frac{10\sqrt{3}}{3}&lt;/math&gt; and &lt;math&gt;\overline{BF} = \frac{20\sqrt{3}}{3}&lt;/math&gt;. Additionally, &lt;cmath&gt;\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}&lt;/cmath&gt;Now, substituting in the obtained values, we get &lt;math&gt;\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}&lt;/math&gt; and &lt;math&gt;\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}&lt;/math&gt;. Substituting the first equation into the second yields &lt;math&gt;\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}&lt;/math&gt;, so &lt;math&gt;\overline{DE} = 10\sqrt{3}&lt;/math&gt;. Because &lt;math&gt;\triangle ADE&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, &lt;math&gt;\overline{AE} = \boxed{\textbf{(E)}~20}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Trigonometry)==<br /> By Law of Sines&lt;cmath&gt;\frac{BC}{\sin 75^\circ}=\frac{EC}{\sin15^\circ}\rightarrow\frac{10}{\frac{\sqrt{2}+\sqrt{6}}{4}}=\frac{EC}{\frac{\sqrt{6}-\sqrt{2}}{4}}\rightarrow\frac{10}{EC}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\rightarrow EC=\frac{10}{2+\sqrt{3}}=20-10\sqrt{3}.&lt;/cmath&gt;Thus, &lt;math&gt;DE=20-(20-10\sqrt{3})=10\sqrt{3}.&lt;/math&gt;<br /> <br /> We see that &lt;math&gt;\triangle{ADE}&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, leaving &lt;math&gt;\overline{AE}=\boxed{\textbf{(E)}~20}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Measuring) ==<br /> If we draw rectangle &lt;math&gt;ABCD&lt;/math&gt; and whip out a protractor, we can draw a perfect &lt;math&gt;\overline{BE}&lt;/math&gt;, almost perfectly &lt;math&gt;15^\circ&lt;/math&gt; degrees off of &lt;math&gt;\overline{BC}&lt;/math&gt;. Then we can draw &lt;math&gt;\overline{AE}&lt;/math&gt;, and use a ruler to measure it. <br /> We can clearly see that the &lt;math&gt;\overline{AE}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(E)}~20}&lt;/math&gt;.<br /> <br /> NOTE: this method is a last resort, and is pretty risky. Answer choice &lt;math&gt;\textbf{(D)}~11\sqrt{3}&lt;/math&gt; is also very close to &lt;math&gt;\textbf{(E)}~20&lt;/math&gt;, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of &lt;math&gt;\triangle ADE&lt;/math&gt;, we can clearly see that it is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, which verifies our answer of &lt;math&gt;\boxed{20}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_22&diff=89716 2014 AMC 10A Problems/Problem 22 2018-01-13T05:11:57Z <p>Pulusona: /* Solution 4 (Measuring) */</p> <hr /> <div>==Problem==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;\overline{AB}=20&lt;/math&gt; and &lt;math&gt;\overline{BC}=10&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be a point on &lt;math&gt;\overline{CD}&lt;/math&gt; such that &lt;math&gt;\angle CBE=15^\circ&lt;/math&gt;. What is &lt;math&gt;\overline{AE}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution (Trigonometry)==<br /> Note that &lt;math&gt;\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3&lt;/math&gt;. (If you do not know the tangent half-angle formula, it is &lt;math&gt;\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}&lt;/math&gt;). Therefore, we have &lt;math&gt;DE=10\sqrt 3&lt;/math&gt;. Since &lt;math&gt;ADE&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, &lt;math&gt;AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}&lt;/math&gt;<br /> <br /> ==Solution 2 (Without Trigonometry)==<br /> <br /> Let &lt;math&gt;F&lt;/math&gt; be a point on line &lt;math&gt;\overline{CD}&lt;/math&gt; such that points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are distinct and that &lt;math&gt;\angle EBF = 15^\circ&lt;/math&gt;. By the angle bisector theorem, &lt;math&gt;\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}&lt;/math&gt;. Since &lt;math&gt;\triangle BFC&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle, &lt;math&gt;\overline{CF} = \frac{10\sqrt{3}}{3}&lt;/math&gt; and &lt;math&gt;\overline{BF} = \frac{20\sqrt{3}}{3}&lt;/math&gt;. Additionally, &lt;cmath&gt;\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}&lt;/cmath&gt;Now, substituting in the obtained values, we get &lt;math&gt;\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}&lt;/math&gt; and &lt;math&gt;\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}&lt;/math&gt;. Substituting the first equation into the second yields &lt;math&gt;\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}&lt;/math&gt;, so &lt;math&gt;\overline{DE} = 10\sqrt{3}&lt;/math&gt;. Because &lt;math&gt;\triangle ADE&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, &lt;math&gt;\overline{AE} = \boxed{\textbf{(E)}~20}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Trigonometry)==<br /> By Law of Sines&lt;cmath&gt;\frac{BC}{\sin 75^\circ}=\frac{EC}{\sin15^\circ}\rightarrow\frac{10}{\frac{\sqrt{2}+\sqrt{6}}{4}}=\frac{EC}{\frac{\sqrt{6}-\sqrt{2}}{4}}\rightarrow\frac{10}{EC}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\rightarrow EC=\frac{10}{2+\sqrt{3}}=20-10\sqrt{3}.&lt;/cmath&gt;Thus, &lt;math&gt;DE=20-(20-10\sqrt{3})=10\sqrt{3}.&lt;/math&gt;<br /> <br /> We see that &lt;math&gt;\triangle{ADE}&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, leaving &lt;math&gt;\overline{AE}=\boxed{\textbf{(E)}~20}.&lt;/math&gt;<br /> <br /> ==Solution 4 (Measuring) ==<br /> If we draw rectangle &lt;math&gt;ABCD&lt;/math&gt; and whip out a protractor, we can draw a perfect &lt;math&gt;\overline{BE}&lt;/math&gt;, almost perfectly &lt;math&gt;15^\circ&lt;/math&gt; degrees off of &lt;math&gt;\overline{BC}&lt;/math&gt;. Then we can draw &lt;math&gt;\overline{AE}&lt;/math&gt;, and use a ruler to measure it. <br /> We can clearly see that the &lt;math&gt;\overline{AE}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(E)}~20}&lt;/math&gt;.<br /> <br /> NOTE: this method is a last resort, and is pretty risky. Answer choice &lt;math&gt;\textbf{(D)}~11\sqrt{3}&lt;/math&gt; is also very close to &lt;math&gt;\textbf{(E)}~20&lt;/math&gt;, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of &lt;math&gt;\triangle ADE, we can clearly see that it is a &lt;/math&gt;30-60-90&lt;math&gt; triangle, which verifies our answer of &lt;/math&gt;20.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_22&diff=89714 2015 AMC 10A Problems/Problem 22 2018-01-13T04:44:47Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #17]] and [[2015 AMC 10A Problems|2015 AMC 10A #22]]}}<br /> ==Problem==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt;\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. We casework on how many people are standing.<br /> <br /> Case &lt;math&gt;1:&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; people are standing. This yields &lt;math&gt;1&lt;/math&gt; arrangement.<br /> <br /> Case &lt;math&gt;2:&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; person is standing. This yields &lt;math&gt;8&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;3:&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; people are standing. This yields &lt;math&gt;\dbinom{8}{2} - 8 = 20&lt;/math&gt; arrangements, because the two people cannot be next to each other.<br /> <br /> Case &lt;math&gt;4:&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding &lt;math&gt;2&lt;/math&gt; possible arrangements.<br /> <br /> More difficult is:<br /> <br /> Case &lt;math&gt;5:&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; people are standing. First, choose the location of the first person standing (&lt;math&gt;8&lt;/math&gt; choices). Next, choose &lt;math&gt;2&lt;/math&gt; of the remaining people in the remaining &lt;math&gt;5&lt;/math&gt; legal seats to stand, amounting to &lt;math&gt;6&lt;/math&gt; arrangements considering that these two people cannot stand next to each other. However, we have to divide by &lt;math&gt;3,&lt;/math&gt; because there are &lt;math&gt;3&lt;/math&gt; ways to choose the first person given any three. This yields &lt;math&gt;\dfrac{8 \cdot 6}{3} = 16&lt;/math&gt; arrangements for Case &lt;math&gt;5.&lt;/math&gt;<br /> <br /> Alternate Case &lt;math&gt;5:&lt;/math&gt; Use complementary counting. Total number of ways to choose 3 people from 8 which is &lt;math&gt;\dbinom{8}{3}&lt;/math&gt;. Sub-case &lt;math&gt;1:&lt;/math&gt; three people are next to each other which is &lt;math&gt;\dbinom{8}{1}&lt;/math&gt;. Sub-case &lt;math&gt;2:&lt;/math&gt; two people are next to each other and the third person is not &lt;math&gt;\dbinom{8}{1}&lt;/math&gt; &lt;math&gt;\dbinom{4}{1}&lt;/math&gt;. This yields &lt;math&gt;\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16&lt;/math&gt; <br /> <br /> Summing gives &lt;math&gt;1 + 8 + 20 + 2 + 16 = 47,&lt;/math&gt; and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We will count how many valid standing arrangements there are counting rotations as distinct and divide by &lt;math&gt;256&lt;/math&gt; at the end.<br /> Line up all &lt;math&gt;8&lt;/math&gt; people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires &lt;math&gt;2&lt;/math&gt; spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.<br /> <br /> If there are &lt;math&gt;4&lt;/math&gt; standing, there are &lt;math&gt;{4 \choose 4}=1&lt;/math&gt; ways to place them.<br /> For &lt;math&gt;3,&lt;/math&gt; there are &lt;math&gt;{3+2 \choose 3}=10&lt;/math&gt; ways.<br /> etc.<br /> Summing, we get &lt;math&gt;{4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34&lt;/math&gt; ways.<br /> <br /> Now we consider that the far right person can be standing as well, so we have<br /> &lt;math&gt;{3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13&lt;/math&gt; ways<br /> <br /> Together we have &lt;math&gt;34+13=47&lt;/math&gt;, and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only &lt;math&gt;7&lt;/math&gt; people. If they stand, we count the arrangements with &lt;math&gt;6&lt;/math&gt; instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with &lt;math&gt;1&lt;/math&gt; person there are two ways and with &lt;math&gt;2&lt;/math&gt; people there are three ways. Carrying out the Fibonacci recursion until we get to &lt;math&gt;8&lt;/math&gt; people, we find there are &lt;math&gt;55&lt;/math&gt; standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for &lt;math&gt;4&lt;/math&gt; people to stand in a line, which is &lt;math&gt;8&lt;/math&gt; from our sequence. Therefore our probability is &lt;math&gt;\frac{55 - 8}{256} = \boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2015|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2015|ab=A|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&diff=89690 2013 AMC 10B Problems/Problem 24 2018-01-12T03:39:28Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> A positive integer &lt;math&gt;n&lt;/math&gt; is ''nice'' if there is a positive integer &lt;math&gt;m&lt;/math&gt; with exactly four positive divisors (including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt;) such that the sum of the four divisors is equal to &lt;math&gt;n&lt;/math&gt;. How many numbers in the set &lt;math&gt;\{ 2010,2011,2012,\dotsc,2019 \}&lt;/math&gt; are nice?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> A positive integer with only four positive divisors has its prime factorization in the form of &lt;math&gt;a*b&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both prime positive integers or &lt;math&gt;c^3&lt;/math&gt; where &lt;math&gt;c&lt;/math&gt; is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of &lt;math&gt;a*b&lt;/math&gt;. The four factors of this number would be &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;ab&lt;/math&gt;. The sum of these would be &lt;math&gt;ab+a+b+1&lt;/math&gt;, which can be factored into the form &lt;math&gt;(a+1)(b+1)&lt;/math&gt;. Easily we can see that now we can take cases again. <br /> <br /> Case 1: Either &lt;math&gt;a&lt;/math&gt; or &lt;math&gt;b&lt;/math&gt; is 2. <br /> <br /> If this is true then we have to have that one of &lt;math&gt;(a+1)&lt;/math&gt; or &lt;math&gt;(b+1)&lt;/math&gt; is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either &lt;math&gt;\frac{2016}{3} - 1&lt;/math&gt; or &lt;math&gt;\frac{2010}{3} - 1&lt;/math&gt; is a prime. We see that in this case none of them work.<br /> <br /> Case 2: Both &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are odd primes. <br /> <br /> This implies that both &lt;math&gt;(a+1)&lt;/math&gt; and &lt;math&gt;(b+1)&lt;/math&gt; are even which implies that in this case the number must be divisible by &lt;math&gt;4&lt;/math&gt;. This leaves only &lt;math&gt;2012&lt;/math&gt; and &lt;math&gt;2016&lt;/math&gt;.<br /> &lt;math&gt;2012=4*503&lt;/math&gt; so either &lt;math&gt;(a+1)&lt;/math&gt; or &lt;math&gt;(b+1)&lt;/math&gt; both has a factor of &lt;math&gt;2&lt;/math&gt; or one has a factor of &lt;math&gt;4&lt;/math&gt;. If it was the first case, then &lt;math&gt;a&lt;/math&gt; or &lt;math&gt;b&lt;/math&gt; will equal &lt;math&gt;1&lt;/math&gt;. That means that either &lt;math&gt;(a+1)&lt;/math&gt; or &lt;math&gt;(b+1)&lt;/math&gt; has a factor of &lt;math&gt;4&lt;/math&gt;. That means that &lt;math&gt;a&lt;/math&gt; or &lt;math&gt;b&lt;/math&gt; is &lt;math&gt;502&lt;/math&gt; which isn't a prime, so 2012 does not work. &lt;math&gt;2016 = 4 * 504&lt;/math&gt; so we have &lt;math&gt;(503 + 1)(3 + 1)&lt;/math&gt;. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 1}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> If &lt;math&gt;m&lt;/math&gt; has four divisors, then its divisors would be 1, &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;ab&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are prime. Therefore, the sum of the divisors of &lt;math&gt;m&lt;/math&gt; is &lt;math&gt;1+a+b+ab=(a+1)(b+1)&lt;/math&gt;. <br /> <br /> If either &lt;math&gt;a+1&lt;/math&gt; or &lt;math&gt;b+1&lt;/math&gt; are odd, then &lt;math&gt;a&lt;/math&gt; or &lt;math&gt;b&lt;/math&gt; are even. Therefore, &lt;math&gt;a+1&lt;/math&gt; and &lt;math&gt;b+1&lt;/math&gt; are even, so &lt;math&gt;m&lt;/math&gt; is a multiple of 4. The only two numbers from the range &lt;math&gt;2010-2019&lt;/math&gt; that are multiples of 4 are 2012 and 2016.<br /> <br /> Factoring 2012, we get &lt;math&gt;2^2*503&lt;/math&gt;. To make &lt;math&gt;a+1&lt;/math&gt; and &lt;math&gt;b+1&lt;/math&gt; even, WLOG, we have &lt;math&gt;a+1=2&lt;/math&gt; and &lt;math&gt;b+1=1006&lt;/math&gt;. However, if &lt;math&gt;a&lt;/math&gt; was 1, then &lt;math&gt;a&lt;/math&gt; is not prime, so 2012 is not nice.<br /> <br /> Factoring 2016, we get &lt;math&gt;2^5*3^2*7&lt;/math&gt;. WLOG, we have &lt;math&gt;a&lt;b&lt;/math&gt;.<br /> <br /> Testing for the lowest &lt;math&gt;a&lt;/math&gt;, we get &lt;math&gt;a+1=4&lt;/math&gt; and &lt;math&gt;b+1=504&lt;/math&gt;. Therefore, &lt;math&gt;a=3&lt;/math&gt;, and &lt;math&gt;b=503&lt;/math&gt;, so &lt;math&gt;n=2016&lt;/math&gt; is nice, with &lt;math&gt;m=1509&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 1}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_23&diff=89689 2013 AMC 10A Problems/Problem 23 2018-01-12T03:30:22Z <p>Pulusona: /* Solution 2 (Stewart's Theorem) */</p> <hr /> <div>{{duplicate|[[2013 AMC 12A Problems|2013 AMC 12A #19]] and [[2013 AMC 10A Problems|2013 AMC 10A #23]]}}<br /> <br /> ==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ===Solution 1 (Power of a Point)===<br /> <br /> Let &lt;math&gt;BX = q&lt;/math&gt;, &lt;math&gt;CX = p&lt;/math&gt;, and &lt;math&gt;AC&lt;/math&gt; meets the circle at &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;, with &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;AZ = AY = 86&lt;/math&gt;. Using the Power of a Point (Secant-Secant Power Theorem), we get that &lt;math&gt;p(p+q) = 11(183) = 11 * 3 * 61&lt;/math&gt;. We know that &lt;math&gt;p+q&gt;p&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is either 3,11, or 33. We also know that &lt;math&gt;p&gt;11&lt;/math&gt; by the triangle inequality on &lt;math&gt;\triangle ACX&lt;/math&gt;. &lt;math&gt;p&lt;/math&gt; is 33. Thus, we get that &lt;math&gt;BC = p+q = \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 2 (Stewart's Theorem)===<br /> Stewart's Theorem: https://en.wikipedia.org/wiki/Stewart%27s_theorem.<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; represent &lt;math&gt;CX&lt;/math&gt;, and let &lt;math&gt;y&lt;/math&gt; represent &lt;math&gt;BX&lt;/math&gt;. Since the circle goes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;AB = AX = 86&lt;/math&gt;.<br /> Then by Stewart's Theorem,<br /> <br /> &lt;math&gt;xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 + xy + 86^2 = 97^2&lt;/math&gt;<br /> <br /> (Since &lt;math&gt;y&lt;/math&gt; cannot be equal to &lt;math&gt;0&lt;/math&gt;, dividing both sides of the equation by &lt;math&gt;y&lt;/math&gt; is allowed.)<br /> <br /> &lt;math&gt;x(x+y) = (97+86)(97-86)&lt;/math&gt;<br /> <br /> &lt;math&gt;x(x+y) = 2013&lt;/math&gt;<br /> <br /> The prime factors of &lt;math&gt;2013&lt;/math&gt; are &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt;, and &lt;math&gt;61&lt;/math&gt;. Obviously, &lt;math&gt;x &lt; x+y&lt;/math&gt;. In addition, by the Triangle Inequality, &lt;math&gt;BC &lt; AB + AC&lt;/math&gt;, so &lt;math&gt;x+y &lt; 183&lt;/math&gt;. Therefore, &lt;math&gt;x&lt;/math&gt; must equal &lt;math&gt;33&lt;/math&gt;, and &lt;math&gt;x+y&lt;/math&gt; must equal &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;CX=x, BX=y&lt;/math&gt;. Let the circle intersect &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and the diameter including &lt;math&gt;AD&lt;/math&gt; intersect the circle again at &lt;math&gt;E&lt;/math&gt;.<br /> Use power of a point on point C to the circle centered at A.<br /> <br /> So &lt;math&gt;CX*CB=CD*CE=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=(97-86)(97+86)=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=3*11*61&lt;/math&gt;.<br /> <br /> Obviously &lt;math&gt;x+y&gt;x&lt;/math&gt; so we have three solution pairs for &lt;math&gt;(x,x+y)=(1,2013),(3,671),(11,183),(33,61)&lt;/math&gt;.<br /> By the Triangle Inequality, only&lt;math&gt; x+y=61&lt;/math&gt; yields a possible length of &lt;math&gt;BX+CX=BC&lt;/math&gt;.<br /> <br /> Therefore, the answer is &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_23&diff=89688 2013 AMC 10A Problems/Problem 23 2018-01-12T03:30:05Z <p>Pulusona: /* Solution 2 (Stewart's Theorem) */</p> <hr /> <div>{{duplicate|[[2013 AMC 12A Problems|2013 AMC 12A #19]] and [[2013 AMC 10A Problems|2013 AMC 10A #23]]}}<br /> <br /> ==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ===Solution 1 (Power of a Point)===<br /> <br /> Let &lt;math&gt;BX = q&lt;/math&gt;, &lt;math&gt;CX = p&lt;/math&gt;, and &lt;math&gt;AC&lt;/math&gt; meets the circle at &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;, with &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;AZ = AY = 86&lt;/math&gt;. Using the Power of a Point (Secant-Secant Power Theorem), we get that &lt;math&gt;p(p+q) = 11(183) = 11 * 3 * 61&lt;/math&gt;. We know that &lt;math&gt;p+q&gt;p&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is either 3,11, or 33. We also know that &lt;math&gt;p&gt;11&lt;/math&gt; by the triangle inequality on &lt;math&gt;\triangle ACX&lt;/math&gt;. &lt;math&gt;p&lt;/math&gt; is 33. Thus, we get that &lt;math&gt;BC = p+q = \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 2 (Stewart's Theorem)===<br /> Stewart's Theorem: https://en.wikipedia.org/wiki/Stewart%27s_theorem<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; represent &lt;math&gt;CX&lt;/math&gt;, and let &lt;math&gt;y&lt;/math&gt; represent &lt;math&gt;BX&lt;/math&gt;. Since the circle goes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;AB = AX = 86&lt;/math&gt;.<br /> Then by Stewart's Theorem,<br /> <br /> &lt;math&gt;xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 + xy + 86^2 = 97^2&lt;/math&gt;<br /> <br /> (Since &lt;math&gt;y&lt;/math&gt; cannot be equal to &lt;math&gt;0&lt;/math&gt;, dividing both sides of the equation by &lt;math&gt;y&lt;/math&gt; is allowed.)<br /> <br /> &lt;math&gt;x(x+y) = (97+86)(97-86)&lt;/math&gt;<br /> <br /> &lt;math&gt;x(x+y) = 2013&lt;/math&gt;<br /> <br /> The prime factors of &lt;math&gt;2013&lt;/math&gt; are &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt;, and &lt;math&gt;61&lt;/math&gt;. Obviously, &lt;math&gt;x &lt; x+y&lt;/math&gt;. In addition, by the Triangle Inequality, &lt;math&gt;BC &lt; AB + AC&lt;/math&gt;, so &lt;math&gt;x+y &lt; 183&lt;/math&gt;. Therefore, &lt;math&gt;x&lt;/math&gt; must equal &lt;math&gt;33&lt;/math&gt;, and &lt;math&gt;x+y&lt;/math&gt; must equal &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;CX=x, BX=y&lt;/math&gt;. Let the circle intersect &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and the diameter including &lt;math&gt;AD&lt;/math&gt; intersect the circle again at &lt;math&gt;E&lt;/math&gt;.<br /> Use power of a point on point C to the circle centered at A.<br /> <br /> So &lt;math&gt;CX*CB=CD*CE=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=(97-86)(97+86)=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=3*11*61&lt;/math&gt;.<br /> <br /> Obviously &lt;math&gt;x+y&gt;x&lt;/math&gt; so we have three solution pairs for &lt;math&gt;(x,x+y)=(1,2013),(3,671),(11,183),(33,61)&lt;/math&gt;.<br /> By the Triangle Inequality, only&lt;math&gt; x+y=61&lt;/math&gt; yields a possible length of &lt;math&gt;BX+CX=BC&lt;/math&gt;.<br /> <br /> Therefore, the answer is &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_23&diff=89687 2013 AMC 10A Problems/Problem 23 2018-01-12T03:29:48Z <p>Pulusona: /* Solution 2 (Stewart's Theorem) */</p> <hr /> <div>{{duplicate|[[2013 AMC 12A Problems|2013 AMC 12A #19]] and [[2013 AMC 10A Problems|2013 AMC 10A #23]]}}<br /> <br /> ==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ===Solution 1 (Power of a Point)===<br /> <br /> Let &lt;math&gt;BX = q&lt;/math&gt;, &lt;math&gt;CX = p&lt;/math&gt;, and &lt;math&gt;AC&lt;/math&gt; meets the circle at &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;, with &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;AZ = AY = 86&lt;/math&gt;. Using the Power of a Point (Secant-Secant Power Theorem), we get that &lt;math&gt;p(p+q) = 11(183) = 11 * 3 * 61&lt;/math&gt;. We know that &lt;math&gt;p+q&gt;p&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is either 3,11, or 33. We also know that &lt;math&gt;p&gt;11&lt;/math&gt; by the triangle inequality on &lt;math&gt;\triangle ACX&lt;/math&gt;. &lt;math&gt;p&lt;/math&gt; is 33. Thus, we get that &lt;math&gt;BC = p+q = \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 2 (Stewart's Theorem)===<br /> Stewart's Theorem: https://en.wikipedia.org/wiki/Stewart%27s_theorem<br /> <br /> <br /> Let &lt;math&gt;x&lt;/math&gt; represent &lt;math&gt;CX&lt;/math&gt;, and let &lt;math&gt;y&lt;/math&gt; represent &lt;math&gt;BX&lt;/math&gt;. Since the circle goes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;AB = AX = 86&lt;/math&gt;.<br /> Then by Stewart's Theorem,<br /> <br /> &lt;math&gt;xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 + xy + 86^2 = 97^2&lt;/math&gt;<br /> <br /> (Since &lt;math&gt;y&lt;/math&gt; cannot be equal to &lt;math&gt;0&lt;/math&gt;, dividing both sides of the equation by &lt;math&gt;y&lt;/math&gt; is allowed.)<br /> <br /> &lt;math&gt;x(x+y) = (97+86)(97-86)&lt;/math&gt;<br /> <br /> &lt;math&gt;x(x+y) = 2013&lt;/math&gt;<br /> <br /> The prime factors of &lt;math&gt;2013&lt;/math&gt; are &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt;, and &lt;math&gt;61&lt;/math&gt;. Obviously, &lt;math&gt;x &lt; x+y&lt;/math&gt;. In addition, by the Triangle Inequality, &lt;math&gt;BC &lt; AB + AC&lt;/math&gt;, so &lt;math&gt;x+y &lt; 183&lt;/math&gt;. Therefore, &lt;math&gt;x&lt;/math&gt; must equal &lt;math&gt;33&lt;/math&gt;, and &lt;math&gt;x+y&lt;/math&gt; must equal &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;CX=x, BX=y&lt;/math&gt;. Let the circle intersect &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and the diameter including &lt;math&gt;AD&lt;/math&gt; intersect the circle again at &lt;math&gt;E&lt;/math&gt;.<br /> Use power of a point on point C to the circle centered at A.<br /> <br /> So &lt;math&gt;CX*CB=CD*CE=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=(97-86)(97+86)=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=3*11*61&lt;/math&gt;.<br /> <br /> Obviously &lt;math&gt;x+y&gt;x&lt;/math&gt; so we have three solution pairs for &lt;math&gt;(x,x+y)=(1,2013),(3,671),(11,183),(33,61)&lt;/math&gt;.<br /> By the Triangle Inequality, only&lt;math&gt; x+y=61&lt;/math&gt; yields a possible length of &lt;math&gt;BX+CX=BC&lt;/math&gt;.<br /> <br /> Therefore, the answer is &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_23&diff=89686 2013 AMC 10A Problems/Problem 23 2018-01-12T03:28:55Z <p>Pulusona: /* Solution 2 (Stewart's Theorem) */</p> <hr /> <div>{{duplicate|[[2013 AMC 12A Problems|2013 AMC 12A #19]] and [[2013 AMC 10A Problems|2013 AMC 10A #23]]}}<br /> <br /> ==Problem==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ===Solution 1 (Power of a Point)===<br /> <br /> Let &lt;math&gt;BX = q&lt;/math&gt;, &lt;math&gt;CX = p&lt;/math&gt;, and &lt;math&gt;AC&lt;/math&gt; meets the circle at &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;, with &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;AZ = AY = 86&lt;/math&gt;. Using the Power of a Point (Secant-Secant Power Theorem), we get that &lt;math&gt;p(p+q) = 11(183) = 11 * 3 * 61&lt;/math&gt;. We know that &lt;math&gt;p+q&gt;p&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is either 3,11, or 33. We also know that &lt;math&gt;p&gt;11&lt;/math&gt; by the triangle inequality on &lt;math&gt;\triangle ACX&lt;/math&gt;. &lt;math&gt;p&lt;/math&gt; is 33. Thus, we get that &lt;math&gt;BC = p+q = \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 2 (Stewart's Theorem)===<br /> Stewart's Theorem: https://en.wikipedia.org/wiki/Stewart%27s_theorem<br /> Let &lt;math&gt;x&lt;/math&gt; represent &lt;math&gt;CX&lt;/math&gt;, and let &lt;math&gt;y&lt;/math&gt; represent &lt;math&gt;BX&lt;/math&gt;. Since the circle goes through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;AB = AX = 86&lt;/math&gt;.<br /> Then by Stewart's Theorem,<br /> <br /> &lt;math&gt;xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x&lt;/math&gt;<br /> <br /> &lt;math&gt;x^2 + xy + 86^2 = 97^2&lt;/math&gt;<br /> <br /> (Since &lt;math&gt;y&lt;/math&gt; cannot be equal to &lt;math&gt;0&lt;/math&gt;, dividing both sides of the equation by &lt;math&gt;y&lt;/math&gt; is allowed.)<br /> <br /> &lt;math&gt;x(x+y) = (97+86)(97-86)&lt;/math&gt;<br /> <br /> &lt;math&gt;x(x+y) = 2013&lt;/math&gt;<br /> <br /> The prime factors of &lt;math&gt;2013&lt;/math&gt; are &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt;, and &lt;math&gt;61&lt;/math&gt;. Obviously, &lt;math&gt;x &lt; x+y&lt;/math&gt;. In addition, by the Triangle Inequality, &lt;math&gt;BC &lt; AB + AC&lt;/math&gt;, so &lt;math&gt;x+y &lt; 183&lt;/math&gt;. Therefore, &lt;math&gt;x&lt;/math&gt; must equal &lt;math&gt;33&lt;/math&gt;, and &lt;math&gt;x+y&lt;/math&gt; must equal &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> Let &lt;math&gt;CX=x, BX=y&lt;/math&gt;. Let the circle intersect &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and the diameter including &lt;math&gt;AD&lt;/math&gt; intersect the circle again at &lt;math&gt;E&lt;/math&gt;.<br /> Use power of a point on point C to the circle centered at A.<br /> <br /> So &lt;math&gt;CX*CB=CD*CE=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=(97-86)(97+86)=&gt;&lt;/math&gt;<br /> &lt;math&gt;x(x+y)=3*11*61&lt;/math&gt;.<br /> <br /> Obviously &lt;math&gt;x+y&gt;x&lt;/math&gt; so we have three solution pairs for &lt;math&gt;(x,x+y)=(1,2013),(3,671),(11,183),(33,61)&lt;/math&gt;.<br /> By the Triangle Inequality, only&lt;math&gt; x+y=61&lt;/math&gt; yields a possible length of &lt;math&gt;BX+CX=BC&lt;/math&gt;.<br /> <br /> Therefore, the answer is &lt;math&gt; \boxed{\textbf{(D) }61}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_8&diff=89653 2015 AMC 12B Problems/Problem 8 2018-01-11T05:23:12Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;math&gt;(625^{\log_5 2015})^{\frac{1}{4}}&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt{5^{2015}}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;(625^{\log_5 2015})^\frac{1}{4}<br /> = ((5^4)^{\log_5 2015})^\frac{1}{4}<br /> = (5^{4 \cdot \log_5 2015})^\frac{1}{4}<br /> = (5^{\log_5 2015 \cdot 4})^\frac{1}{4}<br /> = ((5^{\log_5 2015})^4)^\frac{1}{4}<br /> = (2015^4)^\frac{1}{4}<br /> = \boxed{\textbf{(D)}\; 2015}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can rewrite &lt;math&gt;\log_5 2015&lt;/math&gt; as as &lt;math&gt;5^x = 2015&lt;/math&gt;. Thus, &lt;math&gt;625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=9|num-b=7}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_8&diff=89652 2015 AMC 12B Problems/Problem 8 2018-01-11T05:22:26Z <p>Pulusona: </p> <hr /> <div>==Problem==<br /> What is the value of &lt;math&gt;(625^{\log_5 2015})^{\frac{1}{4}}&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt{5^{2015}}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt;(625^{\log_5 2015})^\frac{1}{4}<br /> = ((5^4)^{\log_5 2015})^\frac{1}{4}<br /> = (5^{4 \cdot \log_5 2015})^\frac{1}{4}<br /> = (5^{\log_5 2015 \cdot 4})^\frac{1}{4}<br /> = ((5^{\log_5 2015})^4)^\frac{1}{4}<br /> = (2015^4)^\frac{1}{4}<br /> = \boxed{\textbf{(D)}\; 2015}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We can rewrite &lt;math&gt;\log_5 2015&lt;/math&gt; as as &lt;math&gt;5^x = 2015&lt;/math&gt;. Thus, &lt;math&gt;625^{x \cdot \frac{1}{4}} = 5^x = 2015&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2015|ab=B|num-a=9|num-b=7}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_16&diff=89556 2011 AMC 10A Problems/Problem 16 2018-01-07T22:51:53Z <p>Pulusona: /* Solution */</p> <hr /> <div>==Problem 16==<br /> Which of the following is equal to &lt;math&gt;\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We find the answer by squaring, then square rooting the expression.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\ = \ &amp;\sqrt{\left(\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\right)^2}\\ = \ &amp;\sqrt{9-6\sqrt{2}+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}+9+6\sqrt{2}}\\ = \ &amp;\sqrt{18+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}}\\ = \ &amp;\sqrt{18+2\sqrt{9^2-(6\sqrt{2})^2}}\\ = \ &amp;\sqrt{18+2\sqrt{81-72}}\\ = \ &amp;\sqrt{18+2\sqrt{9}}\\ = \ &amp;\sqrt{18+6}\\= \ &amp;\sqrt{24}\\ = \ &amp;\boxed{2\sqrt{6} \ \mathbf{(B)}}.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_8&diff=89555 2011 AMC 10A Problems/Problem 8 2018-01-07T22:46:54Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>== Problem 8 ==<br /> <br /> Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> === Solution 1 ===<br /> <br /> 75% of the total birds were not swans. Out of that 75%, there was &lt;math&gt;30\% / 75\% = \boxed{40\%(C)}&lt;/math&gt; of the birds that were not swans that were geese.<br /> <br /> === Solution 2 ===<br /> <br /> WLOG, suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. &lt;math&gt;100-25 = 75&lt;/math&gt; of the birds are not swans and 30 of these are geese, so the answer is &lt;math&gt;\frac{30}{75} \times 100 = \boxed{40 \%(C)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2011|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=User:Pulusona&diff=89521 User:Pulusona 2018-01-07T05:15:45Z <p>Pulusona: </p> <hr /> <div>Random musings of mine: A firefly is the opposite of a waterfall. Why do people care about wasting water if there is a water cycle? When cleaning a vacuum cleaner, you are not a vacuum cleaner. Why don't people all use only one language?</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=User:Pulusona&diff=89520 User:Pulusona 2018-01-07T05:15:15Z <p>Pulusona: </p> <hr /> <div>Random musings of mine: A firefly is the opposite of a waterfall. Why do people care about wasting water if there is a water cycle? When cleaning a vacuum cleaner, you are not a vacuum cleaner.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=User:Pulusona&diff=89519 User:Pulusona 2018-01-07T05:14:33Z <p>Pulusona: </p> <hr /> <div>Random musings of mine:<br /> A firefly is the opposite of a waterfall; <br /> Why do people care about wasting water if there is a water cycle?; <br /> When cleaning a vacuum cleaner, you are not a vacuum cleaner;</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=User:Pulusona&diff=89518 User:Pulusona 2018-01-07T05:13:57Z <p>Pulusona: </p> <hr /> <div>Random musings of mine:<br /> A firefly is the opposite of a waterfall. ||<br /> Why do people care about wasting water if there is a water cycle? ||<br /> When cleaning a vacuum cleaner, you are not a vacuum cleaner.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=User:Pulusona&diff=89517 User:Pulusona 2018-01-07T05:13:26Z <p>Pulusona: </p> <hr /> <div>Random musings of mine:<br /> A firefly is the opposite of a waterfall. //<br /> Why do people care about wasting water if there is a water cycle? //<br /> When cleaning a vacuum cleaner, you are not a vacuum cleaner.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=User:Pulusona&diff=89516 User:Pulusona 2018-01-07T05:12:51Z <p>Pulusona: Created page with &quot;Random musings of mine: A firefly is the opposite of a waterfall. Why do people care about wasting water if there is a water cycle? When cleaning a vacuum cleaner, you are not...&quot;</p> <hr /> <div>Random musings of mine:<br /> A firefly is the opposite of a waterfall.<br /> Why do people care about wasting water if there is a water cycle?<br /> When cleaning a vacuum cleaner, you are not a vacuum cleaner.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_25&diff=89515 2010 AMC 10A Problems/Problem 25 2018-01-07T05:04:40Z <p>Pulusona: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Jim starts with a positive integer &lt;math&gt;n&lt;/math&gt; and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with &lt;math&gt;n = 55&lt;/math&gt;, then his sequence contains &lt;math&gt;5&lt;/math&gt; numbers:<br /> <br /> <br /> &lt;cmath&gt;\begin{array}{ccccc}<br /> {}&amp;{}&amp;{}&amp;{}&amp;55\\<br /> 55&amp;-&amp;7^2&amp;=&amp;6\\<br /> 6&amp;-&amp;2^2&amp;=&amp;2\\<br /> 2&amp;-&amp;1^2&amp;=&amp;1\\<br /> 1&amp;-&amp;1^2&amp;=&amp;0\\<br /> \end{array}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;N&lt;/math&gt; be the smallest number for which Jim’s sequence has &lt;math&gt;8&lt;/math&gt; numbers. What is the units digit of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 1<br /> \qquad<br /> \mathrm{(B)}\ 3<br /> \qquad<br /> \mathrm{(C)}\ 5<br /> \qquad<br /> \mathrm{(D)}\ 7<br /> \qquad<br /> \mathrm{(E)}\ 9<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We can find the answer by working backwards. We begin with &lt;math&gt;1-1^2=0&lt;/math&gt; on the bottom row, then the &lt;math&gt;1&lt;/math&gt; goes to the right of the equal's sign in the row above. We find the smallest value &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;x-1^2=1&lt;/math&gt; and &lt;math&gt;x&gt;1^2&lt;/math&gt;, which is &lt;math&gt;x=2&lt;/math&gt;.<br /> <br /> We repeat the same procedure except with &lt;math&gt;x-1^2=1&lt;/math&gt; for the next row and &lt;math&gt;x-1^2=2&lt;/math&gt; for the row after that. However, at the fourth row, we see that solving &lt;math&gt;x-1^2=3&lt;/math&gt; yields &lt;math&gt;x=4&lt;/math&gt;, in which case it would be incorrect since &lt;math&gt;1^2=1&lt;/math&gt; is not the greatest perfect square less than or equal to &lt;math&gt;x&lt;/math&gt; . So we make it a &lt;math&gt;2^2&lt;/math&gt; and solve &lt;math&gt;x-2^2=3&lt;/math&gt;. We continue on using this same method where we increase the perfect square until &lt;math&gt;x&lt;/math&gt; can be made bigger than it. When we repeat this until we have &lt;math&gt;8&lt;/math&gt; rows, we get:<br /> <br /> &lt;cmath&gt; \begin{array}{ccccc}{}&amp;{}&amp;{}&amp;{}&amp;7223\\ 7223&amp;-&amp;84^{2}&amp;=&amp;167\\ 167&amp;-&amp;12^{2}&amp;=&amp;23\\ 23&amp;-&amp;4^{2}&amp;=&amp;7\\ 7&amp;-&amp;2^{2}&amp;=&amp;3\\ 3&amp;-&amp;1^{2}&amp;=&amp;2\\ 2&amp;-&amp;1^{2}&amp;=&amp;1\\ 1&amp;-&amp;1^{2}&amp;=&amp;0\\ \end{array} &lt;/cmath&gt;<br /> <br /> Hence the solution is the last digit of &lt;math&gt;7223&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)}\ 3}&lt;/math&gt;.<br /> <br /> Note: We can go up to &lt;math&gt;167&lt;/math&gt;, and then notice the pattern of units digits alternating between &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt;, so we do not need to calculate &lt;math&gt;7223&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=24|after=Last Question|ab=A}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_10&diff=89437 2010 AMC 10B Problems/Problem 10 2018-01-05T06:22:09Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Shelby drives her scooter at a speed of &lt;math&gt;30&lt;/math&gt; miles per hour if it is not raining, and &lt;math&gt;20&lt;/math&gt; miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of &lt;math&gt;16&lt;/math&gt; miles in &lt;math&gt;40&lt;/math&gt; minutes. How many minutes did she drive in the rain?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We know that &lt;math&gt;d = vt&lt;/math&gt;<br /> <br /> Since we know that she drove both when it was raining and when it was not and that her total distance traveled is &lt;math&gt;16&lt;/math&gt; miles.<br /> <br /> We also know that she drove a total of &lt;math&gt;40&lt;/math&gt; minutes which is &lt;math&gt;\dfrac{2}{3}&lt;/math&gt; of an hour.<br /> <br /> We get the following system of equations, where &lt;math&gt;x&lt;/math&gt; is the time traveled when it was not raining and &lt;math&gt;y&lt;/math&gt; is the time traveled when it was raining:<br /> <br /> &lt;math&gt;\left\{\begin{array}{ccc} 30x + 20y &amp; = &amp; 16 \\x + y &amp; = &amp; \dfrac{2}{3} \end{array} \right.&lt;/math&gt;<br /> <br /> Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:<br /> <br /> &lt;math&gt;-10y = -4 \Leftrightarrow y = \dfrac{2}{5}&lt;/math&gt;<br /> <br /> We know now that the time traveled in rain was &lt;math&gt;\dfrac{2}{5}&lt;/math&gt; of an hour, which is &lt;math&gt;\dfrac{2}{5}*60 = 24&lt;/math&gt; minutes<br /> <br /> So, our answer is &lt;math&gt; \boxed{\textbf{(C)}\ 24} &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We let &lt;math&gt;t&lt;/math&gt; be the time Shelby drives in the rain. This gives us the equation &lt;math&gt; 20t + 30\left(\frac{2}{3} - t \right ) = 16&lt;/math&gt;. Expanding and rearranging gives us &lt;math&gt;10t = 4&lt;/math&gt;, or &lt;math&gt;t = 0.4&lt;/math&gt; hours. we multiply &lt;math&gt;0.4&lt;/math&gt; by &lt;math&gt;60&lt;/math&gt;, which gives us &lt;math&gt; \boxed{\textbf{(C)}\ 24} &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_10&diff=89436 2010 AMC 10B Problems/Problem 10 2018-01-05T06:21:13Z <p>Pulusona: </p> <hr /> <div>==Problem==<br /> Shelby drives her scooter at a speed of &lt;math&gt;30&lt;/math&gt; miles per hour if it is not raining, and &lt;math&gt;20&lt;/math&gt; miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of &lt;math&gt;16&lt;/math&gt; miles in &lt;math&gt;40&lt;/math&gt; minutes. How many minutes did she drive in the rain?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We know that &lt;math&gt;d = vt&lt;/math&gt;<br /> <br /> Since we know that she drove both when it was raining and when it was not and that her total distance traveled is &lt;math&gt;16&lt;/math&gt; miles.<br /> <br /> We also know that she drove a total of &lt;math&gt;40&lt;/math&gt; minutes which is &lt;math&gt;\dfrac{2}{3}&lt;/math&gt; of an hour.<br /> <br /> We get the following system of equations, where &lt;math&gt;x&lt;/math&gt; is the time traveled when it was not raining and &lt;math&gt;y&lt;/math&gt; is the time traveled when it was raining:<br /> <br /> &lt;math&gt;\left\{\begin{array}{ccc} 30x + 20y &amp; = &amp; 16 \\x + y &amp; = &amp; \dfrac{2}{3} \end{array} \right.&lt;/math&gt;<br /> <br /> Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:<br /> <br /> &lt;math&gt;-10y = -4 \Leftrightarrow y = \dfrac{2}{5}&lt;/math&gt;<br /> <br /> We know now that the time traveled in rain was &lt;math&gt;\dfrac{2}{5}&lt;/math&gt; of an hour, which is &lt;math&gt;\dfrac{2}{5}*60 = 24&lt;/math&gt; minutes<br /> <br /> So, our answer is &lt;math&gt; \boxed{\textbf{(C)}\ 24} &lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We let &lt;math&gt;t&lt;/math&gt; be the time Shelby drives in the rain. This gives us the equation &lt;math&gt; 20t + 30(\frac{2}{3} - t) = 16&lt;/math&gt;. Expanding and rearranging gives us &lt;math&gt;10t = 4&lt;/math&gt;, or &lt;math&gt;t = 0.4&lt;/math&gt; hours. we multiply &lt;math&gt;0.4&lt;/math&gt; by &lt;math&gt;60&lt;/math&gt;, which gives us &lt;math&gt; \boxed{\textbf{(C)}\ 24} &lt;/math&gt;.<br /> ==See Also==<br /> {{AMC10 box|year=2010|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=Talk:2016_AMC_10B_Problems/Problem_19&diff=89404 Talk:2016 AMC 10B Problems/Problem 19 2018-01-04T06:04:21Z <p>Pulusona: Created page with &quot;pulusona: For solution 1, how do we know that PQ is not a multiple of 7?&quot;</p> <hr /> <div>pulusona: For solution 1, how do we know that PQ is not a multiple of 7?</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=Talk:2016_AMC_10B_Problems/Problem_14&diff=89403 Talk:2016 AMC 10B Problems/Problem 14 2018-01-04T06:00:55Z <p>Pulusona: </p> <hr /> <div>pulusona: Is graphing the only solution?</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=Talk:2016_AMC_10B_Problems/Problem_14&diff=89402 Talk:2016 AMC 10B Problems/Problem 14 2018-01-04T05:59:12Z <p>Pulusona: Created page with &quot;Pulusona: Is graphing the only solution?&quot;</p> <hr /> <div>Pulusona: Is graphing the only solution?</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=89401 2016 AMC 10B Problems/Problem 17 2018-01-04T05:57:31Z <p>Pulusona: /* Warning */</p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad<br /> \textbf{(B)}\ 343 \qquad<br /> \textbf{(C)}\ 625 \qquad<br /> \textbf{(D)}\ 729 \qquad<br /> \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let us call the six sides of our cube &lt;math&gt;a,b,c,d,e,&lt;/math&gt; and &lt;math&gt;f&lt;/math&gt; (where &lt;math&gt;a&lt;/math&gt; is opposite &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; is opposite &lt;math&gt;e&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; is opposite &lt;math&gt;f&lt;/math&gt;.<br /> Thus, for the eight vertices, we have the following products: &lt;math&gt;abc&lt;/math&gt;,&lt;math&gt;abe&lt;/math&gt;,&lt;math&gt;bcd&lt;/math&gt;,&lt;math&gt;bde&lt;/math&gt;,&lt;math&gt;acf&lt;/math&gt;,&lt;math&gt;cdf&lt;/math&gt;,&lt;math&gt;cef&lt;/math&gt;, and &lt;math&gt;def&lt;/math&gt;.<br /> Let us find the sum of these products:<br /> &lt;math&gt;abc+abe+bcd+bde+acf+cdf+aef+def&lt;/math&gt;<br /> We notice &lt;math&gt;b&lt;/math&gt; is a factor of the first four terms, and &lt;math&gt;f&lt;/math&gt; is factor is the last four terms.<br /> &lt;math&gt;b(ac+ae+cd+de)+f(ac+ae+cd+de)&lt;/math&gt;<br /> Now, we can factor even more:<br /> &lt;math&gt;(b+f)(ac+ae+cd+de)&lt;/math&gt;<br /> &lt;math&gt;(b+f)(a(c+e)+d(c+e))&lt;/math&gt;<br /> &lt;math&gt;(b+f)(a+d)(c+e)&lt;/math&gt;<br /> We have the product. Notice how the factors are sums of opposite faces. The best sum for this is to make &lt;math&gt;(7+2)&lt;/math&gt;,&lt;math&gt;(6+3)&lt;/math&gt;, and &lt;math&gt;(5+4)&lt;/math&gt; all factors.<br /> &lt;math&gt; (7+2)(6+3)(5+4)&lt;/math&gt;<br /> &lt;math&gt;9&lt;/math&gt; &lt;math&gt; *&lt;/math&gt; &lt;math&gt; 9&lt;/math&gt; &lt;math&gt; *&lt;/math&gt; &lt;math&gt;9&lt;/math&gt;<br /> &lt;math&gt; 729 &lt;/math&gt;<br /> Thus our answer is &lt;math&gt;\textbf{(D)}\ 729&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> We first find the factorization &lt;math&gt;(b+f)(a+d)(c+e)&lt;/math&gt; using the method in Solution 1. By using AM-GM, we get, &lt;math&gt;(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3&lt;/math&gt;. To maximize, the factorization, we get the answer is &lt;math&gt;\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}&lt;/math&gt;<br /> <br /> ==Solution 3 (Cheap Solution)==<br /> ===Warning===<br /> Only use this if you are stuck on the problem or are low on time, and if you don't want to get the correct answer.<br /> <br /> ===Solution===<br /> Create a pairing that seems to intuitively seem to be optimal value. So put a number and it's complement(the number that's the difference of 9 and this number). &lt;math&gt;1680&lt;/math&gt; is way too high using reasonability after you do this so you put &lt;math&gt;\boxed{\textbf{D}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_13&diff=89338 2015 AMC 10B Problems/Problem 13 2018-01-01T22:01:24Z <p>Pulusona: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The line &lt;math&gt;12x+5y=60&lt;/math&gt; forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?<br /> <br /> &lt;math&gt;\textbf{(A) } 20 \qquad\textbf{(B) } \dfrac{360}{17} \qquad\textbf{(C) } \dfrac{107}{5} \qquad\textbf{(D) } \dfrac{43}{2} \qquad\textbf{(E) } \dfrac{281}{13} &lt;/math&gt;<br /> <br /> ==Solution==<br /> We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If &lt;math&gt;x=0&lt;/math&gt;, then &lt;math&gt;y=12&lt;/math&gt;. If &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;0&lt;/math&gt;, then &lt;math&gt;x=5&lt;/math&gt;. Our three vertices are &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;(5,0)&lt;/math&gt;, and &lt;math&gt;(0,12)&lt;/math&gt;. Two of our altitudes are &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;12&lt;/math&gt;, and since it is a 5-12-13 right triangle, the hypotenuse is &lt;math&gt;13&lt;/math&gt;. Since the area of the triangle is &lt;math&gt;30&lt;/math&gt;, so our final altitude is &lt;math&gt;\frac{30(2)}{13}=\frac{60}{13}&lt;/math&gt;. The sum of our altitudes is &lt;math&gt;\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}&lt;/math&gt;. Note that there is no need to calculate the final answer after we know that the third altitude has length &lt;math&gt;\frac{60}{13}&lt;/math&gt; since &lt;math&gt;E&lt;/math&gt; is the only choice with a denominator of &lt;math&gt;13&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=89337 2015 AMC 10B Problems/Problem 22 2018-01-01T21:58:31Z <p>Pulusona: /* Solution 2(Trigonometry) */</p> <hr /> <div>==Problem==<br /> In the figure shown below, &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon and &lt;math&gt;AG=1&lt;/math&gt;. What is &lt;math&gt;FG + JH + CD&lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br /> //(0,0) is a convenient point<br /> //E1 to prevent conflict with direction E(ast)<br /> pair F=intersectionpoints(D--A,E1--B), G=intersectionpoints(A--C,E1--B), H=intersectionpoints(B--D,A--C), I=intersectionpoints(C--E1,D--B), J=intersectionpoints(E1--C,D--A);<br /> draw(A--B--C--D--E1--A);<br /> draw(A--D--B--E1--C--A);<br /> draw(F--I--G--J--H--F);<br /> label(&quot;A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,NW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,S);<br /> label(&quot;$J$&quot;,J,W);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Triangle &lt;math&gt;AFG&lt;/math&gt; is isosceles, so &lt;math&gt;AG=AF=1&lt;/math&gt;. &lt;math&gt;FJ = FG&lt;/math&gt; since &lt;math&gt;\triangle FGJ&lt;/math&gt; is also isosceles. Using the symmetry of pentagon &lt;math&gt;FGHIJ&lt;/math&gt;, notice that &lt;math&gt;\triangle JHG \cong \triangle AFG&lt;/math&gt;. Therefore, &lt;math&gt;JH=AF=1&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle AJH \sim \triangle AFG&lt;/math&gt;, <br /> &lt;cmath&gt;\frac{JH}{AF+FJ}=\frac{FG}{FA}&lt;/cmath&gt;.<br /> &lt;cmath&gt;\frac{1}{1+FG} = \frac{FG}1&lt;/cmath&gt;<br /> &lt;cmath&gt;1 = FG^2 + FG&lt;/cmath&gt;<br /> &lt;cmath&gt;FG^2+FG-1 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;FG = \frac{-1 \pm \sqrt{5} }{2}&lt;/cmath&gt;<br /> <br /> However, &lt;math&gt;FG=\frac{-1 + \sqrt{5}}{2}&lt;/math&gt; since &lt;math&gt;FG&lt;/math&gt; must be greater than 0.<br /> <br /> Notice that &lt;math&gt;CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==Solution 2 (Trigonometry)==<br /> Note that since &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon, all of its interior angles are &lt;math&gt;108^\circ&lt;/math&gt;. We can say that pentagon &lt;math&gt;FGHIJ&lt;/math&gt; is also regular by symmetry. So, all of the interior angles of &lt;math&gt;FGHIJ&lt;/math&gt; are &lt;math&gt;108^\circ&lt;/math&gt;. Now, we can angle chase and use trigonometry to get that &lt;math&gt;FG=2\sin18^\circ&lt;/math&gt;, &lt;math&gt;JH=2\sin18^\circ*(2\sin18^\circ+1)&lt;/math&gt;, and &lt;math&gt;DC=2\sin18^\circ*(2\sin18^\circ+2)&lt;/math&gt;. Adding these together, we get that &lt;math&gt;FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;. Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to &lt;math&gt;8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;, but we can find that this is closest to &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=89336 2015 AMC 10B Problems/Problem 22 2018-01-01T21:58:14Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> In the figure shown below, &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon and &lt;math&gt;AG=1&lt;/math&gt;. What is &lt;math&gt;FG + JH + CD&lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br /> //(0,0) is a convenient point<br /> //E1 to prevent conflict with direction E(ast)<br /> pair F=intersectionpoints(D--A,E1--B), G=intersectionpoints(A--C,E1--B), H=intersectionpoints(B--D,A--C), I=intersectionpoints(C--E1,D--B), J=intersectionpoints(E1--C,D--A);<br /> draw(A--B--C--D--E1--A);<br /> draw(A--D--B--E1--C--A);<br /> draw(F--I--G--J--H--F);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,E);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,NW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,E);<br /> label(&quot;$I$&quot;,I,S);<br /> label(&quot;$J\$&quot;,J,W);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Triangle &lt;math&gt;AFG&lt;/math&gt; is isosceles, so &lt;math&gt;AG=AF=1&lt;/math&gt;. &lt;math&gt;FJ = FG&lt;/math&gt; since &lt;math&gt;\triangle FGJ&lt;/math&gt; is also isosceles. Using the symmetry of pentagon &lt;math&gt;FGHIJ&lt;/math&gt;, notice that &lt;math&gt;\triangle JHG \cong \triangle AFG&lt;/math&gt;. Therefore, &lt;math&gt;JH=AF=1&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;\triangle AJH \sim \triangle AFG&lt;/math&gt;, <br /> &lt;cmath&gt;\frac{JH}{AF+FJ}=\frac{FG}{FA}&lt;/cmath&gt;.<br /> &lt;cmath&gt;\frac{1}{1+FG} = \frac{FG}1&lt;/cmath&gt;<br /> &lt;cmath&gt;1 = FG^2 + FG&lt;/cmath&gt;<br /> &lt;cmath&gt;FG^2+FG-1 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;FG = \frac{-1 \pm \sqrt{5} }{2}&lt;/cmath&gt;<br /> <br /> However, &lt;math&gt;FG=\frac{-1 + \sqrt{5}}{2}&lt;/math&gt; since &lt;math&gt;FG&lt;/math&gt; must be greater than 0.<br /> <br /> Notice that &lt;math&gt;CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;<br /> <br /> ==Solution 2(Trigonometry)==<br /> Note that since &lt;math&gt;ABCDE&lt;/math&gt; is a regular pentagon, all of its interior angles are &lt;math&gt;108^\circ&lt;/math&gt;. We can say that pentagon &lt;math&gt;FGHIJ&lt;/math&gt; is also regular by symmetry. So, all of the interior angles of &lt;math&gt;FGHIJ&lt;/math&gt; are &lt;math&gt;108^\circ&lt;/math&gt;. Now, we can angle chase and use trigonometry to get that &lt;math&gt;FG=2\sin18^\circ&lt;/math&gt;, &lt;math&gt;JH=2\sin18^\circ*(2\sin18^\circ+1)&lt;/math&gt;, and &lt;math&gt;DC=2\sin18^\circ*(2\sin18^\circ+2)&lt;/math&gt;. Adding these together, we get that &lt;math&gt;FG+JH+CD=2\sin18^\circ*(4+4\sin18^\circ)=8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;. Because calculators were not permitted in the 2015 AMC 10B, we can not use a calculator to find out which of the options is equal to &lt;math&gt;8\sin18^\circ*(1+\sin18^\circ)&lt;/math&gt;, but we can find that this is closest to &lt;math&gt;\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_24&diff=89335 2015 AMC 10A Problems/Problem 24 2018-01-01T21:50:29Z <p>Pulusona: /* Solution */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #19]] and [[2015 AMC 10A Problems|2015 AMC 10A #24]]}}<br /> ==Problem==<br /> For some positive integers &lt;math&gt;p&lt;/math&gt;, there is a quadrilateral &lt;math&gt;ABCD&lt;/math&gt; with positive integer side lengths, perimeter &lt;math&gt;p&lt;/math&gt;, right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;AB=2&lt;/math&gt;, and &lt;math&gt;CD=AD&lt;/math&gt;. How many different values of &lt;math&gt;p&lt;2015&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63&lt;/math&gt;<br /> <br /> ==Solution ==<br /> Let &lt;math&gt;BC = x&lt;/math&gt; and &lt;math&gt;CD = AD = y&lt;/math&gt; be positive integers. Drop a perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;CD&lt;/math&gt; to show that, using the Pythagorean Theorem, that<br /> &lt;cmath&gt;x^2 + (y - 2)^2 = y^2.&lt;/cmath&gt;<br /> Simplifying yields &lt;math&gt;x^2 - 4y + 4 = 0&lt;/math&gt;, so &lt;math&gt;x^2 = 4(y - 1)&lt;/math&gt;. Thus, &lt;math&gt;y&lt;/math&gt; is one more than a perfect square.<br /> <br /> The perimeter &lt;math&gt;p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2&lt;/math&gt; must be less than 2015. Simple calculations demonstrate that &lt;math&gt;y = 31^2 + 1 = 962&lt;/math&gt; is valid, but &lt;math&gt;y = 32^2 + 1 = 1025&lt;/math&gt; is not. On the lower side, &lt;math&gt;y = 1&lt;/math&gt; does not work (because &lt;math&gt;x &gt; 0&lt;/math&gt;), but &lt;math&gt;y = 1^2 + 1&lt;/math&gt; does work. Hence, there are 31 valid &lt;math&gt;y&lt;/math&gt; (all &lt;math&gt;y&lt;/math&gt; such that &lt;math&gt;y = n^2 + 1&lt;/math&gt; for &lt;math&gt;1 \le n \le 31&lt;/math&gt;), and so our answer is &lt;math&gt;\boxed{\textbf{(B) } 31}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2015|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2015|ab=A|num-b=18|num-a=20}}<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_20&diff=89334 2015 AMC 10A Problems/Problem 20 2018-01-01T21:45:58Z <p>Pulusona: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A rectangle with positive integer side lengths in &lt;math&gt;\text{cm}&lt;/math&gt; has area &lt;math&gt;A&lt;/math&gt; &lt;math&gt;\text{cm}^2&lt;/math&gt; and perimeter &lt;math&gt;P&lt;/math&gt; &lt;math&gt;\text{cm}&lt;/math&gt;. Which of the following numbers cannot equal &lt;math&gt;A+P&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the rectangle's length be &lt;math&gt;a&lt;/math&gt; and its width be &lt;math&gt;b&lt;/math&gt;. Its area is &lt;math&gt;ab&lt;/math&gt; and the perimeter is &lt;math&gt;2a+2b&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;A + P = ab + 2a + 2b&lt;/math&gt;. Factoring, we have &lt;math&gt;(a + 2)(b + 2) - 4&lt;/math&gt;.<br /> <br /> The only one of the answer choices that cannot be expressed in this form is &lt;math&gt;102&lt;/math&gt;, as &lt;math&gt;102 + 4&lt;/math&gt; is twice a prime. There would then be no way to express &lt;math&gt;106&lt;/math&gt; as &lt;math&gt;(a + 2)(b + 2)&lt;/math&gt;, keeping &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; as positive integers.<br /> <br /> Our answer is then &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> Note: The original problem only stated that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Algebra Problems]]</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=User:Datboiiiiiiii&diff=89332 User:Datboiiiiiiii 2018-01-01T21:35:27Z <p>Pulusona: Created page with &quot;bot.&quot;</p> <hr /> <div>bot.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_15&diff=89327 2014 AMC 10B Problems/Problem 15 2018-01-01T21:06:53Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;DC = 2CB&lt;/math&gt; and points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;\overline{ED}&lt;/math&gt; and &lt;math&gt;\overline{FD}&lt;/math&gt; trisect &lt;math&gt;\angle ADC&lt;/math&gt; as shown. What is the ratio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to the area of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle);<br /> draw((0, 0)--(sqrt(3)/3, 1));<br /> draw((0, 0)--(sqrt(3), 1));<br /> label(&quot;A&quot;, (0, 1), N);<br /> label(&quot;B&quot;, (2, 1), N);<br /> label(&quot;C&quot;, (2, 0), S);<br /> label(&quot;D&quot;, (0, 0), S);<br /> label(&quot;E&quot;, (sqrt(3)/3, 1), N);<br /> label(&quot;F&quot;, (sqrt(3), 1), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let the length of &lt;math&gt;AD&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;, so that the length of &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;2x&lt;/math&gt; and &lt;math&gt;\text{[}ABCD\text{]}=2x^2&lt;/math&gt;. <br /> <br /> Because &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;\angle ADC=90^{\circ}&lt;/math&gt;, and so &lt;math&gt;\angle ADE=\angle EDF=\angle FDC=30^{\circ}&lt;/math&gt;. Thus &lt;math&gt;\triangle DAE&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle; this implies that &lt;math&gt;\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}&lt;/math&gt;. Now drop the altitude from &lt;math&gt;E&lt;/math&gt; of &lt;math&gt;\triangle DEF&lt;/math&gt;, forming two &lt;math&gt;30-60-90&lt;/math&gt; triangles. <br /> <br /> Because the length of &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt;, from the properties of a &lt;math&gt;30-60-90&lt;/math&gt; triangle the length of &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\frac{x\sqrt{3}}{3}&lt;/math&gt; and the length of &lt;math&gt;DE&lt;/math&gt; is thus &lt;math&gt;\frac{2x\sqrt{3}}{3}&lt;/math&gt;. Thus the altitude of &lt;math&gt;\triangle DEF&lt;/math&gt; is &lt;math&gt;\frac{x\sqrt{3}}{3}&lt;/math&gt;, and its base is &lt;math&gt;2x&lt;/math&gt;, so its area is &lt;math&gt;\frac{1}{2}(2x)\left(\frac{x\sqrt{3}}{3}\right)=\frac{x^2\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> To finish, &lt;math&gt;\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> WLOG, let &lt;math&gt;AD = 1&lt;/math&gt; and &lt;math&gt;BC = 2&lt;/math&gt;. Furthermore, drop an an altitude from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;CD&lt;/math&gt;, which meets &lt;math&gt;CD&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;. Since &lt;math&gt;\angle ADC&lt;/math&gt; is right and has been trisected, it follows that &lt;math&gt;\triangle ADE&lt;/math&gt; and &lt;math&gt;\triangle DXF&lt;/math&gt; are both &lt;math&gt;30-60-90&lt;/math&gt; triangles. Therefore, &lt;math&gt;AE = \frac{\sqrt{3}}{3}&lt;/math&gt;, and &lt;math&gt;DX = AF = \sqrt{3}&lt;/math&gt;. Hence, it follows that &lt;math&gt;EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}&lt;/math&gt;. Since &lt;math&gt;\angle ADE&lt;/math&gt; is right, the height and base of &lt;math&gt;\triangle DEF&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt; \frac{2\sqrt{3}}{3}&lt;/math&gt;, respectively. Thus, the area of &lt;math&gt;\triangle DEF&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt;, and the area of rectengle &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;, so the ratio beween the area of &lt;math&gt;\triangle DEF&lt;/math&gt; and &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}&lt;/math&gt;. Note that we are able to assume that &lt;math&gt;AD=1&lt;/math&gt; and &lt;math&gt;BC = 2&lt;/math&gt; because we were asked to find the ratio between two areas.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_15&diff=89326 2014 AMC 10B Problems/Problem 15 2018-01-01T21:05:28Z <p>Pulusona: </p> <hr /> <div>==Problem==<br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;DC = 2CB&lt;/math&gt; and points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;\overline{ED}&lt;/math&gt; and &lt;math&gt;\overline{FD}&lt;/math&gt; trisect &lt;math&gt;\angle ADC&lt;/math&gt; as shown. What is the ratio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to the area of rectangle &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle);<br /> draw((0, 0)--(sqrt(3)/3, 1));<br /> draw((0, 0)--(sqrt(3), 1));<br /> label(&quot;A&quot;, (0, 1), N);<br /> label(&quot;B&quot;, (2, 1), N);<br /> label(&quot;C&quot;, (2, 0), S);<br /> label(&quot;D&quot;, (0, 0), S);<br /> label(&quot;E&quot;, (sqrt(3)/3, 1), N);<br /> label(&quot;F&quot;, (sqrt(3), 1), N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> Let the length of &lt;math&gt;AD&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;, so that the length of &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;2x&lt;/math&gt; and &lt;math&gt;\text{[}ABCD\text{]}=2x^2&lt;/math&gt;. <br /> <br /> Because &lt;math&gt;ABCD&lt;/math&gt; is a rectangle, &lt;math&gt;\angle ADC=90^{\circ}&lt;/math&gt;, and so &lt;math&gt;\angle ADE=\angle EDF=\angle FDC=30^{\circ}&lt;/math&gt;. Thus &lt;math&gt;\triangle DAE&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; right triangle; this implies that &lt;math&gt;\angle DEF=180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;, so &lt;math&gt;\angle EFD=180^{\circ}-(120^{\circ}+30^{\circ})=30^{\circ}&lt;/math&gt;. Now drop the altitude from &lt;math&gt;E&lt;/math&gt; of &lt;math&gt;\triangle DEF&lt;/math&gt;, forming two &lt;math&gt;30-60-90&lt;/math&gt; triangles. <br /> <br /> Because the length of &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt;, from the properties of a &lt;math&gt;30-60-90&lt;/math&gt; triangle the length of &lt;math&gt;AE&lt;/math&gt; is &lt;math&gt;\frac{x\sqrt{3}}{3}&lt;/math&gt; and the length of &lt;math&gt;DE&lt;/math&gt; is thus &lt;math&gt;\frac{2x\sqrt{3}}{3}&lt;/math&gt;. Thus the altitude of &lt;math&gt;\triangle DEF&lt;/math&gt; is &lt;math&gt;\frac{x\sqrt{3}}{3}&lt;/math&gt;, and its base is &lt;math&gt;2x&lt;/math&gt;, so its area is &lt;math&gt;\frac{1}{2}(2x)\left(\frac{x\sqrt{3}}{3}\right)=\frac{x^2\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> To finish, &lt;math&gt;\frac{\text{[}\triangle DEF\text{]}}{\text{[}ABCD\text{]}}=\frac{\frac{x^2\sqrt{3}}{3}}{2x^2}=\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> WLOG, let &lt;math&gt;AD = 1&lt;/math&gt; and &lt;math&gt;BC = 2&lt;/math&gt;. Furthermore, drop an an altitude from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;CD&lt;/math&gt;, which meets &lt;math&gt;CD&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;. Since &lt;math&gt;\angle ADC&lt;/math&gt; is right and has been trisected, it follows that &lt;math&gt;\triangle ADE&lt;/math&gt; and &lt;math&gt;\triangle DXF&lt;/math&gt; are both &lt;math&gt;30-60-90&lt;/math&gt; triangles. Therefore, &lt;math&gt;AE = \frac{\sqrt{3}}{3}&lt;/math&gt;, and &lt;math&gt;DX = AF = \sqrt{3}&lt;/math&gt;. Hence, it follows that &lt;math&gt;EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}&lt;/math&gt;. Since &lt;math&gt;\angle ADE&lt;/math&gt; is right, the height and base of &lt;math&gt;\triangle DEF&lt;/math&gt; are &lt;math&gt;1&lt;/math&gt; and &lt;math&gt; \frac{2\sqrt{3}}{3}&lt;/math&gt;, respectively. Thus, the area of &lt;math&gt;\triangle DEF&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{3}&lt;/math&gt;, and the area of rectengle &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;, so the ratio beween the area of &lt;math&gt;\triangle DEF&lt;/math&gt; and &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}&lt;/math&gt;. <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=Talk:2013_AMC_12B_Problems/Problem_10&diff=89300 Talk:2013 AMC 12B Problems/Problem 10 2017-12-31T20:55:52Z <p>Pulusona: </p> <hr /> <div>pulusona: I think it would be easier to go to the blue booth first in solution 1 because you aren't left with a remainder.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=Talk:2013_AMC_12B_Problems/Problem_10&diff=89299 Talk:2013 AMC 12B Problems/Problem 10 2017-12-31T20:55:04Z <p>Pulusona: Created page with &quot;I think it would be easier to go to the blue booth first in solution 1 because you aren't left with a remainder.&quot;</p> <hr /> <div>I think it would be easier to go to the blue booth first in solution 1 because you aren't left with a remainder.</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_12&diff=89298 2013 AMC 10B Problems/Problem 12 2017-12-31T20:50:06Z <p>Pulusona: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt; S &lt;/math&gt; be the set of sides and diagonals of a regular pentagon. A pair of elements of &lt;math&gt; S &lt;/math&gt; are selected at random without replacement. What is the probability that the two chosen segments have the same length?<br /> <br /> &lt;math&gt; \textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is &lt;math&gt;\boxed{\textbf{(B) }\frac{4}{9}}&lt;/math&gt;. <br /> ===Solution 2===<br /> Alternatively, we can divide this problem into two cases.<br /> Case 1: Side;<br /> In this case, there is a &lt;math&gt;\frac{5}{10}&lt;/math&gt; chance of picking a side, and a &lt;math&gt;\frac{4}{9}&lt;/math&gt; chance of picking another side.<br /> Case 2: Diagonal;<br /> This case is similar to the first, for again, there is a &lt;math&gt;\frac{5}{10}&lt;/math&gt; chance of picking a diagonal, and a &lt;math&gt;\frac{4}{9}&lt;/math&gt; chance of picking another diagonal.<br /> <br /> Summing these cases up gives us a probability of &lt;math&gt;\boxed{\textbf{(B) }\frac{4}{9}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_12&diff=89297 2013 AMC 10B Problems/Problem 12 2017-12-31T20:49:31Z <p>Pulusona: /* Solution */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt; S &lt;/math&gt; be the set of sides and diagonals of a regular pentagon. A pair of elements of &lt;math&gt; S &lt;/math&gt; are selected at random without replacement. What is the probability that the two chosen segments have the same length?<br /> <br /> &lt;math&gt; \textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is &lt;math&gt;\boxed{\textbf{(B) }\frac{4}{9}}&lt;/math&gt;. <br /> ===Solution 2===<br /> Alternatively, we can divide this problem into two cases.<br /> Case 1: Side<br /> In this case, there is a &lt;math&gt;\frac{5}{10}&lt;/math&gt; chance of picking a side, and a &lt;math&gt;\frac{4}{9}&lt;/math&gt; chance of picking another side.<br /> Case 2: Diagonal<br /> This case is similar to the first, for again, there is a &lt;math&gt;\frac{5}{10}&lt;/math&gt; chance of picking a diagonal, and a &lt;math&gt;\frac{4}{9}&lt;/math&gt; chance of picking another diagonal.<br /> <br /> Summing these cases up gives us a probability of &lt;math&gt;\boxed{\textbf{(B) }\frac{4}{9}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=89296 2013 AMC 12B Problems/Problem 4 2017-12-31T20:38:22Z <p>Pulusona: /* Solution */</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br /> <br /> ==Problem==<br /> Ray's car averages &lt;math&gt;40&lt;/math&gt; miles per gallon of gasoline, and Tom's car averages &lt;math&gt;10&lt;/math&gt; miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?&lt;br \&gt;<br /> &lt;math&gt;\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> ==Solution==<br /> WLOG, let Ray and Tom drive 40 miles. Ray's car would require &lt;math&gt;\frac{40}{40}=1&lt;/math&gt; gallon of gas and Tom's car would require &lt;math&gt;\frac{40}{10}=4&lt;/math&gt; gallons of gas. They would have driven a total of &lt;math&gt;40+40=80&lt;/math&gt; miles, on &lt;math&gt;1+4=5&lt;/math&gt; gallons of gas, for a combined rate of &lt;math&gt;\frac{80}{5}=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) }16}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br /> {{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=89259 2003 AMC 10A Problems/Problem 23 2017-12-30T03:54:57Z <p>Pulusona: </p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> [[Image:2003amc10a23.gif]]<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1===<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> We see that the bottom row of &lt;math&gt;2003&lt;/math&gt; small triangles are formed from &lt;math&gt;1002&lt;/math&gt; upward-facing triangles and &lt;math&gt;1001&lt;/math&gt; downward-facing triangles. Since each upward-facing triangle uses three distinct toothpicks, and since the total number of upward-facing triangles is &lt;math&gt;1002+1001+...+1=\frac{1003\cdot1002}{2}=502503&lt;/math&gt;, we have that the total number of toothpicks is &lt;math&gt;3\cdot 502503=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> Experimenting a bit we find that the number of toothpicks needs a triangle with &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; rows is &lt;math&gt;1\cdot{3}&lt;/math&gt;, &lt;math&gt;3\cdot{3}&lt;/math&gt; and &lt;math&gt;6\cdot{3}&lt;/math&gt; respectively. Since &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; are triangular numbers we know that depending on how many rows there are in the triangle, the number we multiply by &lt;math&gt;3&lt;/math&gt; to find total no.toothpicks is the corresponding triangular number. Since the triangle in question has &lt;math&gt;2n-1=2003\implies{n=1002}&lt;/math&gt; rows, we can use &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt; to find the triangular number for that row and multiply by &lt;math&gt;3&lt;/math&gt;, hence finding the total no.toothpicks; this is just &lt;math&gt;\frac{3\cdot{1002}\cdot{1003}}{2}=3\cdot{501}\cdot{1003}=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt;.<br /> ===Note===<br /> In the final step of the problem, we know that the units digit of the answer must &lt;math&gt;9&lt;/math&gt;, so the only answer choice applicable must be &lt;math&gt;\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt;. This saves you the time it takes to compute &lt;math&gt;3\cdot{501}\cdot{1003}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Answer_Key&diff=89258 2003 AMC 10A Answer Key 2017-12-30T03:45:37Z <p>Pulusona: </p> <hr /> <div>#D<br /> #B<br /> #D<br /> #A<br /> #B<br /> #C<br /> #B<br /> #E<br /> #A<br /> #E<br /> #E<br /> #A<br /> #A<br /> #A<br /> #C<br /> #C<br /> #B<br /> #B<br /> #C<br /> #E<br /> #D<br /> #B<br /> #C<br /> #E<br /> #B</div> Pulusona https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Answer_Key&diff=89257 2003 AMC 10A Answer Key 2017-12-30T03:45:25Z <p>Pulusona: </p> <hr /> <div>#E<br /> #B<br /> #D<br /> #A<br /> #B<br /> #C<br /> #B<br /> #E<br /> #A<br /> #E<br /> #E<br /> #A<br /> #A<br /> #A<br /> #C<br /> #C<br /> #B<br /> #B<br /> #C<br /> #E<br /> #D<br /> #B<br /> #C<br /> #E<br /> #B</div> Pulusona