https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Matongxu&feedformat=atom AoPS Wiki - User contributions [en] 2021-01-23T10:38:52Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_15&diff=92204 2015 AIME II Problems/Problem 15 2018-02-22T02:57:47Z <p>Matongxu: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; have radii &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, respectively, and are externally tangent at point &lt;math&gt;A&lt;/math&gt;. Point &lt;math&gt;B&lt;/math&gt; is on &lt;math&gt;\mathcal{P}&lt;/math&gt; and point &lt;math&gt;C&lt;/math&gt; is on &lt;math&gt;\mathcal{Q}&lt;/math&gt; so that line &lt;math&gt;BC&lt;/math&gt; is a common external tangent of the two circles. A line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; again at &lt;math&gt;D&lt;/math&gt; and intersects &lt;math&gt;\mathcal{Q}&lt;/math&gt; again at &lt;math&gt;E&lt;/math&gt;. Points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on the same side of &lt;math&gt;\ell&lt;/math&gt;, and the areas of &lt;math&gt;\triangle DBA&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equal. This common area is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> import cse5;<br /> pathpen=black; pointpen=black;<br /> size(6cm);<br /> <br /> pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689);<br /> <br /> filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7));<br /> filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7));<br /> D(CR((0,1),1)); D(CR((4,4),4,150,390));<br /> D(L(MP(&quot;D&quot;,D(D),N),MP(&quot;A&quot;,D((0.8,1.6)),NE),1,5.5));<br /> D((-1.2,0)--MP(&quot;B&quot;,D((0,0)),S)--MP(&quot;C&quot;,D((4,0)),S)--(8,0));<br /> D(MP(&quot;E&quot;,E,N));<br /> &lt;/asy&gt;<br /> <br /> ==Hint==<br /> This is a #15 on an AIME, so it must be difficult. Indeed, although there are several possible approaches, all of them are very computational.<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> unitsize(35);<br /> draw(Circle((-1,0),1));<br /> draw(Circle((4,0),4));<br /> pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;<br /> A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);<br /> label(&quot;$A$&quot;,A,NE);label(&quot;$O_1$&quot;,O_1,NE);label(&quot;$O_2$&quot;,O_2,NE);label(&quot;$B$&quot;,B,SW);label(&quot;$C$&quot;,C,SW);label(&quot;$D$&quot;,D,NE);label(&quot;$E$&quot;,E,NE);label(&quot;$N$&quot;,N,W);label(&quot;$K$&quot;,(-24/15,0.2));label(&quot;$L$&quot;,(24/15,0.2));label(&quot;$n$&quot;,(-0.8,-0.12));label(&quot;$p$&quot;,((29/15,-48/15)));label(&quot;$\mathcal{P}$&quot;,(-1.6,1.1));label(&quot;$\mathcal{Q}$&quot;,(6,4));<br /> draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);<br /> dot(O_1);dot(O_2);<br /> draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);<br /> &lt;/asy&gt;<br /> <br /> Call &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; the centers of circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt;, respectively, and extend &lt;math&gt;CB&lt;/math&gt; and &lt;math&gt;O_2O_1&lt;/math&gt; to meet at point &lt;math&gt;N&lt;/math&gt;. Call &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;O_1N&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;O_2N&lt;/math&gt;, respectively. Using the fact that &lt;math&gt;\triangle{O_1BN} \sim \triangle{O_2CN}&lt;/math&gt; and setting &lt;math&gt;NO_1 = k&lt;/math&gt;, we have that &lt;math&gt;\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}&lt;/math&gt;. We can do some more length chasing using triangles similar to &lt;math&gt;OBN&lt;/math&gt; to get that &lt;math&gt;AK = AL = \frac{24}{15}&lt;/math&gt;, &lt;math&gt;BK = \frac{12}{15}&lt;/math&gt;, and &lt;math&gt;CL = \frac{48}{15}&lt;/math&gt;. Now, consider the circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; on the coordinate plane, where &lt;math&gt;A&lt;/math&gt; is the origin. If the line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; then &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;. To verify this, notice that &lt;math&gt;\triangle{AO_1D} \sim \triangle{EO_2A}&lt;/math&gt; from the fact that both triangles are isosceles with &lt;math&gt;\angle{O_1AD} \cong \angle{O_2AE}&lt;/math&gt;, which are corresponding angles. Since &lt;math&gt;O_2A = 4\cdot O_1A&lt;/math&gt;, we can conclude that &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;.<br /> <br /> <br /> Hence, we need to find the slope &lt;math&gt;m&lt;/math&gt; of line &lt;math&gt;\ell&lt;/math&gt; such that the perpendicular distance &lt;math&gt;n&lt;/math&gt; from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is four times the perpendicular distance &lt;math&gt;p&lt;/math&gt; from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AE&lt;/math&gt;. This will mean that the product of the bases and heights of triangles &lt;math&gt;ACE&lt;/math&gt; and &lt;math&gt;DBA&lt;/math&gt; will be equal, which in turn means that their areas will be equal. Let the line &lt;math&gt;\ell&lt;/math&gt; have the equation &lt;math&gt;y = -mx \implies mx + y = 0&lt;/math&gt;, and let &lt;math&gt;m&lt;/math&gt; be a positive real number so that the negative slope of &lt;math&gt;\ell&lt;/math&gt; is preserved. Setting &lt;math&gt;A = (0,0)&lt;/math&gt;, the coordinates of &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)&lt;/math&gt;, and the coordinates of &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)&lt;/math&gt;. Using the point-to-line distance formula and the condition &lt;math&gt;n = 4p&lt;/math&gt;, we have &lt;cmath&gt;\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}&lt;/cmath&gt; &lt;cmath&gt;\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.&lt;/cmath&gt; If &lt;math&gt;m &gt; 2&lt;/math&gt;, then clearly &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; would not lie on the same side of &lt;math&gt;\ell&lt;/math&gt;. Thus since &lt;math&gt;m &gt; 0&lt;/math&gt;, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have &lt;cmath&gt;\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.&lt;/cmath&gt; Thus, the equation of &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt;.<br /> <br /> <br /> Then we can find the coordinates of &lt;math&gt;D&lt;/math&gt; by finding the point &lt;math&gt;(x,y)&lt;/math&gt; other than &lt;math&gt;A = (0,0)&lt;/math&gt; where the circle &lt;math&gt;\mathcal{P}&lt;/math&gt; intersects &lt;math&gt;\ell&lt;/math&gt;. &lt;math&gt;\mathcal{P}&lt;/math&gt; can be represented with the equation &lt;math&gt;(x + 1)^2 + y^2 = 1&lt;/math&gt;, and substituting &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt; into this equation yields &lt;math&gt;x = 0, -\frac{8}{13}&lt;/math&gt; as solutions. Discarding &lt;math&gt;x = 0&lt;/math&gt;, the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}&lt;/math&gt;. The distance from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; is then &lt;math&gt;\frac{4}{\sqrt{13}}.&lt;/math&gt; The perpendicular distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; or the height of &lt;math&gt;\triangle{DBA}&lt;/math&gt; is &lt;math&gt;\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.&lt;/math&gt; Finally, the common area is &lt;math&gt;\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}&lt;/math&gt;, and &lt;math&gt;m + n = 64 + 65 = \boxed{129}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> By [[homothety]], we deduce that &lt;math&gt;AE = 4 AD&lt;/math&gt;. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt; is four times that from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;. Let the distance from &lt;math&gt;C&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; and the distance from &lt;math&gt;B&lt;/math&gt; be &lt;math&gt;4x&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the centers of their respective circles. Then dropping a perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; creates a &lt;math&gt;3-4-5&lt;/math&gt; right triangle, from which &lt;math&gt;BC = 4&lt;/math&gt; and, if &lt;math&gt;\alpha = \angle{AQC}&lt;/math&gt;, that &lt;math&gt;\cos \alpha = \dfrac{3}{5}&lt;/math&gt;. Then &lt;math&gt;\angle{BPA} = 180^\circ - \alpha&lt;/math&gt;, and the Law of Cosines on triangles &lt;math&gt;APB&lt;/math&gt; and &lt;math&gt;AQC&lt;/math&gt; gives &lt;math&gt;AB = \dfrac{4}{\sqrt{5}}&lt;/math&gt; and &lt;math&gt;AC = \dfrac{8}{\sqrt{5}}.&lt;/math&gt;<br /> <br /> Now, using the Pythagorean Theorem to express the length of the projection of &lt;math&gt;BC&lt;/math&gt; onto line &lt;math&gt;l&lt;/math&gt; gives<br /> &lt;cmath&gt;\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.&lt;/cmath&gt;<br /> Squaring and simplifying gives<br /> &lt;cmath&gt;\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,&lt;/cmath&gt;<br /> and squaring and solving gives &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}.&lt;/math&gt;<br /> <br /> By the Law of Sines on triangle &lt;math&gt;ABD&lt;/math&gt;, we have<br /> &lt;cmath&gt;\frac{BD}{\sin A} = 2.&lt;/cmath&gt;<br /> But we know &lt;math&gt;\sin A = \dfrac{4x}{AB}&lt;/math&gt;, and so a small computation gives &lt;math&gt;BD = \dfrac{16}{\sqrt{65}}.&lt;/math&gt; The Pythagorean Theorem now gives<br /> &lt;cmath&gt;AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},&lt;/cmath&gt;<br /> and so the common area is &lt;math&gt;\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{129}.&lt;/math&gt;<br /> <br /> ==Alternate Path to x==<br /> Call the intersection of lines &lt;math&gt;l&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;E&lt;/math&gt;.You can use similar triangles to find that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt; is four times the distance from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. Then draw a perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call the point &lt;math&gt;F&lt;/math&gt;. &lt;math&gt;AF = \frac{8}{5}&lt;/math&gt; and &lt;math&gt;FE = FC + CE = \frac{16}{5} + \frac{4}{3} = \frac{68}{15}&lt;/math&gt;, so by the Pythagorean Theorem, &lt;math&gt;AE = \dfrac{20\sqrt{13}}{5}&lt;/math&gt;. You can now use similar triangles to find that &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}&lt;/math&gt; and continue on like in solution 2.<br /> <br /> ==Solution 3==<br /> &lt;math&gt;DE&lt;/math&gt; goes through &lt;math&gt;A&lt;/math&gt;, the point of tangency of both circles. So &lt;math&gt;DE&lt;/math&gt; intercepts equal arcs in circle &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;: [[homothety]]. Hence, &lt;math&gt;AE=4AD&lt;/math&gt;. We will use such similarity later.<br /> <br /> The diagonal distance between the centers of the circles is &lt;math&gt;4+1=5&lt;/math&gt;. The difference in heights is &lt;math&gt;4-1=3&lt;/math&gt;. So &lt;math&gt;BC=\sqrt{5^2-3^2}=4&lt;/math&gt;.<br /> <br /> The triangle connecting the centers with a side parallel to &lt;math&gt;BC&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; right triangle. Since &lt;math&gt;O_PA=1&lt;/math&gt;, the height of &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;1+3/5=8/5&lt;/math&gt;. Drop an altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call it &lt;math&gt;I&lt;/math&gt;: &lt;math&gt;IB=4/5&lt;/math&gt; and &lt;math&gt;IC=4-4/5=32/5&lt;/math&gt;. Since right &lt;math&gt;\triangle AIB\sim\triangle CIB&lt;/math&gt;, &lt;math&gt;ABC&lt;/math&gt; is a right triangle also; &lt;math&gt;IB:IA:IC&lt;/math&gt; form a geometric progression &lt;math&gt;\times 2&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;BA&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;G&lt;/math&gt; on the other side of &lt;math&gt;\circ Q&lt;/math&gt;. By [[homothety]], &lt;math&gt;\triangle DAB\sim\triangle EAG&lt;/math&gt;. By angle chasing &lt;math&gt;\triangle DAB&lt;/math&gt; through right triangle &lt;math&gt;ABC&lt;/math&gt;, we deduce that &lt;math&gt;\angle CEG&lt;/math&gt; is a right angle. Since &lt;math&gt;ACEG&lt;/math&gt; is cyclic, &lt;math&gt;\angle GAC&lt;/math&gt; is also right. So &lt;math&gt;CG&lt;/math&gt; is a diameter of &lt;math&gt;\circ G&lt;/math&gt;. Because of this, &lt;math&gt;CG \perp BC&lt;/math&gt;, the tangent line. &lt;math&gt;\triangle BCG&lt;/math&gt; is right and &lt;math&gt;\triangle BCG\sim\triangle ABC\sim\triangle CAG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5&lt;/math&gt; so &lt;math&gt;AG=2AC=16\sqrt{5}/5&lt;/math&gt; and &lt;math&gt;[\triangle CAG]=64/5&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;[\triangle DAB]=[\triangle ACE]&lt;/math&gt;, the common area is &lt;math&gt;[ACEG]/17&lt;/math&gt;. &lt;math&gt;16[\triangle DAB]=[\triangle GAE]&lt;/math&gt; because the triangles are similar with a ratio of &lt;math&gt;1:4&lt;/math&gt;. So we only need to find &lt;math&gt;[\triangle CEG]&lt;/math&gt; now.<br /> <br /> <br /> Extend &lt;math&gt;DE&lt;/math&gt; through &lt;math&gt;E&lt;/math&gt; to intersect the tangent at &lt;math&gt;F&lt;/math&gt;. Because &lt;math&gt;4DA=AE&lt;/math&gt;, the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; times the height from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;EA&lt;/math&gt;. So &lt;math&gt;BC=3/4BF&lt;/math&gt; and &lt;math&gt;BF=16/3&lt;/math&gt;. We look at right triangle &lt;math&gt;\triangle AIF&lt;/math&gt;. &lt;math&gt;IF=68/15&lt;/math&gt; and &lt;math&gt;AI=8/5&lt;/math&gt;. &lt;math&gt;\triangle AIF&lt;/math&gt; is a &lt;math&gt;17-6-5\sqrt{13}&lt;/math&gt; right triangle. Hypotenuse &lt;math&gt;AF&lt;/math&gt; intersects &lt;math&gt;CG&lt;/math&gt; at a point, we call it &lt;math&gt;H&lt;/math&gt;. &lt;math&gt;CH=4/3\div 68/15\cdot 8/5=8/17&lt;/math&gt;. So &lt;math&gt;HG=8-8/17=128/17&lt;/math&gt;.<br /> <br /> <br /> By [[Power of a Point Theorem|Power of a Point]], &lt;math&gt;CH\cdot HG=AH\cdot HE&lt;/math&gt;. &lt;math&gt;AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17.&lt;/math&gt; So &lt;math&gt;HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})&lt;/math&gt;. The height from &lt;math&gt;E&lt;/math&gt; to &lt;math&gt;CG&lt;/math&gt; is &lt;math&gt;17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;[\triangle CEG]=64/65\cdot 8\div 2=256/65&lt;/math&gt;. The area of the whole cyclic quadrilateral is &lt;math&gt;64/5+256/65=(832+256)/65=1088/65&lt;/math&gt;. Lastly, the common area is &lt;math&gt;1/17&lt;/math&gt; the area of the quadrilateral, or &lt;math&gt;64/65&lt;/math&gt;. So &lt;math&gt;64+65=\boxed{129}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_15&diff=92170 2015 AIME II Problems/Problem 15 2018-02-21T19:04:23Z <p>Matongxu: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; have radii &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, respectively, and are externally tangent at point &lt;math&gt;A&lt;/math&gt;. Point &lt;math&gt;B&lt;/math&gt; is on &lt;math&gt;\mathcal{P}&lt;/math&gt; and point &lt;math&gt;C&lt;/math&gt; is on &lt;math&gt;\mathcal{Q}&lt;/math&gt; so that line &lt;math&gt;BC&lt;/math&gt; is a common external tangent of the two circles. A line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; again at &lt;math&gt;D&lt;/math&gt; and intersects &lt;math&gt;\mathcal{Q}&lt;/math&gt; again at &lt;math&gt;E&lt;/math&gt;. Points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on the same side of &lt;math&gt;\ell&lt;/math&gt;, and the areas of &lt;math&gt;\triangle DBA&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equal. This common area is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> import cse5;<br /> pathpen=black; pointpen=black;<br /> size(6cm);<br /> <br /> pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689);<br /> <br /> filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7));<br /> filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7));<br /> D(CR((0,1),1)); D(CR((4,4),4,150,390));<br /> D(L(MP(&quot;D&quot;,D(D),N),MP(&quot;A&quot;,D((0.8,1.6)),NE),1,5.5));<br /> D((-1.2,0)--MP(&quot;B&quot;,D((0,0)),S)--MP(&quot;C&quot;,D((4,0)),S)--(8,0));<br /> D(MP(&quot;E&quot;,E,N));<br /> &lt;/asy&gt;<br /> <br /> ==Hint==<br /> This is a #15 on an AIME, so it must be difficult. Indeed, although there are several possible approaches, all of them are very computational.<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> unitsize(35);<br /> draw(Circle((-1,0),1));<br /> draw(Circle((4,0),4));<br /> pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;<br /> A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);<br /> label(&quot;$A$&quot;,A,NE);label(&quot;$O_1$&quot;,O_1,NE);label(&quot;$O_2$&quot;,O_2,NE);label(&quot;$B$&quot;,B,SW);label(&quot;$C$&quot;,C,SW);label(&quot;$D$&quot;,D,NE);label(&quot;$E$&quot;,E,NE);label(&quot;$N$&quot;,N,W);label(&quot;$K$&quot;,(-24/15,0.2));label(&quot;$L$&quot;,(24/15,0.2));label(&quot;$n$&quot;,(-0.8,-0.12));label(&quot;$p$&quot;,((29/15,-48/15)));label(&quot;$\mathcal{P}$&quot;,(-1.6,1.1));label(&quot;$\mathcal{Q}$&quot;,(6,4));<br /> draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);<br /> dot(O_1);dot(O_2);<br /> draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);<br /> &lt;/asy&gt;<br /> <br /> Call &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; the centers of circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt;, respectively, and extend &lt;math&gt;CB&lt;/math&gt; and &lt;math&gt;O_2O_1&lt;/math&gt; to meet at point &lt;math&gt;N&lt;/math&gt;. Call &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;O_1N&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;O_2N&lt;/math&gt;, respectively. Using the fact that &lt;math&gt;\triangle{O_1BN} \sim \triangle{O_2CN}&lt;/math&gt; and setting &lt;math&gt;NO_1 = k&lt;/math&gt;, we have that &lt;math&gt;\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}&lt;/math&gt;. We can do some more length chasing using triangles similar to &lt;math&gt;OBN&lt;/math&gt; to get that &lt;math&gt;AK = AL = \frac{24}{15}&lt;/math&gt;, &lt;math&gt;BK = \frac{12}{15}&lt;/math&gt;, and &lt;math&gt;CL = \frac{48}{15}&lt;/math&gt;. Now, consider the circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; on the coordinate plane, where &lt;math&gt;A&lt;/math&gt; is the origin. If the line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; then &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;. To verify this, notice that &lt;math&gt;\triangle{AO_1D} \sim \triangle{EO_2A}&lt;/math&gt; from the fact that both triangles are isosceles with &lt;math&gt;\angle{O_1AD} \cong \angle{O_2AE}&lt;/math&gt;, which are corresponding angles. Since &lt;math&gt;O_2A = 4\cdot O_1A&lt;/math&gt;, we can conclude that &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;.<br /> <br /> <br /> Hence, we need to find the slope &lt;math&gt;m&lt;/math&gt; of line &lt;math&gt;\ell&lt;/math&gt; such that the perpendicular distance &lt;math&gt;n&lt;/math&gt; from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is four times the perpendicular distance &lt;math&gt;p&lt;/math&gt; from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AE&lt;/math&gt;. This will mean that the product of the bases and heights of triangles &lt;math&gt;ACE&lt;/math&gt; and &lt;math&gt;DBA&lt;/math&gt; will be equal, which in turn means that their areas will be equal. Let the line &lt;math&gt;\ell&lt;/math&gt; have the equation &lt;math&gt;y = -mx \implies mx + y = 0&lt;/math&gt;, and let &lt;math&gt;m&lt;/math&gt; be a positive real number so that the negative slope of &lt;math&gt;\ell&lt;/math&gt; is preserved. Setting &lt;math&gt;A = (0,0)&lt;/math&gt;, the coordinates of &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)&lt;/math&gt;, and the coordinates of &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)&lt;/math&gt;. Using the point-to-line distance formula and the condition &lt;math&gt;n = 4p&lt;/math&gt;, we have &lt;cmath&gt;\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}&lt;/cmath&gt; &lt;cmath&gt;\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.&lt;/cmath&gt; If &lt;math&gt;m &gt; 2&lt;/math&gt;, then clearly &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; would not lie on the same side of &lt;math&gt;\ell&lt;/math&gt;. Thus since &lt;math&gt;m &gt; 0&lt;/math&gt;, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have &lt;cmath&gt;\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.&lt;/cmath&gt; Thus, the equation of &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt;.<br /> <br /> <br /> Then we can find the coordinates of &lt;math&gt;D&lt;/math&gt; by finding the point &lt;math&gt;(x,y)&lt;/math&gt; other than &lt;math&gt;A = (0,0)&lt;/math&gt; where the circle &lt;math&gt;\mathcal{P}&lt;/math&gt; intersects &lt;math&gt;\ell&lt;/math&gt;. &lt;math&gt;\mathcal{P}&lt;/math&gt; can be represented with the equation &lt;math&gt;(x + 1)^2 + y^2 = 1&lt;/math&gt;, and substituting &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt; into this equation yields &lt;math&gt;x = 0, -\frac{8}{13}&lt;/math&gt; as solutions. Discarding &lt;math&gt;x = 0&lt;/math&gt;, the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}&lt;/math&gt;. The distance from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; is then &lt;math&gt;\frac{4}{\sqrt{13}}.&lt;/math&gt; The perpendicular distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; or the height of &lt;math&gt;\triangle{DBA}&lt;/math&gt; is &lt;math&gt;\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.&lt;/math&gt; Finally, the common area is &lt;math&gt;\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}&lt;/math&gt;, and &lt;math&gt;m + n = 64 + 65 = \boxed{129}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> By [[homothety]], we deduce that &lt;math&gt;AE = 4 AD&lt;/math&gt;. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt; is four times that from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;. Let the distance from &lt;math&gt;C&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; and the distance from &lt;math&gt;B&lt;/math&gt; be &lt;math&gt;4x&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the centers of their respective circles. Then dropping a perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; creates a &lt;math&gt;3-4-5&lt;/math&gt; right triangle, from which &lt;math&gt;BC = 4&lt;/math&gt; and, if &lt;math&gt;\alpha = \angle{AQC}&lt;/math&gt;, that &lt;math&gt;\cos \alpha = \dfrac{3}{5}&lt;/math&gt;. Then &lt;math&gt;\angle{BPA} = 180^\circ - \alpha&lt;/math&gt;, and the Law of Cosines on triangles &lt;math&gt;APB&lt;/math&gt; and &lt;math&gt;AQC&lt;/math&gt; gives &lt;math&gt;AB = \dfrac{4}{\sqrt{5}}&lt;/math&gt; and &lt;math&gt;AC = \dfrac{8}{\sqrt{5}}.&lt;/math&gt;<br /> <br /> Now, using the Pythagorean Theorem to express the length of the projection of &lt;math&gt;BC&lt;/math&gt; onto line &lt;math&gt;l&lt;/math&gt; gives<br /> &lt;cmath&gt;\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.&lt;/cmath&gt;<br /> Squaring and simplifying gives<br /> &lt;cmath&gt;\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,&lt;/cmath&gt;<br /> and squaring and solving gives &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}.&lt;/math&gt;<br /> <br /> By the Law of Sines on triangle &lt;math&gt;ABD&lt;/math&gt;, we have<br /> &lt;cmath&gt;\frac{BD}{\sin A} = 2.&lt;/cmath&gt;<br /> But we know &lt;math&gt;\sin A = \dfrac{4x}{AB}&lt;/math&gt;, and so a small computation gives &lt;math&gt;BD = \dfrac{16}{\sqrt{65}}.&lt;/math&gt; The Pythagorean Theorem now gives<br /> &lt;cmath&gt;AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},&lt;/cmath&gt;<br /> and so the common area is &lt;math&gt;\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{129}.&lt;/math&gt;<br /> <br /> ==Alternate Path to x==<br /> Call the intersection of lines &lt;math&gt;l&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;E&lt;/math&gt;.You can use similar triangles to find that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt; is four times the distance from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. Then draw a perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call the point &lt;math&gt;F&lt;/math&gt;. &lt;math&gt;AF = \frac{8}{5}&lt;/math&gt; and &lt;math&gt;FE = FC + CE = \frac{16}{5} + \frac{4}{3} = \frac{68}{15}&lt;/math&gt;, so by the Pythagorean Theorem, &lt;math&gt;AE = \dfrac{20\sqrt{13}}{5}&lt;/math&gt;. You can now use similar triangles to find that &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}&lt;/math&gt; and continue on like in solution 2.<br /> <br /> ==Solution 3==<br /> &lt;math&gt;DE&lt;/math&gt; goes through &lt;math&gt;A&lt;/math&gt;, the point of tangency of both circles. So &lt;math&gt;DE&lt;/math&gt; intercepts equal arcs in circle &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;: [[homothety]]. Hence, &lt;math&gt;AE=4AD&lt;/math&gt;. We will use such similarity later.<br /> <br /> The diagonal distance between the centers of the circles is &lt;math&gt;4+1=5&lt;/math&gt;. The difference in heights is &lt;math&gt;4-1=3&lt;/math&gt;. So &lt;math&gt;BC=\sqrt{5^2-3^2}=4&lt;/math&gt;.<br /> <br /> The triangle connecting the centers with a side parallel to &lt;math&gt;BC&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; right triangle. Since &lt;math&gt;O_PA=1&lt;/math&gt;, the height of &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;1+3/5=8/5&lt;/math&gt;. Drop an altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call it &lt;math&gt;I&lt;/math&gt;: &lt;math&gt;IB=4/5&lt;/math&gt; and &lt;math&gt;IC=4-4/5=32/5&lt;/math&gt;. Since right &lt;math&gt;\triangle AIB\sim\triangle CIB&lt;/math&gt;, &lt;math&gt;ABC&lt;/math&gt; is a right triangle also; &lt;math&gt;IB:IA:IC&lt;/math&gt; form a geometric progression &lt;math&gt;\times 2&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;BA&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;G&lt;/math&gt; on the other side of &lt;math&gt;\circ Q&lt;/math&gt;. By [[homothety]], &lt;math&gt;\triangle DAB\sim\triangle EAG&lt;/math&gt;. By angle chasing &lt;math&gt;\triangle DAB&lt;/math&gt; through right triangle &lt;math&gt;ABC&lt;/math&gt;, we deduce that &lt;math&gt;\angle CEG&lt;/math&gt; is a right angle. Since &lt;math&gt;ACEG&lt;/math&gt; is cyclic, &lt;math&gt;\angle GAC&lt;/math&gt; is also right. So &lt;math&gt;CG&lt;/math&gt; is a diameter of &lt;math&gt;\circ G&lt;/math&gt;. Because of this, &lt;math&gt;CG \perp BC&lt;/math&gt;, the tangent line. &lt;math&gt;\triangle BCG&lt;/math&gt; is right and &lt;math&gt;\triangle BCG\sim\triangle ABC\sim\triangle CAG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5&lt;/math&gt; so &lt;math&gt;AG=2AC=16\sqrt{5}/5&lt;/math&gt; and &lt;math&gt;[\triangle CAG]=64/5&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;[\triangle DAB]=[\triangle ACE]&lt;/math&gt;, the common area is &lt;math&gt;[ACEG]/17&lt;/math&gt;. &lt;math&gt;16[\triangle DAB]=[\triangle GAE]&lt;/math&gt; because the triangles are similar with a ratio of &lt;math&gt;1:4&lt;/math&gt;. So we only need to find &lt;math&gt;[\triangle CEG]&lt;/math&gt; now.<br /> <br /> <br /> Extend &lt;math&gt;DE&lt;/math&gt; through &lt;math&gt;E&lt;/math&gt; to intersect the tangent at &lt;math&gt;F&lt;/math&gt;. Because &lt;math&gt;4DA=AE&lt;/math&gt;, the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; times the height from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;EA&lt;/math&gt;. So &lt;math&gt;BC=3/4BF&lt;/math&gt; and &lt;math&gt;BF=16/3&lt;/math&gt;. We look at right triangle &lt;math&gt;\triangle AIF&lt;/math&gt;. &lt;math&gt;IF=68/15&lt;/math&gt; and &lt;math&gt;AI=8/5&lt;/math&gt;. &lt;math&gt;\triangle AIF&lt;/math&gt; is a &lt;math&gt;17-6-5\sqrt{13}&lt;/math&gt; right triangle. Hypotenuse &lt;math&gt;AF&lt;/math&gt; intersects &lt;math&gt;CG&lt;/math&gt; at a point, we call it &lt;math&gt;H&lt;/math&gt;. &lt;math&gt;CH=4/3\div 68/15\cdot 8/5=8/17&lt;/math&gt;. So &lt;math&gt;HG=8-8/17=128/17&lt;/math&gt;.<br /> <br /> <br /> By [[Power of a Point Theorem|Power of a Point]], &lt;math&gt;CH\cdot HG=AH\cdot HE&lt;/math&gt;. &lt;math&gt;AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17&lt;/math&gt;. So &lt;math&gt;HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})&lt;/math&gt;. The height from &lt;math&gt;E&lt;/math&gt; to &lt;math&gt;CG&lt;/math&gt; is &lt;math&gt;17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;[\triangle CEG]=64/65\cdot 8\div 2=256/65&lt;/math&gt;. The area of the whole cyclic quadrilateral is &lt;math&gt;64/5+256/65=(832+256)/65=1088/65&lt;/math&gt;. Lastly, the common area is &lt;math&gt;1/17&lt;/math&gt; the area of the quadrilateral, or &lt;math&gt;64/65&lt;/math&gt;. So &lt;math&gt;64+65=\boxed{129}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=AIME_Problems_and_Solutions&diff=77349 AIME Problems and Solutions 2016-03-05T00:46:10Z <p>Matongxu: </p> <hr /> <div>This is a list of all [[AIME]] exams in the AoPSWiki. Some of them contain complete questions and solutions, others complete questions, and others are lacking both questions and solutions. Many of these problems and solutions are also available in the [http://www.artofproblemsolving.com/Forum/resources.php?c=182 AoPS Resources] section. If you find problems that are in the Resources section which are not in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome.<br /> The [[2016 AIME I]] was held on Thursday, March 3, 2016. <br /> The [[2016 AIME II]] will be held on Wednesday, March 16, 2016.<br /> {| class=&quot;wikitable&quot;<br /> |-<br /> ! Year || Test I || Test II<br /> |-<br /> | 2016 || [[2016 AIME I | AIME I]] || [[2016 AIME II | AIME II]]<br /> |-<br /> | 2015 || [[2015 AIME I | AIME I]] || [[2015 AIME II | AIME II]]<br /> |-<br /> | 2014 || [[2014 AIME I | AIME I]] || [[2014 AIME II | AIME II]]<br /> |-<br /> | 2013 || [[2013 AIME I | AIME I]] || [[2013 AIME II | AIME II]]<br /> |-<br /> | 2012 || [[2012 AIME I | AIME I]] || [[2012 AIME II | AIME II]]<br /> |-<br /> | 2011 || [[2011 AIME I | AIME I]] || [[2011 AIME II | AIME II]]<br /> |-<br /> | 2010 || [[2010 AIME I | AIME I]] || [[2010 AIME II | AIME II]]<br /> |-<br /> | 2009 || [[2009 AIME I | AIME I]] || [[2009 AIME II | AIME II]]<br /> |-<br /> | 2008 || [[2008 AIME I | AIME I]] || [[2008 AIME II | AIME II]]<br /> |-<br /> | 2007 || [[2007 AIME I | AIME I]] || [[2007 AIME II | AIME II]]<br /> |-<br /> | 2006 || [[2006 AIME I | AIME I]] || [[2006 AIME II | AIME II]]<br /> |-<br /> | 2005 || [[2005 AIME I | AIME I]] || [[2005 AIME II | AIME II]]<br /> |-<br /> | 2004 || [[2004 AIME I | AIME I]] || [[2004 AIME II | AIME II]]<br /> |-<br /> | 2003 || [[2003 AIME I | AIME I]] || [[2003 AIME II | AIME II]]<br /> |-<br /> | 2002 || [[2002 AIME I | AIME I]] || [[2002 AIME II | AIME II]]<br /> |-<br /> | 2001 || [[2001 AIME I | AIME I]] || [[2001 AIME II | AIME II]]<br /> |-<br /> | 2000 || [[2000 AIME I | AIME I]] || [[2000 AIME II | AIME II]]<br /> |-<br /> | 1999 || [[1999 AIME | AIME]] || <br /> |-<br /> | 1998 || [[1998 AIME | AIME]] || <br /> |-<br /> | 1997 || [[1997 AIME | AIME]] || <br /> |-<br /> | 1996 || [[1996 AIME | AIME]] || <br /> |-<br /> | 1995 || [[1995 AIME | AIME]] || <br /> |-<br /> | 1994 || [[1994 AIME | AIME]] || <br /> |-<br /> | 1993 || [[1993 AIME | AIME]] || <br /> |-<br /> | 1992 || [[1992 AIME | AIME]] || <br /> |-<br /> | 1991 || [[1991 AIME | AIME]] || <br /> |-<br /> | 1990 || [[1990 AIME | AIME]] || <br /> |-<br /> | 1989 || [[1989 AIME | AIME]] || <br /> |-<br /> | 1988 || [[1988 AIME | AIME]] || <br /> |-<br /> | 1987 || [[1987 AIME | AIME]] || <br /> |-<br /> | 1986 || [[1986 AIME | AIME]] || <br /> |-<br /> | 1985 || [[1985 AIME | AIME]] || <br /> |-<br /> | 1984 || [[1984 AIME | AIME]] || <br /> |-<br /> | 1983 || [[1983 AIME | AIME]] || <br /> |}<br /> <br /> == Resources ==<br /> * [[American Mathematics Competitions]]<br /> * [[AMC Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> <br /> [[Category:Math Contest Problems]]</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I&diff=77222 2016 AIME I 2016-03-04T15:00:26Z <p>Matongxu: Verb Tense</p> <hr /> <div>'''2016 [[American Invitational Mathematics Examination|AIME]] I''' problems and solutions. The test was held on Thursday, March 3, 2016. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br /> <br /> * [[2016 AIME I Problems|Entire Test]]<br /> * [[2016 AIME I Answer Key|Answer Key]]<br /> ** [[2016 AIME I Problems/Problem 1|Problem 1]] <br /> ** [[2016 AIME I Problems/Problem 2|Problem 2]]<br /> ** [[2016 AIME I Problems/Problem 3|Problem 3]]<br /> ** [[2016 AIME I Problems/Problem 4|Problem 4]]<br /> ** [[2016 AIME I Problems/Problem 5|Problem 5]]<br /> ** [[2016 AIME I Problems/Problem 6|Problem 6]]<br /> ** [[2016 AIME I Problems/Problem 7|Problem 7]]<br /> ** [[2016 AIME I Problems/Problem 8|Problem 8]]<br /> ** [[2016 AIME I Problems/Problem 9|Problem 9]]<br /> ** [[2016 AIME I Problems/Problem 10|Problem 10]]<br /> ** [[2016 AIME I Problems/Problem 11|Problem 11]]<br /> ** [[2016 AIME I Problems/Problem 12|Problem 12]]<br /> ** [[2016 AIME I Problems/Problem 13|Problem 13]]<br /> ** [[2016 AIME I Problems/Problem 14|Problem 14]]<br /> ** [[2016 AIME I Problems/Problem 15|Problem 15]]</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=AIME_Problems_and_Solutions&diff=77221 AIME Problems and Solutions 2016-03-04T14:59:40Z <p>Matongxu: Added 2016 AIME I+II pages, did not add contests</p> <hr /> <div>This is a list of all [[AIME]] exams in the AoPSWiki. Some of them contain complete questions and solutions, others complete questions, and others are lacking both questions and solutions. Many of these problems and solutions are also available in the [http://www.artofproblemsolving.com/Forum/resources.php?c=182 AoPS Resources] section. If you find problems that are in the Resources section which are not in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome.<br /> The [2016 AIME I] was held on Thursday, March 3, 2016. <br /> The [2016 AIME II] will be held on Wednesday, March 16, 2016.<br /> {| class=&quot;wikitable&quot;<br /> |-<br /> ! Year || Test I || Test II<br /> |-<br /> | 2016 || [[2016 AIME I | AIME I]] || [[2016 AIME II | AIME II]]<br /> |-<br /> | 2015 || [[2015 AIME I | AIME I]] || [[2015 AIME II | AIME II]]<br /> |-<br /> | 2014 || [[2014 AIME I | AIME I]] || [[2014 AIME II | AIME II]]<br /> |-<br /> | 2013 || [[2013 AIME I | AIME I]] || [[2013 AIME II | AIME II]]<br /> |-<br /> | 2012 || [[2012 AIME I | AIME I]] || [[2012 AIME II | AIME II]]<br /> |-<br /> | 2011 || [[2011 AIME I | AIME I]] || [[2011 AIME II | AIME II]]<br /> |-<br /> | 2010 || [[2010 AIME I | AIME I]] || [[2010 AIME II | AIME II]]<br /> |-<br /> | 2009 || [[2009 AIME I | AIME I]] || [[2009 AIME II | AIME II]]<br /> |-<br /> | 2008 || [[2008 AIME I | AIME I]] || [[2008 AIME II | AIME II]]<br /> |-<br /> | 2007 || [[2007 AIME I | AIME I]] || [[2007 AIME II | AIME II]]<br /> |-<br /> | 2006 || [[2006 AIME I | AIME I]] || [[2006 AIME II | AIME II]]<br /> |-<br /> | 2005 || [[2005 AIME I | AIME I]] || [[2005 AIME II | AIME II]]<br /> |-<br /> | 2004 || [[2004 AIME I | AIME I]] || [[2004 AIME II | AIME II]]<br /> |-<br /> | 2003 || [[2003 AIME I | AIME I]] || [[2003 AIME II | AIME II]]<br /> |-<br /> | 2002 || [[2002 AIME I | AIME I]] || [[2002 AIME II | AIME II]]<br /> |-<br /> | 2001 || [[2001 AIME I | AIME I]] || [[2001 AIME II | AIME II]]<br /> |-<br /> | 2000 || [[2000 AIME I | AIME I]] || [[2000 AIME II | AIME II]]<br /> |-<br /> | 1999 || [[1999 AIME | AIME]] || <br /> |-<br /> | 1998 || [[1998 AIME | AIME]] || <br /> |-<br /> | 1997 || [[1997 AIME | AIME]] || <br /> |-<br /> | 1996 || [[1996 AIME | AIME]] || <br /> |-<br /> | 1995 || [[1995 AIME | AIME]] || <br /> |-<br /> | 1994 || [[1994 AIME | AIME]] || <br /> |-<br /> | 1993 || [[1993 AIME | AIME]] || <br /> |-<br /> | 1992 || [[1992 AIME | AIME]] || <br /> |-<br /> | 1991 || [[1991 AIME | AIME]] || <br /> |-<br /> | 1990 || [[1990 AIME | AIME]] || <br /> |-<br /> | 1989 || [[1989 AIME | AIME]] || <br /> |-<br /> | 1988 || [[1988 AIME | AIME]] || <br /> |-<br /> | 1987 || [[1987 AIME | AIME]] || <br /> |-<br /> | 1986 || [[1986 AIME | AIME]] || <br /> |-<br /> | 1985 || [[1985 AIME | AIME]] || <br /> |-<br /> | 1984 || [[1984 AIME | AIME]] || <br /> |-<br /> | 1983 || [[1983 AIME | AIME]] || <br /> |}<br /> <br /> == Resources ==<br /> * [[American Mathematics Competitions]]<br /> * [[AMC Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> <br /> [[Category:Math Contest Problems]]</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=73540 2015 AMC 8 Problems/Problem 25 2015-12-05T16:20:11Z <p>Matongxu: /* See Also */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) \ } 9\qquad \textbf{(B) \ } 12\frac{1}{2}\qquad \textbf{(C) \ } 15\qquad \textbf{(D) \ } 15\frac{1}{2}\qquad \textbf{(E) \ } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the &lt;math&gt;4&lt;/math&gt; big triangles by &lt;math&gt;AA.&lt;/math&gt; Let the height of a big triangle be &lt;math&gt;x&lt;/math&gt; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;<br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> This the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> We draw a square as shown:<br /> &lt;asy&gt;<br /> pair Q,R,S,T;<br /> Q=(1.381966,0);<br /> R=(5,1.381966);<br /> S=(3.618034,5);<br /> T=(0,3.618034);<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> draw(Q--R--S--T--cycle);<br /> draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed);<br /> &lt;/asy&gt;<br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=\boxed{{\textbf{(C)}}~15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt;<br /> <br /> Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = \boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=73539 2015 AMC 8 Problems/Problem 25 2015-12-05T16:19:23Z <p>Matongxu: /* See Also */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) \ } 9\qquad \textbf{(B) \ } 12\frac{1}{2}\qquad \textbf{(C) \ } 15\qquad \textbf{(D) \ } 15\frac{1}{2}\qquad \textbf{(E) \ } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the &lt;math&gt;4&lt;/math&gt; big triangles by &lt;math&gt;AA.&lt;/math&gt; Let the height of a big triangle be &lt;math&gt;x&lt;/math&gt; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;<br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> This the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> We draw a square as shown:<br /> &lt;asy&gt;<br /> pair Q,R,S,T;<br /> Q=(1.381966,0);<br /> R=(5,1.381966);<br /> S=(3.618034,5);<br /> T=(0,3.618034);<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> draw(Q--R--S--T--cycle);<br /> draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed);<br /> &lt;/asy&gt;<br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=\boxed{{\textbf{(C)}}~15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt;<br /> <br /> Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = \boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=73538 2015 AMC 8 Problems/Problem 25 2015-12-05T16:19:03Z <p>Matongxu: /* See Also */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) \ } 9\qquad \textbf{(B) \ } 12\frac{1}{2}\qquad \textbf{(C) \ } 15\qquad \textbf{(D) \ } 15\frac{1}{2}\qquad \textbf{(E) \ } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the &lt;math&gt;4&lt;/math&gt; big triangles by &lt;math&gt;AA.&lt;/math&gt; Let the height of a big triangle be &lt;math&gt;x&lt;/math&gt; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;<br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> This the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> We draw a square as shown:<br /> &lt;asy&gt;<br /> pair Q,R,S,T;<br /> Q=(1.381966,0);<br /> R=(5,1.381966);<br /> S=(3.618034,5);<br /> T=(0,3.618034);<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> draw(Q--R--S--T--cycle);<br /> draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed);<br /> &lt;/asy&gt;<br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=\boxed{{\textbf{(C)}}~15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt;<br /> <br /> Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = \boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=73537 2015 AMC 8 Problems/Problem 25 2015-12-05T16:18:49Z <p>Matongxu: /* See Also */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) \ } 9\qquad \textbf{(B) \ } 12\frac{1}{2}\qquad \textbf{(C) \ } 15\qquad \textbf{(D) \ } 15\frac{1}{2}\qquad \textbf{(E) \ } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the &lt;math&gt;4&lt;/math&gt; big triangles by &lt;math&gt;AA.&lt;/math&gt; Let the height of a big triangle be &lt;math&gt;x&lt;/math&gt; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;<br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> This the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> We draw a square as shown:<br /> &lt;asy&gt;<br /> pair Q,R,S,T;<br /> Q=(1.381966,0);<br /> R=(5,1.381966);<br /> S=(3.618034,5);<br /> T=(0,3.618034);<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> draw(Q--R--S--T--cycle);<br /> draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed);<br /> &lt;/asy&gt;<br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=\boxed{{\textbf{(C)}}~15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt;<br /> <br /> Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = \boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after=}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=73536 2015 AMC 8 Problems/Problem 25 2015-12-05T16:18:13Z <p>Matongxu: /* Solution 1 */</p> <hr /> <div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;math&gt;\textbf{(A) \ } 9\qquad \textbf{(B) \ } 12\frac{1}{2}\qquad \textbf{(C) \ } 15\qquad \textbf{(D) \ } 15\frac{1}{2}\qquad \textbf{(E) \ } 17&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> &lt;/asy&gt;<br /> <br /> ===Solution 1===<br /> We can draw a diagram as shown.<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt; <br /> Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the &lt;math&gt;4&lt;/math&gt; big triangles by &lt;math&gt;AA.&lt;/math&gt; Let the height of a big triangle be &lt;math&gt;x&lt;/math&gt; then &lt;math&gt;\tfrac{x}{x-1}=\tfrac{5-x}{1}&lt;/math&gt;.<br /> &lt;cmath&gt;x=-x^2+6x-5&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2-5x+5=0&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;x=\dfrac{5\pm \sqrt{5}}{2}&lt;/cmath&gt;<br /> Thus &lt;math&gt;x=\dfrac{5-\sqrt{5}}{2}&lt;/math&gt;<br /> This means the area of each triangle is &lt;math&gt;\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}&lt;/math&gt;<br /> This the area of the square is &lt;math&gt;25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ===Solution 2=== <br /> <br /> We draw a square as shown:<br /> &lt;asy&gt;<br /> pair Q,R,S,T;<br /> Q=(1.381966,0);<br /> R=(5,1.381966);<br /> S=(3.618034,5);<br /> T=(0,3.618034);<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> draw(Q--R--S--T--cycle);<br /> draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed);<br /> &lt;/asy&gt;<br /> <br /> We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so the combined area of the four triangles is &lt;math&gt;4 \cdot \frac 32=6&lt;/math&gt;. The area of the smaller square is &lt;math&gt;9&lt;/math&gt;. We add these to see that the area of the large square is &lt;math&gt;9+6=\boxed{{\textbf{(C)}}~15}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> &lt;asy&gt;<br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> path arc = arc((2.5,4),1.5,0,90);<br /> pair P = intersectionpoint(arc,(0,5)--(5,5));<br /> pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br /> draw(P--Pp--Ppp--Pppp--cycle);<br /> &lt;/asy&gt;<br /> <br /> Let's find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length &lt;math&gt;1&lt;/math&gt;, and let's label the other legs &lt;math&gt;x&lt;/math&gt; for one of the triangles and &lt;math&gt;y&lt;/math&gt; for the other. Note that &lt;math&gt;x + y = 3&lt;/math&gt;.<br /> The area of each of the triangles is &lt;math&gt;\frac{x}{2}&lt;/math&gt; and &lt;math&gt;\frac{y}{2}&lt;/math&gt;, and there are &lt;math&gt;4&lt;/math&gt; of each. So now we need to find &lt;math&gt;4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;(4)\frac{x}{2} + (4)\frac{y}{2}&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)&lt;/math&gt;<br /> &lt;math&gt;\Rightarrow~~4\left(\frac{x+y}{2}\right)&lt;/math&gt;<br /> Remember that &lt;math&gt;x+y=3&lt;/math&gt;, so substituting this in we find that the area of all of the triangles is &lt;math&gt;4\left(\frac{3}{2}\right) = 6&lt;/math&gt;. <br /> The area of the &lt;math&gt;4&lt;/math&gt; unit squares is &lt;math&gt;4&lt;/math&gt;, so the area of the square we need is &lt;math&gt;25- (4+6) = \boxed{\textbf{(C)}~15}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=24|after= }}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_23&diff=73135 2015 AMC 8 Problems/Problem 23 2015-11-25T21:41:53Z <p>Matongxu: Solution presented on the problems page, modified to fit standards</p> <hr /> <div>Tom has twelve slips of paper which he wants to put into five cups labeled &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. The numbers on the papers are &lt;math&gt;2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,&lt;/math&gt; and &lt;math&gt;4.5&lt;/math&gt;. If a slip with &lt;math&gt;2&lt;/math&gt; goes into cup &lt;math&gt;E&lt;/math&gt; and a slip with &lt;math&gt;3&lt;/math&gt; goes into cup &lt;math&gt;B&lt;/math&gt;, then the slip with &lt;math&gt;3.5&lt;/math&gt; must go into what cup?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } A \qquad<br /> \textbf{(B) } B \qquad<br /> \textbf{(C) } C \qquad<br /> \textbf{(D) } D \qquad<br /> \textbf{(E) } E<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The numbers have a sum of &lt;math&gt;6+5+12+4+8=35&lt;/math&gt;, which averages to &lt;math&gt;7&lt;/math&gt;, which means &lt;math&gt;A, B, C, D, E&lt;/math&gt; will have values &lt;math&gt;5, 6, 7, 8, 9&lt;/math&gt;, respectively. Now it's process of elimination: Cup &lt;math&gt;A&lt;/math&gt; will have a sum of &lt;math&gt;5&lt;/math&gt;, so putting a &lt;math&gt;3.5&lt;/math&gt; slip in the cup will leave &lt;math&gt;5-3.5=1.5&lt;/math&gt;; However, all of our slips are bigger than &lt;math&gt;1.5&lt;/math&gt;, so this is impossible. Cup &lt;math&gt;B&lt;/math&gt; has a sum of &lt;math&gt;6&lt;/math&gt;, but we are told that it already has a &lt;math&gt;3&lt;/math&gt; slip, leaving &lt;math&gt;6-3=3&lt;/math&gt;, which is too small for the &lt;math&gt;3.5&lt;/math&gt; slip. Cup &lt;math&gt;C&lt;/math&gt; is a little bit trickier, but still manageable. It must have a value of &lt;math&gt;7&lt;/math&gt;, so adding the &lt;math&gt;3.5&lt;/math&gt; slip leaves room for &lt;math&gt;7-3.5=3.5&lt;/math&gt;. This looks good at first, as we do have slips smaller than that, but upon closer inspection, we see that no slip fits exactly, and the smallest sum of 2 slips is &lt;math&gt;2+2=4&lt;/math&gt;, which is too big, so this case is also impossible. Cup &lt;math&gt;E&lt;/math&gt; has a sum of &lt;math&gt;9&lt;/math&gt;, but we are told it already has a &lt;math&gt;2&lt;/math&gt; slip, so we are left with &lt;math&gt;9-2=7&lt;/math&gt;, which is identical to the Cup C case, and thus also impossible. With all other choices removed, we are left with the answer: Cup &lt;math&gt;\boxed{D}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_23&diff=73132 2015 AMC 8 Problems/Problem 23 2015-11-25T21:37:51Z <p>Matongxu: /* See Also */</p> <hr /> <div>Tom has twelve slips of paper which he wants to put into five cups labeled &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. The numbers on the papers are &lt;math&gt;2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,&lt;/math&gt; and &lt;math&gt;4.5&lt;/math&gt;. If a slip with &lt;math&gt;2&lt;/math&gt; goes into cup &lt;math&gt;E&lt;/math&gt; and a slip with &lt;math&gt;3&lt;/math&gt; goes into cup &lt;math&gt;B&lt;/math&gt;, then the slip with &lt;math&gt;3.5&lt;/math&gt; must go into what cup?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } A \qquad<br /> \textbf{(B) } B \qquad<br /> \textbf{(C) } C \qquad<br /> \textbf{(D) } D \qquad<br /> \textbf{(E) } E<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems&diff=73130 2015 AMC 8 Problems 2015-11-25T21:36:44Z <p>Matongxu: /* Problem 23 */</p> <hr /> <div>==Problem 1==<br /> <br /> How many square yards of carpet are required to cover a rectangular floor that is &lt;math&gt;12&lt;/math&gt; feet long and &lt;math&gt;9&lt;/math&gt; feet wide? (There are 3 feet in a yard.)<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Point &lt;math&gt;O&lt;/math&gt; is the center of the regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; is the midpoint of the side &lt;math&gt;\overline{AB}.&lt;/math&gt; What fraction of the area of the octagon is shaded?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G,H,O,X;<br /> A=dir(45);<br /> B=dir(90);<br /> C=dir(135);<br /> D=dir(180);<br /> E=dir(-135);<br /> F=dir(-90);<br /> G=dir(-45);<br /> H=dir(0);<br /> O=(0,0);<br /> X=midpoint(A--B);<br /> <br /> fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br /> draw(A--B--C--D--E--F--G--H--cycle);<br /> <br /> dot(&quot;$A$&quot;,A,dir(45));<br /> dot(&quot;$B$&quot;,B,dir(90));<br /> dot(&quot;$C$&quot;,C,dir(135));<br /> dot(&quot;$D$&quot;,D,dir(180));<br /> dot(&quot;$E$&quot;,E,dir(-135));<br /> dot(&quot;$F$&quot;,F,dir(-90));<br /> dot(&quot;$G$&quot;,G,dir(-45));<br /> dot(&quot;$H$&quot;,H,dir(0));<br /> dot(&quot;$X$&quot;,X,dir(135/2));<br /> dot(&quot;$O$&quot;,O,dir(0));<br /> draw(E--O--X);<br /> &lt;/asy&gt;<br /> [[2015 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of &lt;math&gt;10&lt;/math&gt; miles per hour. Jack walks to the pool at a constant speed of &lt;math&gt;4&lt;/math&gt; miles per hour. How many minutes before Jack does Jill arrive?<br /> <br /> &lt;math&gt;\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Billy's basketball team scored the following points over the course of the first 11 games of the season: &lt;cmath&gt;42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73&lt;/cmath&gt; If his team scores 40 in the 12th game, which of the following statistics will show an increase?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{range} \qquad \textbf{(B) } \text{median} \qquad \textbf{(C) } \text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> In &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, &lt;math&gt;AB=BC=29&lt;/math&gt;, and &lt;math&gt;AC=42&lt;/math&gt;. What is the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Each of two boxes contains three chips numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8== <br /> <br /> What is the smallest whole number larger than the perimeter of any triangle with a side of length &lt;math&gt;5&lt;/math&gt; and a side of length &lt;math&gt;19&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working &lt;math&gt;20&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> How many integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt; have four distinct digits?<br /> <br /> &lt;math&gt;\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read &quot;AMC8&quot;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> How many pairs of parallel edges, such as &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{GH}&lt;/math&gt; or &lt;math&gt;\overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt;, does a cube have?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }6 \quad\textbf{(B) }12 \quad\textbf{(C) } 18 \quad\textbf{(D) } 24 \quad \textbf{(E) } 36&lt;/math&gt; &lt;asy&gt; import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label(&quot;$D$&quot;,(0,0,0),S); label(&quot;$A$&quot;,(0,0,1),N); label(&quot;$H$&quot;,(0,1,0),S); label(&quot;$E$&quot;,(0,1,1),N); label(&quot;$C$&quot;,(1,0,0),S); label(&quot;$B$&quot;,(1,0,1),N); label(&quot;$G$&quot;,(1,1,0),S); label(&quot;$F$&quot;,(1,1,1),N); &lt;/asy&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many subsets of two elements can be removed from the set &lt;math&gt;\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}&lt;/math&gt; so that the mean (average) of the remaining numbers is &lt;math&gt;6&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\qquad\textbf{(B)}\text{ 40}\qquad\textbf{(C)}\text{ 72}\qquad\textbf{(D)}\text{ 100}\qquad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> At Euler Middle School, &lt;math&gt;198&lt;/math&gt; students voted on two issues in a school referendum with the following results: &lt;math&gt;149&lt;/math&gt; voted in favor of the first issue and &lt;math&gt;119&lt;/math&gt; voted in favor of the second issue. If there were exactly &lt;math&gt;29&lt;/math&gt; students who voted against both issues, how many students voted in favor of both issues?<br /> <br /> &lt;math&gt;\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; of all the ninth graders are paired with &lt;math&gt;\tfrac{2}{5}&lt;/math&gt; of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{2}{15} \qquad<br /> \textbf{(B) } \frac{4}{11} \qquad<br /> \textbf{(C) } \frac{11}{30} \qquad<br /> \textbf{(D) } \frac{3}{8} \qquad<br /> \textbf{(E) } \frac{11}{15}<br /> &lt;/math&gt;<br /> <br /> ==Problem 17==<br /> <br /> Jeremy's father drives him to school in rush hour traffic in &lt;math&gt;20&lt;/math&gt; minutes. One day there is no traffic, so his father can drive him &lt;math&gt;18&lt;/math&gt; miles per hour faster and gets him to school in &lt;math&gt;12&lt;/math&gt; minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, &lt;math&gt;2,5,8,11,14&lt;/math&gt; is an arithmetic sequence with five terms, in which the first term is &lt;math&gt;2&lt;/math&gt; and the constant added is &lt;math&gt;3&lt;/math&gt;. Each row and each column in this &lt;math&gt;5\times5&lt;/math&gt; array is an arithmetic sequence with five terms. What is the value of &lt;math&gt;X&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42&lt;/math&gt;<br /> <br /> &lt;asy&gt; size(3.85cm); label(&quot;$X$&quot;,(2.5,2.1),N); for (int i=0; i&lt;=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j&lt;=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, &quot; &quot;); foo(3, 4, &quot; &quot;); foo(0, 3, &quot; &quot;); foo(2, 3, &quot; &quot;); foo(1, 2, &quot; &quot;); foo(3, 2, &quot; &quot;); foo(1, 1, &quot; &quot;); foo(2, 1, &quot; &quot;); foo(3, 1, &quot; &quot;); foo(4, 1, &quot; &quot;); foo(2, 0, &quot; &quot;); foo(3, 0, &quot; &quot;); foo(0, 1, &quot; &quot;); foo(0, 2, &quot; &quot;); foo(1, 0, &quot; &quot;); foo(1, 3, &quot; &quot;); foo(1, 4, &quot; &quot;); foo(3, 3, &quot; &quot;); foo(4, 2, &quot; &quot;); foo(4, 3, &quot; &quot;); &lt;/asy&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> A triangle with vertices as &lt;math&gt;A=(1,3)&lt;/math&gt;, &lt;math&gt;B=(5,1)&lt;/math&gt;, and &lt;math&gt;C=(4,4)&lt;/math&gt; is plotted on a &lt;math&gt;6\times5&lt;/math&gt; grid. What fraction of the grid is covered by the triangle?<br /> <br /> &lt;asy&gt;<br /> <br /> draw((1,0)--(1,5),linewidth(.5));<br /> draw((2,0)--(2,5),linewidth(.5));<br /> draw((3,0)--(3,5),linewidth(.5));<br /> draw((4,0)--(4,5),linewidth(.5));<br /> draw((5,0)--(5,5),linewidth(.5));<br /> draw((6,0)--(6,5),linewidth(.5));<br /> draw((0,1)--(6,1),linewidth(.5));<br /> draw((0,2)--(6,2),linewidth(.5));<br /> draw((0,3)--(6,3),linewidth(.5));<br /> draw((0,4)--(6,4),linewidth(.5));<br /> draw((0,5)--(6,5),linewidth(.5)); <br /> draw((0,0)--(0,6),EndArrow);<br /> draw((0,0)--(7,0),EndArrow);<br /> draw((1,3)--(4,4)--(5,1)--cycle);<br /> label(&quot;$y$&quot;,(0,6),W); label(&quot;$x$&quot;,(7,0),S);<br /> label(&quot;$A$&quot;,(1,3),dir(210)); label(&quot;$B$&quot;,(5,1),SE); label(&quot;$C$&quot;,(4,4),dir(100));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Ralph went to the store and bought 12 pairs of socks for a total of \$24. Some of the socks he bought cost \$1 a pair, some of the socks he bought cost \$3 a pair, and some of the socks he bought cost \$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy?<br /> <br /> &lt;math&gt;\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> In the given figure hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is equiangular, &lt;math&gt;ABJI&lt;/math&gt; and &lt;math&gt;FEHG&lt;/math&gt; are squares with areas &lt;math&gt;18&lt;/math&gt; and &lt;math&gt;32&lt;/math&gt; respectively, &lt;math&gt;\triangle JBK&lt;/math&gt; is equilateral and &lt;math&gt;FE=BC&lt;/math&gt;. What is the area of &lt;math&gt;\triangle KBC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }6\sqrt{2}\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }9\sqrt{2}\qquad\textbf{(E) }32&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> <br /> draw((-4,6*sqrt(2))--(4,6*sqrt(2)));<br /> draw((-4,-6*sqrt(2))--(4,-6*sqrt(2)));<br /> draw((-8,0)--(-4,6*sqrt(2)));<br /> draw((-8,0)--(-4,-6*sqrt(2)));<br /> draw((4,6*sqrt(2))--(8,0));<br /> draw((8,0)--(4,-6*sqrt(2)));<br /> draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle);<br /> draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle);<br /> label(&quot;$I$&quot;,(-4,8+6*sqrt(2)),dir(100)); label(&quot;$J$&quot;,(4,8+6*sqrt(2)),dir(80));<br /> label(&quot;$A$&quot;,(-4,6*sqrt(2)),dir(280)); label(&quot;$B$&quot;,(4,6*sqrt(2)),dir(250));<br /> label(&quot;$C$&quot;,(8,0),W); label(&quot;$D$&quot;,(4,-6*sqrt(2)),NW); label(&quot;$E$&quot;,(-4,-6*sqrt(2)),NE); label(&quot;$F$&quot;,(-8,0),E);<br /> draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle);<br /> label(&quot;$K$&quot;,(4+4*sqrt(3),4+6*sqrt(2)),E);<br /> draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed);<br /> label(&quot;$H$&quot;,(-4-6*sqrt(2),-4-6*sqrt(2)),S);<br /> label(&quot;$G$&quot;,(-8-6*sqrt(2),-4),W);<br /> label(&quot;$32$&quot;,(-10,-8),N);<br /> label(&quot;$18$&quot;,(0,6*sqrt(2)+2),N);<br /> <br /> &lt;/asy&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br /> <br /> &lt;math&gt;\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Tom has twelve slips of paper which he wants to put into five cups labeled &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. The numbers on the papers are &lt;math&gt;2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,&lt;/math&gt; and &lt;math&gt;4.5&lt;/math&gt;. If a slip with 2 goes into cup &lt;math&gt;E&lt;/math&gt; and a slip with 3 goes into cup &lt;math&gt;B&lt;/math&gt;, then the slip with 3.5 must go into what cup?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } A \qquad<br /> \textbf{(B) } B \qquad<br /> \textbf{(C) } C \qquad<br /> \textbf{(D) } D \qquad<br /> \textbf{(E) } E<br /> &lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> A baseball league consists of two four-team divisions. Each team plays every other team in its division &lt;math&gt;N&lt;/math&gt; games. Each team plays every team in the other division &lt;math&gt;M&lt;/math&gt; games with &lt;math&gt;N&gt;2M&lt;/math&gt; and &lt;math&gt;M&gt;4&lt;/math&gt;. Each team plays a 76 game schedule. How many games does a team play within its own division?<br /> <br /> &lt;math&gt;\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;asy&gt;<br /> <br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 25|Solution]]</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems&diff=73128 2015 AMC 8 Problems 2015-11-25T21:35:02Z <p>Matongxu: /* Problem 23 */</p> <hr /> <div>==Problem 1==<br /> <br /> How many square yards of carpet are required to cover a rectangular floor that is &lt;math&gt;12&lt;/math&gt; feet long and &lt;math&gt;9&lt;/math&gt; feet wide? (There are 3 feet in a yard.)<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Point &lt;math&gt;O&lt;/math&gt; is the center of the regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt;, and &lt;math&gt;X&lt;/math&gt; is the midpoint of the side &lt;math&gt;\overline{AB}.&lt;/math&gt; What fraction of the area of the octagon is shaded?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G,H,O,X;<br /> A=dir(45);<br /> B=dir(90);<br /> C=dir(135);<br /> D=dir(180);<br /> E=dir(-135);<br /> F=dir(-90);<br /> G=dir(-45);<br /> H=dir(0);<br /> O=(0,0);<br /> X=midpoint(A--B);<br /> <br /> fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br /> draw(A--B--C--D--E--F--G--H--cycle);<br /> <br /> dot(&quot;$A$&quot;,A,dir(45));<br /> dot(&quot;$B$&quot;,B,dir(90));<br /> dot(&quot;$C$&quot;,C,dir(135));<br /> dot(&quot;$D$&quot;,D,dir(180));<br /> dot(&quot;$E$&quot;,E,dir(-135));<br /> dot(&quot;$F$&quot;,F,dir(-90));<br /> dot(&quot;$G$&quot;,G,dir(-45));<br /> dot(&quot;$H$&quot;,H,dir(0));<br /> dot(&quot;$X$&quot;,X,dir(135/2));<br /> dot(&quot;$O$&quot;,O,dir(0));<br /> draw(E--O--X);<br /> &lt;/asy&gt;<br /> [[2015 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of &lt;math&gt;10&lt;/math&gt; miles per hour. Jack walks to the pool at a constant speed of &lt;math&gt;4&lt;/math&gt; miles per hour. How many minutes before Jack does Jill arrive?<br /> <br /> &lt;math&gt;\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Billy's basketball team scored the following points over the course of the first 11 games of the season: &lt;cmath&gt;42, 47, 53, 53, 58, 58, 58, 61, 64, 65, 73&lt;/cmath&gt; If his team scores 40 in the 12th game, which of the following statistics will show an increase?<br /> <br /> &lt;math&gt;\textbf{(A) } \text{range} \qquad \textbf{(B) } \text{median} \qquad \textbf{(C) } \text{mean} \qquad \textbf{(D) } \text{mode} \qquad \textbf{(E) } \text{mid-range}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> In &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, &lt;math&gt;AB=BC=29&lt;/math&gt;, and &lt;math&gt;AC=42&lt;/math&gt;. What is the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Each of two boxes contains three chips numbered &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8== <br /> <br /> What is the smallest whole number larger than the perimeter of any triangle with a side of length &lt;math&gt;5&lt;/math&gt; and a side of length &lt;math&gt;19&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working &lt;math&gt;20&lt;/math&gt; days?<br /> <br /> &lt;math&gt;\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> How many integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt; have four distinct digits?<br /> <br /> &lt;math&gt;\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read &quot;AMC8&quot;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> How many pairs of parallel edges, such as &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{GH}&lt;/math&gt; or &lt;math&gt;\overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{FG}&lt;/math&gt;, does a cube have?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }6 \quad\textbf{(B) }12 \quad\textbf{(C) } 18 \quad\textbf{(D) } 24 \quad \textbf{(E) } 36&lt;/math&gt; &lt;asy&gt; import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label(&quot;$D$&quot;,(0,0,0),S); label(&quot;$A$&quot;,(0,0,1),N); label(&quot;$H$&quot;,(0,1,0),S); label(&quot;$E$&quot;,(0,1,1),N); label(&quot;$C$&quot;,(1,0,0),S); label(&quot;$B$&quot;,(1,0,1),N); label(&quot;$G$&quot;,(1,1,0),S); label(&quot;$F$&quot;,(1,1,1),N); &lt;/asy&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many subsets of two elements can be removed from the set &lt;math&gt;\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}&lt;/math&gt; so that the mean (average) of the remaining numbers is &lt;math&gt;6&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Which of the following integers cannot be written as the sum of four consecutive odd integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\text{ 16}\qquad\textbf{(B)}\text{ 40}\qquad\textbf{(C)}\text{ 72}\qquad\textbf{(D)}\text{ 100}\qquad\textbf{(E)}\text{ 200}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> At Euler Middle School, &lt;math&gt;198&lt;/math&gt; students voted on two issues in a school referendum with the following results: &lt;math&gt;149&lt;/math&gt; voted in favor of the first issue and &lt;math&gt;119&lt;/math&gt; voted in favor of the second issue. If there were exactly &lt;math&gt;29&lt;/math&gt; students who voted against both issues, how many students voted in favor of both issues?<br /> <br /> &lt;math&gt;\textbf{(A) }49\qquad\textbf{(B) }70\qquad\textbf{(C) }79\qquad\textbf{(D) }99\qquad \textbf{(E) }149&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; of all the ninth graders are paired with &lt;math&gt;\tfrac{2}{5}&lt;/math&gt; of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } \frac{2}{15} \qquad<br /> \textbf{(B) } \frac{4}{11} \qquad<br /> \textbf{(C) } \frac{11}{30} \qquad<br /> \textbf{(D) } \frac{3}{8} \qquad<br /> \textbf{(E) } \frac{11}{15}<br /> &lt;/math&gt;<br /> <br /> ==Problem 17==<br /> <br /> Jeremy's father drives him to school in rush hour traffic in &lt;math&gt;20&lt;/math&gt; minutes. One day there is no traffic, so his father can drive him &lt;math&gt;18&lt;/math&gt; miles per hour faster and gets him to school in &lt;math&gt;12&lt;/math&gt; minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, &lt;math&gt;2,5,8,11,14&lt;/math&gt; is an arithmetic sequence with five terms, in which the first term is &lt;math&gt;2&lt;/math&gt; and the constant added is &lt;math&gt;3&lt;/math&gt;. Each row and each column in this &lt;math&gt;5\times5&lt;/math&gt; array is an arithmetic sequence with five terms. What is the value of &lt;math&gt;X&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42&lt;/math&gt;<br /> <br /> &lt;asy&gt; size(3.85cm); label(&quot;$X$&quot;,(2.5,2.1),N); for (int i=0; i&lt;=5; ++i) draw((i,0)--(i,5), linewidth(.5)); for (int j=0; j&lt;=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num) { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); } draw_num((0,0), 17); draw_num((4, 0), 81); draw_num((0, 4), 1); draw_num((4,4), 25); void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, &quot; &quot;); foo(3, 4, &quot; &quot;); foo(0, 3, &quot; &quot;); foo(2, 3, &quot; &quot;); foo(1, 2, &quot; &quot;); foo(3, 2, &quot; &quot;); foo(1, 1, &quot; &quot;); foo(2, 1, &quot; &quot;); foo(3, 1, &quot; &quot;); foo(4, 1, &quot; &quot;); foo(2, 0, &quot; &quot;); foo(3, 0, &quot; &quot;); foo(0, 1, &quot; &quot;); foo(0, 2, &quot; &quot;); foo(1, 0, &quot; &quot;); foo(1, 3, &quot; &quot;); foo(1, 4, &quot; &quot;); foo(3, 3, &quot; &quot;); foo(4, 2, &quot; &quot;); foo(4, 3, &quot; &quot;); &lt;/asy&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> A triangle with vertices as &lt;math&gt;A=(1,3)&lt;/math&gt;, &lt;math&gt;B=(5,1)&lt;/math&gt;, and &lt;math&gt;C=(4,4)&lt;/math&gt; is plotted on a &lt;math&gt;6\times5&lt;/math&gt; grid. What fraction of the grid is covered by the triangle?<br /> <br /> &lt;asy&gt;<br /> <br /> draw((1,0)--(1,5),linewidth(.5));<br /> draw((2,0)--(2,5),linewidth(.5));<br /> draw((3,0)--(3,5),linewidth(.5));<br /> draw((4,0)--(4,5),linewidth(.5));<br /> draw((5,0)--(5,5),linewidth(.5));<br /> draw((6,0)--(6,5),linewidth(.5));<br /> draw((0,1)--(6,1),linewidth(.5));<br /> draw((0,2)--(6,2),linewidth(.5));<br /> draw((0,3)--(6,3),linewidth(.5));<br /> draw((0,4)--(6,4),linewidth(.5));<br /> draw((0,5)--(6,5),linewidth(.5)); <br /> draw((0,0)--(0,6),EndArrow);<br /> draw((0,0)--(7,0),EndArrow);<br /> draw((1,3)--(4,4)--(5,1)--cycle);<br /> label(&quot;$y$&quot;,(0,6),W); label(&quot;$x$&quot;,(7,0),S);<br /> label(&quot;$A$&quot;,(1,3),dir(210)); label(&quot;$B$&quot;,(5,1),SE); label(&quot;$C$&quot;,(4,4),dir(100));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Ralph went to the store and bought 12 pairs of socks for a total of \$24. Some of the socks he bought cost \$1 a pair, some of the socks he bought cost \$3 a pair, and some of the socks he bought cost \$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy?<br /> <br /> &lt;math&gt;\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> In the given figure hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is equiangular, &lt;math&gt;ABJI&lt;/math&gt; and &lt;math&gt;FEHG&lt;/math&gt; are squares with areas &lt;math&gt;18&lt;/math&gt; and &lt;math&gt;32&lt;/math&gt; respectively, &lt;math&gt;\triangle JBK&lt;/math&gt; is equilateral and &lt;math&gt;FE=BC&lt;/math&gt;. What is the area of &lt;math&gt;\triangle KBC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }6\sqrt{2}\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }9\sqrt{2}\qquad\textbf{(E) }32&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> <br /> draw((-4,6*sqrt(2))--(4,6*sqrt(2)));<br /> draw((-4,-6*sqrt(2))--(4,-6*sqrt(2)));<br /> draw((-8,0)--(-4,6*sqrt(2)));<br /> draw((-8,0)--(-4,-6*sqrt(2)));<br /> draw((4,6*sqrt(2))--(8,0));<br /> draw((8,0)--(4,-6*sqrt(2)));<br /> draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle);<br /> draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle);<br /> label(&quot;$I$&quot;,(-4,8+6*sqrt(2)),dir(100)); label(&quot;$J$&quot;,(4,8+6*sqrt(2)),dir(80));<br /> label(&quot;$A$&quot;,(-4,6*sqrt(2)),dir(280)); label(&quot;$B$&quot;,(4,6*sqrt(2)),dir(250));<br /> label(&quot;$C$&quot;,(8,0),W); label(&quot;$D$&quot;,(4,-6*sqrt(2)),NW); label(&quot;$E$&quot;,(-4,-6*sqrt(2)),NE); label(&quot;$F$&quot;,(-8,0),E);<br /> draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle);<br /> label(&quot;$K$&quot;,(4+4*sqrt(3),4+6*sqrt(2)),E);<br /> draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed);<br /> label(&quot;$H$&quot;,(-4-6*sqrt(2),-4-6*sqrt(2)),S);<br /> label(&quot;$G$&quot;,(-8-6*sqrt(2),-4),W);<br /> label(&quot;$32$&quot;,(-10,-8),N);<br /> label(&quot;$18$&quot;,(0,6*sqrt(2)+2),N);<br /> <br /> &lt;/asy&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br /> <br /> &lt;math&gt;\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Tom has twelve slips of paper which he wants to put into five cups labeled &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. The numbers on the papers are &lt;math&gt;2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,&lt;/math&gt; and &lt;math&gt;4.5&lt;/math&gt;. If a slip with 2 goes into cup &lt;math&gt;E&lt;/math&gt; and a slip with 3 goes into cup &lt;math&gt;B&lt;/math&gt;, then the slip with 3.5 must go into what cup?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } A \qquad<br /> \textbf{(B) } B \qquad<br /> \textbf{(C) } C \qquad<br /> \textbf{(D) } D \qquad<br /> \textbf{(E) } E<br /> &lt;/math&gt;<br /> <br /> ==Problem 24==<br /> <br /> A baseball league consists of two four-team divisions. Each team plays every other team in its division &lt;math&gt;N&lt;/math&gt; games. Each team plays every team in the other division &lt;math&gt;M&lt;/math&gt; games with &lt;math&gt;N&gt;2M&lt;/math&gt; and &lt;math&gt;M&gt;4&lt;/math&gt;. Each team plays a 76 game schedule. How many games does a team play within its own division?<br /> <br /> &lt;math&gt;\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br /> <br /> &lt;asy&gt;<br /> <br /> draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br /> filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br /> filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br /> filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17&lt;/math&gt;<br /> <br /> [[2015 AMC 8 Problems/Problem 25|Solution]]</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=AMC_8_Problems_and_Solutions&diff=72952 AMC 8 Problems and Solutions 2015-11-23T21:17:47Z <p>Matongxu: Change of grammatical tense</p> <hr /> <div>The 2015 AMC 8 was held on November 17th, 2015.<br /> <br /> [[AMC 8]] / [[AJHSME]] problems and solutions.<br /> <br /> * [[2014 AMC 8]]<br /> * [[2013 AMC 8]]<br /> * [[2012 AMC 8]]<br /> * [[2011 AMC 8]]<br /> * [[2010 AMC 8]]<br /> * [[2009 AMC 8]]<br /> * [[2008 AMC 8]]<br /> * [[2007 AMC 8]]<br /> * [[2006 AMC 8]]<br /> * [[2005 AMC 8]]<br /> * [[2004 AMC 8]]<br /> * [[2003 AMC 8]]<br /> * [[2002 AMC 8]]<br /> * [[2001 AMC 8]]<br /> * [[2000 AMC 8]]<br /> * [[1999 AMC 8]]<br /> * [[1998 AJHSME]]<br /> * [[1997 AJHSME]]<br /> * [[1996 AJHSME]]<br /> * [[1995 AJHSME]]<br /> * [[1994 AJHSME]]<br /> * [[1993 AJHSME]]<br /> * [[1992 AJHSME]]<br /> * [[1991 AJHSME]]<br /> * [[1990 AJHSME]]<br /> * [[1989 AJHSME]]<br /> * [[1988 AJHSME]]<br /> * [[1987 AJHSME]]<br /> * [[1986 AJHSME]]<br /> * [[1985 AJHSME]]<br /> <br /> == Resources ==<br /> * [[American Mathematics Competitions]]<br /> * [[AMC Problems and Solutions]]<br /> * [[Mathematics competition resources]]</div> Matongxu https://artofproblemsolving.com/wiki/index.php?title=Community_Service_Scholarships&diff=71422 Community Service Scholarships 2015-08-01T19:21:02Z <p>Matongxu: /* National community service scholarships */</p> <hr /> <div>The following list of '''community service scholarships''' is primarily for American students. This list can be reorganized to incorporate scholarship programs in other countries. Just make that reorganization as clear and as clean as possible.<br /> <br /> Additions to this list are welcomed and encouraged. Please don't be stingy about letting students in on how they can finance their educations! If you are unfamiliar with how to edit a Wiki and don't have time to learn, please contact [[Mathew Crawford]] using the email crawford@artofproblemsolving.com with the scholarship listing you wish to contribute.<br /> <br /> === National community service scholarships ===<br /> * Prudential Spirit of Community Awards [http://spirit.prudential.com/ website]<br /> * Kohl's Kids Who Care Scholarship Program [http://www.kohlscorporation.com/communityrelations/scholarship/index.asp website]<br /> * Comcast Leaders and Achievers scholarship (requires nomination by the principal or guidance counselor) [http://www.comcast.com/Corporate/About/InTheCommunity/Partners/LeadersAndAchievers.html website]<br /> * [http://www.womeninservice.org/index.php?option=com_content&amp;view=article&amp;id=7&amp;Itemid=8 Women in Service to Appalachia Scholarships] of \$500 for female undergraduate and graduate students who are regular involved in community service.<br /> * [http://www.aaiusa.org/foundation/154/student-resource-center#ylawards Helen Abbott Community Service Awards] for high school and undergraduate students and student organizations devoted to community service.<br /> * [http://www.goodtidings.org/ Good Tidings Foundation Community Service Scholarships] for high school seniors in the Greater Bay Area.<br /> * [http://www.barronprize.org/ The Gloria Barron Prize for Young Heroes], recognizing outstanding young leaders nominated by adults.<br /> * [http://www.bonner.org/campus/bsp/howtoapply.htm Bonner Scholars Program] for with high financial need and commitment to service to attend one of 27 participating colleges and universities.<br /> <br /> === State-specific ===<br /> * [http://www.youthlaunch.org/programs/scholarship.php YouthLaunch Scholarship for Outstanding Service]. A &lt;dollar/&gt;3,000 scholarship for Texas students who demonstrate commitment to their community service projects.<br /> * [http://publicaffairs.disneyland.com/education/youth-scholarships/youth_scholarships.html The Disneyland Resort Scholarship Program] for Orange County, California, high school seniors who have demonstrated excellence in volunteerism and leadership.<br /> * [http://jlrockford.org/scholarships.asp Junior League of Rockford Outstanding Volunteer Scholarship] for a high school senior in Winnebago County, Illinois.<br /> <br /> == See also ==<br /> * [[Academic scholarships]]<br /> * [[Academic competitions]]</div> Matongxu