https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathgl2018&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-06T04:36:50Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=164045 2004 AMC 10A Problems/Problem 20 2021-10-24T17:49:44Z <p>Mathgl2018: /* Solution 5 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are located on square &lt;math&gt;ABCD&lt;/math&gt; so that &lt;math&gt;\triangle BEF&lt;/math&gt; is equilateral. What is the ratio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to that of &lt;math&gt;\triangle ABE&lt;/math&gt;?<br /> <br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> unitsize(3 cm);<br /> <br /> pair A, B, C, D, E, F;<br /> <br /> A = (0,0);<br /> B = (1,0);<br /> C = (1,1);<br /> D = (0,1);<br /> E = (0,Tan(15));<br /> F = (1 - Tan(15),1);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(B--E--F--cycle);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&quot;, C, NE);<br /> label(&quot;$D$&quot;, D, NW);<br /> label(&quot;$E$&quot;, E, W);<br /> label(&quot;$F$&quot;, F, N);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since triangle &lt;math&gt;BEF&lt;/math&gt; is equilateral, &lt;math&gt;EA=FC&lt;/math&gt;, and &lt;math&gt;EAB&lt;/math&gt; and &lt;math&gt;FCB&lt;/math&gt; are &lt;math&gt;SAS&lt;/math&gt; congruent. Thus, triangle &lt;math&gt;DEF&lt;/math&gt; is an isosceles right triangle. So we let &lt;math&gt;DE=x&lt;/math&gt;. Thus &lt;math&gt;EF=EB=FB=x\sqrt{2}&lt;/math&gt;. If we go angle chasing, we find out that &lt;math&gt;\angle AEB=75^{\circ}&lt;/math&gt;, thus &lt;math&gt;\angle ABE=15^{\circ}&lt;/math&gt;. &lt;math&gt;\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;. Thus &lt;math&gt;\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;, or &lt;math&gt;AE=\frac{x(\sqrt{3}-1)}{2}&lt;/math&gt;. Thus &lt;math&gt;AB=\frac{x(\sqrt{3}+1)}{2}&lt;/math&gt;, and &lt;math&gt;[ABE]=\frac{x^2}{4}&lt;/math&gt;, and &lt;math&gt;[DEF]=\frac{x^2}{2}&lt;/math&gt;. Thus the ratio of the areas is &lt;math&gt;\boxed{\mathrm{(D)}\ 2}&lt;/math&gt;<br /> <br /> ==Solution 2 (Non-trig) ==<br /> WLOG, let the side length of &lt;math&gt;ABCD&lt;/math&gt; be 1. Let &lt;math&gt;DE = x&lt;/math&gt;. It suffices that &lt;math&gt;AE = 1 - x&lt;/math&gt;. Then triangles &lt;math&gt;ABE&lt;/math&gt; and &lt;math&gt;CBF&lt;/math&gt; are congruent by HL, so &lt;math&gt;CF = AE&lt;/math&gt; and &lt;math&gt;DE = DF&lt;/math&gt;. We find that &lt;math&gt;BE = EF = x \sqrt{2}&lt;/math&gt;, and so, by the Pythagorean Theorem, we have<br /> &lt;math&gt;(1 - x)^2 + 1 = 2x^2.&lt;/math&gt; This yields &lt;math&gt;x^2 + 2x = 2&lt;/math&gt;, so &lt;math&gt;x^2 = 2 - 2x&lt;/math&gt;. Thus, the desired ratio of areas is<br /> &lt;cmath&gt;\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;\angle EBF = 60^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle EBA = \angle FBC&lt;/math&gt; so they must each be &lt;math&gt;15^{\circ}&lt;/math&gt;. Then let &lt;math&gt;BE=EF=FB=1&lt;/math&gt;, which gives &lt;math&gt;EA=\sin{15^{\circ}}&lt;/math&gt; and &lt;math&gt;AB=\cos{15^{\circ}}&lt;/math&gt;. <br /> The area of &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is then &lt;math&gt;\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}&lt;/math&gt;. <br /> &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; is an isosceles right triangle with hypotenuse 1, so &lt;math&gt;DE=DF=\frac{1}{\sqrt{2}}&lt;/math&gt; and therefore its area is &lt;math&gt;\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}&lt;/math&gt;. <br /> The ratio of areas is then &lt;math&gt;\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}&lt;/math&gt;<br /> <br /> ==Solution 4 (System of Equations)==<br /> Assume &lt;math&gt;AB=1&lt;/math&gt;. Then, &lt;math&gt;FC&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt; is &lt;math&gt;1-x&lt;/math&gt;. We see that using &lt;math&gt;HL&lt;/math&gt;, &lt;math&gt;FCB&lt;/math&gt; is congruent to EAB. Using Pythagoras of triangles &lt;math&gt;FCB&lt;/math&gt; and &lt;math&gt;FDE&lt;/math&gt; we get &lt;math&gt;2{(1-x)}^2=x^2+1&lt;/math&gt;. Expanding, we get &lt;math&gt;2x^2-4x+2=x^2+1&lt;/math&gt;. Simplifying gives &lt;math&gt;x^2-4x+1=0&lt;/math&gt; solving using completing the square (or other methods) gives 2 answers: &lt;math&gt;2-\sqrt{3}&lt;/math&gt; and &lt;math&gt;2+\sqrt{3}&lt;/math&gt;. Because &lt;math&gt;x &lt; 1&lt;/math&gt;, &lt;math&gt;x=2-\sqrt{3}&lt;/math&gt;. Using the areas, the answer is &lt;math&gt;\boxed{\text{(D) }2}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> First, since &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral and &lt;math&gt;ABCD&lt;/math&gt; is a square, by the Hypothenuse Leg Theorem, &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is congruent to &lt;math&gt;\bigtriangleup CBF&lt;/math&gt;. Then, assume length &lt;math&gt;AB = BC = x&lt;/math&gt; and length &lt;math&gt;DE = DF = y&lt;/math&gt;, then &lt;math&gt;AE = FC = x - y&lt;/math&gt;. &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;EF = EB&lt;/math&gt; and &lt;math&gt;EB^2 = EF^2&lt;/math&gt;, it is given that &lt;math&gt;ABCD&lt;/math&gt; is a square and &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; and &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; are right triangles. Then we use the Pythagorean theorem to prove that &lt;math&gt;AB^2 + AE^2 = EB^2&lt;/math&gt; and since we know that &lt;math&gt;EB^2 = EF^2&lt;/math&gt; and &lt;math&gt;EF^2 = DE^2 + DF^2&lt;/math&gt;, which means &lt;math&gt;AB^2 + AE^2 = DE^2 + DF^2&lt;/math&gt;. Now we plug in the variables and the equation becomes &lt;math&gt;x^2 + (x+y)^2 = 2y^2&lt;/math&gt;, expand and simplify and you get &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;. We want the ratio of area of &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; to &lt;math&gt;\bigtriangleup ABE&lt;/math&gt;. Expressed in our variables, the ratio of the area is &lt;math&gt;\frac{y^2}{x^2 - xy}&lt;/math&gt; and we know &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;, so the ratio must be &lt;math&gt;2&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\text{(D) }2}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/hwHIHRukYMk<br /> <br /> Education, the Study of Everything<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/BFKo9h8GhLY <br /> <br /> ~IceMatrix<br /> <br /> ==See also==<br /> {{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=164044 2004 AMC 10A Problems/Problem 20 2021-10-24T17:49:21Z <p>Mathgl2018: /* Solution 5 */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are located on square &lt;math&gt;ABCD&lt;/math&gt; so that &lt;math&gt;\triangle BEF&lt;/math&gt; is equilateral. What is the ratio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to that of &lt;math&gt;\triangle ABE&lt;/math&gt;?<br /> <br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> unitsize(3 cm);<br /> <br /> pair A, B, C, D, E, F;<br /> <br /> A = (0,0);<br /> B = (1,0);<br /> C = (1,1);<br /> D = (0,1);<br /> E = (0,Tan(15));<br /> F = (1 - Tan(15),1);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(B--E--F--cycle);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&quot;, C, NE);<br /> label(&quot;$D$&quot;, D, NW);<br /> label(&quot;$E$&quot;, E, W);<br /> label(&quot;$F$&quot;, F, N);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since triangle &lt;math&gt;BEF&lt;/math&gt; is equilateral, &lt;math&gt;EA=FC&lt;/math&gt;, and &lt;math&gt;EAB&lt;/math&gt; and &lt;math&gt;FCB&lt;/math&gt; are &lt;math&gt;SAS&lt;/math&gt; congruent. Thus, triangle &lt;math&gt;DEF&lt;/math&gt; is an isosceles right triangle. So we let &lt;math&gt;DE=x&lt;/math&gt;. Thus &lt;math&gt;EF=EB=FB=x\sqrt{2}&lt;/math&gt;. If we go angle chasing, we find out that &lt;math&gt;\angle AEB=75^{\circ}&lt;/math&gt;, thus &lt;math&gt;\angle ABE=15^{\circ}&lt;/math&gt;. &lt;math&gt;\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;. Thus &lt;math&gt;\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;, or &lt;math&gt;AE=\frac{x(\sqrt{3}-1)}{2}&lt;/math&gt;. Thus &lt;math&gt;AB=\frac{x(\sqrt{3}+1)}{2}&lt;/math&gt;, and &lt;math&gt;[ABE]=\frac{x^2}{4}&lt;/math&gt;, and &lt;math&gt;[DEF]=\frac{x^2}{2}&lt;/math&gt;. Thus the ratio of the areas is &lt;math&gt;\boxed{\mathrm{(D)}\ 2}&lt;/math&gt;<br /> <br /> ==Solution 2 (Non-trig) ==<br /> WLOG, let the side length of &lt;math&gt;ABCD&lt;/math&gt; be 1. Let &lt;math&gt;DE = x&lt;/math&gt;. It suffices that &lt;math&gt;AE = 1 - x&lt;/math&gt;. Then triangles &lt;math&gt;ABE&lt;/math&gt; and &lt;math&gt;CBF&lt;/math&gt; are congruent by HL, so &lt;math&gt;CF = AE&lt;/math&gt; and &lt;math&gt;DE = DF&lt;/math&gt;. We find that &lt;math&gt;BE = EF = x \sqrt{2}&lt;/math&gt;, and so, by the Pythagorean Theorem, we have<br /> &lt;math&gt;(1 - x)^2 + 1 = 2x^2.&lt;/math&gt; This yields &lt;math&gt;x^2 + 2x = 2&lt;/math&gt;, so &lt;math&gt;x^2 = 2 - 2x&lt;/math&gt;. Thus, the desired ratio of areas is<br /> &lt;cmath&gt;\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;\angle EBF = 60^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle EBA = \angle FBC&lt;/math&gt; so they must each be &lt;math&gt;15^{\circ}&lt;/math&gt;. Then let &lt;math&gt;BE=EF=FB=1&lt;/math&gt;, which gives &lt;math&gt;EA=\sin{15^{\circ}}&lt;/math&gt; and &lt;math&gt;AB=\cos{15^{\circ}}&lt;/math&gt;. <br /> The area of &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is then &lt;math&gt;\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}&lt;/math&gt;. <br /> &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; is an isosceles right triangle with hypotenuse 1, so &lt;math&gt;DE=DF=\frac{1}{\sqrt{2}}&lt;/math&gt; and therefore its area is &lt;math&gt;\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}&lt;/math&gt;. <br /> The ratio of areas is then &lt;math&gt;\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}&lt;/math&gt;<br /> <br /> ==Solution 4 (System of Equations)==<br /> Assume &lt;math&gt;AB=1&lt;/math&gt;. Then, &lt;math&gt;FC&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt; is &lt;math&gt;1-x&lt;/math&gt;. We see that using &lt;math&gt;HL&lt;/math&gt;, &lt;math&gt;FCB&lt;/math&gt; is congruent to EAB. Using Pythagoras of triangles &lt;math&gt;FCB&lt;/math&gt; and &lt;math&gt;FDE&lt;/math&gt; we get &lt;math&gt;2{(1-x)}^2=x^2+1&lt;/math&gt;. Expanding, we get &lt;math&gt;2x^2-4x+2=x^2+1&lt;/math&gt;. Simplifying gives &lt;math&gt;x^2-4x+1=0&lt;/math&gt; solving using completing the square (or other methods) gives 2 answers: &lt;math&gt;2-\sqrt{3}&lt;/math&gt; and &lt;math&gt;2+\sqrt{3}&lt;/math&gt;. Because &lt;math&gt;x &lt; 1&lt;/math&gt;, &lt;math&gt;x=2-\sqrt{3}&lt;/math&gt;. Using the areas, the answer is &lt;math&gt;\boxed{\text{(D) }2}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> First, since &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral and &lt;math&gt;ABCD&lt;/math&gt; is a square, by the Hypothenuse Leg Theorem, &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is congruent to &lt;math&gt;\bigtriangleup CBF&lt;/math&gt;. Then, assume length &lt;math&gt;AB = BC = x&lt;/math&gt; and length &lt;math&gt;DE = DF = y&lt;/math&gt;, then &lt;math&gt;AE = FC = x - y&lt;/math&gt;. &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;EF = EB&lt;/math&gt; and &lt;math&gt;EB^2 = EF^2&lt;/math&gt;, it is given that &lt;math&gt;ABCD&lt;/math&gt; is a square and &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; and &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; are right triangles. Then we use the Pythagorean theorem to prove that &lt;math&gt;AB^2 + AE^2 = EB^2&lt;/math&gt; and since we know that &lt;math&gt;EB^2 = EF^2&lt;/math&gt; and &lt;math&gt;EF^2 = DE^2 + DF^2&lt;/math&gt;, which means &lt;math&gt;AB^2 + AE^2 = DE^2 + DF^2&lt;/math&gt;. Now we plug in the variables and the equation becomes &lt;math&gt;x^2 + (x+y)^2 = 2y^2&lt;/math&gt;, expand and simplify and you get &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;. We want the ratio of area of &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; to &lt;math&gt;\bigtriangleup ABE&lt;/math&gt;. Expressed in our variables, the ratio of the area is &lt;math&gt;\frac{y^2}{x^2 - xy}&lt;/math&gt; and we know &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;, so the ratio must be &lt;math&gt;2&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\text{(D )}2}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/hwHIHRukYMk<br /> <br /> Education, the Study of Everything<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/BFKo9h8GhLY <br /> <br /> ~IceMatrix<br /> <br /> ==See also==<br /> {{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=164043 2004 AMC 10A Problems/Problem 20 2021-10-24T17:48:03Z <p>Mathgl2018: /* Solution 4 (System of Equations) */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are located on square &lt;math&gt;ABCD&lt;/math&gt; so that &lt;math&gt;\triangle BEF&lt;/math&gt; is equilateral. What is the ratio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to that of &lt;math&gt;\triangle ABE&lt;/math&gt;?<br /> <br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> unitsize(3 cm);<br /> <br /> pair A, B, C, D, E, F;<br /> <br /> A = (0,0);<br /> B = (1,0);<br /> C = (1,1);<br /> D = (0,1);<br /> E = (0,Tan(15));<br /> F = (1 - Tan(15),1);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(B--E--F--cycle);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&quot;, C, NE);<br /> label(&quot;$D$&quot;, D, NW);<br /> label(&quot;$E$&quot;, E, W);<br /> label(&quot;$F$&quot;, F, N);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since triangle &lt;math&gt;BEF&lt;/math&gt; is equilateral, &lt;math&gt;EA=FC&lt;/math&gt;, and &lt;math&gt;EAB&lt;/math&gt; and &lt;math&gt;FCB&lt;/math&gt; are &lt;math&gt;SAS&lt;/math&gt; congruent. Thus, triangle &lt;math&gt;DEF&lt;/math&gt; is an isosceles right triangle. So we let &lt;math&gt;DE=x&lt;/math&gt;. Thus &lt;math&gt;EF=EB=FB=x\sqrt{2}&lt;/math&gt;. If we go angle chasing, we find out that &lt;math&gt;\angle AEB=75^{\circ}&lt;/math&gt;, thus &lt;math&gt;\angle ABE=15^{\circ}&lt;/math&gt;. &lt;math&gt;\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;. Thus &lt;math&gt;\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;, or &lt;math&gt;AE=\frac{x(\sqrt{3}-1)}{2}&lt;/math&gt;. Thus &lt;math&gt;AB=\frac{x(\sqrt{3}+1)}{2}&lt;/math&gt;, and &lt;math&gt;[ABE]=\frac{x^2}{4}&lt;/math&gt;, and &lt;math&gt;[DEF]=\frac{x^2}{2}&lt;/math&gt;. Thus the ratio of the areas is &lt;math&gt;\boxed{\mathrm{(D)}\ 2}&lt;/math&gt;<br /> <br /> ==Solution 2 (Non-trig) ==<br /> WLOG, let the side length of &lt;math&gt;ABCD&lt;/math&gt; be 1. Let &lt;math&gt;DE = x&lt;/math&gt;. It suffices that &lt;math&gt;AE = 1 - x&lt;/math&gt;. Then triangles &lt;math&gt;ABE&lt;/math&gt; and &lt;math&gt;CBF&lt;/math&gt; are congruent by HL, so &lt;math&gt;CF = AE&lt;/math&gt; and &lt;math&gt;DE = DF&lt;/math&gt;. We find that &lt;math&gt;BE = EF = x \sqrt{2}&lt;/math&gt;, and so, by the Pythagorean Theorem, we have<br /> &lt;math&gt;(1 - x)^2 + 1 = 2x^2.&lt;/math&gt; This yields &lt;math&gt;x^2 + 2x = 2&lt;/math&gt;, so &lt;math&gt;x^2 = 2 - 2x&lt;/math&gt;. Thus, the desired ratio of areas is<br /> &lt;cmath&gt;\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;\angle EBF = 60^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle EBA = \angle FBC&lt;/math&gt; so they must each be &lt;math&gt;15^{\circ}&lt;/math&gt;. Then let &lt;math&gt;BE=EF=FB=1&lt;/math&gt;, which gives &lt;math&gt;EA=\sin{15^{\circ}}&lt;/math&gt; and &lt;math&gt;AB=\cos{15^{\circ}}&lt;/math&gt;. <br /> The area of &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is then &lt;math&gt;\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}&lt;/math&gt;. <br /> &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; is an isosceles right triangle with hypotenuse 1, so &lt;math&gt;DE=DF=\frac{1}{\sqrt{2}}&lt;/math&gt; and therefore its area is &lt;math&gt;\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}&lt;/math&gt;. <br /> The ratio of areas is then &lt;math&gt;\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}&lt;/math&gt;<br /> <br /> ==Solution 4 (System of Equations)==<br /> Assume &lt;math&gt;AB=1&lt;/math&gt;. Then, &lt;math&gt;FC&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt; is &lt;math&gt;1-x&lt;/math&gt;. We see that using &lt;math&gt;HL&lt;/math&gt;, &lt;math&gt;FCB&lt;/math&gt; is congruent to EAB. Using Pythagoras of triangles &lt;math&gt;FCB&lt;/math&gt; and &lt;math&gt;FDE&lt;/math&gt; we get &lt;math&gt;2{(1-x)}^2=x^2+1&lt;/math&gt;. Expanding, we get &lt;math&gt;2x^2-4x+2=x^2+1&lt;/math&gt;. Simplifying gives &lt;math&gt;x^2-4x+1=0&lt;/math&gt; solving using completing the square (or other methods) gives 2 answers: &lt;math&gt;2-\sqrt{3}&lt;/math&gt; and &lt;math&gt;2+\sqrt{3}&lt;/math&gt;. Because &lt;math&gt;x &lt; 1&lt;/math&gt;, &lt;math&gt;x=2-\sqrt{3}&lt;/math&gt;. Using the areas, the answer is &lt;math&gt;\boxed{\text{(D) }2}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> First, since &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral and &lt;math&gt;ABCD&lt;/math&gt; is a square, by the Hypothenuse Leg Theorem, &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is congruent to &lt;math&gt;\bigtriangleup CBF&lt;/math&gt;. Then, assume length &lt;math&gt;AB = BC = x&lt;/math&gt; and length &lt;math&gt;DE = DF = y&lt;/math&gt;, then &lt;math&gt;AE = FC = x - y&lt;/math&gt;. &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;EF = EB&lt;/math&gt; and &lt;math&gt;EB^2 = EF^2&lt;/math&gt;, it is given that &lt;math&gt;ABCD&lt;/math&gt; is a square and &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; and &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; are right triangles. Then we use the Pythagorean theorem to prove that &lt;math&gt;AB^2 + AE^2 = EB^2&lt;/math&gt; and since we know that &lt;math&gt;EB^2 = EF^2&lt;/math&gt; and &lt;math&gt;EF^2 = DE^2 + DF^2&lt;/math&gt;, which means &lt;math&gt;AB^2 + AE^2 = DE^2 + DF^2&lt;/math&gt;. Now we plug in the variables and the equation becomes &lt;math&gt;x^2 + (x+y)^2 = 2y^2&lt;/math&gt;, expand and simplify and you get &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;. We want the ratio of area of &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; to &lt;math&gt;\bigtriangleup ABE&lt;/math&gt;. Expressed in our variables, the ratio of the area is &lt;math&gt;\frac{y^2}{x^2 - xy}&lt;/math&gt; and we know &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;, so the ratio must be 2. Choice D<br /> <br /> ==Video Solution==<br /> https://youtu.be/hwHIHRukYMk<br /> <br /> Education, the Study of Everything<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/BFKo9h8GhLY <br /> <br /> ~IceMatrix<br /> <br /> ==See also==<br /> {{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=164042 2004 AMC 10A Problems/Problem 20 2021-10-24T17:47:31Z <p>Mathgl2018: /* Solution 4(system of equations) */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are located on square &lt;math&gt;ABCD&lt;/math&gt; so that &lt;math&gt;\triangle BEF&lt;/math&gt; is equilateral. What is the ratio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to that of &lt;math&gt;\triangle ABE&lt;/math&gt;?<br /> <br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> unitsize(3 cm);<br /> <br /> pair A, B, C, D, E, F;<br /> <br /> A = (0,0);<br /> B = (1,0);<br /> C = (1,1);<br /> D = (0,1);<br /> E = (0,Tan(15));<br /> F = (1 - Tan(15),1);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(B--E--F--cycle);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&quot;, C, NE);<br /> label(&quot;$D$&quot;, D, NW);<br /> label(&quot;$E$&quot;, E, W);<br /> label(&quot;$F$&quot;, F, N);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since triangle &lt;math&gt;BEF&lt;/math&gt; is equilateral, &lt;math&gt;EA=FC&lt;/math&gt;, and &lt;math&gt;EAB&lt;/math&gt; and &lt;math&gt;FCB&lt;/math&gt; are &lt;math&gt;SAS&lt;/math&gt; congruent. Thus, triangle &lt;math&gt;DEF&lt;/math&gt; is an isosceles right triangle. So we let &lt;math&gt;DE=x&lt;/math&gt;. Thus &lt;math&gt;EF=EB=FB=x\sqrt{2}&lt;/math&gt;. If we go angle chasing, we find out that &lt;math&gt;\angle AEB=75^{\circ}&lt;/math&gt;, thus &lt;math&gt;\angle ABE=15^{\circ}&lt;/math&gt;. &lt;math&gt;\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;. Thus &lt;math&gt;\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;, or &lt;math&gt;AE=\frac{x(\sqrt{3}-1)}{2}&lt;/math&gt;. Thus &lt;math&gt;AB=\frac{x(\sqrt{3}+1)}{2}&lt;/math&gt;, and &lt;math&gt;[ABE]=\frac{x^2}{4}&lt;/math&gt;, and &lt;math&gt;[DEF]=\frac{x^2}{2}&lt;/math&gt;. Thus the ratio of the areas is &lt;math&gt;\boxed{\mathrm{(D)}\ 2}&lt;/math&gt;<br /> <br /> ==Solution 2 (Non-trig) ==<br /> WLOG, let the side length of &lt;math&gt;ABCD&lt;/math&gt; be 1. Let &lt;math&gt;DE = x&lt;/math&gt;. It suffices that &lt;math&gt;AE = 1 - x&lt;/math&gt;. Then triangles &lt;math&gt;ABE&lt;/math&gt; and &lt;math&gt;CBF&lt;/math&gt; are congruent by HL, so &lt;math&gt;CF = AE&lt;/math&gt; and &lt;math&gt;DE = DF&lt;/math&gt;. We find that &lt;math&gt;BE = EF = x \sqrt{2}&lt;/math&gt;, and so, by the Pythagorean Theorem, we have<br /> &lt;math&gt;(1 - x)^2 + 1 = 2x^2.&lt;/math&gt; This yields &lt;math&gt;x^2 + 2x = 2&lt;/math&gt;, so &lt;math&gt;x^2 = 2 - 2x&lt;/math&gt;. Thus, the desired ratio of areas is<br /> &lt;cmath&gt;\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;\angle EBF = 60^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle EBA = \angle FBC&lt;/math&gt; so they must each be &lt;math&gt;15^{\circ}&lt;/math&gt;. Then let &lt;math&gt;BE=EF=FB=1&lt;/math&gt;, which gives &lt;math&gt;EA=\sin{15^{\circ}}&lt;/math&gt; and &lt;math&gt;AB=\cos{15^{\circ}}&lt;/math&gt;. <br /> The area of &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is then &lt;math&gt;\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}&lt;/math&gt;. <br /> &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; is an isosceles right triangle with hypotenuse 1, so &lt;math&gt;DE=DF=\frac{1}{\sqrt{2}}&lt;/math&gt; and therefore its area is &lt;math&gt;\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}&lt;/math&gt;. <br /> The ratio of areas is then &lt;math&gt;\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}&lt;/math&gt;<br /> <br /> ==Solution 4 (System of Equations)==<br /> Assume &lt;math&gt;AB=1&lt;/math&gt;. Then, &lt;math&gt;FC&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt; is &lt;math&gt;1-x&lt;/math&gt; then we see that using &lt;math&gt;HL&lt;/math&gt;, &lt;math&gt;FCB&lt;/math&gt; is congruent to EAB. Using Pythagoras of triangles &lt;math&gt;FCB&lt;/math&gt; and &lt;math&gt;FDE&lt;/math&gt; we get &lt;math&gt;2{(1-x)}^2=x^2+1&lt;/math&gt;. Expanding, we get &lt;math&gt;2x^2-4x+2=x^2+1&lt;/math&gt;. Simplifying gives &lt;math&gt;x^2-4x+1=0&lt;/math&gt; solving using completing the square (or other methods) gives 2 answers: &lt;math&gt;2-\sqrt{3}&lt;/math&gt; and &lt;math&gt;2+\sqrt{3}&lt;/math&gt;. Because &lt;math&gt;x &lt; 1&lt;/math&gt;, &lt;math&gt;x=2-\sqrt{3}&lt;/math&gt;. Using the areas, the answer is &lt;math&gt;\boxed{\text{(D) }2}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> First, since &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral and &lt;math&gt;ABCD&lt;/math&gt; is a square, by the Hypothenuse Leg Theorem, &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is congruent to &lt;math&gt;\bigtriangleup CBF&lt;/math&gt;. Then, assume length &lt;math&gt;AB = BC = x&lt;/math&gt; and length &lt;math&gt;DE = DF = y&lt;/math&gt;, then &lt;math&gt;AE = FC = x - y&lt;/math&gt;. &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;EF = EB&lt;/math&gt; and &lt;math&gt;EB^2 = EF^2&lt;/math&gt;, it is given that &lt;math&gt;ABCD&lt;/math&gt; is a square and &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; and &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; are right triangles. Then we use the Pythagorean theorem to prove that &lt;math&gt;AB^2 + AE^2 = EB^2&lt;/math&gt; and since we know that &lt;math&gt;EB^2 = EF^2&lt;/math&gt; and &lt;math&gt;EF^2 = DE^2 + DF^2&lt;/math&gt;, which means &lt;math&gt;AB^2 + AE^2 = DE^2 + DF^2&lt;/math&gt;. Now we plug in the variables and the equation becomes &lt;math&gt;x^2 + (x+y)^2 = 2y^2&lt;/math&gt;, expand and simplify and you get &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;. We want the ratio of area of &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; to &lt;math&gt;\bigtriangleup ABE&lt;/math&gt;. Expressed in our variables, the ratio of the area is &lt;math&gt;\frac{y^2}{x^2 - xy}&lt;/math&gt; and we know &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;, so the ratio must be 2. Choice D<br /> <br /> ==Video Solution==<br /> https://youtu.be/hwHIHRukYMk<br /> <br /> Education, the Study of Everything<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/BFKo9h8GhLY <br /> <br /> ~IceMatrix<br /> <br /> ==See also==<br /> {{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_20&diff=164041 2004 AMC 10A Problems/Problem 20 2021-10-24T17:47:12Z <p>Mathgl2018: /* Solution 4(system of equations) */</p> <hr /> <div>==Problem==<br /> Points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are located on square &lt;math&gt;ABCD&lt;/math&gt; so that &lt;math&gt;\triangle BEF&lt;/math&gt; is equilateral. What is the ratio of the area of &lt;math&gt;\triangle DEF&lt;/math&gt; to that of &lt;math&gt;\triangle ABE&lt;/math&gt;?<br /> <br /> &lt;center&gt;<br /> &lt;asy&gt;<br /> unitsize(3 cm);<br /> <br /> pair A, B, C, D, E, F;<br /> <br /> A = (0,0);<br /> B = (1,0);<br /> C = (1,1);<br /> D = (0,1);<br /> E = (0,Tan(15));<br /> F = (1 - Tan(15),1);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(B--E--F--cycle);<br /> <br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&quot;, C, NE);<br /> label(&quot;$D$&quot;, D, NW);<br /> label(&quot;$E$&quot;, E, W);<br /> label(&quot;$F$&quot;, F, N);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 1+\sqrt{3} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since triangle &lt;math&gt;BEF&lt;/math&gt; is equilateral, &lt;math&gt;EA=FC&lt;/math&gt;, and &lt;math&gt;EAB&lt;/math&gt; and &lt;math&gt;FCB&lt;/math&gt; are &lt;math&gt;SAS&lt;/math&gt; congruent. Thus, triangle &lt;math&gt;DEF&lt;/math&gt; is an isosceles right triangle. So we let &lt;math&gt;DE=x&lt;/math&gt;. Thus &lt;math&gt;EF=EB=FB=x\sqrt{2}&lt;/math&gt;. If we go angle chasing, we find out that &lt;math&gt;\angle AEB=75^{\circ}&lt;/math&gt;, thus &lt;math&gt;\angle ABE=15^{\circ}&lt;/math&gt;. &lt;math&gt;\frac{AE}{EB}=\sin{15^{\circ}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;. Thus &lt;math&gt;\frac{AE}{x\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}&lt;/math&gt;, or &lt;math&gt;AE=\frac{x(\sqrt{3}-1)}{2}&lt;/math&gt;. Thus &lt;math&gt;AB=\frac{x(\sqrt{3}+1)}{2}&lt;/math&gt;, and &lt;math&gt;[ABE]=\frac{x^2}{4}&lt;/math&gt;, and &lt;math&gt;[DEF]=\frac{x^2}{2}&lt;/math&gt;. Thus the ratio of the areas is &lt;math&gt;\boxed{\mathrm{(D)}\ 2}&lt;/math&gt;<br /> <br /> ==Solution 2 (Non-trig) ==<br /> WLOG, let the side length of &lt;math&gt;ABCD&lt;/math&gt; be 1. Let &lt;math&gt;DE = x&lt;/math&gt;. It suffices that &lt;math&gt;AE = 1 - x&lt;/math&gt;. Then triangles &lt;math&gt;ABE&lt;/math&gt; and &lt;math&gt;CBF&lt;/math&gt; are congruent by HL, so &lt;math&gt;CF = AE&lt;/math&gt; and &lt;math&gt;DE = DF&lt;/math&gt;. We find that &lt;math&gt;BE = EF = x \sqrt{2}&lt;/math&gt;, and so, by the Pythagorean Theorem, we have<br /> &lt;math&gt;(1 - x)^2 + 1 = 2x^2.&lt;/math&gt; This yields &lt;math&gt;x^2 + 2x = 2&lt;/math&gt;, so &lt;math&gt;x^2 = 2 - 2x&lt;/math&gt;. Thus, the desired ratio of areas is<br /> &lt;cmath&gt;\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{(D) }2}.&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;\angle EBF = 60^{\circ}&lt;/math&gt;, and &lt;math&gt;\angle EBA = \angle FBC&lt;/math&gt; so they must each be &lt;math&gt;15^{\circ}&lt;/math&gt;. Then let &lt;math&gt;BE=EF=FB=1&lt;/math&gt;, which gives &lt;math&gt;EA=\sin{15^{\circ}}&lt;/math&gt; and &lt;math&gt;AB=\cos{15^{\circ}}&lt;/math&gt;. <br /> The area of &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is then &lt;math&gt;\frac{1}{2}\sin{15^{\circ}}\cos{15^{\circ}}=\frac{1}{4}\sin{30^{\circ}}=\frac{1}{8}&lt;/math&gt;. <br /> &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; is an isosceles right triangle with hypotenuse 1, so &lt;math&gt;DE=DF=\frac{1}{\sqrt{2}}&lt;/math&gt; and therefore its area is &lt;math&gt;\frac{1}{2}\left(\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}\right)=\frac{1}{4}&lt;/math&gt;. <br /> The ratio of areas is then &lt;math&gt;\frac{\frac{1}{4}}{\frac{1}{8}}=\framebox{(D) 2}&lt;/math&gt;<br /> <br /> ==Solution 4(system of equations)==<br /> Assume &lt;math&gt;AB=1&lt;/math&gt;. Then, &lt;math&gt;FC&lt;/math&gt; is &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt; is &lt;math&gt;1-x&lt;/math&gt; then we see that using &lt;math&gt;HL&lt;/math&gt;, &lt;math&gt;FCB&lt;/math&gt; is congruent to EAB. Using Pythagoras of triangles &lt;math&gt;FCB&lt;/math&gt; and &lt;math&gt;FDE&lt;/math&gt; we get &lt;math&gt;2{(1-x)}^2=x^2+1&lt;/math&gt;. Expanding, we get &lt;math&gt;2x^2-4x+2=x^2+1&lt;/math&gt;. Simplifying gives &lt;math&gt;x^2-4x+1=0&lt;/math&gt; solving using completing the square (or other methods) gives 2 answers: &lt;math&gt;2-\sqrt{3}&lt;/math&gt; and &lt;math&gt;2+\sqrt{3}&lt;/math&gt;. Because &lt;math&gt;x &lt; 1&lt;/math&gt;, &lt;math&gt;x=2-\sqrt{3}&lt;/math&gt;. Using the areas, the answer is &lt;math&gt;\boxed{\text{(D) }2}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> First, since &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral and &lt;math&gt;ABCD&lt;/math&gt; is a square, by the Hypothenuse Leg Theorem, &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; is congruent to &lt;math&gt;\bigtriangleup CBF&lt;/math&gt;. Then, assume length &lt;math&gt;AB = BC = x&lt;/math&gt; and length &lt;math&gt;DE = DF = y&lt;/math&gt;, then &lt;math&gt;AE = FC = x - y&lt;/math&gt;. &lt;math&gt;\bigtriangleup BEF&lt;/math&gt; is equilateral, so &lt;math&gt;EF = EB&lt;/math&gt; and &lt;math&gt;EB^2 = EF^2&lt;/math&gt;, it is given that &lt;math&gt;ABCD&lt;/math&gt; is a square and &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; and &lt;math&gt;\bigtriangleup ABE&lt;/math&gt; are right triangles. Then we use the Pythagorean theorem to prove that &lt;math&gt;AB^2 + AE^2 = EB^2&lt;/math&gt; and since we know that &lt;math&gt;EB^2 = EF^2&lt;/math&gt; and &lt;math&gt;EF^2 = DE^2 + DF^2&lt;/math&gt;, which means &lt;math&gt;AB^2 + AE^2 = DE^2 + DF^2&lt;/math&gt;. Now we plug in the variables and the equation becomes &lt;math&gt;x^2 + (x+y)^2 = 2y^2&lt;/math&gt;, expand and simplify and you get &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;. We want the ratio of area of &lt;math&gt;\bigtriangleup DEF&lt;/math&gt; to &lt;math&gt;\bigtriangleup ABE&lt;/math&gt;. Expressed in our variables, the ratio of the area is &lt;math&gt;\frac{y^2}{x^2 - xy}&lt;/math&gt; and we know &lt;math&gt;2x^2 - 2xy = y^2&lt;/math&gt;, so the ratio must be 2. Choice D<br /> <br /> ==Video Solution==<br /> https://youtu.be/hwHIHRukYMk<br /> <br /> Education, the Study of Everything<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/BFKo9h8GhLY <br /> <br /> ~IceMatrix<br /> <br /> ==See also==<br /> {{AMC10 box|year=2004|ab=A|num-b=19|num-a=21}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_15&diff=162780 2002 AMC 12A Problems/Problem 15 2021-09-27T02:24:03Z <p>Mathgl2018: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #15]] and [[2002 AMC 10A Problems|2002 AMC 10A #21]]}}<br /> <br /> == Problem ==<br /> <br /> The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is <br /> <br /> &lt;math&gt;<br /> \text{(A) }11<br /> \qquad<br /> \text{(B) }12<br /> \qquad<br /> \text{(C) }13<br /> \qquad<br /> \text{(D) }14<br /> \qquad<br /> \text{(E) }15<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> As the unique mode is &lt;math&gt;8&lt;/math&gt;, there are at least two &lt;math&gt;8&lt;/math&gt;s.<br /> <br /> As the range is &lt;math&gt;8&lt;/math&gt; and one of the numbers is &lt;math&gt;8&lt;/math&gt;, the largest one can be at most &lt;math&gt;16&lt;/math&gt;.<br /> <br /> If the largest one is &lt;math&gt;16&lt;/math&gt;, then the smallest one is &lt;math&gt;8&lt;/math&gt;, and thus the mean is strictly larger than &lt;math&gt;8&lt;/math&gt;, which is a contradiction.<br /> <br /> If the largest one is &lt;math&gt;15&lt;/math&gt;, then the smallest one is &lt;math&gt;7&lt;/math&gt;. This means that we already know four of the values: &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;. Since the mean of all the numbers is &lt;math&gt;8&lt;/math&gt;, their sum must be &lt;math&gt;64&lt;/math&gt;. Thus the sum of the missing four numbers is &lt;math&gt;64-8-8-7-15=26&lt;/math&gt;. But if &lt;math&gt;7&lt;/math&gt; is the smallest number, then the sum of the missing numbers must be at least &lt;math&gt;4\cdot 7=28&lt;/math&gt;, which is again a contradiction.<br /> <br /> If the largest number is &lt;math&gt;14&lt;/math&gt;, we can easily find the solution &lt;math&gt;(6,6,6,8,8,8,8,14)&lt;/math&gt;. Hence, our answer is &lt;math&gt;\boxed{\text{(D)}\ 14 }&lt;/math&gt;.<br /> <br /> ===Note===<br /> <br /> The solution for &lt;math&gt;14&lt;/math&gt; is, in fact, unique. As the median must be &lt;math&gt;8&lt;/math&gt;, this means that both the &lt;math&gt;4^\text{th}&lt;/math&gt; and the &lt;math&gt;5^\text{th}&lt;/math&gt; number, when ordered by size, must be &lt;math&gt;8&lt;/math&gt;s. This gives the partial solution &lt;math&gt;(6,a,b,8,8,c,d,14)&lt;/math&gt;. For the mean to be &lt;math&gt;8&lt;/math&gt; each missing variable must be replaced by the smallest allowed value.<br /> The solution that works is &lt;math&gt;(6,6,6,8,8,8,8,14)&lt;/math&gt;<br /> <br /> ==Solution 2 ==<br /> Let the 8 numbers be &lt;math&gt;a, b, c, d, e, f, g, h&lt;/math&gt;, arranged in increasing order. Since the range of the eight numbers is 8, &lt;math&gt;h=a+8&lt;/math&gt;.<br /> <br /> I claim that &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;e&lt;/math&gt; must both be &lt;math&gt;8&lt;/math&gt;. Since the median is 8, the mean of &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; must be 8. Let's assume that &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; aren't &lt;math&gt;8&lt;/math&gt;. The mode of the collection is &lt;math&gt;8&lt;/math&gt;, and if &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; aren't, then &lt;math&gt;8&lt;/math&gt; must be between &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; (i.e. not in the collection). This is a contradiction, so &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; have to be 8.<br /> <br /> Now, we have the eight numbers are &lt;math&gt;a, b, c, 8, 8, f, g, a+8&lt;/math&gt;.<br /> <br /> Since the mean is &lt;math&gt;8&lt;/math&gt;, we have &lt;math&gt;a+b+c+8+8+f+g+a+8=8 \times 8 = 64&lt;/math&gt;, giving us &lt;math&gt;2a+b+c+f+g=40&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;f, g \ge 8&lt;/math&gt;, &lt;math&gt;f+g \ge 16&lt;/math&gt;. Plugging that in, we have &lt;math&gt;2a+b+c \le 24&lt;/math&gt;. Note that we can't do the same for &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, but we can do &lt;math&gt;b+c \ge 2a&lt;/math&gt;, giving us &lt;math&gt;4a \le 24&lt;/math&gt;, which means &lt;math&gt;a\le 6&lt;/math&gt;.<br /> We want to find &lt;math&gt;h=a+8\le 6+8=14&lt;/math&gt;<br /> This is our answer, so &lt;math&gt;\boxed{\text{(D) }14}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2002|ab=A|num-b=14|num-a=16}}<br /> {{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_15&diff=162779 2002 AMC 12A Problems/Problem 15 2021-09-27T02:23:44Z <p>Mathgl2018: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #15]] and [[2002 AMC 10A Problems|2002 AMC 10A #21]]}}<br /> <br /> == Problem ==<br /> <br /> The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is <br /> <br /> &lt;math&gt;<br /> \text{(A) }11<br /> \qquad<br /> \text{(B) }12<br /> \qquad<br /> \text{(C) }13<br /> \qquad<br /> \text{(D) }14<br /> \qquad<br /> \text{(E) }15<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> As the unique mode is &lt;math&gt;8&lt;/math&gt;, there are at least two &lt;math&gt;8&lt;/math&gt;s.<br /> <br /> As the range is &lt;math&gt;8&lt;/math&gt; and one of the numbers is &lt;math&gt;8&lt;/math&gt;, the largest one can be at most &lt;math&gt;16&lt;/math&gt;.<br /> <br /> If the largest one is &lt;math&gt;16&lt;/math&gt;, then the smallest one is &lt;math&gt;8&lt;/math&gt;, and thus the mean is strictly larger than &lt;math&gt;8&lt;/math&gt;, which is a contradiction.<br /> <br /> If the largest one is &lt;math&gt;15&lt;/math&gt;, then the smallest one is &lt;math&gt;7&lt;/math&gt;. This means that we already know four of the values: &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;. Since the mean of all the numbers is &lt;math&gt;8&lt;/math&gt;, their sum must be &lt;math&gt;64&lt;/math&gt;. Thus the sum of the missing four numbers is &lt;math&gt;64-8-8-7-15=26&lt;/math&gt;. But if &lt;math&gt;7&lt;/math&gt; is the smallest number, then the sum of the missing numbers must be at least &lt;math&gt;4\cdot 7=28&lt;/math&gt;, which is again a contradiction.<br /> <br /> If the largest number is &lt;math&gt;14&lt;/math&gt;, we can easily find the solution &lt;math&gt;(6,6,6,8,8,8,8,14)&lt;/math&gt;. Hence, our answer is &lt;math&gt;\boxed{\text{(D)}\ 14 }&lt;/math&gt;.<br /> <br /> ===Note===<br /> <br /> The solution for &lt;math&gt;14&lt;/math&gt; is, in fact, unique. As the median must be &lt;math&gt;8&lt;/math&gt;, this means that both the &lt;math&gt;4^\text{th}&lt;/math&gt; and the &lt;math&gt;5^\text{th}&lt;/math&gt; number, when ordered by size, must be &lt;math&gt;8&lt;/math&gt;s. This gives the partial solution &lt;math&gt;(6,a,b,8,8,c,d,14)&lt;/math&gt;. For the mean to be &lt;math&gt;8&lt;/math&gt; each missing variable must be replaced by the smallest allowed value.<br /> The solution that works is &lt;math&gt;(6,6,6,8,8,8,8,14)&lt;/math&gt;<br /> <br /> ==Solution 2 ==<br /> Let the 8 numbers be &lt;math&gt;a, b, c, d, e, f, g, h&lt;/math&gt;, arranged in increasing order. Since the range of the eight numbers is 8, &lt;math&gt;h=a+8&lt;/math&gt;.<br /> <br /> I claim that &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;e&lt;/math&gt; must both be &lt;math&gt;8&lt;/math&gt;. Since the median is 8, the mean of &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; must be 8. Let's assume that &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; aren't &lt;math&gt;8&lt;/math&gt;. The mode of the collection is &lt;math&gt;8&lt;/math&gt;, and if &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; aren't, then &lt;math&gt;8&lt;/math&gt; must be between &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; (i.e. not in the collection). This is a contradiction, so &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; have to be 8.<br /> <br /> Now, we have the eight numbers are &lt;math&gt;a, b, c, 8, 8, f, g, a+8&lt;/math&gt;.<br /> <br /> Since the mean is &lt;math&gt;8&lt;/math&gt;, we have &lt;math&gt;a+b+c+8+8+f+g+a+8=8 \times 8 = 64&lt;/math&gt;, giving us &lt;math&gt;2a+b+c+f+g=40&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;f, g \ge 8&lt;/math&gt;, &lt;math&gt;f+g \ge 16&lt;/math&gt;. Plugging that in, we have &lt;math&gt;2a+b+c \le 24&lt;/math&gt;. Note that we can't do the same for &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, but we can do &lt;math&gt;b+c \ge 2a&lt;/math&gt;, giving us &lt;math&gt;4a \le 24&lt;/math&gt;, which means &lt;math&gt;a\le 6&lt;/math&gt;.<br /> We want to find &lt;math&gt;h=a+8\le 6+8=14&lt;/math&gt;<br /> This is our answer, so &lt;math&gt;\boxed{\text{(D)}14}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2002|ab=A|num-b=14|num-a=16}}<br /> {{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_15&diff=162777 2002 AMC 12A Problems/Problem 15 2021-09-27T02:22:25Z <p>Mathgl2018: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #15]] and [[2002 AMC 10A Problems|2002 AMC 10A #21]]}}<br /> <br /> == Problem ==<br /> <br /> The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is <br /> <br /> &lt;math&gt;<br /> \text{(A) }11<br /> \qquad<br /> \text{(B) }12<br /> \qquad<br /> \text{(C) }13<br /> \qquad<br /> \text{(D) }14<br /> \qquad<br /> \text{(E) }15<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> As the unique mode is &lt;math&gt;8&lt;/math&gt;, there are at least two &lt;math&gt;8&lt;/math&gt;s.<br /> <br /> As the range is &lt;math&gt;8&lt;/math&gt; and one of the numbers is &lt;math&gt;8&lt;/math&gt;, the largest one can be at most &lt;math&gt;16&lt;/math&gt;.<br /> <br /> If the largest one is &lt;math&gt;16&lt;/math&gt;, then the smallest one is &lt;math&gt;8&lt;/math&gt;, and thus the mean is strictly larger than &lt;math&gt;8&lt;/math&gt;, which is a contradiction.<br /> <br /> If the largest one is &lt;math&gt;15&lt;/math&gt;, then the smallest one is &lt;math&gt;7&lt;/math&gt;. This means that we already know four of the values: &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;. Since the mean of all the numbers is &lt;math&gt;8&lt;/math&gt;, their sum must be &lt;math&gt;64&lt;/math&gt;. Thus the sum of the missing four numbers is &lt;math&gt;64-8-8-7-15=26&lt;/math&gt;. But if &lt;math&gt;7&lt;/math&gt; is the smallest number, then the sum of the missing numbers must be at least &lt;math&gt;4\cdot 7=28&lt;/math&gt;, which is again a contradiction.<br /> <br /> If the largest number is &lt;math&gt;14&lt;/math&gt;, we can easily find the solution &lt;math&gt;(6,6,6,8,8,8,8,14)&lt;/math&gt;. Hence, our answer is &lt;math&gt;\boxed{\text{(D)}\ 14 }&lt;/math&gt;.<br /> <br /> ===Note===<br /> <br /> The solution for &lt;math&gt;14&lt;/math&gt; is, in fact, unique. As the median must be &lt;math&gt;8&lt;/math&gt;, this means that both the &lt;math&gt;4^\text{th}&lt;/math&gt; and the &lt;math&gt;5^\text{th}&lt;/math&gt; number, when ordered by size, must be &lt;math&gt;8&lt;/math&gt;s. This gives the partial solution &lt;math&gt;(6,a,b,8,8,c,d,14)&lt;/math&gt;. For the mean to be &lt;math&gt;8&lt;/math&gt; each missing variable must be replaced by the smallest allowed value.<br /> The solution that works is &lt;math&gt;(6,6,6,8,8,8,8,14)&lt;/math&gt;<br /> <br /> ==Solution 2 ==<br /> Let the 8 numbers be &lt;math&gt;a, b, c, d, e, f, g, h&lt;/math&gt;, arranged in increasing order. Since the range of the eight numbers is 8, &lt;math&gt;h=a+8&lt;/math&gt;.<br /> <br /> I claim that &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;e&lt;/math&gt; must both be &lt;math&gt;8&lt;/math&gt;. Since the median is 8, the mean of &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; must be 8. Let's assume that &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; aren't &lt;math&gt;8&lt;/math&gt;. The mode of the collection is &lt;math&gt;8&lt;/math&gt;, and if &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; aren't, then &lt;math&gt;8&lt;/math&gt; must be between &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; (i.e. not in the collection). This is a contradiction, so &lt;math&gt;d&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; have to be 8.<br /> <br /> Now, we have the eight numbers are &lt;math&gt;a, b, c, 8, 8, f, g, a+8&lt;/math&gt;.<br /> <br /> Since the mean is &lt;math&gt;8&lt;/math&gt;, we have &lt;math&gt;a+b+c+8+8+f+g+a+8=8 \times 8 = 64&lt;/math&gt;, giving us &lt;math&gt;2a+b+c+f+g=40&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;f, g \ge 8&lt;/math&gt;, &lt;math&gt;f+g \ge 16&lt;/math&gt;. Plugging that in, we have &lt;math&gt;2a+b+c \le 24&lt;/math&gt;. Note that we can't do the same for &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, but we can do &lt;math&gt;b+c \ge 2a&lt;/math&gt;, giving us &lt;math&gt;4a \le 24&lt;/math&gt;, which means &lt;math&gt;a\le 6&lt;/math&gt;.<br /> We want to find &lt;math&gt;h=a+8\le 6+8=14&lt;/math&gt;<br /> This is our answer, so &lt;math&gt;\boxed{(D)14}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2002|ab=A|num-b=14|num-a=16}}<br /> {{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_23&diff=135401 2006 AMC 8 Problems/Problem 23 2020-10-19T20:08:03Z <p>Mathgl2018: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The counting numbers that leave a remainder of &lt;math&gt;4&lt;/math&gt; when divided by &lt;math&gt;6&lt;/math&gt; are<br /> &lt;math&gt;4, 10, 16, 22, 28, 34, \cdots&lt;/math&gt; The counting numbers that leave a remainder of &lt;math&gt;3&lt;/math&gt; when<br /> divided by &lt;math&gt;5&lt;/math&gt; are &lt;math&gt;3,8,13,18,23,28,33, \cdots&lt;/math&gt; So &lt;math&gt;28&lt;/math&gt; is the smallest possible number<br /> of coins that meets both conditions. Because &lt;math&gt;4\cdot 7 = 28&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left<br /> when they are divided among seven people.<br /> <br /> ===Solution 2===<br /> <br /> If there were two more coins in the box, the number of coins would be divisible<br /> by both &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. The smallest number that is divisible by &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; is &lt;math&gt;30&lt;/math&gt;, so the<br /> smallest possible number of coins in the box is &lt;math&gt;28&lt;/math&gt; and the remainder when divided by &lt;math&gt;7&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can set up a system of modular congruencies:<br /> &lt;cmath&gt;g\equiv 4 \pmod{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;g\equiv 3 \pmod{5}&lt;/cmath&gt;<br /> We can use the division algorithm to say &lt;math&gt;g=6n+4&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;6n\equiv 4 \pmod{5}&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;. If we plug the division algorithm in again, we get &lt;math&gt;n=5q+4&lt;/math&gt;. This means that &lt;math&gt;g=30q+28&lt;/math&gt;, which means that &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. From this, we can see that &lt;math&gt;28&lt;/math&gt; is our smallest possible integer satisfying &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. &lt;math&gt;28&lt;/math&gt; &lt;math&gt;\div&lt;/math&gt; &lt;math&gt;7=4&lt;/math&gt;, making our remainder &lt;math&gt;0&lt;/math&gt;. This means that there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left over when equally divided amongst &lt;math&gt;7&lt;/math&gt; people.<br /> <br /> ~Champion1234<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/g1PLxYVZE_U<br /> -Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_23&diff=135400 2006 AMC 8 Problems/Problem 23 2020-10-19T20:06:55Z <p>Mathgl2018: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The counting numbers that leave a remainder of 4 when divided by 6 are<br /> &lt;math&gt;4, 10, 16, 22, 28, 34, \cdots&lt;/math&gt; The counting numbers that leave a remainder of 3 when<br /> divided by 5 are &lt;math&gt;3,8,13,18,23,28,33, \cdots&lt;/math&gt; So 28 is the smallest possible number<br /> of coins that meets both conditions. Because &lt;math&gt;4\cdot 7 = 28&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left<br /> when they are divided among seven people.<br /> <br /> ===Solution 2===<br /> <br /> If there were two more coins in the box, the number of coins would be divisible<br /> by both &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. The smallest number that is divisible by &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; is &lt;math&gt;30&lt;/math&gt;, so the<br /> smallest possible number of coins in the box is &lt;math&gt;28&lt;/math&gt; and the remainder when divided by &lt;math&gt;7&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can set up a system of modular congruencies:<br /> &lt;cmath&gt;g\equiv 4 \pmod{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;g\equiv 3 \pmod{5}&lt;/cmath&gt;<br /> We can use the division algorithm to say &lt;math&gt;g=6n+4&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;6n\equiv 4 \pmod{5}&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;. If we plug the division algorithm in again, we get &lt;math&gt;n=5q+4&lt;/math&gt;. This means that &lt;math&gt;g=30q+28&lt;/math&gt;, which means that &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. From this, we can see that &lt;math&gt;28&lt;/math&gt; is our smallest possible integer satisfying &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. &lt;math&gt;28&lt;/math&gt; &lt;math&gt;\div&lt;/math&gt; &lt;math&gt;7=4&lt;/math&gt;, making our remainder &lt;math&gt;0&lt;/math&gt;. This means that there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left over when equally divided amongst &lt;math&gt;7&lt;/math&gt; people.<br /> <br /> ~Champion1234<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/g1PLxYVZE_U<br /> -Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_23&diff=135399 2006 AMC 8 Problems/Problem 23 2020-10-19T20:05:16Z <p>Mathgl2018: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The counting numbers that leave a remainder of 4 when divided by 6 are<br /> &lt;math&gt;4, 10, 16, 22, 28, 34, \cdots&lt;/math&gt; The counting numbers that leave a remainder of 3 when<br /> divided by 5 are &lt;math&gt;3,8,13,18,23,28,33, \cdots&lt;/math&gt; So 28 is the smallest possible number<br /> of coins that meets both conditions. Because &lt;math&gt;4\cdot 7 = 28&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left<br /> when they are divided among seven people.<br /> <br /> ===Solution 2===<br /> <br /> If there were two more coins in the box, the number of coins would be divisible<br /> by both 6 and 5. The smallest number that is divisible by 6 and 5 is &lt;math&gt;30&lt;/math&gt;, so the<br /> smallest possible number of coins in the box is &lt;math&gt;28&lt;/math&gt; and the remainder when divided by 7 is &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can set up a system of modular congruencies:<br /> &lt;cmath&gt;g\equiv 4 \pmod{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;g\equiv 3 \pmod{5}&lt;/cmath&gt;<br /> We can use the division algorithm to say &lt;math&gt;g=6n+4&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;6n\equiv 4 \pmod{5}&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;. If we plug the division algorithm in again, we get &lt;math&gt;n=5q+4&lt;/math&gt;. This means that &lt;math&gt;g=30q+28&lt;/math&gt;, which means that &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. From this, we can see that &lt;math&gt;28&lt;/math&gt; is our smallest possible integer satisfying &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. &lt;math&gt;28&lt;/math&gt; &lt;math&gt;\div&lt;/math&gt; &lt;math&gt;7=4&lt;/math&gt;, making our remainder &lt;math&gt;0&lt;/math&gt;. This means that there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left over when equally divided amongst &lt;math&gt;7&lt;/math&gt; people.<br /> <br /> ~Champion1234<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/g1PLxYVZE_U<br /> -Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_23&diff=135398 2006 AMC 8 Problems/Problem 23 2020-10-19T20:03:24Z <p>Mathgl2018: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The counting numbers that leave a remainder of 4 when divided by 6 are<br /> &lt;math&gt;4, 10, 16, 22, 28, 34, \cdots&lt;/math&gt; The counting numbers that leave a remainder of 3 when<br /> divided by 5 are &lt;math&gt;3,8,13,18,23,28,33, \cdots&lt;/math&gt; So 28 is the smallest possible number<br /> of coins that meets both conditions. Because &lt;math&gt;4\cdot 7 = 28&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left<br /> when they are divided among seven people.<br /> <br /> ===Solution 2===<br /> <br /> If there were two more coins in the box, the number of coins would be divisible<br /> by both 6 and 5. The smallest number that is divisible by 6 and 5 is &lt;math&gt;30&lt;/math&gt;, so the<br /> smallest possible number of coins in the box is &lt;math&gt;28&lt;/math&gt; and the remainder when divided by 7 is &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can set up a system of modular congruencies:<br /> &lt;cmath&gt;g\equiv 4 \pmod{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;g\equiv 3 \pmod{5}&lt;/cmath&gt;<br /> We can use the division algorithm to say g=6n+4 &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;6n\equiv 4 \pmod{5}&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;. If we plug the division algorithm in again, we get &lt;math&gt;n=5q+4&lt;/math&gt;. This means that &lt;math&gt;g=30q+28&lt;/math&gt;, which means that &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. From this, we can see that 28 is our smallest possible integer satisfying &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. 28 &lt;math&gt;\div&lt;/math&gt; 7=4, making our remainder 0. This means that there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left over when equally divided amongst 7 people.<br /> <br /> ~Champion1234<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/g1PLxYVZE_U<br /> -Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_23&diff=135397 2006 AMC 8 Problems/Problem 23 2020-10-19T20:01:59Z <p>Mathgl2018: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> The counting numbers that leave a remainder of 4 when divided by 6 are<br /> &lt;math&gt;4, 10, 16, 22, 28, 34, \cdots&lt;/math&gt; The counting numbers that leave a remainder of 3 when<br /> divided by 5 are &lt;math&gt;3,8,13,18,23,28,33, \cdots&lt;/math&gt; So 28 is the smallest possible number<br /> of coins that meets both conditions. Because &lt;math&gt;4\cdot 7 = 28&lt;/math&gt;, there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left<br /> when they are divided among seven people.<br /> <br /> ===Solution 2===<br /> <br /> If there were two more coins in the box, the number of coins would be divisible<br /> by both 6 and 5. The smallest number that is divisible by 6 and 5 is &lt;math&gt;30&lt;/math&gt;, so the<br /> smallest possible number of coins in the box is &lt;math&gt;28&lt;/math&gt; and the remainder when divided by 7 is &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> <br /> We can set up a system of modular congruencies:<br /> &lt;cmath&gt;g\equiv 4 \pmod{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;g\equiv 3 \pmod{5}&lt;/cmath&gt;<br /> We can use the division algorithm to say g=6n+4 &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;6n\equiv 4 \pmod{5}&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;. If we plug the division algorithm in again, we get &lt;math&gt;n=5q+4&lt;/math&gt;. This means that g=30q+28, which means that &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. From this, we can see that 28 is our smallest possible integer satisfying &lt;math&gt;g\equiv 28 \pmod{30}&lt;/math&gt;. 28 &lt;math&gt;\div&lt;/math&gt; 7=4, making our remainder 0. This means there are &lt;math&gt;\boxed{\textbf{(A)}\ 0}&lt;/math&gt; coins left over when equally divided amongst 7 people.<br /> <br /> ~Champion1234<br /> <br /> ==Video Solution==<br /> <br /> https://youtu.be/g1PLxYVZE_U<br /> -Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_22&diff=135396 2006 AMC 8 Problems/Problem 22 2020-10-19T19:57:39Z <p>Mathgl2018: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?<br /> <br /> &lt;asy&gt;<br /> path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> path sw=((0,0)--(1,sqrt(3)));<br /> path se=((5,0)--(4,sqrt(3)));<br /> draw(cell, linewidth(1));<br /> draw(shift(2,0)*cell, linewidth(1));<br /> draw(shift(4,0)*cell, linewidth(1));<br /> draw(shift(1,3)*cell, linewidth(1));<br /> draw(shift(3,3)*cell, linewidth(1));<br /> draw(shift(2,6)*cell, linewidth(1));<br /> draw(shift(0.45,1.125)*sw, EndArrow);<br /> draw(shift(2.45,1.125)*sw, EndArrow);<br /> draw(shift(1.45,4.125)*sw, EndArrow);<br /> draw(shift(-0.45,1.125)*se, EndArrow);<br /> draw(shift(-2.45,1.125)*se, EndArrow);<br /> draw(shift(-1.45,4.125)*se, EndArrow);<br /> label(&quot;$+$&quot;, (1.5,1.5));<br /> label(&quot;$+$&quot;, (3.5,1.5));<br /> label(&quot;$+$&quot;, (2.5,4.5));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If the lower cells contain &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, then the second row will contain<br /> &lt;math&gt;A + B&lt;/math&gt; and &lt;math&gt;B + C&lt;/math&gt;, and the top cell will contain &lt;math&gt;A + 2B + C&lt;/math&gt;. To obtain the<br /> smallest sum, place &lt;math&gt;1&lt;/math&gt; in the center cell and &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; in the outer ones. The top<br /> number will be &lt;math&gt;7&lt;/math&gt;. For the largest sum, place &lt;math&gt;9&lt;/math&gt; in the center cell and &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; in<br /> the outer ones. This top number will be &lt;math&gt;33&lt;/math&gt;. The difference is &lt;math&gt;33 - 7 = \boxed{\textbf{(D)}\ 26<br /> }&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_22&diff=135395 2006 AMC 8 Problems/Problem 22 2020-10-19T19:56:45Z <p>Mathgl2018: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?<br /> <br /> &lt;asy&gt;<br /> path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> path sw=((0,0)--(1,sqrt(3)));<br /> path se=((5,0)--(4,sqrt(3)));<br /> draw(cell, linewidth(1));<br /> draw(shift(2,0)*cell, linewidth(1));<br /> draw(shift(4,0)*cell, linewidth(1));<br /> draw(shift(1,3)*cell, linewidth(1));<br /> draw(shift(3,3)*cell, linewidth(1));<br /> draw(shift(2,6)*cell, linewidth(1));<br /> draw(shift(0.45,1.125)*sw, EndArrow);<br /> draw(shift(2.45,1.125)*sw, EndArrow);<br /> draw(shift(1.45,4.125)*sw, EndArrow);<br /> draw(shift(-0.45,1.125)*se, EndArrow);<br /> draw(shift(-2.45,1.125)*se, EndArrow);<br /> draw(shift(-1.45,4.125)*se, EndArrow);<br /> label(&quot;$+$&quot;, (1.5,1.5));<br /> label(&quot;$+$&quot;, (3.5,1.5));<br /> label(&quot;$+$&quot;, (2.5,4.5));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If the lower cells contain &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, then the second row will contain<br /> &lt;math&gt;A + B&lt;/math&gt; and &lt;math&gt;B + C&lt;/math&gt;, and the top cell will contain &lt;math&gt;A + 2B + C&lt;/math&gt;. To obtain the<br /> smallest sum, place &lt;math&gt;1&lt;/math&gt; in the center cell and &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; in the outer ones. The top<br /> number will be &lt;math&gt;7&lt;/math&gt;. For the largest sum, place &lt;math&gt;9&lt;/math&gt; in the center cell and &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; in<br /> the outer ones. This top number will be &lt;math&gt;33&lt;/math&gt;. The difference is &lt;math&gt;33 - 7 = \boxed{\textbf{(D)} 26\ <br /> }&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_22&diff=135394 2006 AMC 8 Problems/Problem 22 2020-10-19T19:56:16Z <p>Mathgl2018: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?<br /> <br /> &lt;asy&gt;<br /> path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br /> path sw=((0,0)--(1,sqrt(3)));<br /> path se=((5,0)--(4,sqrt(3)));<br /> draw(cell, linewidth(1));<br /> draw(shift(2,0)*cell, linewidth(1));<br /> draw(shift(4,0)*cell, linewidth(1));<br /> draw(shift(1,3)*cell, linewidth(1));<br /> draw(shift(3,3)*cell, linewidth(1));<br /> draw(shift(2,6)*cell, linewidth(1));<br /> draw(shift(0.45,1.125)*sw, EndArrow);<br /> draw(shift(2.45,1.125)*sw, EndArrow);<br /> draw(shift(1.45,4.125)*sw, EndArrow);<br /> draw(shift(-0.45,1.125)*se, EndArrow);<br /> draw(shift(-2.45,1.125)*se, EndArrow);<br /> draw(shift(-1.45,4.125)*se, EndArrow);<br /> label(&quot;$+$&quot;, (1.5,1.5));<br /> label(&quot;$+$&quot;, (3.5,1.5));<br /> label(&quot;$+$&quot;, (2.5,4.5));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If the lower cells contain &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, then the second row will contain<br /> &lt;math&gt;A + B&lt;/math&gt; and &lt;math&gt;B + C&lt;/math&gt;, and the top cell will contain &lt;math&gt;A + 2B + C&lt;/math&gt;. To obtain the<br /> smallest sum, place &lt;math&gt;1&lt;/math&gt; in the center cell and &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; in the outer ones. The top<br /> number will be &lt;math&gt;7&lt;/math&gt;. For the largest sum, place &lt;math&gt;9&lt;/math&gt; in the center cell and &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; in<br /> the outer ones. This top number will be &lt;math&gt;33&lt;/math&gt;. The difference is &lt;math&gt;33 - 7 = \boxed{\textbf{(D) 26}\ <br /> }&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_21&diff=135393 2006 AMC 8 Problems/Problem 21 2020-10-19T19:52:49Z <p>Mathgl2018: /* Super Hard Solution */</p> <hr /> <div>==Problem==<br /> An aquarium has a rectangular base that measures &lt;math&gt;100&lt;/math&gt; cm by &lt;math&gt;40&lt;/math&gt; cm and has a height of &lt;math&gt;50&lt;/math&gt; cm. The aquarium is filled with water to a depth of &lt;math&gt;37&lt;/math&gt; cm. A rock with volume &lt;math&gt;1000\text{cm}^3&lt;/math&gt; is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5 &lt;/math&gt;<br /> <br /> ==Super Easy Solution==<br /> The water level will rise &lt;math&gt;1&lt;/math&gt;cm for every &lt;math&gt;100 \cdot 40 = 4000\text{cm}^2&lt;/math&gt;. Since &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; of &lt;math&gt;4000&lt;/math&gt;, the water will rise &lt;math&gt;\frac{1}{4}\cdot1 = \boxed{\textbf{(A)}\ 0.25}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_21&diff=135392 2006 AMC 8 Problems/Problem 21 2020-10-19T19:52:34Z <p>Mathgl2018: /* Super Easy Solution */</p> <hr /> <div>==Problem==<br /> An aquarium has a rectangular base that measures &lt;math&gt;100&lt;/math&gt; cm by &lt;math&gt;40&lt;/math&gt; cm and has a height of &lt;math&gt;50&lt;/math&gt; cm. The aquarium is filled with water to a depth of &lt;math&gt;37&lt;/math&gt; cm. A rock with volume &lt;math&gt;1000\text{cm}^3&lt;/math&gt; is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5 &lt;/math&gt;<br /> <br /> ==Super Hard Solution==<br /> The water level will rise &lt;math&gt;1&lt;/math&gt;cm for every &lt;math&gt;100 \cdot 40 = 4000\text{cm}^2&lt;/math&gt;. Since &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;\frac{1}{4}&lt;/math&gt; of &lt;math&gt;4000&lt;/math&gt;, the water will rise &lt;math&gt;\frac{1}{4}\cdot1 = \boxed{\textbf{(A)}\ 0.25}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|n=II|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Mathgl2018 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems&diff=135045 2002 AMC 8 Problems 2020-10-14T19:28:37Z <p>Mathgl2018: /* Problem 9 */</p> <hr /> <div>==Problem 1==<br /> <br /> A [[circle]] and two distinct [[Line|lines]] are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?<br /> <br /> &lt;math&gt;\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad {(C)}\ 4 \qquad {(D)}\ 5 \qquad {(E)}\ 6&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> <br /> How many different combinations of \$5 bills and \$2 bills can be used to make a total of \$17? Order does not matter in this problem.<br /> <br /> &lt;math&gt;\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> <br /> What is the smallest possible average of four distinct positive even integers?<br /> <br /> &lt;math&gt;\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?<br /> <br /> &lt;math&gt;\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Wednesday} \qquad \text{(C)}\ \text{Friday} \qquad \text{(D)}\ \text{Saturday} \qquad \text{(E)}\ \text{Sunday}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 5 | Solution]]<br /> <br /> ==Problem 6==<br /> <br /> A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it?<br /> <br /> [[Image:2002amc8prob6graph.png|center]]<br /> <br /> &lt;math&gt;\text{(A)}\ \text{A} \qquad \text{(B)}\ \text{B} \qquad \text{(C)}\ \text{C} \qquad \text{(D)}\ \text{D} \qquad \text{(E)}\ \text{E}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 6 | Solution]]<br /> <br /> ==Problem 7==<br /> <br /> The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?<br /> <br /> &lt;asy&gt;<br /> real[] r={6, 8, 4, 2, 5};<br /> int i;<br /> for(i=0; i&lt;5; i=i+1) {<br /> filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black);<br /> }<br /> draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9));<br /> for(i=1; i&lt;8; i=i+1) {<br /> draw((0,2i)--(19,2i));<br /> }<br /> label(&quot;$0$&quot;, (0,2*0), W);<br /> label(&quot;$1$&quot;, (0,2*1), W);<br /> label(&quot;$2$&quot;, (0,2*2), W);<br /> label(&quot;$3$&quot;, (0,2*3), W);<br /> label(&quot;$4$&quot;, (0,2*4), W);<br /> label(&quot;$5$&quot;, (0,2*5), W);<br /> label(&quot;$6$&quot;, (0,2*6), W);<br /> label(&quot;$7$&quot;, (0,2*7), W);<br /> label(&quot;$8$&quot;, (0,2*8), W);<br /> label(&quot;$A$&quot;, (0*4+1.5, 0), S);<br /> label(&quot;$B$&quot;, (1*4+1.5, 0), S);<br /> label(&quot;$C$&quot;, (2*4+1.5, 0), S);<br /> label(&quot;$D$&quot;, (3*4+1.5, 0), S);<br /> label(&quot;$E$&quot;, (4*4+1.5, 0), S);<br /> label(&quot;SWEET TOOTH&quot;, (9.5,18), N);<br /> label(&quot;Kinds of candy&quot;, (9.5,-2), S);<br /> label(rotate(90)*&quot;Number of students&quot;, (-2,8), W);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 7 | Solution]]<br /> <br /> ==Juan's Old Stamping Grounds==<br /> <br /> Problems 8,9 and 10 use the data found in the accompanying paragraph and table:<br /> <br /> &lt;center&gt;<br /> Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and<br /> France, 6 cents each, Peru 4 cents each, and Spain 5 cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.)<br /> &lt;/center&gt;<br /> <br /> &lt;asy&gt;<br /> /* AMC8 2002 #8, 9, 10 Problem */<br /> size(3inch, 1.5inch);<br /> for ( int y = 0; y &amp;lt;= 5; ++y )<br /> {<br /> draw((0,y)--(18,y));<br /> }<br /> draw((0,0)--(0,5));<br /> draw((6,0)--(6,5));<br /> draw((9,0)--(9,5));<br /> draw((12,0)--(12,5));<br /> draw((15,0)--(15,5));<br /> draw((18,0)--(18,5));<br /> draw(scale(0.8)*&quot;50s&quot;, (7.5,4.5));<br /> draw(scale(0.8)*&quot;4&quot;, (7.5,3.5));<br /> draw(scale(0.8)*&quot;8&quot;, (7.5,2.5));<br /> draw(scale(0.8)*&quot;6&quot;, (7.5,1.5));<br /> draw(scale(0.8)*&quot;3&quot;, (7.5,0.5));<br /> draw(scale(0.8)*&quot;60s&quot;, (10.5,4.5));<br /> draw(scale(0.8)*&quot;7&quot;, (10.5,3.5));<br /> draw(scale(0.8)*&quot;4&quot;, (10.5,2.5));<br /> draw(scale(0.8)*&quot;4&quot;, (10.5,1.5));<br /> draw(scale(0.8)*&quot;9&quot;, (10.5,0.5));<br /> draw(scale(0.8)*&quot;70s&quot;, (13.5,4.5));<br /> draw(scale(0.8)*&quot;12&quot;, (13.5,3.5));<br /> draw(scale(0.8)*&quot;12&quot;, (13.5,2.5));<br /> draw(scale(0.8)*&quot;6&quot;, (13.5,1.5));<br /> draw(scale(0.8)*&quot;13&quot;, (13.5,0.5));<br /> draw(scale(0.8)*&quot;80s&quot;, (16.5,4.5));<br /> draw(scale(0.8)*&quot;8&quot;, (16.5,3.5));<br /> draw(scale(0.8)*&quot;15&quot;, (16.5,2.5));<br /> draw(scale(0.8)*&quot;10&quot;, (16.5,1.5));<br /> draw(scale(0.8)*&quot;9&quot;, (16.5,0.5));<br /> label(scale(0.8)*&quot;Country&quot;, (3,4.5));<br /> label(scale(0.8)*&quot;Brazil&quot;, (3,3.5));<br /> label(scale(0.8)*&quot;France&quot;, (3,2.5));<br /> label(scale(0.8)*&quot;Peru&quot;, (3,1.5));<br /> label(scale(0.8)*&quot;Spain&quot;, (3,0.5));<br /> label(scale(0.9)*&quot;Juan's Stamp Collection&quot;, (9,0), S);<br /> label(scale(0.9)*&quot;Number of Stamps by Decade&quot;, (9,5), N);&lt;/asy&gt;<br /> <br /> ===Problem 8===<br /> <br /> How many of his European stamps were issued in the '80s? <br /> <br /> &lt;math&gt;\text{(A)}\ 9 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 42&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 8 | Solution]]<br /> <br /> ===Problem 9===<br /> <br /> His South American stamps issued before the '70s did it cost him?<br /> <br /> &lt;math&gt;\text{(A)}\ \textdollar 0.40 \qquad \text{(B)}\ \textdollar 1.06 \qquad \text{(C)}\ \textdollar 1.80 \qquad \text{(D)}\ \textdollar 2.38 \qquad \text{(E)}\ \textdollar 2.64&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 9 | Solution]]<br /> <br /> ===Problem 10===<br /> <br /> The average price of his '70s stamps is closest to<br /> <br /> &lt;math&gt;\text{(A)}\ 3.5 \text{ cents} \qquad \text{(B)}\ 4 \text{ cents} \qquad \text{(C)}\ 4.5 \text{ cents} \qquad \text{(D)}\ 5 \text{ cents} \qquad \text{(E)}\ 5.4 \text{ cents}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 10 | Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?<br /> <br /> &lt;asy&gt;<br /> path p=origin--(1,0)--(1,1)--(0,1)--cycle;<br /> draw(p);<br /> draw(shift(3,0)*p);<br /> draw(shift(3,1)*p);<br /> draw(shift(4,0)*p);<br /> draw(shift(4,1)*p);<br /> draw(shift(7,0)*p);<br /> draw(shift(7,1)*p);<br /> draw(shift(7,2)*p);<br /> draw(shift(8,0)*p);<br /> draw(shift(8,1)*p);<br /> draw(shift(8,2)*p);<br /> draw(shift(9,0)*p);<br /> draw(shift(9,1)*p);<br /> draw(shift(9,2)*p);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 11 | Solution]]<br /> <br /> ==Problem 12==<br /> <br /> A board game spinner is divided into three regions labeled &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. The probability of the arrow stopping on region &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;\frac{1}{3}&lt;/math&gt; and on region &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. The probability of the arrow stopping on region &lt;math&gt;C&lt;/math&gt; is<br /> <br /> &lt;math&gt;\text{(A)}\ \frac{1}{12} \qquad \text{(B)}\ \frac{1}{6} \qquad \text{(C)}\ \frac{1}{5} \qquad \text{(D)}\ \frac{1}{3} \qquad \text{(E)}\ \frac{2}{5}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 12 | Solution]]<br /> <br /> ==Problem 13==<br /> <br /> For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?<br /> <br /> &lt;math&gt;\text{(A)}\ 250 \qquad \text{(B)}\ 500 \qquad \text{(C)}\ 625 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1000&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 13 | Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A merchant offers a large group of items at 30% off. Later, the merchant takes 20% off these sale prices and claims that the final price of these items is 50% off the original price. The total discount is<br /> <br /> &lt;math&gt;\text{(A)}\ 35\% \qquad \text{(B)}\ 44\% \qquad \text{(C)}\ 50\% \qquad \text{(D)}\ 56\% \qquad \text{(E)}\ 60\%&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 14 | Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Which of the following polygons has the largest area?<br /> <br /> &lt;asy&gt;<br /> size(330);<br /> int i,j,k;<br /> for(i=0;i&lt;5; i=i+1) {<br /> for(j=0;j&lt;5;j=j+1) {<br /> for(k=0;k&lt;5;k=k+1) {<br /> dot((6i+j, k));<br /> }}}<br /> draw((0,0)--(4,0)--(3,1)--(3,3)--(2,3)--(2,1)--(1,1)--cycle);<br /> draw(shift(6,0)*((0,0)--(4,0)--(4,1)--(3,1)--(3,2)--(2,1)--(1,1)--(0,2)--cycle));<br /> draw(shift(12,0)*((0,1)--(1,0)--(3,2)--(3,3)--(1,1)--(1,3)--(0,4)--cycle));<br /> draw(shift(18,0)*((0,1)--(2,1)--(3,0)--(3,3)--(2,2)--(1,3)--(1,2)--(0,2)--cycle));<br /> draw(shift(24,0)*((1,0)--(2,1)--(2,3)--(3,2)--(3,4)--(0,4)--(1,3)--cycle));<br /> label(&quot;$A$&quot;, (0*6+2, 0), S);<br /> label(&quot;$B$&quot;, (1*6+2, 0), S);<br /> label(&quot;$C$&quot;, (2*6+2, 0), S);<br /> label(&quot;$D$&quot;, (3*6+2, 0), S);<br /> label(&quot;$E$&quot;, (4*6+2, 0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)} \text{A} \qquad \text{(B)}\ \text{B} \qquad \text{(C)}\ \text{C} \qquad \text{(D)}\ \text{D} \qquad \text{(E)}\ \text{E}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 15 | Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?<br /> <br /> &lt;asy&gt;/* AMC8 2002 #16 Problem */<br /> draw((0,0)--(4,0)--(4,3)--cycle);<br /> draw((4,3)--(-4,4)--(0,0));<br /> draw((-0.15,0.1)--(0,0.25)--(.15,0.1));<br /> draw((0,0)--(4,-4)--(4,0));<br /> draw((4,0.2)--(3.8,0.2)--(3.8,-0.2)--(4,-0.2));<br /> draw((4,0)--(7,3)--(4,3));<br /> draw((4,2.8)--(4.2,2.8)--(4.2,3));<br /> label(scale(0.8)*&quot;$Z$&quot;, (0, 3), S);<br /> label(scale(0.8)*&quot;$Y$&quot;, (3,-2));<br /> label(scale(0.8)*&quot;$X$&quot;, (5.5, 2.5));<br /> label(scale(0.8)*&quot;$W$&quot;, (2.6,1));<br /> label(scale(0.65)*&quot;5&quot;, (2,2));<br /> label(scale(0.65)*&quot;4&quot;, (2.3,-0.4));<br /> label(scale(0.65)*&quot;3&quot;, (4.3,1.5));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ X + Z = W + Y \qquad \text{(B)}\ W + X = Z \qquad \text{(C)}\ 3X + 4Y = 5Z&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(D)}\ X +W = \frac{1}{2} (Y + Z) \qquad \text{(E)}\ X + Y = Z&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 16 | Solution]]<br /> <br /> ==Problem 17==<br /> <br /> In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 17 | Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each day for 3 days. How long would he have to skate the ninth day in order to average 85 minutes of skating each day for the entire time?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{1 hr} \qquad \text{(B)}\ \text{1 hr 10 min} \qquad \text{(C)}\ \text{1 hr 20 min} \qquad \text{(D)}\ \text{1 hr 40 min} \qquad \text{(E)}\ \text{2 hr}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 18 | Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many whole numbers between 99 and 999 contain exactly one 0?<br /> <br /> &lt;math&gt;\text{(A)}\ 72 \qquad \text{(B)}\ 90 \qquad \text{(C)}\ 144 \qquad \text{(D)}\ 162 \qquad \text{(E)}\ 180&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 19 | Solution]]<br /> <br /> ==Problem 20==<br /> <br /> The area of triangle &lt;math&gt;XYZ&lt;/math&gt; is 8 square inches. Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are midpoints of congruent segments &lt;math&gt;\overline{XY}&lt;/math&gt; and &lt;math&gt;\overline{XZ}&lt;/math&gt;. Altitude &lt;math&gt;\overline{XC}&lt;/math&gt; bisects &lt;math&gt;\overline{YZ}&lt;/math&gt;. The area (in square inches) of the shaded region is<br /> <br /> &lt;asy&gt;/* AMC8 2002 #20 Problem */<br /> fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey);<br /> draw((0,0)--(10,0)--(5,4)--cycle);<br /> draw((2.5,2)--(7.5,2));<br /> draw((5,4)--(5,0));<br /> label(scale(0.8)*&quot;$X$&quot;, (5,4), N);<br /> label(scale(0.8)*&quot;$Y$&quot;, (0,0), W);<br /> label(scale(0.8)*&quot;$Z$&quot;, (10,0), E);<br /> label(scale(0.8)*&quot;$A$&quot;, (2.5,2.2), W);<br /> label(scale(0.8)*&quot;$B$&quot;, (7.5,2.2), E);<br /> label(scale(0.8)*&quot;$C\$&quot;, (5,0), S);<br /> fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 1\frac{1}{2} \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 2\frac{1}{2} \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 3\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 20 | Solution]]<br /> <br /> ==Problem 21==<br /> <br /> Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is<br /> <br /> &lt;math&gt;\text{(A)}\ \frac{5}{16} \qquad \text{(B)}\ \frac{3}{8} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{5}{8} \qquad \text{(E)}\ \frac{11}{16}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 21 | Solution]]<br /> <br /> ==Problem 22==<br /> <br /> Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom and sides.<br /> <br /> &lt;asy&gt;/* AMC8 2002 #22 Problem */<br /> draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br /> draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1));<br /> draw((1,0)--(1.5,0.5)--(1.5,1.5));<br /> draw((0.5,1.5)--(1,2)--(1.5,2));<br /> draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5));<br /> draw((1.5,3.5)--(2.5,3.5));<br /> draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5));<br /> draw((3,4)--(3,3)--(2.5,2.5));<br /> draw((3,3)--(4,3)--(4,2)--(3.5,1.5));<br /> draw((4,3)--(3.5,2.5));<br /> draw((2.5,.5)--(3,1)--(3,1.5));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ 18 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 22 | Solution]]<br /> <br /> ==Problem 23==<br /> <br /> A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of<br /> darker tiles?<br /> <br /> &lt;asy&gt;/* AMC8 2002 #23 Problem */<br /> fill((0,2)--(1,3)--(2,3)--(2,4)--(3,5)--(4,4)--(4,3)--(5,3)--(6,2)--(5,1)--(4,1)--(4,0)--(2,0)--(2,1)--(1,1)--cycle, mediumgrey);<br /> fill((7,1)--(6,2)--(7,3)--(8,3)--(8,4)--(9,5)--(10,4)--(7,0)--cycle, mediumgrey);<br /> fill((3,5)--(2,6)--(2,7)--(1,7)--(0,8)--(1,9)--(2,9)--(2,10)--(3,11)--(4,10)--(4,9)--(5,9)--(6,8)--(5,7)--(4,7)--(4,6)--cycle, mediumgrey);<br /> fill((6,8)--(7,9)--(8,9)--(8,10)--(9,11)--(10,10)--(10,9)--(11,9)--(11,7)--(10,7)--(10,6)--(9,5)--(8,6)--(8,7)--(7,7)--cycle, mediumgrey);<br /> <br /> draw((0,0)--(0,11)--(11,11));<br /> for ( int x = 1; x &amp;lt; 11; ++x )<br /> {<br /> draw((x,11)--(x,0), linetype(&quot;4 4&quot;));<br /> }<br /> <br /> for ( int y = 1; y &amp;lt; 11; ++y )<br /> {<br /> draw((0,y)--(11,y), linetype(&quot;4 4&quot;));<br /> }<br /> clip((0,0)--(0,11)--(11,11)--(11,5)--(4,1)--cycle);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{4}{9} \qquad \text{(C)}\ \frac{1}{2} \qquad \text{(D)}\ \frac{5}{9} \qquad \text{(E)}\ \frac{5}{8}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 23 | Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?<br /> <br /> &lt;math&gt;\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 70&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 24 | Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?<br /> <br /> &lt;math&gt;\text{(A)}\ \frac{1}{10} \qquad \text{(B)}\ \frac{1}{4} \qquad \text{(C)}\ \frac{1}{3} \qquad \text{(D)}\ \frac{2}{5} \qquad \text{(E)}\ \frac{1}{2}&lt;/math&gt;<br /> <br /> [[2002 AMC 8 Problems/Problem 25 | Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2002|before=[[2001 AMC 8 Problems|2001 AMC 8]]|after=[[2003 AMC 8 Problems|2003 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathgl2018