https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Snow52&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-27T07:22:21Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_25&diff=133553 2015 AMC 10B Problems/Problem 25 2020-09-12T21:41:37Z <p>Snow52: /* Solution 3 (Basically the exact same as Solution 1) */</p> <hr /> <div>==Problem==<br /> A rectangular box measures &lt;math&gt;a \times b \times c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers and &lt;math&gt;1\leq a \leq b \leq c&lt;/math&gt;. The volume and the surface area of the box are numerically equal. How many ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, the volume is &lt;math&gt;abc&lt;/math&gt;, so &lt;math&gt;2(ab+bc+ca)=abc&lt;/math&gt;.<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt;, we have: &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> <br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\ge 3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c\ge b\ge a&gt;0&lt;/math&gt;, we have &lt;math&gt;\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}&lt;/math&gt;.<br /> Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a\le 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;.<br /> <br /> From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b\le \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b\le \frac{2}{k}&lt;/math&gt;<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, \cdots, 12&lt;/math&gt;. The solutions we find are &lt;math&gt;(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. The solutions we find are &lt;math&gt;(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{(B)}\;10}&lt;/math&gt;<br /> <br /> ==Simplification of Solution 1==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, the volume is &lt;math&gt;abc&lt;/math&gt;, so &lt;math&gt;2(ab+bc+ca)=abc&lt;/math&gt;.<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt;, we have: &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\ge 3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c\ge b\ge a&gt;0&lt;/math&gt;, we have &lt;math&gt;\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}&lt;/math&gt;. Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a\le 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> <br /> We can say &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{q}&lt;/math&gt;, where &lt;math&gt;\frac{1}{q} = \frac{1}{2}-\frac{1}{a}&lt;/math&gt;.<br /> <br /> Notice &lt;math&gt;immediately&lt;/math&gt; that &lt;math&gt;b, c &gt; q&lt;/math&gt; This is our key step.<br /> Then we can say &lt;math&gt;b=q+d&lt;/math&gt;, &lt;math&gt;c=q+e&lt;/math&gt;. If we clear the fraction about b and c (do the math), our immediate result is that &lt;math&gt;de = q^2&lt;/math&gt;. Realize also that &lt;math&gt;d \leq e&lt;/math&gt;. <br /> <br /> Now go through cases for &lt;math&gt;a&lt;/math&gt; and you end up with the same result. However, now you don't have to guess solutions. For example, when &lt;math&gt;a=3&lt;/math&gt;, then &lt;math&gt;de = 36&lt;/math&gt; and &lt;math&gt;d=1, 2, 3, 4, 6&lt;/math&gt;.<br /> <br /> - minor edit by Williamgolly, minor edit by Tiblis<br /> <br /> ==Solution 2==<br /> We need&lt;cmath&gt;abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).&lt;/cmath&gt;Since &lt;math&gt;ab, ac \le bc&lt;/math&gt;, we get &lt;math&gt;abc \le 6bc&lt;/math&gt;. Thus &lt;math&gt;a\le 6&lt;/math&gt;. From the second equation we see that &lt;math&gt;a &gt; 2&lt;/math&gt;. Thus &lt;math&gt;a\in \{3, 4, 5, 6\}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a=3&lt;/math&gt; we need &lt;math&gt;bc = 6(b+c) \Rightarrow (b-6)(c-6)=36&lt;/math&gt;. We get five roots &lt;math&gt;\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.&lt;/math&gt;<br /> If &lt;math&gt;a=4&lt;/math&gt; we need &lt;math&gt;bc = 4(b+c) \Rightarrow (b-4)(c-4)=16&lt;/math&gt;. We get three roots &lt;math&gt;\{(4,5,20), (4,6,12), (4,8,8)\}&lt;/math&gt;.<br /> If &lt;math&gt;a=5&lt;/math&gt; we need &lt;math&gt;3bc = 10(b+c)&lt;/math&gt;, which is the same as &lt;math&gt;9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100&lt;/math&gt;. We get only one root (corresponding to &lt;math&gt;100=5\cdot 20&lt;/math&gt;) &lt;math&gt;(5,5,10)&lt;/math&gt;.<br /> If &lt;math&gt;a=6&lt;/math&gt; we need &lt;math&gt;4bc = 12(b+c)&lt;/math&gt;. Then &lt;math&gt;(b-3)(c-3)=9&lt;/math&gt;. We get one root &lt;math&gt;(6,6,6)&lt;/math&gt;.<br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==Solution 3 (Basically the exact same as Solution 1)==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, and the volume is &lt;math&gt;abc&lt;/math&gt;, so equating the two yields<br /> <br /> &lt;cmath&gt;2(ab+bc+ca)=abc.&lt;/cmath&gt;<br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt; to obtain&lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\geqslant3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c \geq b \geq a &gt; 0&lt;/math&gt;, hence &lt;math&gt;\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}&lt;/math&gt;. Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a \leq 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;. From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b \leq \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b \leq \frac{2}{k}&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, 9, 10, 11, 12&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(3, 7, 42)&lt;/math&gt;, &lt;math&gt;(3, 8, 24)&lt;/math&gt;, &lt;math&gt;(3, 9, 18)&lt;/math&gt;, &lt;math&gt;(3, 10, 15)&lt;/math&gt;, &lt;math&gt;(3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(4, 5, 20)&lt;/math&gt;, &lt;math&gt;(4, 6, 12)&lt;/math&gt;, &lt;math&gt;(4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> Minor Edit by Snow52<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Snow52 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_25&diff=133552 2015 AMC 10B Problems/Problem 25 2020-09-12T21:40:32Z <p>Snow52: /* Solution 3 (Basically the exact same as Solution 1) */</p> <hr /> <div>==Problem==<br /> A rectangular box measures &lt;math&gt;a \times b \times c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers and &lt;math&gt;1\leq a \leq b \leq c&lt;/math&gt;. The volume and the surface area of the box are numerically equal. How many ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, the volume is &lt;math&gt;abc&lt;/math&gt;, so &lt;math&gt;2(ab+bc+ca)=abc&lt;/math&gt;.<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt;, we have: &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> <br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\ge 3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c\ge b\ge a&gt;0&lt;/math&gt;, we have &lt;math&gt;\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}&lt;/math&gt;.<br /> Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a\le 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;.<br /> <br /> From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b\le \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b\le \frac{2}{k}&lt;/math&gt;<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, \cdots, 12&lt;/math&gt;. The solutions we find are &lt;math&gt;(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. The solutions we find are &lt;math&gt;(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{(B)}\;10}&lt;/math&gt;<br /> <br /> ==Simplification of Solution 1==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, the volume is &lt;math&gt;abc&lt;/math&gt;, so &lt;math&gt;2(ab+bc+ca)=abc&lt;/math&gt;.<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt;, we have: &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\ge 3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c\ge b\ge a&gt;0&lt;/math&gt;, we have &lt;math&gt;\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}&lt;/math&gt;. Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a\le 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> <br /> We can say &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{q}&lt;/math&gt;, where &lt;math&gt;\frac{1}{q} = \frac{1}{2}-\frac{1}{a}&lt;/math&gt;.<br /> <br /> Notice &lt;math&gt;immediately&lt;/math&gt; that &lt;math&gt;b, c &gt; q&lt;/math&gt; This is our key step.<br /> Then we can say &lt;math&gt;b=q+d&lt;/math&gt;, &lt;math&gt;c=q+e&lt;/math&gt;. If we clear the fraction about b and c (do the math), our immediate result is that &lt;math&gt;de = q^2&lt;/math&gt;. Realize also that &lt;math&gt;d \leq e&lt;/math&gt;. <br /> <br /> Now go through cases for &lt;math&gt;a&lt;/math&gt; and you end up with the same result. However, now you don't have to guess solutions. For example, when &lt;math&gt;a=3&lt;/math&gt;, then &lt;math&gt;de = 36&lt;/math&gt; and &lt;math&gt;d=1, 2, 3, 4, 6&lt;/math&gt;.<br /> <br /> - minor edit by Williamgolly, minor edit by Tiblis<br /> <br /> ==Solution 2==<br /> We need&lt;cmath&gt;abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).&lt;/cmath&gt;Since &lt;math&gt;ab, ac \le bc&lt;/math&gt;, we get &lt;math&gt;abc \le 6bc&lt;/math&gt;. Thus &lt;math&gt;a\le 6&lt;/math&gt;. From the second equation we see that &lt;math&gt;a &gt; 2&lt;/math&gt;. Thus &lt;math&gt;a\in \{3, 4, 5, 6\}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a=3&lt;/math&gt; we need &lt;math&gt;bc = 6(b+c) \Rightarrow (b-6)(c-6)=36&lt;/math&gt;. We get five roots &lt;math&gt;\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.&lt;/math&gt;<br /> If &lt;math&gt;a=4&lt;/math&gt; we need &lt;math&gt;bc = 4(b+c) \Rightarrow (b-4)(c-4)=16&lt;/math&gt;. We get three roots &lt;math&gt;\{(4,5,20), (4,6,12), (4,8,8)\}&lt;/math&gt;.<br /> If &lt;math&gt;a=5&lt;/math&gt; we need &lt;math&gt;3bc = 10(b+c)&lt;/math&gt;, which is the same as &lt;math&gt;9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100&lt;/math&gt;. We get only one root (corresponding to &lt;math&gt;100=5\cdot 20&lt;/math&gt;) &lt;math&gt;(5,5,10)&lt;/math&gt;.<br /> If &lt;math&gt;a=6&lt;/math&gt; we need &lt;math&gt;4bc = 12(b+c)&lt;/math&gt;. Then &lt;math&gt;(b-3)(c-3)=9&lt;/math&gt;. We get one root &lt;math&gt;(6,6,6)&lt;/math&gt;.<br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==Solution 3 (Basically the exact same as Solution 1)==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, and the volume is &lt;math&gt;abc&lt;/math&gt;, so equating the two yields<br /> <br /> &lt;cmath&gt;2(ab+bc+ca)=abc.&lt;/cmath&gt;<br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt; to obtain&lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\geqslant3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c \geq b \geq a &gt; 0&lt;/math&gt;, hence &lt;math&gt;\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}&lt;/math&gt;. Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a \leq 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;. From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b \leq \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b \leq \frac{2}{k}&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, 9, 10, 11, 12&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(3, 7, 42)&lt;/math&gt;, &lt;math&gt;(3, 8, 24)&lt;/math&gt;, &lt;math&gt;(3, 9, 18)&lt;/math&gt;, &lt;math&gt;(3, 10, 15)&lt;/math&gt;, &lt;math&gt;(3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(4, 5, 20)&lt;/math&gt;, &lt;math&gt;(4, 6, 12)&lt;/math&gt;, &lt;math&gt;(4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> Minor edit by Snow52<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Snow52 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_25&diff=133551 2015 AMC 10B Problems/Problem 25 2020-09-12T21:39:59Z <p>Snow52: /* Solution 3 (Practically the exact same as Solution 1) */</p> <hr /> <div>==Problem==<br /> A rectangular box measures &lt;math&gt;a \times b \times c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers and &lt;math&gt;1\leq a \leq b \leq c&lt;/math&gt;. The volume and the surface area of the box are numerically equal. How many ordered triples &lt;math&gt;(a,b,c)&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, the volume is &lt;math&gt;abc&lt;/math&gt;, so &lt;math&gt;2(ab+bc+ca)=abc&lt;/math&gt;.<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt;, we have: &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> <br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\ge 3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c\ge b\ge a&gt;0&lt;/math&gt;, we have &lt;math&gt;\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}&lt;/math&gt;.<br /> Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a\le 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;.<br /> <br /> From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b\le \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b\le \frac{2}{k}&lt;/math&gt;<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, \cdots, 12&lt;/math&gt;. The solutions we find are &lt;math&gt;(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. The solutions we find are &lt;math&gt;(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{(B)}\;10}&lt;/math&gt;<br /> <br /> ==Simplification of Solution 1==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, the volume is &lt;math&gt;abc&lt;/math&gt;, so &lt;math&gt;2(ab+bc+ca)=abc&lt;/math&gt;.<br /> <br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt;, we have: &lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\ge 3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c\ge b\ge a&gt;0&lt;/math&gt;, we have &lt;math&gt;\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}&lt;/math&gt;. Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a\le 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> <br /> We can say &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{q}&lt;/math&gt;, where &lt;math&gt;\frac{1}{q} = \frac{1}{2}-\frac{1}{a}&lt;/math&gt;.<br /> <br /> Notice &lt;math&gt;immediately&lt;/math&gt; that &lt;math&gt;b, c &gt; q&lt;/math&gt; This is our key step.<br /> Then we can say &lt;math&gt;b=q+d&lt;/math&gt;, &lt;math&gt;c=q+e&lt;/math&gt;. If we clear the fraction about b and c (do the math), our immediate result is that &lt;math&gt;de = q^2&lt;/math&gt;. Realize also that &lt;math&gt;d \leq e&lt;/math&gt;. <br /> <br /> Now go through cases for &lt;math&gt;a&lt;/math&gt; and you end up with the same result. However, now you don't have to guess solutions. For example, when &lt;math&gt;a=3&lt;/math&gt;, then &lt;math&gt;de = 36&lt;/math&gt; and &lt;math&gt;d=1, 2, 3, 4, 6&lt;/math&gt;.<br /> <br /> - minor edit by Williamgolly, minor edit by Tiblis<br /> <br /> ==Solution 2==<br /> We need&lt;cmath&gt;abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).&lt;/cmath&gt;Since &lt;math&gt;ab, ac \le bc&lt;/math&gt;, we get &lt;math&gt;abc \le 6bc&lt;/math&gt;. Thus &lt;math&gt;a\le 6&lt;/math&gt;. From the second equation we see that &lt;math&gt;a &gt; 2&lt;/math&gt;. Thus &lt;math&gt;a\in \{3, 4, 5, 6\}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a=3&lt;/math&gt; we need &lt;math&gt;bc = 6(b+c) \Rightarrow (b-6)(c-6)=36&lt;/math&gt;. We get five roots &lt;math&gt;\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.&lt;/math&gt;<br /> If &lt;math&gt;a=4&lt;/math&gt; we need &lt;math&gt;bc = 4(b+c) \Rightarrow (b-4)(c-4)=16&lt;/math&gt;. We get three roots &lt;math&gt;\{(4,5,20), (4,6,12), (4,8,8)\}&lt;/math&gt;.<br /> If &lt;math&gt;a=5&lt;/math&gt; we need &lt;math&gt;3bc = 10(b+c)&lt;/math&gt;, which is the same as &lt;math&gt;9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100&lt;/math&gt;. We get only one root (corresponding to &lt;math&gt;100=5\cdot 20&lt;/math&gt;) &lt;math&gt;(5,5,10)&lt;/math&gt;.<br /> If &lt;math&gt;a=6&lt;/math&gt; we need &lt;math&gt;4bc = 12(b+c)&lt;/math&gt;. Then &lt;math&gt;(b-3)(c-3)=9&lt;/math&gt;. We get one root &lt;math&gt;(6,6,6)&lt;/math&gt;.<br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==Solution 3 (Basically the exact same as Solution 1)==<br /> The surface area is &lt;math&gt;2(ab+bc+ca)&lt;/math&gt;, and the volume is &lt;math&gt;abc&lt;/math&gt;, so equating the two yields<br /> <br /> &lt;cmath&gt;2(ab+bc+ca)=abc.&lt;/cmath&gt;<br /> Divide both sides by &lt;math&gt;2abc&lt;/math&gt; to obtain&lt;cmath&gt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.&lt;/cmath&gt;<br /> First consider the bound of the variable &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\frac{1}{a}&lt;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},&lt;/math&gt; we have &lt;math&gt;a&gt;2&lt;/math&gt;, or &lt;math&gt;a\geqslant3&lt;/math&gt;.<br /> <br /> Also note that &lt;math&gt;c \geq b \geq a &gt; 0&lt;/math&gt;, hence &lt;math&gt;\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}&lt;/math&gt;. Thus, &lt;math&gt;\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}&lt;/math&gt;, so &lt;math&gt;a \leq 6&lt;/math&gt;.<br /> <br /> So we have &lt;math&gt;a=3, 4, 5&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;.<br /> <br /> Before the casework, let's consider the possible range for &lt;math&gt;b&lt;/math&gt; if &lt;math&gt;\frac{1}{b}+\frac{1}{c}=k&gt;0&lt;/math&gt;. From &lt;math&gt;\frac{1}{b}&lt;k&lt;/math&gt;, we have &lt;math&gt;b&gt;\frac{1}{k}&lt;/math&gt;. From &lt;math&gt;\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k&lt;/math&gt;, we have &lt;math&gt;b \leq \frac{2}{k}&lt;/math&gt;. Thus &lt;math&gt;\frac{1}{k}&lt;b \leq \frac{2}{k}&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=3&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{6}&lt;/math&gt;, so &lt;math&gt;b=7, 8, 9, 10, 11, 12&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(3, 7, 42)&lt;/math&gt;, &lt;math&gt;(3, 8, 24)&lt;/math&gt;, &lt;math&gt;(3, 9, 18)&lt;/math&gt;, &lt;math&gt;(3, 10, 15)&lt;/math&gt;, &lt;math&gt;(3, 12, 12)&lt;/math&gt;, for a total of &lt;math&gt;5&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=4&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{1}{4}&lt;/math&gt;, so &lt;math&gt;b=5, 6, 7, 8&lt;/math&gt;. We find the solutions &lt;math&gt;(a, b, c)=(4, 5, 20)&lt;/math&gt;, &lt;math&gt;(4, 6, 12)&lt;/math&gt;, &lt;math&gt;(4, 8, 8)&lt;/math&gt;, for a total of &lt;math&gt;3&lt;/math&gt; solutions.<br /> <br /> When &lt;math&gt;a=5&lt;/math&gt;, we get &lt;math&gt;\frac{1}{b}+\frac{1}{c}=\frac{3}{10}&lt;/math&gt;, so &lt;math&gt;b=5, 6&lt;/math&gt;. The only solution in this case is &lt;math&gt;(a, b, c)=(5, 5, 10)&lt;/math&gt;.<br /> <br /> When &lt;math&gt;a=6&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; is forced to be &lt;math&gt;6&lt;/math&gt;, and thus &lt;math&gt;(a, b, c)=(6, 6, 6)&lt;/math&gt;.<br /> <br /> Thus, there are &lt;math&gt;5+3+1+1 = \boxed{\textbf{(B)}\; 10}&lt;/math&gt; solutions.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Geometry Problems]]</div> Snow52 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Answer_Key&diff=112252 2014 AMC 10A Answer Key 2019-11-28T02:02:09Z <p>Snow52: </p> <hr /> <div>#C<br /> #C<br /> #E<br /> #B<br /> #C<br /> #A<br /> #B<br /> #D<br /> #C<br /> #B<br /> #C<br /> #C<br /> #C<br /> #D<br /> #C<br /> #E<br /> #D<br /> #B<br /> #A<br /> #D<br /> #E<br /> #E<br /> #C<br /> #A<br /> #B<br /> <br /> Get Pranked I was able to edit this. If Admins want to know how, pm me and I will explain ~Firebolt360<br /> PPS: I did not change any of the answers please don't ban me. All you have to do is press the &quot;Edit&quot; button in the sidebar. You are supposed to be able to edit it, I think.</div> Snow52 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems&diff=110765 2017 AMC 8 Problems 2019-11-02T17:08:44Z <p>Snow52: /* Problem 16 */</p> <hr /> <div>==Problem 1==<br /> Which of the following values is largest?<br /> <br /> &lt;math&gt;\textbf{(A) }2+0+1+7\qquad\textbf{(B) }2 \times 0 +1+7\qquad\textbf{(C) }2+0 \times 1 + 7\qquad\textbf{(D) }2+0+1 \times 7\qquad\textbf{(E) }2 \times 0 \times 1 \times 7&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 1|Solution<br /> ]]<br /> <br /> ==Problem 2==<br /> <br /> Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received 36 votes, then how many votes were cast all together?<br /> <br /> &lt;asy&gt;<br /> draw((-1,0)--(0,0)--(0,1));<br /> draw((0,0)--(0.309, -0.951));<br /> filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray);<br /> filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray);<br /> draw(arc((0,0), (-1,0), (0.309, -0.951)));<br /> label(&quot;Colby&quot;, (-0.5, 0.5));<br /> label(&quot;25\%&quot;, (-0.5, 0.3));<br /> label(&quot;Alicia&quot;, (0.7, 0.2));<br /> label(&quot;45\%&quot;, (0.7, 0));<br /> label(&quot;Brenda&quot;, (-0.5, -0.4));<br /> label(&quot;30\%&quot;, (-0.5, -0.6));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }100\qquad\textbf{(D) }106\qquad\textbf{(E) }120&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 2|Solution<br /> ]]<br /> <br /> ==Problem 3==<br /> <br /> What is the value of the expression &lt;math&gt;\sqrt{16\sqrt{8\sqrt{4}}}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 3|Solution<br /> ]]<br /> <br /> ==Problem 4==<br /> <br /> When 0.000315 is multiplied by 7,928,564 the product is closest to which of the following?<br /> <br /> &lt;math&gt;\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 4|Solution<br /> ]]<br /> <br /> ==Problem 5==<br /> <br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 5|Solution<br /> ]]<br /> <br /> ==Problem 6==<br /> <br /> If the degree measures of the angles of a triangle are in the ratio &lt;math&gt;3:3:4&lt;/math&gt;, what is the degree measure of the largest angle of the triangle?<br /> <br /> &lt;math&gt;\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 6|Solution<br /> ]]<br /> <br /> ==Problem 7==<br /> <br /> Let &lt;math&gt;Z&lt;/math&gt; be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of &lt;math&gt;Z&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 7|Solution<br /> ]]<br /> <br /> ==Problem 8==<br /> <br /> Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, &quot;My house number has two digits, and exactly three of the following four statements about it are true.&quot;<br /> <br /> (1) It is prime.<br /> <br /> (2) It is even.<br /> <br /> (3) It is divisible by 7.<br /> <br /> (4) One of its digits is 9.<br /> <br /> This information allows Malcolm to determine Isabella's house number. What is its units digit?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 8|Solution<br /> ]]<br /> <br /> ==Problem 9==<br /> <br /> All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles?<br /> <br /> &lt;math&gt;\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 9|Solution<br /> ]]<br /> <br /> ==Problem 10==<br /> <br /> A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 10|Solution<br /> ]]<br /> <br /> ==Problem 11==<br /> <br /> A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?<br /> <br /> &lt;math&gt;\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 11|Solution<br /> ]]<br /> <br /> ==Problem 12==<br /> <br /> The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }2\text{ and }19\qquad\textbf{(B) }20\text{ and }39\qquad\textbf{(C) }40\text{ and }59\qquad\textbf{(D) }60\text{ and }79\qquad\textbf{(E) }80\text{ and }124&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 12|Solution<br /> ]]<br /> <br /> ==Problem 13==<br /> <br /> Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 13|Solution<br /> ]]<br /> <br /> ==Problem 14==<br /> <br /> Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only &lt;math&gt;80\%&lt;/math&gt; of the problems she solved alone, but overall &lt;math&gt;88\%&lt;/math&gt; of her answers were correct. Zoe had correct answers to &lt;math&gt;90\%&lt;/math&gt; of the problems she solved alone. What was Zoe's overall percentage of correct answers?<br /> <br /> &lt;math&gt;\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 14|Solution<br /> ]]<br /> <br /> ==Problem 15==<br /> <br /> In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.<br /> &lt;asy&gt;<br /> fill((0.5, 4.5)--(1.5,4.5)--(1.5,2.5)--(0.5,2.5)--cycle,lightgray);<br /> fill((1.5,3.5)--(2.5,3.5)--(2.5,1.5)--(1.5,1.5)--cycle,lightgray);<br /> label(&quot;$8$&quot;, (1, 0));<br /> label(&quot;$C$&quot;, (2, 0));<br /> label(&quot;$8$&quot;, (3, 0));<br /> label(&quot;$8$&quot;, (0, 1));<br /> label(&quot;$C$&quot;, (1, 1));<br /> label(&quot;$M$&quot;, (2, 1));<br /> label(&quot;$C$&quot;, (3, 1));<br /> label(&quot;$8$&quot;, (4, 1));<br /> label(&quot;$C$&quot;, (0, 2));<br /> label(&quot;$M$&quot;, (1, 2));<br /> label(&quot;$A$&quot;, (2, 2));<br /> label(&quot;$M$&quot;, (3, 2));<br /> label(&quot;$C$&quot;, (4, 2));<br /> label(&quot;$8$&quot;, (0, 3));<br /> label(&quot;$C$&quot;, (1, 3));<br /> label(&quot;$M$&quot;, (2, 3));<br /> label(&quot;$C$&quot;, (3, 3));<br /> label(&quot;$8$&quot;, (4, 3));<br /> label(&quot;$8$&quot;, (1, 4));<br /> label(&quot;$C$&quot;, (2, 4));<br /> label(&quot;$8$&quot;, (3, 4));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 15|Solution<br /> ]]<br /> <br /> ==Problem 16==<br /> <br /> In the figure below, choose point &lt;math&gt;D&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle ABD&lt;/math&gt; have equal perimeters. What is the area of &lt;math&gt;\triangle ABD&lt;/math&gt;?<br /> &lt;asy&gt; draw((0,0)--(4,0)--(0,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,0), ESE);<br /> label(&quot;$C$&quot;, (0, 3), N);<br /> label(&quot;$3$&quot;, (0, 1.5), W);<br /> label(&quot;$4$&quot;, (2, 0), S);<br /> label(&quot;$5$&quot;, (2, 1.5), NE);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 16|Solution<br /> ]]<br /> <br /> ==Problem 17==<br /> Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 17|Solution<br /> ]]<br /> <br /> ==Problem 18==<br /> <br /> In the non-convex quadrilateral &lt;math&gt;ABCD&lt;/math&gt; shown below, &lt;math&gt;\angle BCD&lt;/math&gt; is a right angle, &lt;math&gt;AB=12&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;CD=3&lt;/math&gt;, and &lt;math&gt;AD=13&lt;/math&gt;.<br /> &lt;asy&gt;draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0));<br /> label(&quot;$B$&quot;, (0, 0), SW);<br /> label(&quot;$A$&quot;, (12, 0), ESE);<br /> label(&quot;$C$&quot;, (2.4, 3.6), SE);<br /> label(&quot;$D$&quot;, (0, 5), N);&lt;/asy&gt;<br /> What is the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 18|Solution<br /> ]]<br /> <br /> ==Problem 19==<br /> <br /> For any positive integer &lt;math&gt;M&lt;/math&gt;, the notation &lt;math&gt;M!&lt;/math&gt; denotes the product of the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;M&lt;/math&gt;. What is the largest integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;5^n&lt;/math&gt; is a factor of the sum &lt;math&gt;98!+99!+100!&lt;/math&gt; ?<br /> <br /> &lt;math&gt;\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 19|Solution<br /> ]]<br /> <br /> ==Problem 20==<br /> <br /> An integer between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{14}{75}\qquad\textbf{(B) }\frac{56}{225}\qquad\textbf{(C) }\frac{107}{400}\qquad\textbf{(D) }\frac{7}{25}\qquad\textbf{(E) }\frac{9}{25}&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 20|Solution<br /> ]]<br /> <br /> ==Problem 21==<br /> <br /> Suppose &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are nonzero real numbers, and &lt;math&gt;a+b+c=0&lt;/math&gt;. What are the possible value(s) for &lt;math&gt;\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 21|Solution<br /> ]]<br /> <br /> ==Problem 22==<br /> <br /> In the right triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC=12&lt;/math&gt;, &lt;math&gt;BC=5&lt;/math&gt;, and angle &lt;math&gt;C&lt;/math&gt; is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?<br /> &lt;asy&gt;<br /> draw((0,0)--(12,0)--(12,5)--(0,0));<br /> draw(arc((8.67,0),(12,0),(5.33,0)));<br /> label(&quot;$A$&quot;, (0,0), W);<br /> label(&quot;$C$&quot;, (12,0), E);<br /> label(&quot;$B$&quot;, (12,5), NE);<br /> label(&quot;$12$&quot;, (6, 0), S);<br /> label(&quot;$5$&quot;, (12, 2.5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 22|Solution<br /> ]]<br /> <br /> <br /> [[2017 AMC 8 Problems/Problem 23|Solution<br /> ]]<br /> <br /> ==Problem 23==<br /> <br /> Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 23|Solution<br /> ]]<br /> <br /> ==Problem 24==<br /> <br /> Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?<br /> <br /> &lt;math&gt;\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 24|Solution<br /> ]]<br /> <br /> ==Problem 25==<br /> <br /> In the figure shown, &lt;math&gt;\overline{US}&lt;/math&gt; and &lt;math&gt;\overline{UT}&lt;/math&gt; are line segments each of length 2, and &lt;math&gt;m\angle TUS = 60^\circ&lt;/math&gt;. Arcs &lt;math&gt;\overarc{TR}&lt;/math&gt; and &lt;math&gt;\overarc{SR}&lt;/math&gt; are each one-sixth of a circle with radius 2. What is the area of the region shown?<br /> &lt;asy&gt;draw((1,1.732)--(2,3.464)--(3,1.732));<br /> draw(arc((0,0),(2,0),(1,1.732)));<br /> draw(arc((4,0),(3,1.732),(2,0)));<br /> label(&quot;$U$&quot;, (2,3.464), N);<br /> label(&quot;$S$&quot;, (1,1.732), W);<br /> label(&quot;$T$&quot;, (3,1.732), E);<br /> label(&quot;$R$&quot;, (2,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> [[2017 AMC 8 Problems/Problem 25|Solution<br /> ]]<br /> <br /> {{MAA Notice}}</div> Snow52 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&diff=110637 2018 AMC 8 Problems/Problem 23 2019-10-27T23:20:19Z <p>Snow52: /* Solution 2 */</p> <hr /> <div>From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> We will use constructive counting to solve this. There are &lt;math&gt;2&lt;/math&gt; cases: Either all &lt;math&gt;3&lt;/math&gt; points are adjacent, or exactly &lt;math&gt;2&lt;/math&gt; points are adjacent.<br /> If all &lt;math&gt;3&lt;/math&gt; points are adjacent, then we have &lt;math&gt;8&lt;/math&gt; choices. If we have exactly &lt;math&gt;2&lt;/math&gt; adjacent points, then we will have &lt;math&gt;8&lt;/math&gt; places to put the adjacent points and also &lt;math&gt;4&lt;/math&gt; places to put the remaining point, so we have &lt;math&gt;8\cdot4&lt;/math&gt; choices. The total amount of choices is &lt;math&gt;{8 \choose 3} = 8\cdot7&lt;/math&gt;.<br /> Thus our answer is &lt;math&gt;\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ==Solution 2 (Complementary)==<br /> <br /> We can decide &lt;math&gt;2&lt;/math&gt; adjacent points with &lt;math&gt;8&lt;/math&gt; choices. The remaining point will have &lt;math&gt;6&lt;/math&gt; choices. However, we have counted the case with &lt;math&gt;3&lt;/math&gt; adjacent points twice, so we need to subtract this case once. The case with the &lt;math&gt;3&lt;/math&gt; adjacent points has &lt;math&gt;8&lt;/math&gt; arrangements, so our answer is &lt;math&gt;\frac{8\cdot6-8}{8\cdot7}=\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Snow52 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&diff=110636 2018 AMC 8 Problems/Problem 23 2019-10-27T23:19:41Z <p>Snow52: /* Solution 2 (Complementary) */</p> <hr /> <div>From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> We will use constructive counting to solve this. There are &lt;math&gt;2&lt;/math&gt; cases: Either all &lt;math&gt;3&lt;/math&gt; points are adjacent, or exactly &lt;math&gt;2&lt;/math&gt; points are adjacent.<br /> If all &lt;math&gt;3&lt;/math&gt; points are adjacent, then we have &lt;math&gt;8&lt;/math&gt; choices. If we have exactly &lt;math&gt;2&lt;/math&gt; adjacent points, then we will have &lt;math&gt;8&lt;/math&gt; places to put the adjacent points and also &lt;math&gt;4&lt;/math&gt; places to put the remaining point, so we have &lt;math&gt;8\cdot4&lt;/math&gt; choices. The total amount of choices is &lt;math&gt;{8 \choose 3} = 8\cdot7&lt;/math&gt;.<br /> Thus our answer is &lt;math&gt;\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can decide &lt;math&gt;2&lt;/math&gt; adjacent points with &lt;math&gt;8&lt;/math&gt; choices. The remaining point will have &lt;math&gt;6&lt;/math&gt; choices. However, we have counted the case with &lt;math&gt;3&lt;/math&gt; adjacent points twice, so we need to subtract this case once. The case with the &lt;math&gt;3&lt;/math&gt; adjacent points has &lt;math&gt;8&lt;/math&gt; arrangements, so our answer is &lt;math&gt;\frac{8\cdot6-8}{8\cdot7}=\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Snow52 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=98603 2016 AMC 8 Problems/Problem 25 2018-11-11T21:03:43Z <p>Snow52: /* Solution 6: Area (Even simpler than Solution 5) */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> First, we draw a line perpendicular to the base of the triangle and cut the triangle in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; results in the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 2: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> Notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AC}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> ==Solution 3: Inscribed Circle==<br /> <br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB = BC = CD = DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area &lt;math&gt;[ABCD]&lt;/math&gt; of the rhombus is twice that, which is &lt;math&gt;bh = (16)(15) = 240&lt;/math&gt;.<br /> <br /> The [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34r = 240&lt;/math&gt;. Solving this, &lt;math&gt;r = \frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> Noting that we have a 8-15-17 triangle, we can find &lt;math&gt;AE&lt;/math&gt; and &lt;math&gt;CE.&lt;/math&gt; Let &lt;math&gt;AE=x&lt;/math&gt;, &lt;math&gt;CE=17-x.&lt;/math&gt; Then by similar triangles (or &quot;Altitude on Hypotenuse&quot;) we have &lt;math&gt;15^2=x*17.&lt;/math&gt; Thus, &lt;math&gt;AE=x=225/17, CE=64/17.&lt;/math&gt; Now again by &quot;Altitude on Hypotenuse”, &lt;math&gt;r=\sqrt{AE*CE}.&lt;/math&gt; Therefore &lt;math&gt;r=120/17.&lt;/math&gt;<br /> <br /> ==Solution 5: Simple Trigonometry(10 second solve)==<br /> Denote the bottom left vertex of the isosceles triangle to be &lt;math&gt;A&lt;/math&gt;<br /> <br /> Denote the bottom right vertex of the isosceles triangle to be &lt;math&gt;B&lt;/math&gt;<br /> <br /> Denote the top vertex of the isosceles triangle to be &lt;math&gt;C&lt;/math&gt;<br /> <br /> Drop an altitude from &lt;math&gt;C&lt;/math&gt; to side &lt;math&gt;AB&lt;/math&gt;. Denote the foot of intersection to be &lt;math&gt;D&lt;/math&gt;.<br /> <br /> By the Pythagorean Theorem, &lt;math&gt;AC=17&lt;/math&gt;.<br /> <br /> Now, we see that &lt;math&gt;\sin{A}=\frac{15}{17}&lt;/math&gt;.<br /> <br /> This implies that &lt;math&gt;\sin{A}=\frac{r}{8}&lt;/math&gt; (r=radius of semicircle).<br /> <br /> Hence, &lt;math&gt;r=\boxed{\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 6: Area (Even simpler than Solution 5)==<br /> <br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Credits for Asymptote go to whoever wrote this diagram up in Solution 2. <br /> <br /> There are two ways to find the area of &lt;math&gt;\triangle ABC&lt;/math&gt;. The first way is the most obvious, and that is to multiply the base times the height (&lt;math&gt;16\cdot15&lt;/math&gt;) and then divide it by two. The second way is, in a way, a little more complex. Note that &lt;math&gt;\triangle ACD&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt; are congruent. This means that if we find the area of one triangle, we can just multiply it's area by 2 and we've found the area of the larger triangle &lt;math&gt;ABC&lt;/math&gt;. But since we always divide by two, as it is in the formula for finding the area of a triangle, the multiply by two and divide by two cancel out, giving us that if we just multiply base times height of &lt;math&gt;\triangle ACD&lt;/math&gt;, we will get the area of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> Now you might think: &quot;But what is the other way of finding the area&quot;. Well that is &lt;math&gt;AC&lt;/math&gt;, which would be the base, times &lt;math&gt;DE&lt;/math&gt;, the radius, which would be the height. <br /> <br /> The first way to find the area gives us the area of the &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\frac{15\cdot16}{2}=120&lt;/math&gt;. This gives us &lt;math&gt;120=AC\cdot r&lt;/math&gt; (&lt;math&gt;r&lt;/math&gt; signifies radius). We can find &lt;math&gt;AC&lt;/math&gt; using the Pythagorean Theorem on &lt;math&gt;\triangle ACD&lt;/math&gt;. The two legs are &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;, which gives us that the hypotenuse, &lt;math&gt;AC&lt;/math&gt;, is equal to &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now that we have an equation with only one variable for the radius, &lt;math&gt;120=17r&lt;/math&gt;, we can just solve for the radius. We get &lt;math&gt;r=\frac{120}{17}&lt;/math&gt;. <br /> <br /> This may seem like a lengthy explanation, but doing it yourself when you know what to do, it actually takes very little time. Try it yourself!</div> Snow52