https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hong2021&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T11:17:27ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_15&diff=1057761997 AIME Problems/Problem 152019-05-12T01:14:48Z<p>Hong2021: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
The sides of [[rectangle]] <math>ABCD</math> have lengths <math>10</math> and <math>11</math>. An [[equilateral triangle]] is drawn so that no point of the triangle lies outside <math>ABCD</math>. The maximum possible [[area]] of such a triangle can be written in the form <math>p\sqrt{q}-r</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are positive integers, and <math>q</math> is not divisible by the square of any prime number. Find <math>p+q+r</math>.<br />
<br />
__TOC__<br />
=== Solution 1 (Coordinate Bash)===<br />
Consider points on the [[complex plane]] <math>A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)</math>. Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one [[vertex]] of the triangle at <math>A</math>, and the other two points <math>E</math> and <math>F</math> on <math>BC</math> and <math>CD</math>, respectively. Let <math>E (11,a)</math> and <math>F (b, 10)</math>. Since it's equilateral, then <math>E\cdot\text{cis}60^{\circ} = F</math>, so <math>(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i</math>, and expanding we get <math>\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i</math>.<br />
<br />
[[Image:1997 AIME-15a.PNG|center]]<br />
<br />
We can then set the real and imaginary parts equal, and solve for <math>(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})</math>. Hence a side <math>s</math> of the equilateral triangle can be found by <math>s^2 = AE^2 = a^2 + AB^2 = 884 - 440\sqrt{3}</math>. Using the area formula <math>\frac{s^2\sqrt{3}}{4}</math>, the area of the equilateral triangle is <math>\frac{(884-440\sqrt{3})\sqrt{3}}{4} = 221\sqrt{3} - 330</math>. Thus <math>p + q + r = 221 + 3 + 330 = \boxed{554}</math>.<br />
<br />
===Solution 2===<br />
This is a trigonometric re-statement of the above. Let <math>x = \angle EAB</math>; by alternate interior angles, <math>\angle DFA=60+x</math>. Let <math>a = EB</math> and the side of the equilateral triangle be <math>s</math>, so <math>s= \sqrt{a^2+121}</math> by the [[Pythagorean Theorem]]. Now <math>\frac{10}{s} = \sin(60+x)= \sin {60} \cos x+ \cos {60} \sin x = \left(\frac{\sqrt{3}}2\right)\left(\frac{11}s\right)+\left(\frac 12\right)\left( \frac as \right)</math>. This reduces to <math>a=20-11\sqrt{3}</math>. <br />
<br />
Thus, the area of the triangle is <math>\frac{s^2\sqrt{3}}{4} =(a^2+121)\frac{\sqrt{3}}{4}</math>, which yields the same answer as above.<br />
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===Solution 3===<br />
<br />
Since <math>\angle{BAD}=90</math> and <math>\angle{EAF}=60</math>, it follows that <math>\angle{DAF}+\angle{BAE}=90-60=30</math>. Rotate triangle <math>ADF</math> <math>60</math> degrees clockwise. Note that the image of <math>AF</math> is <math>AE</math>. Let the image of <math>D</math> be <math>D'</math>. Since angles are preserved under rotation, <math>\angle{DAF}=\angle{D'AE}</math>. It follows that <math>\angle{D'AE}+\angle{BAE}=\angle{D'AB}=30</math>. Since <math>\angle{ADF}=\angle{ABE}=90</math>, it follows that quadrilateral <math>ABED'</math> is cyclic with circumdiameter <math>AE=s</math> and thus circumradius <math>\frac{s}{2}</math>. Let <math>O</math> be its circumcenter. By Inscribed Angles, <math>\angle{BOD'}=2\angle{BAD}=60</math>. By the definition of circle, <math>OB=OD'</math>. It follows that triangle <math>OBD'</math> is equilateral. Therefore, <math>BD'=r=\frac{s}{2}</math>. Applying the Law of Cosines to triangle <math>ABD'</math>, <math>\frac{s}{2}=\sqrt{10^2+11^2-(2)(10)(11)(\cos{30})}</math>. Squaring and multiplying by <math>\sqrt{3}</math> yields <math>\frac{s^2\sqrt{3}}{4}=221\sqrt{3}-330\implies{p+q+r=221+3+330=\boxed{554}}</math><br />
<br />
-Solution by '''thecmd999'''<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_7&diff=1056861995 AIME Problems/Problem 72019-05-05T23:13:27Z<p>Hong2021: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Given that <math>(1+\sin t)(1+\cos t)=5/4</math> and<br />
:<math>(1-\sin t)(1-\cos t)=\frac mn-\sqrt{k},</math><br />
where <math>k, m,</math> and <math>n_{}</math> are [[positive integer]]s with <math>m_{}</math> and <math>n_{}</math> [[relatively prime]], find <math>k+m+n.</math><br />
<br />
== Solution ==<br />
From the givens, <br />
<math>2\sin t \cos t + 2 \sin t + 2 \cos t = \frac{1}{2}</math>, and adding <math>\sin^2 t + \cos^2t = 1</math> to both sides gives <math>(\sin t + \cos t)^2 + 2(\sin t + \cos t) = \frac{3}{2}</math>. Completing the square on the left in the variable <math>(\sin t + \cos t)</math> gives <math>\sin t + \cos t = -1 \pm \sqrt{\frac{5}{2}}</math>. Since <math>|\sin t + \cos t| \leq \sqrt 2 < 1 + \sqrt{\frac{5}{2}}</math>, we have <math>\sin t + \cos t = \sqrt{\frac{5}{2}} - 1</math>. Subtracting twice this from our original equation gives <math>(\sin t - 1)(\cos t - 1) = \sin t \cos t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}</math>, so the answer is <math>13 + 4 + 10 = \boxed{027}</math>.<br />
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== Solution 2 ==<br />
Let <math>(1 - \sin t)(1 - \cos t) = x</math>. Multiplying <math>x</math> with the given equation, <math>\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t</math>, and <math>\frac{\sqrt{5x}}{2} = \sin t \cos t</math>. Simplifying and rearranging the given equation, <math>\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}</math>. Notice that <math>(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x</math>, and substituting, <math>x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}</math>. Rearranging and squaring, <math>5x = x^2 - \frac{3}{2} x + \frac{9}{16}</math>, so <math>x^2 - \frac{13}{2} x + \frac{9}{16} = 0</math>, and <math>x = \frac{13}{4} \pm \sqrt{10}</math>, but clearly, <math>0 \leq x < 4</math>. Therefore, <math>x = \frac{13}{4} - \sqrt{10}</math>, and the answer is <math> 13 + 4 + 10 = \boxed{027}</math>.<br />
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== Solution 3 == <br />
<br />
(-synergy)<br />
<br />
We have <math>1+\sin x\cos x+\sin x+\cos x = \frac{5}{4}</math><br />
<br />
We want to find <math>1+\sin x \cos x-\sin x-\cos x</math><br />
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If we find <math>\sin x+\cos x</math>, we will be done with the problem.<br />
<br />
Let <math>y = \sin x+\cos x</math><br />
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Squaring, we have <math>y^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x</math><br />
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From this we have <math>\sin x \cos x = \frac{y^2-1}{2}</math> and <math>\sin x + \cos x = y</math><br />
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Substituting this into the first equation we have <math>2y^2+4y-3=0</math>, <math>y = \frac {-2 \pm \sqrt{10}}{2}</math><br />
<br />
<math>\frac{5}{4} - 2(\frac{-2+\sqrt{10}}{2}) = \frac{13}{4}-\sqrt{10} \rightarrow 13+10+4=\boxed{027}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1995|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_9&diff=1056821983 AIME Problems/Problem 92019-05-05T18:24:50Z<p>Hong2021: /* Solution */</p>
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<div>== Problem ==<br />
Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.<br />
<br />
== Solution 1 ==<br />
Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>.<br />
<br />
Since <math>x>0</math>, and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. So we can apply [[AM-GM]]:<br />
<br />
<cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath><br />
<br />
The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>.<br />
<br />
Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when we have <math>x \sin{x} = \frac{2}{3}</math> in the original equation (since <math>x\sin x</math> is continuous and increasing on the interval <math>0 \le x \le \frac{\pi}{2}</math>, and its range on that interval is from <math>0 \le x\sin x \le \frac{\pi}{2}</math>, this value of <math>\frac{2}{3}</math> is attainable by the [[Intermediate Value Theorem]]).<br />
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== Solution 2 ==<br />
We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>.<br />
<br />
This results in <math>\dfrac{(3x \sin x-2)^2}{x \sin x}+12</math>.<br />
<br />
Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.<br />
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== Solution 3 (uses calculus) ==<br />
<br />
Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero. <br />
<br />
The derivative of <math>f(y)</math>, using the Power Rule, is<br />
<br />
<math>f'(y)</math> = <math>9 - 4y^{-2}</math><br />
<br />
<math>f'(y)</math> is zero only when <math>y = \frac{2}{3}</math> or <math>y = -\frac{2}{3}</math>. It can further be verified that <math>\frac{2}{3}</math> and <math>-\frac{2}{3}</math> are relative minima by finding the derivatives at other points near the critical points. However, since <math>x \sin{x}</math> is always positive in the given domain, <math>y = \frac{2}{3}</math>. Therefore, <math>x\sin{x}</math> = <math>\frac{2}{3}</math>, and the answer is <math>\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}</math>.<br />
<br />
== Solution 4 (also uses calculus) ==<br />
<br />
As above, let <math>y = x\sin{x}</math>. Add <math>\frac{12y}{y}</math> to the expression and subtract <math>12</math>, giving <math>f(x) = \frac{(3y+2)^2}{y} - 12</math>. Taking the [[derivative]] of <math>f(x)</math> using the [[Chain Rule]] and [[Quotient Rule]], we have <math>\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to <math>0</math>. Simplifying, we have <math>6y(3y+2) = (3y+2)^2</math> and <math>y = \pm{\frac{2}{3}} = x\sin{x}</math>. Since both <math>x</math> and <math>\sin{x}</math> are positive on the given interval, we can ignore the negative root. Plugging <math>y = \frac{2}{3}</math> into our expression for <math>f(x)</math>, we have <math>\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}</math>.<br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_39&diff=1040371953 AHSME Problems/Problem 392019-03-05T01:45:42Z<p>Hong2021: /Solution/</p>
<hr />
<div><cmath>a^x=b</cmath><br />
<cmath>b^y=a</cmath><br />
<cmath>{a^x}^y=a</cmath><br />
<cmath>xy=1</cmath><br />
<cmath>\log_a b\log_b a=1</cmath><br />
As a result, the answer should be 1 \boxed{A}.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=Helios_Hong&diff=104036Helios Hong2019-03-05T01:30:53Z<p>Hong2021: MATH GOD</p>
<hr />
<div>One of the greatest mathematician of all time.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_21&diff=1040352015 AMC 12A Problems/Problem 212019-03-05T01:29:17Z<p>Hong2021: /* Solution 3 (Partial) */</p>
<hr />
<div>==Problem==<br />
<br />
A circle of radius <math>r</math> passes through both foci of, and exactly four points on, the ellipse with equation <math>x^2+16y^2=16.</math> The set of all possible values of <math>r</math> is an interval <math>[a,b).</math> What is <math>a+b?</math><br />
<br />
<math> \textbf{(A)}\ 5\sqrt{2}+4\qquad\textbf{(B)}\ \sqrt{17}+7\qquad\textbf{(C)}\ 6\sqrt{2}+3\qquad\textbf{(D)}\ \sqrt{15}+8\qquad\textbf{(E)}\ 12 </math><br />
<br />
==Solution==<br />
<br />
We can graph the ellipse by seeing that the center is <math>(0, 0)</math> and finding the ellipse's intercepts. The points where the ellipse intersects the coordinate axes are <math>(0, 1), (0, -1), (4, 0)</math>, and <math>(-4, 0)</math>. Recall that the two foci lie on the major axis of the ellipse and are a distance of <math>c</math> away from the center of the ellipse, where <math>c^2 = a^2 - b^2</math>, with <math>a</math> being half the length of the major (longer) axis and <math>b</math> being half the minor (shorter) axis of the ellipse. We have that <math>c^2 = 4^2 - 1^2 \implies</math> <math>c^2 = 15 \implies c = \pm \sqrt{15}</math>. Hence, the coordinates of both of our foci are <math>(\sqrt{15}, 0)</math> and <math>(-\sqrt{15}, 0)</math>. In order for a circle to pass through both of these foci, we must have that the center of this circle lies on the y-axis.<br />
<br />
The minimum possible value of <math>r</math> belongs to the circle whose diameter's endpoints are the foci of this ellipse, so <math>a = \sqrt{15}</math>. The value for <math>b</math> is achieved when the circle passes through the foci and only three points on the ellipse, which is possible when the circle touches <math>(0, 1)</math> or <math>(0, -1)</math>. Which point we use does not change what value of <math>b</math> is attained, so we use <math>(0, -1)</math>. Here, we must find the point <math>(0, y)</math> such that the distance from <math>(0, y)</math> to both foci and <math>(0, -1)</math> is the same. Now, we have the two following equations. <cmath>(\sqrt{15})^2 + (y)^2 = b^2</cmath> <cmath>y + 1 = b \implies y = b - 1</cmath> Substituting for <math>y</math>, we have that <cmath>15 + (b - 1)^2 = b^2 \implies -2b + 16 = 0.</cmath><br />
<br />
Solving the above simply yields that <math>b = 8</math>, so our answer is <math>a + b = \sqrt{15} + 8 \textbf{ (D)}</math>.<br />
<br />
==Solution 2==<br />
As above, we can show that the foci of the ellipse are <math>(\sqrt{15},0), (-\sqrt{15},0).</math><br />
<br />
To obtain the lower bound, note that the smallest circle is when the diameter is on the line segment formed by the two foci. We can check that this indeed passes through four points on the ellipse since <math>\sqrt{15}>1,</math> so <math>a=\sqrt{15}.</math><br />
<br />
To get the upper bound, note that the circle must go through either <math>(0,1)</math> or <math>(0,-1).</math> WLOG, let let the circle go through <math>(0,1).</math> We know that the circle must go through the foci of the ellipse <math>(\sqrt{15},0), (-\sqrt{15},0),</math> So we can apply power of a point to find the diameter. Let <math>x</math> denote the length of the line segment from the origin to the lower point on the circle. Note that <math>x</math> lies on the diameter. Then by POP, we have<br />
<math>x \cdot 1 = \sqrt{15} \cdot \sqrt{15},</math> yielding <math>x=15</math>, and so the radius of the circle is <math>(15+1)/2=8,</math> so <math>b=8.</math><br />
Thus <math>a+b=\sqrt{15}+8 \textbf{ (D)}</math>. <br />
<br />
~ ccx09 (Roy Boy Apple Short Long)<br />
<br />
== Solution 3 (Partial, If running out of time) ==<br />
Note, this is not a full solution but can be used if one is running out of time.<br />
<br />
As above, the minimum radius of the circle is <math>\sqrt{15}</math> which is the <math>a</math> value. The only answer that contains <math>\sqrt{15}</math> is <math>\boxed{\textbf{(D)}}</math><br />
<br />
== See Also ==<br />
{{AMC12 box|year=2015|ab=A|num-b=20|num-a=22}}<br />
<br />
<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_USAMO_Problems/Problem_1&diff=998512018 USAMO Problems/Problem 12018-12-30T19:42:07Z<p>Hong2021: /* Solution 1*/</p>
<hr />
<div>==Problem 1==<br />
Let <math>a,b,c</math> be positive real numbers such that <math>a+b+c=4\sqrt[3]{abc}</math>. Prove that <cmath>2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.</cmath><br />
<br />
<br />
==Solution 1==<br />
<br />
WLOG let <math>a \leq b \leq c</math>. Add <math>2(ab+bc+ca)</math> to both sides of the inequality and factor to get: <cmath>4(a(a+b+c)+bc) \geq (a+b+c)^2</cmath> <cmath>\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}</cmath><br />
<br />
The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_13&diff=994472008 AMC 10B Problems/Problem 132018-12-15T02:57:43Z<p>Hong2021: /* Solution 2(Using Answer Choices) */</p>
<hr />
<div>==Problem==<br />
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008^{\text{th}}</math> term of the sequence?<br />
<br />
<math>\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}</math><br />
<br />
==Solution==<br />
Since the mean of the first <math>n</math> terms is <math>n</math>, the sum of the first <math>n</math> terms is <math>n^2</math>. Thus, the sum of the first <math>2007</math> terms is <math>2007^2</math> and the sum of the first <math>2008</math> terms is <math>2008^2</math>. Hence, the 2008th term is <math>2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\text{(B)}}</math><br />
<br />
Note that <math>n^2</math> is the sum of the first n odd numbers.<br />
<br />
<br />
==Solution 2 (Using Answer Choices)==<br />
From inspection, we see that the sum of the sequence is basically <math>n^2</math>. We also notice that <math>n^2</math> Is the sum of the first <math>n</math> ODD integers. Because <math>4015</math> is the only odd integer, <math>\boxed{B}</math> is the answer.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_20&diff=994461999 AHSME Problems/Problem 202018-12-15T01:45:15Z<p>Hong2021: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The sequence <math>a_{1},a_{2},a_{3},\ldots</math> statisfies <math>a_{1} = 19,a_{9} = 99</math>, and, for all <math>n\geq 3</math>, <math>a_{n}</math> is the arithmetic mean of the first <math>n - 1</math> terms. Find <math>a_2</math>.<br />
<br />
== Solution ==<br />
<br />
Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>.<br />
<br />
By definition, <math>a_3=m</math>.<br />
<br />
Next, <math>a_4</math> is the mean of <math>m-x</math>, <math>m+x</math> and <math>m</math>, which is again <math>m</math>. <br />
<br />
Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>.<br />
<br />
It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \boxed{(E) 179}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1999|num-b=19|num-a=21}}<br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_20&diff=994451999 AHSME Problems/Problem 202018-12-15T01:45:01Z<p>Hong2021: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The sequence <math>a_{1},a_{2},a_{3},\ldots</math> statisfies <math>a_{1} = 19,a_{9} = 99</math>, and, for all <math>n\geq 3</math>, <math>a_{n}</math> is the arithmetic mean of the first <math>n - 1</math> terms. Find <math>a_2</math>.<br />
<br />
== Solution ==<br />
<br />
Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>.<br />
<br />
By definition, <math>a_3=m</math>.<br />
<br />
Next, <math>a_4</math> is the mean of <math>m-x</math>, <math>m+x</math> and <math>m</math>, which is again <math>m</math>. <br />
<br />
Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>.<br />
<br />
It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \boxed{(E) 179}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1999|num-b=19|num-a=21}}<br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_20&diff=994441999 AHSME Problems/Problem 202018-12-15T01:44:43Z<p>Hong2021: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The sequence <math>a_{1},a_{2},a_{3},\ldots</math> statisfies <math>a_{1} = 19,a_{9} = 99</math>, and, for all <math>n\geq 3</math>, <math>a_{n}</math> is the arithmetic mean of the first <math>n - 1</math> terms. Find <math>a_2</math>.<br />
<br />
== Solution ==<br />
<br />
Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>.<br />
<br />
By definition, <math>a_3=m</math>.<br />
<br />
Next, <math>a_4</math> is the mean of <math>m-x</math>, <math>m+x</math> and <math>m</math>, which is again <math>m</math>. <br />
<br />
Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>.<br />
<br />
It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \ (E) boxed{179}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1999|num-b=19|num-a=21}}<br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_20&diff=994431999 AHSME Problems/Problem 202018-12-15T01:44:22Z<p>Hong2021: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The sequence <math>a_{1},a_{2},a_{3},\ldots</math> statisfies <math>a_{1} = 19,a_{9} = 99</math>, and, for all <math>n\geq 3</math>, <math>a_{n}</math> is the arithmetic mean of the first <math>n - 1</math> terms. Find <math>a_2</math>.<br />
<br />
== Solution ==<br />
<br />
Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>.<br />
<br />
By definition, <math>a_3=m</math>.<br />
<br />
Next, <math>a_4</math> is the mean of <math>m-x</math>, <math>m+x</math> and <math>m</math>, which is again <math>m</math>. <br />
<br />
Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>.<br />
<br />
It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \(E)boxed{179}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1999|num-b=19|num-a=21}}<br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=Perfect_cube&diff=99386Perfect cube2018-12-12T00:50:13Z<p>Hong2021: /* Introductory Problems */</p>
<hr />
<div>A '''perfect cube''' is an [[integer]] that is equal to some other integer raised to the third power. We refer to raising a [[number]] to the third power as ''cubing'' the number.<br />
<br />
For example, 125 is a perfect cube because <math>5^3 = 125</math>. However, 121 is not a perfect cube because there is no integer <math>n</math> such that <math>n^3 = 121</math>.<br />
<br />
== Example Problems ==<br />
=== Introductory Problems ===<br />
* [[2005_AMC_10A_Problems/Problem_15 | 2005 AMC 10A Problem 15]]<br />
* [[2018_AMC_8_Problems/Problem_25 | 2018 AMC 8 Problem 25]]<br />
<br />
== See also ==<br />
* [[Number theory]]<br />
* [[Perfect square]]<br />
* [[Perfect power]]</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_17&diff=993262012 AMC 10B Problems/Problem 172018-12-08T14:54:14Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Problem==<br />
Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br />
<br />
<math>\mathbf{(A)}</math> <math>\dfrac{1}{8}</math> <math>\space</math> <math>\mathbf{(B)}</math> <math>\dfrac{1}{4}</math> <math>\space</math> <math>\mathbf{(C)}</math> <math>\dfrac{\sqrt{10}}{10}</math> <math>\space</math> <math>\mathbf{(D)}</math> <math>\dfrac{\sqrt{5}}{6}</math> <math>\space</math> <math>\mathbf{(E)}</math> <math>\dfrac{\sqrt{5}}{5}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is <math>24\pi</math>, so the circumference of the smaller cone would be <math>\dfrac{120}{360} \times 24\pi = 8\pi</math>. This means that the radius of the smaller cone is <math>4</math>. Since the radius of the paper disk is <math>12</math>, the slant height of the smaller cone would be <math>12</math>. By the Pythagorean Theorem, the height of the cone is <math>\sqrt{12^2-4^2}=8\sqrt{2}</math>. Thus, the volume of the smaller cone is <math>\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi</math>.<br />
<br />
Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is <math>8</math> and the slant height is <math>12</math>. By the Pythagorean Theorem again, the height is <math>\sqrt{12^2-8^2}=4\sqrt{5}</math>. Thus, the volume of the larger cone is <math>\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi</math>.<br />
<br />
The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find <math>\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}</math> after simplifying, or <math>\boxed{(C) \frac{\sqrt{10}}{10}}</math> .<br />
<br />
*A side note<br />
<br />
We can first simplify the volume ratio: <math>\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.</math> Now we can find the GENERAL formulas for <math>r</math> and <math>h</math> based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us <math>C.</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_17&diff=993252012 AMC 10B Problems/Problem 172018-12-08T14:53:51Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Problem==<br />
Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br />
<br />
<math>\mathbf{(A)}</math> <math>\dfrac{1}{8}</math> <math>\space</math> <math>\mathbf{(B)}</math> <math>\dfrac{1}{4}</math> <math>\space</math> <math>\mathbf{(C)}</math> <math>\dfrac{\sqrt{10}}{10}</math> <math>\space</math> <math>\mathbf{(D)}</math> <math>\dfrac{\sqrt{5}}{6}</math> <math>\space</math> <math>\mathbf{(E)}</math> <math>\dfrac{\sqrt{5}}{5}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is <math>24\pi</math>, so the circumference of the smaller cone would be <math>\dfrac{120}{360} \times 24\pi = 8\pi</math>. This means that the radius of the smaller cone is <math>4</math>. Since the radius of the paper disk is <math>12</math>, the slant height of the smaller cone would be <math>12</math>. By the Pythagorean Theorem, the height of the cone is <math>\sqrt{12^2-4^2}=8\sqrt{2}</math>. Thus, the volume of the smaller cone is <math>\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi</math>.<br />
<br />
Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is <math>8</math> and the slant height is <math>12</math>. By the Pythagorean Theorem again, the height is <math>\sqrt{12^2-8^2}=4\sqrt{5}</math>. Thus, the volume of the larger cone is <math>\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi</math>.<br />
<br />
The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find <math>\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}</math> after simplifying, or <math>\boxed{C \frac{\sqrt{10}}{10}}</math> .<br />
<br />
*A side note<br />
<br />
We can first simplify the volume ratio: <math>\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.</math> Now we can find the GENERAL formulas for <math>r</math> and <math>h</math> based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us <math>C.</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_17&diff=993242012 AMC 10B Problems/Problem 172018-12-08T14:53:29Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Problem==<br />
Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br />
<br />
<math>\mathbf{(A)}</math> <math>\dfrac{1}{8}</math> <math>\space</math> <math>\mathbf{(B)}</math> <math>\dfrac{1}{4}</math> <math>\space</math> <math>\mathbf{(C)}</math> <math>\dfrac{\sqrt{10}}{10}</math> <math>\space</math> <math>\mathbf{(D)}</math> <math>\dfrac{\sqrt{5}}{6}</math> <math>\space</math> <math>\mathbf{(E)}</math> <math>\dfrac{\sqrt{5}}{5}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
Let's find the volume of the smaller cone first. We know that the circumference of the paper disk is <math>24\pi</math>, so the circumference of the smaller cone would be <math>\dfrac{120}{360} \times 24\pi = 8\pi</math>. This means that the radius of the smaller cone is <math>4</math>. Since the radius of the paper disk is <math>12</math>, the slant height of the smaller cone would be <math>12</math>. By the Pythagorean Theorem, the height of the cone is <math>\sqrt{12^2-4^2}=8\sqrt{2}</math>. Thus, the volume of the smaller cone is <math>\dfrac{1}{3} \times 4^2\pi \times 8\sqrt{2}=\dfrac{128\sqrt{2}}{3}\pi</math>.<br />
<br />
Now, we need to find the volume of the larger cone. Using the same reason as above, we get that the radius is <math>8</math> and the slant height is <math>12</math>. By the Pythagorean Theorem again, the height is <math>\sqrt{12^2-8^2}=4\sqrt{5}</math>. Thus, the volume of the larger cone is <math>\dfrac{1}{3} \times 8^2\pi \times 4\sqrt{5} = \dfrac{256\sqrt{5}}{3}\pi</math>.<br />
<br />
The question asked for the ratio of the volume of the smaller cone to the larger cone. We need to find <math>\dfrac{\dfrac{128\sqrt{2}}{3}\pi}{\dfrac{256\sqrt{5}}{3}\pi}=\dfrac{\sqrt{10}}{10}</math> after simplifying, or <math>\boxed{C} \frac{\sqrt{10}}{10}</math> .<br />
<br />
*A side note<br />
<br />
We can first simplify the volume ratio: <math>\frac{V_1}{V_2} = \frac{(r_1)^2 \cdot h_1}{(r_2)^2 \cdot h_2}.</math> Now we can find the GENERAL formulas for <math>r</math> and <math>h</math> based on the original circle radius and angle to cut out, then we can substitute the appropriate numbers, which gives us <math>C.</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=991052018 AMC 10B Problems/Problem 102018-11-27T02:44:55Z<p>Hong2021: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
In the rectangular parallelpiped shown, <math>AB</math> = <math>3</math>, <math>BC</math> = <math>1</math>, and <math>CG</math> = <math>2</math>. Point <math>M</math> is the midpoint of <math>\overline{FG}</math>. What is the volume of the rectangular pyramid with base <math>BCHE</math> and apex <math>M</math>?<br />
<br />
<br />
<asy><br />
size(250);<br />
defaultpen(fontsize(10pt));<br />
pair A =origin;<br />
pair B = (4.75,0);<br />
pair E1=(0,3);<br />
pair F = (4.75,3);<br />
pair G = (5.95,4.2);<br />
pair C = (5.95,1.2);<br />
pair D = (1.2,1.2);<br />
pair H= (1.2,4.2);<br />
pair M = ((4.75+5.95)/2,3.6);<br />
draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br />
draw(B--C);<br />
draw(F--G);<br />
draw(A--D--H--C--D,dashed);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,E);<br />
label("$D$",D,W);<br />
label("$E$",E1,W);<br />
label("$F$",F,SW);<br />
label("$G$",G,NE);<br />
label("$H$",H,NW);<br />
label("$M$",M,N);<br />
dot(A);<br />
dot(B);<br />
dot(E1);<br />
dot(F);<br />
dot(G);<br />
dot(C);<br />
dot(D);<br />
dot(H);<br />
dot(M);<br />
label("3",A/2+B/2,S);<br />
label("2",C/2+G/2,E);<br />
label("1",C/2+B/2,SE);<br />
</asy><br />
<br />
<math>\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2</math><br />
<br />
==Solution 1==<br />
Consider the cross-sectional plane and label its area <math>b</math>. Note that the volume of the triangular prism that encloses the pyramid is <math>\frac{bh}{2}=3</math>, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is <math>\frac{bh}{3}</math>, so the answer is <math>\boxed{2}</math>. (AOPS12142015)<br />
<br />
==Solution 2==<br />
We can start by finding the total volume of the parallelepiped. It is <math>2 \cdot 3 \cdot 1 = 6</math>, because a rectangular parallelepiped is a rectangular prism.<br />
<br />
Next, we can consider the wedge-shaped section made when the plane <math>BCHE</math> cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is <math>\frac{1}{2} \cdot 2 \cdot 3 = 3</math>. Since BC is given to be <math>1</math>, we have that FM is <math>\frac{1}{2}</math>. Using the formula for the volume of a triangular pyramid, we have <math>V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}</math>. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume <math>\frac{1}{2}</math> as well.<br />
<br />
The original wedge we considered in the last step has volume <math>3</math>, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have <math>3 - \frac{1}{2} \cdot 2 = 2</math>. Thus, the volume of the figure we are trying to find is <math>2</math>. This means that the correct answer choice is <math>\boxed{E}</math>.<br />
<br />
Written by: Archimedes15<br />
<br />
NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says "rectangular parallelepiped." If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br />
<br />
==Solution 3==<br />
If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is <math>\frac{1}{3}Bh</math>, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is <math>\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}</math>. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br />
<math>(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= </math> <math>\boxed{2}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:3D Geometry Problems]]<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_18&diff=990692018 AMC 8 Problems/Problem 182018-11-26T01:11:20Z<p>Hong2021: /* Problem 18 */</p>
<hr />
<div>==Problem 18==<br />
How many positive factors does 23,232 have?<br />
<br />
<math>\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42</math><br />
<br />
==Solution==<br />
We can first find the prime factorization of <math>23,232</math>, which is <math>2^6\cdot3^1\cdot11^2</math>. Now, we just add one to our powers and multiply. Therefore, the answer is <math>(1+6)*(1+1)*(1+2)=7*2*3=\boxed{42}, \textbf{(E)}</math><br />
==Solution 2==<br />
<br />
Observe that <math>69696 = 264^2</math>, so this is <math>\frac{1}{3}</math> of <math>264^2</math> which is <math>88 \cdot 264 = 11^2 \cdot 8^2 \cdot 3 = 11^2 \cdot 2^6 \cdot 3</math>, which has <math>3 \cdot 7 \cdot 2 = 42</math> factors. The answer is <math>\boxed{\textbf{(E)42}}</math>.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2018|num-b=17|num-a=19}}<br />
<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=988231981 AHSME Problems/Problem 32018-11-21T21:05:41Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math><br />
<br />
The answer is <math>\boxed{\left(D\right) \frac{11}{6x}}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=988221981 AHSME Problems/Problem 32018-11-21T21:03:51Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math><br />
<br />
The answer is <math>\boxed\left(D\right)</math> <math>\frac{11}{6x}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=988211981 AHSME Problems/Problem 32018-11-21T21:02:55Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math><br />
<br />
The answer is <math>\boxed{\left(D\right)</math> <math>\frac{11}{6x}}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_2&diff=988202018 AMC 8 Problems/Problem 22018-11-21T21:00:52Z<p>Hong2021: /* Problem 2 (easy) */</p>
<hr />
<div>==Problem 2==<br />
What is the value of the product<cmath>\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?</cmath><br />
<br />
<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8</math><br />
<br />
==Solution==<br />
<br />
By adding up the numbers in each parentheses, we have: <math>\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}</math>.<br />
<br />
Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus the answer would be <math>\boxed{(D) 7 }</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2018|num-b=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_2&diff=988192018 AMC 8 Problems/Problem 22018-11-21T21:00:30Z<p>Hong2021: /* Problem 2 */</p>
<hr />
<div>==Problem 2 (easy)==<br />
What is the value of the product<cmath>\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?</cmath><br />
<br />
<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8</math><br />
<br />
==Solution==<br />
<br />
By adding up the numbers in each parentheses, we have: <math>\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}</math>.<br />
<br />
Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus the answer would be <math>\boxed{(D) 7 }</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2018|num-b=1|num-a=3}}<br />
<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=987951981 AHSME Problems/Problem 32018-11-21T20:24:54Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math><br />
<br />
The answer is <math>\left(D\right)</math> <math>\frac{11}{6x}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=987941981 AHSME Problems/Problem 32018-11-21T20:24:30Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math><br />
<br />
The answer is <cmath>\left(D\right)</cmath> <math>\frac{11}{6x}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=987891981 AHSME Problems/Problem 32018-11-21T20:08:02Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>\displaystyle{\frac{1}{x}}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math><br />
<br />
The answer is (D) <math>\frac{11}{6x}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Answer_Key&diff=987432018 AMC 8 Answer Key2018-11-21T16:27:41Z<p>Hong2021: Created page with "Not available yet."</p>
<hr />
<div>Not available yet.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2020_USAMO&diff=987412020 USAMO2018-11-21T16:20:23Z<p>Hong2021: Created page with "Not happening yet ..."</p>
<hr />
<div>Not happening yet ...</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=987181981 AHSME Problems/Problem 32018-11-21T15:59:55Z<p>Hong2021: /* Solution */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>\frac{1}{x}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math> = <math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math> = <math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math> = <math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math> + <math>\frac{3}{6x}</math> + <math>\frac{2}{6x}</math> = <math>\frac{11}{6x}</math><br />
<br />
The answer is (D) <math>\frac{11}{6x}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_3&diff=987171981 AHSME Problems/Problem 32018-11-21T15:59:23Z<p>Hong2021: /* Problem 3 */</p>
<hr />
<div>==Solution==<br />
<br />
The least common multiple of <math>\frac{1}{x}</math>, <math>\frac{1}{2x}</math>, and <math>\frac{1}{3x}</math> is <math>\frac{1}{6x}</math>. <br />
<br />
<math>\frac{1}{x}</math>=<math>\frac{6}{6x}</math>, <math>\frac{1}{2x}</math>=<math>\frac{3}{6x}</math>, <math>\frac{1}{3x}</math>=<math>\frac{2}{6x}</math>.<br />
<br />
<math>\frac{6}{6x}</math>+<math>\frac{3}{6x}</math>+<math>\frac{2}{6x}</math>=<math>\frac{11}{6x}</math><br />
<br />
The answer is (D) <math>\frac{11}{6x}</math>.</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems&diff=986981985 AJHSME Problems2018-11-21T14:54:44Z<p>Hong2021: /* Problem 1 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
<math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 57}=</math><br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50</math><br />
<br />
== Problem 2 ==<br />
<br />
<math>90+91+92+93+94+95+96+97+98+99=</math><br />
<br />
<br />
<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math><br />
<br />
== Problem 3 ==<br />
<br />
<math>\frac{10^7}{5\times 10^4}=</math><br />
<br />
<br />
<math>\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000</math><br />
<br />
== Problem 4 ==<br />
<br />
The area of polygon <math>ABCDEF</math>, in square units, is<br />
<br />
<math>\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74</math><br />
<br />
<asy><br />
draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle);<br />
label("A",(0,9),NW);<br />
label("B",(6,9),NE);<br />
label("C",(6,0),SE);<br />
label("D",(2,0),SW);<br />
label("E",(2,4),NE);<br />
label("F",(0,4),SW);<br />
label("6",(3,9),N);<br />
label("9",(6,4.5),E);<br />
label("4",(4,0),S);<br />
label("5",(0,6.5),W);<br />
</asy><br />
<br />
== Problem 5 ==<br />
<br />
<asy><br />
unitsize(13);<br />
draw((0,0)--(20,0));<br />
draw((0,0)--(0,15));<br />
draw((0,3)--(-1,3));<br />
draw((0,6)--(-1,6));<br />
draw((0,9)--(-1,9));<br />
draw((0,12)--(-1,12));<br />
draw((0,15)--(-1,15));<br />
fill((2,0)--(2,15)--(3,15)--(3,0)--cycle,black);<br />
fill((4,0)--(4,12)--(5,12)--(5,0)--cycle,black);<br />
fill((6,0)--(6,9)--(7,9)--(7,0)--cycle,black);<br />
fill((8,0)--(8,9)--(9,9)--(9,0)--cycle,black);<br />
fill((10,0)--(10,15)--(11,15)--(11,0)--cycle,black);<br />
label("A",(2.5,-.5),S);<br />
label("B",(4.5,-.5),S);<br />
label("C",(6.5,-.5),S);<br />
label("D",(8.5,-.5),S);<br />
label("F",(10.5,-.5),S);<br />
label("Grade",(15,-.5),S);<br />
label("$1$",(-1,3),W);<br />
label("$2$",(-1,6),W);<br />
label("$3$",(-1,9),W);<br />
label("$4$",(-1,12),W);<br />
label("$5$",(-1,15),W);<br />
</asy><br />
<br />
The bar graph shows the grades in a mathematics class for the last grading period. If A, B, C, and D are satisfactory grades, what fraction of the grades shown in the graph are satisfactory?<br />
<br />
<math>\text{(A)}\ \frac{1}{2} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{4}{5} \qquad \text{(E)}\ \frac{9}{10}</math><br />
<br />
== Problem 6 ==<br />
<br />
A ream of paper containing <math>500</math> sheets is <math>5</math> cm thick. Approximately how many sheets of this type of paper would there be in a stack <math>7.5</math> cm high?<br />
<br />
<math>\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250</math><br />
<br />
== Problem 7 ==<br />
<br />
A "stair-step" figure is made of alternating black and white squares in each row. Rows <math>1</math> through <math>4</math> are shown. All rows begin and end with a white square. The number of black squares in the <math>37\text{th}</math> row is<br />
<br />
<asy><br />
draw((0,0)--(7,0)--(7,1)--(0,1)--cycle);<br />
draw((1,0)--(6,0)--(6,2)--(1,2)--cycle);<br />
draw((2,0)--(5,0)--(5,3)--(2,3)--cycle);<br />
draw((3,0)--(4,0)--(4,4)--(3,4)--cycle);<br />
fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black);<br />
fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black);<br />
fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black);<br />
fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black);<br />
fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black);<br />
fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black);<br />
</asy><br />
<br />
<math>\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38</math><br />
<br />
== Problem 8 ==<br />
<br />
If <math>a = - 2</math>, the largest number in the set <math> - 3a, 4a, \frac {24}{a}, a^2, 1</math> is<br />
<br />
<math>\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1</math><br />
<br />
== Problem 9 ==<br />
<br />
The product of the 9 factors <math>\Big(1 - \frac12\Big)\Big(1 - \frac13\Big)\Big(1 - \frac14\Big)\cdots\Big(1 - \frac {1}{10}\Big) =</math><br />
<br />
<math>\text{(A)}\ \frac {1}{10} \qquad \text{(B)}\ \frac {1}{9} \qquad \text{(C)}\ \frac {1}{2} \qquad \text{(D)}\ \frac {10}{11} \qquad \text{(E)}\ \frac {11}{2}</math><br />
<br />
== Problem 10 ==<br />
<br />
The fraction halfway between <math>\frac{1}{5}</math> and <math>\frac{1}{3}</math> (on the number line) is<br />
<br />
<asy><br />
unitsize(12);<br />
draw((-1,0)--(20,0),EndArrow);<br />
draw((0,-.75)--(0,.75));<br />
draw((10,-.75)--(10,.75));<br />
draw((17,-.75)--(17,.75));<br />
label("$0$",(0,-.5),S);<br />
label("$\frac{1}{5}$",(10,-.5),S);<br />
label("$\frac{1}{3}$",(17,-.5),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ \frac{1}{4} \qquad \text{(B)}\ \frac{2}{15} \qquad \text{(C)}\ \frac{4}{15} \qquad \text{(D)}\ \frac{53}{200} \qquad \text{(E)}\ \frac{8}{15}</math><br />
<br />
== Problem 11 ==<br />
<br />
A piece of paper containing six joined squares labeled as shown in the diagram is folded along the edges of the squares to form a cube. The label of the face opposite the face labeled <math>\text{X}</math> is <br />
<br />
<asy><br />
draw((0,0)--(0,1)--(2,1)--(2,2)--(3,2)--(3,0)--(2,0)--(2,-2)--(1,-2)--(1,0)--cycle);<br />
draw((1,0)--(1,1));<br />
draw((2,0)--(2,1));<br />
draw((1,0)--(2,0));<br />
draw((1,-1)--(2,-1));<br />
draw((2,1)--(3,1));<br />
label("U",(.5,.3),N);<br />
label("V",(1.5,.3),N);<br />
label("W",(2.5,.3),N);<br />
label("X",(1.5,-.7),N);<br />
label("Y",(2.5,1.3),N);<br />
label("Z",(1.5,-1.7),N);<br />
</asy><br />
<br />
<math>\text{(A)}\ \text{Z} \qquad \text{(B)}\ \text{U} \qquad \text{(C)}\ \text{V} \qquad \text{(D)}\ \ \text{W} \qquad \text{(E)}\ \text{Y}</math><br />
<br />
== Problem 12 ==<br />
<br />
A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are <math>6.2 \text{ cm}</math>, <math>8.3 \text{ cm}</math> and <math>9.5 \text{ cm}</math>. The area of the square is<br />
<br />
<math>\text{(A)}\ 24\text{ cm}^2 \qquad \text{(B)}\ 36\text{ cm}^2 \qquad \text{(C)}\ 48\text{ cm}^2 \qquad \text{(D)}\ 64\text{ cm}^2 \qquad \text{(E)}\ 144\text{ cm}^2</math><br />
<br />
== Problem 13 ==<br />
<br />
If you walk for <math>45</math> minutes at a rate of <math>4 \text{ mph}</math> and then run for <math>30</math> minutes at a rate of <math>10\text{ mph,}</math> how many miles will you have gone at the end of one hour and <math>15</math> minutes?<br />
<br />
<math>\text{(A)}\ 3.5\text{ miles} \qquad \text{(B)}\ 8\text{ miles} \qquad \text{(C)}\ 9\text{ miles} \qquad \text{(D)}\ 25\frac{1}{3}\text{ miles} \qquad \text{(E)}\ 480\text{ miles}</math><br />
<br />
== Problem 14 ==<br />
<br />
The difference between a <math>6.5\% </math> sales tax and a <math>6\% </math> sales tax on an item priced at <math>\$20</math> before tax is<br />
<br />
<math>\text{(A)}</math> <math>\$.01</math> <br />
<br />
<math>\text{(B)}</math> <math>\$.10</math><br />
<br />
<math>\text{(C)}</math> <math>\$ .50</math> <br />
<br />
<math>\text{(D)}</math> <math>\$ 1</math> <br />
<br />
<math>\text{(E)}</math> <math>\$10</math><br />
<br />
== Problem 15 ==<br />
<br />
How many whole numbers between <math>100</math> and <math>400</math> contain the digit <math>2</math>?<br />
<br />
<math>\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148</math><br />
<br />
== Problem 16 ==<br />
<br />
The ratio of boys to girls in Mr. Brown's math class is <math>2:3</math>. If there are <math>30</math> students in the class, how many more girls than boys are in the class?<br />
<br />
<math>\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10</math><br />
<br />
== Problem 17 ==<br />
<br />
If your average score on your first six mathematics tests was <math>84</math> and your average score on your first seven mathematics tests was <math>85</math>, then your score on the seventh test was<br />
<br />
<math>\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92</math><br />
<br />
== Problem 18 ==<br />
<br />
Nine copies of a certain pamphlet cost less than <math>\$10.00</math> while ten copies of the same pamphlet (at the same price) cost more than <math>\$11.00</math>. How much does one copy of this pamphlet cost?<br />
<br />
<math>\text{(A)}</math> <math>\$1.07</math> <br />
<br />
<math>\text{(B)}</math> <math>\$1.08</math><br />
<br />
<math>\text{(C)}</math> <math>\$1.09</math> <br />
<br />
<math>\text{(D)}</math> <math>\$1.10</math> <br />
<br />
<math>\text{(E)}</math> <math>\$1.11</math><br />
<br />
== Problem 19 ==<br />
<br />
If the length and width of a rectangle are each increased by <math>10\% </math>, then the perimeter of the rectangle is increased by<br />
<br />
<math>\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\% </math><br />
<br />
== Problem 20 ==<br />
<br />
In a certain year, January had exactly four Tuesdays and four Saturdays. On what day did January <math>1</math> fall that year?<br />
<br />
<math>\text{(A)}\ \text{Monday} \qquad \text{(B)}\ \text{Tuesday} \qquad \text{(C)}\ \text{Wednesday} \qquad \text{(D)}\ \text{Friday} \qquad \text{(E)}\ \text{Saturday}</math><br />
<br />
== Problem 21 ==<br />
<br />
Mr. Green receives a <math>10\% </math> raise every year. His salary after four such raises has gone up by what percent?<br />
<br />
<math>\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% </math><br />
<br />
== Problem 22 ==<br />
<br />
Assume every 7-digit whole number is a possible telephone number except those that begin with <math>0</math> or <math>1</math>. What fraction of telephone numbers begin with <math>9</math> and end with <math>0</math>?<br />
<br />
<math>\text{(A)}\ \frac{1}{63} \qquad \text{(B)}\ \frac{1}{80} \qquad \text{(C)}\ \frac{1}{81} \qquad \text{(D)}\ \frac{1}{90} \qquad \text{(E)}\ \frac{1}{100}</math><br />
<br />
''Note: All telephone numbers are 7-digit whole numbers.''<br />
<br />
== Problem 23 ==<br />
<br />
King Middle School has <math>1200</math> students. Each student takes <math>5</math> classes a day. Each teacher teaches <math>4</math> classes. Each class has <math>30</math> students and <math>1</math> teacher. How many teachers are there at King Middle School? <br />
<br />
<math>\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50</math><br />
<br />
== Problem 24 ==<br />
<br />
In a magic triangle, each of the six whole numbers <math>10-15</math> is placed in one of the circles so that the sum, <math>S</math>, of the three numbers on each side of the triangle is the same. The largest possible value for <math>S</math> is<br />
<br />
<asy><br />
draw(circle((0,0),1));<br />
draw(dir(60)--6*dir(60));<br />
draw(circle(7*dir(60),1));<br />
draw(8*dir(60)--13*dir(60));<br />
draw(circle(14*dir(60),1));<br />
draw((1,0)--(6,0));<br />
draw(circle((7,0),1));<br />
draw((8,0)--(13,0));<br />
draw(circle((14,0),1));<br />
draw(circle((10.5,6.0621778264910705273460621952706),1));<br />
draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176));<br />
draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788));<br />
</asy><br />
<br />
<math>\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40</math><br />
<br />
== Problem 25 ==<br />
<br />
Five cards are lying on a table as shown.<br />
<br />
<cmath>\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\<br />
\\<br />
\boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}</cmath><br />
<br />
Each card has a letter on one side and a whole number on the other side. Jane said, "If a vowel is on one side of any card, then an even number is on the other side." Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?<br />
<br />
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ \text{P} \qquad \text{(E)}\ \text{Q}</math><br />
<br />
== See Also ==<br />
{{AJHSME box|year=1985|before=First<br>AJHSME|after=[[1986 AJHSME Problems|1986 AJHSME]]}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{MAA Notice}}</div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_10&diff=986812018 AMC 8 Problems/Problem 102018-11-21T14:49:56Z<p>Hong2021: /* Problem 10 */</p>
<hr />
<div>==Problem 10==<br />
The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?<br />
<br />
<math>\textbf{(A) }\frac{3}{7}\qquad\textbf{(B) }\frac{7}{12}\qquad\textbf{(C) }\frac{12}{7}\qquad\textbf{(D) }\frac{7}{4}\qquad \textbf{(E) }\frac{7}{3}</math></div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_5&diff=986802018 AMC 8 Problems/Problem 52018-11-21T14:49:06Z<p>Hong2021: /* Problem 5 */</p>
<hr />
<div>==Problem 5==<br />
What is the value of <math>1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018</math>?<br />
<br />
<math>\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010</math></div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_5&diff=986792018 AMC 8 Problems/Problem 52018-11-21T14:48:47Z<p>Hong2021: /* Problem5 */</p>
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<div>==Problem5==<br />
What is the value of <math>1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018</math>?<br />
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<math>\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010</math></div>Hong2021https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_19&diff=981352007 AMC 12A Problems/Problem 192018-10-14T15:21:36Z<p>Hong2021: /* Problem */</p>
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<div>== Problem ==<br />
[[Triangle]]s <math>ABC</math> and <math>ADE</math> have [[area]]s <math>2007</math> and <math>7002,</math> respectively, with <math>B = (0,0),</math> <math>C = (223,0),</math> <math>D = (680,380),</math> and <math>E = (689,389).</math> What is the sum of all possible x coordinates of <math>A</math>?<br />
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<math>\mathrm{(A)}\ 282 \qquad \mathrm{(B)}\ 300 \qquad \mathrm{(C)}\ 600 \qquad \mathrm{(D)}\ 900 \qquad \mathrm{(E)}\ 1200</math><br />
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__TOC__<br />
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== Solution ==<br />
[[Image:2007_12A_AMC-19.png]]<br />
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=== Solution 1 ===<br />
From <math>k = [ABC] = \frac 12bh</math>, we have that the height of <math>\triangle ABC</math> is <math>h = \frac{2k}{b} = \frac{2007 \cdot 2}{223} = 18</math>. Thus <math>A</math> lies on the lines <math>y = \pm 18 \quad \mathrm{(1)}</math>.<br />
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<math>DE = 9\sqrt{2}</math> using 45-45-90 triangles, so in <math>\triangle ADE</math> we have that <math>h = \frac{2 \cdot 7002}{9\sqrt{2}} = 778\sqrt{2}</math>. The slope of <math>DE</math> is <math>1</math>, so the equation of the line is <math>y = x + b \Longrightarrow b = (380) - (680) = -300 \Longrightarrow y = x - 300</math>. The point <math>A</math> lies on one of two [[parallel]] lines that are <math>778\sqrt{2}</math> units away from <math>\overline{DE}</math>. Now take an arbitrary point on the line <math>\overline{DE}</math> and draw the [[perpendicular]] to one of the parallel lines; then draw a line straight down from the same arbitrary point. These form a 45-45-90 <math>\triangle</math>, so the straight line down has a length of <math>778\sqrt{2} \cdot \sqrt{2} = 1556</math>. Now we note that the [[y-intercept]] of the parallel lines is either <math>1556</math> units above or below the y-intercept of line <math>\overline{DE}</math>; hence the equation of the parallel lines is <math>y = x - 300 \pm 1556 \Longrightarrow x = y + 300 \pm 1556 \quad \mathrm{(2)}</math>.<br />
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We just need to find the intersections of these two lines and sum up the values of the [[x-coordinates]]. Substituting the <math>\mathrm{(1)}</math> into <math>\mathrm{(2)}</math>, we get <math>x = \pm 18 + 300 \pm 1556 = 4(300) = 1200 \Longrightarrow \mathrm{(E)}</math>.<br />
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=== Solution 2 ===<br />
We are finding the intersection of two pairs of [[parallel]] lines, which will form a [[parallelogram]]. The [[centroid]] of this parallelogram is just the intersection of <math>\overline{BC}</math> and <math>\overline{DE}</math>, which can easily be calculated to be <math>(300,0)</math>. Now the sum of the x-coordinates is just <math>4(300) = 1200</math>.<br />
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== See also ==<br />
{{AMC12 box|year=2007|ab=A|num-b=18|num-a=20}}<br />
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[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Hong2021