https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Claudeaops&feedformat=atom AoPS Wiki - User contributions [en] 2022-05-25T12:52:08Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Incircle&diff=157588 Incircle 2021-07-09T02:58:22Z <p>Claudeaops: Remove duplicate text</p> <hr /> <div>{{stub}}<br /> <br /> <br /> An '''incircle''' of a [[convex]] [[polygon]] is a [[circle]] which is inside the figure and [[tangent line | tangent]] to each side. Every [[triangle]] and [[regular polygon]] has a unique incircle, but in general polygons with 4 or more sides (such as non-[[square (geometry) | square]] [[rectangle]]s) do not have an incircle. A quadrilateral that does have an incircle is called a [[Tangential Quadrilateral]]. For a triangle, the center of the incircle is the [[Incenter]], where the [[incircle]] is the largest circle that can be inscribed in the polygon. The [[Incenter]] can be constructed by drawing the intersection of angle bisectors.<br /> <br /> ==Formulas==<br /> *The radius of an incircle of a triangle (the inradius) with sides &lt;math&gt;a,b,c&lt;/math&gt; and area &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;r =&lt;/math&gt; &lt;math&gt;\frac{2A}{a+b+c}.&lt;/math&gt;<br /> *The [[area]] of any [[triangle]] is &lt;math&gt;r * s,&lt;/math&gt; where &lt;math&gt;s&lt;/math&gt; is the [[Semiperimeter]] of the [[triangle]].<br /> *The formula above can be simplified with Heron's Formula, yielding &lt;math&gt;r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.&lt;/math&gt;<br /> *The [[radius]] of an incircle of a right triangle (the inradius) with legs &lt;math&gt;a,b&lt;/math&gt; and hypotenuse &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;r=\frac{ab}{a+b+c}=\frac{a+b-c}{2}&lt;/math&gt;.<br /> *For any polygon with an incircle, &lt;math&gt;A=sr&lt;/math&gt;, where &lt;math&gt;A&lt;/math&gt; is the area, &lt;math&gt;s&lt;/math&gt; is the semi perimeter, and &lt;math&gt;r&lt;/math&gt; is the inradius.<br /> *The coordinates of the incenter (center of incircle) are &lt;math&gt;(\dfrac{aA_x+bB_x+cC_x}{a+b+c}, \dfrac{aA_y+bB_y+cC_y}{a+b+c})&lt;/math&gt;, if the coordinates of each vertex are &lt;math&gt;A(A_x, A_y)&lt;/math&gt;, &lt;math&gt;B(B_x, B_y)&lt;/math&gt;, and &lt;math&gt;C(C_x, C_y)&lt;/math&gt;, the side opposite of &lt;math&gt;A&lt;/math&gt; has length &lt;math&gt;a&lt;/math&gt;, the side opposite of &lt;math&gt;B&lt;/math&gt; has length &lt;math&gt;b&lt;/math&gt;, and the side opposite of &lt;math&gt;C&lt;/math&gt; has length &lt;math&gt;c&lt;/math&gt;.<br /> <br /> *The formula for the [[semiperimeter]] is &lt;math&gt;s=\frac{a+b+c}{2}&lt;/math&gt;.<br /> <br /> *The [[area]] of the [[triangle]] by [[Heron's Formula]] is &lt;math&gt;A=\sqrt{s(s-a)(s-b)(s-c)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> *[[Circumradius]]<br /> *[[Inradius]]<br /> *[[Kimberling center]]<br /> <br /> [[Category:Geometry]]<br /> Click here to learn about the orthocenter, and Line's Tangent</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_15&diff=157587 2010 AIME I Problems/Problem 15 2021-07-09T02:56:13Z <p>Claudeaops: Move sidenote to end of article</p> <hr /> <div>__TOC__<br /> == Problem ==<br /> In &lt;math&gt;\triangle{ABC}&lt;/math&gt; with &lt;math&gt;AB = 12&lt;/math&gt;, &lt;math&gt;BC = 13&lt;/math&gt;, and &lt;math&gt;AC = 15&lt;/math&gt;, let &lt;math&gt;M&lt;/math&gt; be a point on &lt;math&gt;\overline{AC}&lt;/math&gt; such that the [[incircle]]s of &lt;math&gt;\triangle{ABM}&lt;/math&gt; and &lt;math&gt;\triangle{BCM}&lt;/math&gt; have equal [[inradius|radii]]. Then &lt;math&gt;\frac{AM}{CM} = \frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p + q&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt; /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */<br /> import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200);<br /> <br /> /* segments and figures */<br /> draw((0,0)--(15,0));<br /> draw((15,0)--(6.66667,9.97775));<br /> draw((6.66667,9.97775)--(0,0));<br /> draw((7.33333,0)--(6.66667,9.97775));<br /> draw(circle((4.66667,2.49444),2.49444));<br /> draw(circle((9.66667,2.49444),2.49444));<br /> draw((4.66667,0)--(4.66667,2.49444));<br /> draw((9.66667,2.49444)--(9.66667,0));<br /> <br /> /* points and labels */<br /> label(&quot;r&quot;,(10.19662,1.92704),SE);<br /> label(&quot;r&quot;,(5.02391,1.8773),SE);<br /> dot((0,0));<br /> label(&quot;$A$&quot;,(-1.04408,-0.60958),NE);<br /> dot((15,0));<br /> label(&quot;$C$&quot;,(15.41907,-0.46037),NE);<br /> dot((6.66667,9.97775));<br /> label(&quot;$B$&quot;,(6.66525,10.23322),NE);<br /> label(&quot;$15$&quot;,(6.01866,-1.15669),NE);<br /> label(&quot;$13$&quot;,(11.44006,5.50815),NE);<br /> label(&quot;$12$&quot;,(2.28834,5.75684),NE);<br /> dot((7.33333,0));<br /> label(&quot;$M$&quot;,(7.56053,-1.000),NE);<br /> label(&quot;$H_1$&quot;,(3.97942,-1.200),NE);<br /> label(&quot;$H_2$&quot;,(9.54741,-1.200),NE);<br /> dot((4.66667,2.49444));<br /> label(&quot;$I_1$&quot;,(3.97942,2.92179),NE);<br /> dot((9.66667,2.49444));<br /> label(&quot;$I_2$&quot;,(9.54741,2.92179),NE);<br /> clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> === Solution 1 ===<br /> Let &lt;math&gt;AM = x&lt;/math&gt;, then &lt;math&gt;CM = 15 - x&lt;/math&gt;. Also let &lt;math&gt;BM = d&lt;/math&gt; Clearly, &lt;math&gt;\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}&lt;/math&gt;. We can also express each area by the rs formula. Then &lt;math&gt;\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}&lt;/math&gt;. Equating and cross-multiplying yields &lt;math&gt;25x + 2dx = 15d + 180&lt;/math&gt; or &lt;math&gt;d = \frac {25x - 180}{15 - 2x}.&lt;/math&gt; Note that for &lt;math&gt;d&lt;/math&gt; to be positive, we must have &lt;math&gt;7.2 &lt; x &lt; 7.5&lt;/math&gt;.<br /> <br /> By [[Stewart's Theorem]], we have &lt;math&gt;12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)&lt;/math&gt; or &lt;math&gt;432 = 3d^2 + 40x - 3x^2.&lt;/math&gt; Brute forcing by plugging in our previous result for &lt;math&gt;d&lt;/math&gt;, we have &lt;math&gt;432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.&lt;/math&gt; Clearing the fraction and gathering like terms, we get &lt;math&gt;0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.&lt;/math&gt;<br /> <br /> ''Aside: Since &lt;math&gt;x&lt;/math&gt; must be rational in order for our answer to be in the desired form, we can use the [[Rational Root Theorem]] to reveal that &lt;math&gt;12x&lt;/math&gt; is an integer. The only such &lt;math&gt;x&lt;/math&gt; in the above-stated range is &lt;math&gt;\frac {22}3&lt;/math&gt;.''<br /> <br /> Legitimately solving that quartic, note that &lt;math&gt;x = 0&lt;/math&gt; and &lt;math&gt;x = 15&lt;/math&gt; should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get &lt;math&gt;0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).&lt;/math&gt; The only solution in the desired range is thus &lt;math&gt;\frac {22}3&lt;/math&gt;. Then &lt;math&gt;CM = \frac {23}3&lt;/math&gt;, and our desired ratio &lt;math&gt;\frac {AM}{CM} = \frac {22}{23}&lt;/math&gt;, giving us an answer of &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;AM = 2x&lt;/math&gt; and &lt;math&gt;BM = 2y&lt;/math&gt; so &lt;math&gt;CM = 15 - 2x&lt;/math&gt;. Let the [[incenter]]s of &lt;math&gt;\triangle ABM&lt;/math&gt; and &lt;math&gt;\triangle BCM&lt;/math&gt; be &lt;math&gt;I_1&lt;/math&gt; and &lt;math&gt;I_2&lt;/math&gt; respectively, and their equal inradii be &lt;math&gt;r&lt;/math&gt;. From &lt;math&gt;r = \sqrt {(s - a)(s - b)(s - c)/s}&lt;/math&gt;, we find that<br /> <br /> &lt;cmath&gt;\begin{align*}r^2 &amp; = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} &amp; (1) \\<br /> &amp; = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. &amp; (2) \end{align*}&lt;/cmath&gt;<br /> <br /> Let the incircle of &lt;math&gt;\triangle ABM&lt;/math&gt; meet &lt;math&gt;AM&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and the incircle of &lt;math&gt;\triangle BCM&lt;/math&gt; meet &lt;math&gt;CM&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt;. Then note that &lt;math&gt;I_1 P Q I_2&lt;/math&gt; is a rectangle. Also, &lt;math&gt;\angle I_1 M I_2&lt;/math&gt; is right because &lt;math&gt;MI_1&lt;/math&gt; and &lt;math&gt;MI_2&lt;/math&gt; are the angle bisectors of &lt;math&gt;\angle AMB&lt;/math&gt; and &lt;math&gt;\angle CMB&lt;/math&gt; respectively and &lt;math&gt;\angle AMB + \angle CMB = 180^\circ&lt;/math&gt;. By properties of [[tangent (geometry)|tangents]] to [[circle]]s &lt;math&gt;MP = (MA + MB - AB)/2 = x + y - 6&lt;/math&gt; and &lt;math&gt;MQ = (MB + MC - BC)/2 = - x + y + 1&lt;/math&gt;. Now notice that the altitude of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;I_1 I_2&lt;/math&gt; is of length &lt;math&gt;r&lt;/math&gt;, so by similar triangles we find that &lt;math&gt;r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)&lt;/math&gt; (3). Equating (3) with (1) and (2) separately yields<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> 2y^2 - 30 = 2xy + 5x - 7y \\<br /> 2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> and adding these we have<br /> <br /> &lt;cmath&gt;<br /> 4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\<br /> \implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.<br /> &lt;/cmath&gt;<br /> <br /> === Solution 3 ===<br /> Let the incircle of &lt;math&gt;ABM&lt;/math&gt; hit &lt;math&gt;AM&lt;/math&gt;, &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BM&lt;/math&gt; at &lt;math&gt;X_{1},Y_{1},Z_{1}&lt;/math&gt;, and let the incircle of &lt;math&gt;CBM&lt;/math&gt; hit &lt;math&gt;MC&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;BM&lt;/math&gt; at &lt;math&gt;X_{2},Y_{2},Z_{2}&lt;/math&gt;. Draw the incircle of &lt;math&gt;ABC&lt;/math&gt;, and let it be tangent to &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;. Observe that we have a homothety centered at A sending the incircle of &lt;math&gt;ABM&lt;/math&gt; to that of &lt;math&gt;ABC&lt;/math&gt;, and one centered at &lt;math&gt;C&lt;/math&gt; taking the incircle of &lt;math&gt;BCM&lt;/math&gt; to that of &lt;math&gt;ABC&lt;/math&gt;. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is &lt;math&gt;AX_{1}/AX=CX_{2}/CX&lt;/math&gt;.<br /> <br /> By standard computations, &lt;math&gt;AX=\dfrac{AB+AC-BC}{2}=7&lt;/math&gt; and &lt;math&gt;CX=\dfrac{BC+AC-AB}{2}=8&lt;/math&gt;. Now, let &lt;math&gt;AX_{1}=7x&lt;/math&gt; and &lt;math&gt;CX_{2}=8x&lt;/math&gt;. We will now go around and chase lengths. Observe that &lt;math&gt;BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x&lt;/math&gt;. Then, &lt;math&gt;BZ_{1}=12-7x&lt;/math&gt;. We also have &lt;math&gt;CY_{2}=CX_{2}=8x&lt;/math&gt;, so &lt;math&gt;BY_{2}=13-8x&lt;/math&gt; and &lt;math&gt;BZ_{2}=13-8x&lt;/math&gt;.<br /> <br /> Observe now that &lt;math&gt;X_{1}M+MX_{2}=AC-15x=15(1-x)&lt;/math&gt;. Also,&lt;math&gt;X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)&lt;/math&gt;. Solving, we get &lt;math&gt;X_{1}M=8-8x&lt;/math&gt; and &lt;math&gt;MX_{2}=7-7x&lt;/math&gt; (as a side note, note that &lt;math&gt;AX_{1}+MX_{2}=X_{1}M+X_{2}C&lt;/math&gt;, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).<br /> <br /> Now, we get &lt;math&gt;BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x&lt;/math&gt;. To finish, we will compute area ratios. &lt;math&gt;\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}&lt;/math&gt;. Also, since their inradii are equal, we get &lt;math&gt;\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}&lt;/math&gt;. Equating and cross multiplying yields the quadratic &lt;math&gt;3x^{2}-8x+4=0&lt;/math&gt;, so &lt;math&gt;x=2/3,2&lt;/math&gt;. However, observe that &lt;math&gt;AX_{1}+CX_{2}=15x&lt;15&lt;/math&gt;, so we take &lt;math&gt;x=2/3&lt;/math&gt;. Our ratio is therefore &lt;math&gt;\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}&lt;/math&gt;, giving the answer &lt;math&gt;\boxed{045}&lt;/math&gt;.<br /> <br /> Note: Once we have &lt;math&gt;MX_1=8-8x&lt;/math&gt; and &lt;math&gt;MX_2=7-7x&lt;/math&gt;, it's bit easier to use use the right triangle of &lt;math&gt;O_1MO_2&lt;/math&gt; than chasing the area ratio. The inradius of &lt;math&gt;\triangle{ABC}&lt;/math&gt; can be calculated to be &lt;math&gt;r=\sqrt{14}&lt;/math&gt;, and the inradius of &lt;math&gt;ABM&lt;/math&gt; and &lt;math&gt;ACM&lt;/math&gt; are &lt;math&gt;r_1=r_2= xr&lt;/math&gt;, so, <br /> &lt;cmath&gt; O_1O_2^2 = O_1M^2 + O_2M^2 = r_1^2+X_1M^2 + r_2^2 + X_2M^2&lt;/cmath&gt;<br /> or,<br /> &lt;cmath&gt; (15(1-x))^2 = 2(\sqrt{14}x)^2 + (7(1-x))^2 + (8(1-x))^2 &lt;/cmath&gt;<br /> &lt;cmath&gt;112(1-x)^2 = 28x^2&lt;/cmath&gt;<br /> &lt;cmath&gt;4(1-x)^2 = x^2&lt;/cmath&gt;<br /> We get &lt;math&gt;x=\frac{2}{3}&lt;/math&gt; or &lt;math&gt;x=2&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Suppose the incircle of &lt;math&gt;ABM&lt;/math&gt; touches &lt;math&gt;AM&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, and the incircle of &lt;math&gt;CBM&lt;/math&gt; touches &lt;math&gt;CM&lt;/math&gt; at &lt;math&gt;Y&lt;/math&gt;. Then<br /> <br /> &lt;cmath&gt;r = AX \tan(A/2) = CY \tan(C/2)&lt;/cmath&gt;<br /> <br /> We have &lt;math&gt;\cos A = \frac{12^2+15^2-13^2}{2\cdot 12\cdot 15} = \frac{200}{30\cdot 12}=\frac{5}{9}&lt;/math&gt;, &lt;math&gt;\tan(A/2) = \sqrt{\frac{1-\cos A}{1+\cos A}} = \sqrt{\frac{9-5}{9+5}} = \frac{2}{\sqrt{14}}&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos C = \frac{13^2+15^2-12^2}{2\cdot 13\cdot 15} = \frac{250}{30\cdot 13} = \frac{25}{39}&lt;/math&gt;, &lt;math&gt;\tan(C/2) = \sqrt{\frac{39-25}{39+25}}=\frac{\sqrt{14}}{8}&lt;/math&gt;,<br /> <br /> Therefore &lt;math&gt;AX/CY = \tan(C/2)/\tan(A/2) = \frac{14}{2\cdot 8}= \frac{7}{8}.&lt;/math&gt;<br /> <br /> And since &lt;math&gt;AX=\frac{1}{2}(12+AM-BM)&lt;/math&gt;, &lt;math&gt;CY = \frac{1}{2}(13+CM-BM)&lt;/math&gt;, <br /> <br /> &lt;cmath&gt; \frac{12+AM-BM}{13+CM-BM} = \frac{7}{8}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt; 96+8AM-8BM = 91 +7CM-7BM&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;BM= 5 + 8AM-7CM = 5 + 15AM - 7(CM+AM) = 5+15(AM-7) \dots\dots (1)&lt;/cmath&gt;<br /> <br /> Now,<br /> <br /> &lt;math&gt;\frac{AM}{CM} = \frac{[ABM]}{[CBM]} = \frac{\frac{1}{2}(12+AM+BM)r}{\frac{1}{2}(13+CM+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;\frac{AM}{15} = \frac{17+15(AM-7)}{35+30(AM-7)} = \frac{15AM-88}{30AM-175}&lt;/cmath&gt;<br /> &lt;cmath&gt;6AM^2 - 35AM = 45AM-264&lt;/cmath&gt;<br /> &lt;cmath&gt;3AM^2 -40AM+132=0&lt;/cmath&gt;<br /> &lt;cmath&gt;(3AM-22)(AM-6)=0&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;AM=22/3&lt;/math&gt; or &lt;math&gt;6&lt;/math&gt;. But from (1) we know that &lt;math&gt;5+15(AM-7)&gt;0&lt;/math&gt;, or &lt;math&gt;AM&gt;7-1/3&gt;6&lt;/math&gt;, so &lt;math&gt;AM=22/3&lt;/math&gt;, &lt;math&gt;CM=15-22/3=23/3&lt;/math&gt;, &lt;math&gt;AM/CM=22/23&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> Let the common inradius equal r, &lt;math&gt;BM = x&lt;/math&gt;, &lt;math&gt;AM = y&lt;/math&gt;, &lt;math&gt;MC = z&lt;/math&gt;<br /> <br /> From the prespective of &lt;math&gt;\triangle{ABM}&lt;/math&gt; and &lt;math&gt;\triangle{BMC}&lt;/math&gt; we get:<br /> <br /> &lt;math&gt;S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})&lt;/math&gt; &lt;math&gt;S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})&lt;/math&gt;<br /> <br /> Add two triangles up, we get &lt;math&gt;\triangle{ABC}&lt;/math&gt; :<br /> <br /> &lt;math&gt;S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;y + z = 15&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;r = \frac{S_{ABC}}{20 + x}&lt;/math&gt;<br /> <br /> By drawing an altitude from &lt;math&gt;I_1&lt;/math&gt; down to a point &lt;math&gt;H_1&lt;/math&gt; and from &lt;math&gt;I_2&lt;/math&gt; to &lt;math&gt;H_2&lt;/math&gt;, we can get:<br /> <br /> &lt;math&gt;r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2} &lt;/math&gt; and<br /> <br /> &lt;math&gt;r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}&lt;/math&gt;<br /> <br /> Adding these up, we get:<br /> <br /> &lt;math&gt;r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x&lt;/math&gt;<br /> <br /> &lt;math&gt;r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}&lt;/math&gt;<br /> <br /> Now, we have 2 values equal to r, we can set them equal to each other:<br /> <br /> &lt;math&gt;\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}&lt;/math&gt; <br /> <br /> If we let R denote the incircle of ABC, note:<br /> <br /> AC = &lt;math&gt;(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15&lt;/math&gt; and<br /> <br /> &lt;math&gt;S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R&lt;/math&gt;. <br /> <br /> By cross multiplying the equation above, we get:<br /> <br /> &lt;math&gt;400 - x^2 = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot S_{ABC} = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R \cdot 20 = 15 \cdot 20 = 300&lt;/math&gt;<br /> <br /> We can find out x:<br /> <br /> &lt;math&gt;x = 10&lt;/math&gt;.<br /> <br /> Now, we can find ratio of y and z:<br /> <br /> &lt;math&gt;\frac{AM}{CM} = \frac{y}{z} = \frac{S_{ABM}}{S_{BCM}} = \frac{r \cdot \frac{22+y}{2} }{r \cdot \frac{23+z}{2}} = \frac{22+y}{23+z} = \frac{22}{23}&lt;/math&gt; <br /> <br /> The answer is &lt;math&gt;\boxed{045}&lt;/math&gt;. <br /> <br /> -Alexlikemath<br /> <br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=UQVI0Q2PWZw&amp;feature=youtu.be&amp;fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0<br /> <br /> == Sidenote ==<br /> In the problem, &lt;math&gt;BM=10&lt;/math&gt; and the equal inradius of the two triangles happens to be &lt;math&gt; \frac {2\sqrt{14}}{3}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=157586 2013 AMC 10A Problems/Problem 20 2021-07-09T02:53:50Z <p>Claudeaops: Fix typo</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--(0,0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{\sqrt{2}}{2}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram (or a kite with diagonals of &lt;math&gt;(\sqrt{2}-1)&lt;/math&gt; and &lt;math&gt;r \text{ or} \frac{\sqrt{2}}{2}&lt;/math&gt;) with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi \left(\frac{\sqrt{2}}{2}\right)^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;//fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1){path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));draw(arcrot);<br /> fill(arcrot--(0,0)--cycle,grey);}<br /> //draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);}<br /> draw(square^^square2);<br /> //draw((-.5,.5)--(.5,-.5)^^(0,sqrt(.5))--(0,-sqrt(.5)),dotted);draw((.5,.5)--(-.5,-.5),dotted);<br /> &lt;/asy&gt;<br /> (To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.)<br /> &lt;asy&gt;<br /> size(150,Aspect);real r=sqrt(2);real b=2-2/r;<br /> draw((0,0)--(-1,1)--(b-1,1)--(0,r)--cycle);draw((0,1)--(b-1,1)--(b/2-1,1-b/2));draw((0,0)--(b-1,1),dashed);<br /> fill((2,0)--(b+1,1)--(b+2,0)--cycle,lightgray);draw((.5,.5)--(1,.5),EndArrow);<br /> draw((2,0)--(1,1)--(b+1,1)--(b+2,0)--(2,0)^^(b+1,1)--(b/2+1,1-b/2)^^(2,0)--(2+b/2,b/2));<br /> draw((2,0)--(b+1,1),dashed);<br /> &lt;/asy&gt;<br /> Alternatively, you can move the dart-shaped piece to the other side and make a kite.<br /> &lt;asy&gt;<br /> size(75,Aspect);real r=sqrt(2);real b=2-2/r;<br /> draw((r-1,1)--(b-1,1));<br /> draw((0,0)--(b-1,1)--(0,r)--(r-1,1)--cycle);<br /> draw((0,r)--(0,0),dashed);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 2==<br /> [[Image:AMC 10A 2013 20.jpg]]<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> for(int i=0;i&lt;=6;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1));<br /> draw(arcrot);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> &lt;center&gt;&lt;math&gt;\textbf{(high res image; no labels)}&lt;/math&gt;&lt;/center&gt;<br /> <br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the square and &lt;math&gt;C&lt;/math&gt; be the intersection of &lt;math&gt;OB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. The desired area consists of the unit square, plus &lt;math&gt;4&lt;/math&gt; regions congruent to the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; triangular regions congruent to right triangle &lt;math&gt;BCD&lt;/math&gt;. The area of the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt; is &lt;math&gt;\frac{\text{Area of Circle}-\text{Area of Square}}{8}&lt;/math&gt;. Since the circle has radius &lt;math&gt;\dfrac{1}{\sqrt {2}}&lt;/math&gt;, the area of the region is &lt;math&gt;\dfrac{\dfrac{\pi}{2}-1}{8}&lt;/math&gt;, so 4 times the area of that region is &lt;math&gt;\dfrac{\pi}{4}-\dfrac{1}{2}&lt;/math&gt;. Now we find the area of &lt;math&gt;\triangle BCD&lt;/math&gt;. &lt;math&gt;BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}&lt;/math&gt;. Since &lt;math&gt;\triangle BCD&lt;/math&gt; is a &lt;math&gt;45-45-90&lt;/math&gt; right triangle, the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}&lt;/math&gt;, so &lt;math&gt;4&lt;/math&gt; times the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\dfrac{3}{2}-\sqrt {2}&lt;/math&gt;. Finally, the area of the whole region is &lt;math&gt;1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}&lt;/math&gt;, which we can rewrite as &lt;math&gt;\boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_11&diff=157585 2002 AIME II Problems/Problem 11 2021-07-09T02:48:38Z <p>Claudeaops: Add info about first terms and common ratios of series</p> <hr /> <div>== Problem ==<br /> Two distinct, real, infinite geometric series each have a sum of &lt;math&gt;1&lt;/math&gt; and have the same second term. The third term of one of the series is &lt;math&gt;1/8&lt;/math&gt;, and the second term of both series can be written in the form &lt;math&gt;\frac{\sqrt{m}-n}p&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers and &lt;math&gt;m&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;100m+10n+p&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Let the second term of each series be &lt;math&gt;x&lt;/math&gt;. Then, the common ratio is &lt;math&gt;\frac{1}{8x}&lt;/math&gt;, and the first term is &lt;math&gt;8x^2&lt;/math&gt;. <br /> <br /> So, the sum is &lt;math&gt;\frac{8x^2}{1-\frac{1}{8x}}=1&lt;/math&gt;. Thus, &lt;math&gt;64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}&lt;/math&gt;. <br /> <br /> The only solution in the appropriate form is &lt;math&gt;x = \frac{\sqrt{5}-1}{8}&lt;/math&gt;. Therefore, &lt;math&gt;100m+10n+p = \boxed{518}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> <br /> Let the two sequences be &lt;math&gt;a, ar, ar^2 ... \text{ }an^2&lt;/math&gt; and &lt;math&gt;x, xy, xy^2 ... \text{ }xy^n&lt;/math&gt;. We know for a fact that &lt;math&gt;ar = xy&lt;/math&gt;. We also know that the sum of the first sequence = &lt;math&gt;\frac{a}{1-r} = 1&lt;/math&gt;, and the sum of the second sequence = &lt;math&gt;\frac{x}{1-y} = 1&lt;/math&gt;. Therefore we have &lt;cmath&gt;a+r = 1&lt;/cmath&gt;&lt;cmath&gt;x+y = 1&lt;/cmath&gt;&lt;cmath&gt;ar=xy&lt;/cmath&gt; We can then replace &lt;math&gt;r = \frac{xy}{a}&lt;/math&gt; and &lt;math&gt;y = \frac{ar}{x}&lt;/math&gt;. We plug them into the two equations &lt;math&gt;a+r = 1&lt;/math&gt; and &lt;math&gt;x+y = 1&lt;/math&gt;. We then get &lt;cmath&gt;x^2 + ar = x&lt;/cmath&gt;&lt;cmath&gt;a^2 + xy = a&lt;/cmath&gt;We subtract these equations, getting &lt;cmath&gt;x^2 - a^2 + ar - xy = x-a&lt;/cmath&gt;Remember &lt;cmath&gt;ar=xy&lt;/cmath&gt;, so &lt;cmath&gt;(x-a)(x+a-1) = 0&lt;/cmath&gt;Then considering cases, we have either &lt;math&gt;x=a&lt;/math&gt; or &lt;math&gt;y=a&lt;/math&gt;. This suggests that the second sequence is in the form &lt;math&gt;r, ra, ra^2...&lt;/math&gt;, while the first sequence is in the form &lt;math&gt;a, ar, ar^2...&lt;/math&gt; Now we have that &lt;math&gt;ar^2 = \frac18&lt;/math&gt; and we also have that &lt;math&gt;a+r = 1&lt;/math&gt;. We can solve for &lt;math&gt;r&lt;/math&gt; and the only appropriate value for &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\frac{1+\sqrt{5}}{4}&lt;/math&gt;. All we want is the second term, which is &lt;math&gt;ar = \frac{ar^2}{r} = \frac{\frac18}{\frac{1+\sqrt{5}}{4}} = \frac{\sqrt{5} - 1}{8}&lt;/math&gt;<br /> solution by jj_ca888<br /> <br /> == Solution 3 == <br /> <br /> Let's ignore the &quot;two distinct, real, infinite geometric series&quot; part for now and focus on what it means to be a geometric series.<br /> <br /> Let the first term of the series with the third term equal to &lt;math&gt;\frac18&lt;/math&gt; be &lt;math&gt;a,&lt;/math&gt; and the common ratio be &lt;math&gt;r.&lt;/math&gt; Then, we get that &lt;math&gt;\frac{a}{1-r} = 1 \implies a = 1-r,&lt;/math&gt; and &lt;math&gt;ar^2 = \frac18 \implies (1-r)(r^2) = \frac18.&lt;/math&gt;<br /> <br /> We see that this cubic is equivalent to &lt;math&gt;r^3 - r^2 + \frac18 = 0.&lt;/math&gt; Through experimenting, we find that one of the solutions is &lt;math&gt;r = \frac12.&lt;/math&gt; Using synthetic division leads to the quadratic &lt;math&gt;4x^2 - 2x - 1 = 0.&lt;/math&gt; This has roots &lt;math&gt;\dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8},&lt;/math&gt; or, when reduced, &lt;math&gt;\dfrac{1 \pm \sqrt{5}}{4}.&lt;/math&gt;<br /> <br /> It becomes clear that the two geometric series have common ratio &lt;math&gt;\frac{1 + \sqrt{5}}{4}&lt;/math&gt; and &lt;math&gt;\frac{1 - \sqrt{5}}{4}.&lt;/math&gt; Let &lt;math&gt;\frac{1 + \sqrt{5}}{4}&lt;/math&gt; be the ratio that we are inspecting. We see that in this case, &lt;math&gt;a = \dfrac{3 - \sqrt{5}}{4}.&lt;/math&gt;<br /> <br /> Since the second term in the series is &lt;math&gt;ar,&lt;/math&gt; we compute this and have that &lt;cmath&gt;ar = \left(\dfrac{3 - \sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right) = \dfrac{\sqrt{5} - 1}{8},&lt;/cmath&gt;for our answer of &lt;math&gt;100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.&lt;/math&gt;<br /> <br /> Solution by Ilikeapos<br /> <br /> == Sidenote ==<br /> One of the geometric series has first term &lt;math&gt;\frac{3 - \sqrt{5}}{4}&lt;/math&gt; and common ratio &lt;math&gt;\frac{1 + \sqrt{5}}{4}&lt;/math&gt;, and its third term is &lt;math&gt;\frac{1}{8}&lt;/math&gt;. The other series has first term &lt;math&gt;\frac{1 + \sqrt{5}}{4}&lt;/math&gt; and common ratio &lt;math&gt;\frac{3 - \sqrt{5}}{4}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2002|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_15&diff=157584 2002 AIME II Problems/Problem 15 2021-07-09T02:22:58Z <p>Claudeaops: Add remark about centers of circles</p> <hr /> <div>== Problem ==<br /> Circles &lt;math&gt;\mathcal{C}_{1}&lt;/math&gt; and &lt;math&gt;\mathcal{C}_{2}&lt;/math&gt; intersect at two points, one of which is &lt;math&gt;(9,6)&lt;/math&gt;, and the product of the radii is &lt;math&gt;68&lt;/math&gt;. The x-axis and the line &lt;math&gt;y = mx&lt;/math&gt;, where &lt;math&gt;m &gt; 0&lt;/math&gt;, are tangent to both circles. It is given that &lt;math&gt;m&lt;/math&gt; can be written in the form &lt;math&gt;a\sqrt {b}/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 positive integers, &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime, and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime. Find &lt;math&gt;a + b + c&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Let the smaller angle between the &lt;math&gt;x&lt;/math&gt;-axis and the line &lt;math&gt;y=mx&lt;/math&gt; be &lt;math&gt;\theta&lt;/math&gt;. Note that the centers of the two circles lie on the angle bisector of the angle between the &lt;math&gt;x&lt;/math&gt;-axis and the line &lt;math&gt;y=mx&lt;/math&gt;. Also note that if &lt;math&gt;(x,y)&lt;/math&gt; is on said angle bisector, we have that &lt;math&gt;\frac{y}{x}=\tan{\frac{\theta}{2}}&lt;/math&gt;. Let &lt;math&gt;\tan{\frac{\theta}{2}}=m_1&lt;/math&gt;, for convenience. Therefore if &lt;math&gt;(x,y)&lt;/math&gt; is on the angle bisector, then &lt;math&gt;x=\frac{y}{m_1}&lt;/math&gt;. Now let the centers of the two relevant circles be &lt;math&gt;(a/m_1 , a)&lt;/math&gt; and &lt;math&gt;(b/m_1 , b)&lt;/math&gt; for some positive reals &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. These two circles are tangent to the &lt;math&gt;x&lt;/math&gt;-axis, so the radii of the circles are &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively. We know that the point &lt;math&gt;(9,6)&lt;/math&gt; is a point on both circles, so we have that<br /> <br /> &lt;cmath&gt;(9-\frac{a}{m_1})^2+(6-a)^2=a^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;(9-\frac{b}{m_1})^2+(6-b)^2=b^2&lt;/cmath&gt;<br /> <br /> Expanding these and manipulating terms gives<br /> <br /> &lt;cmath&gt;\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{1}{m_1^2}b^2-[(18/m_1)+12]b+117=0&lt;/cmath&gt;<br /> <br /> It follows that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are the roots of the quadratic<br /> <br /> &lt;cmath&gt;\frac{1}{m_1^2}x^2-[(18/m_1)+12]x+117=0&lt;/cmath&gt;<br /> <br /> It follows from Vieta's Formulas that the product of the roots of this quadratic is &lt;math&gt;117m_1^2&lt;/math&gt;, but we were also given that the product of the radii was 68. Therefore &lt;math&gt;68=117m_1^2&lt;/math&gt;, or &lt;math&gt;m_1^2=\frac{68}{117}&lt;/math&gt;. Note that the half-angle formula for tangents is<br /> <br /> &lt;cmath&gt;\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos{\theta}}{1+\cos{\theta}}}&lt;/cmath&gt;<br /> <br /> Therefore<br /> <br /> &lt;cmath&gt;\frac{68}{117}=\frac{1-\cos{\theta}}{1+\cos{\theta}}&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;\cos{\theta}&lt;/math&gt; gives that &lt;math&gt;\cos{\theta}=\frac{49}{185}&lt;/math&gt;. It then follows that &lt;math&gt;\sin{\theta}=\sqrt{1-\cos^2{\theta}}=\frac{12\sqrt{221}}{185}&lt;/math&gt;.<br /> <br /> It then follows that &lt;math&gt;m=\tan{\theta}=\frac{12\sqrt{221}}{49}&lt;/math&gt;. Therefore &lt;math&gt;a=12&lt;/math&gt;, &lt;math&gt;b=221&lt;/math&gt;, and &lt;math&gt;c=49&lt;/math&gt;. The desired answer is then &lt;math&gt;12+221+49=\boxed{282}&lt;/math&gt;.<br /> <br /> == Solution 2 (Alcumus)==<br /> Let &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt; be the radii of the circles. Then the centers of the circles are of the form &lt;math&gt;(kr_1,r_1)&lt;/math&gt; and &lt;math&gt;(kr_2,r_2)&lt;/math&gt; for the same constant &lt;math&gt;k,&lt;/math&gt; since the two centers are collinear with the origin. Since &lt;math&gt;(9,6)&lt;/math&gt; lies on both circles,<br /> &lt;cmath&gt;(kr - 9)^2 + (r - 6)^2 = r^2,&lt;/cmath&gt;where &lt;math&gt;r&lt;/math&gt; represents either radius. Expanding, we get<br /> &lt;cmath&gt;k^2 r^2 - (18k + 12) r + 117 = 0.&lt;/cmath&gt;We are told the product of the circles is 68, so by Vieta's formulas, &lt;math&gt;\frac{117}{k^2} = 68.&lt;/math&gt; Hence, &lt;math&gt;k^2 = \frac{117}{68},&lt;/math&gt; and &lt;math&gt;k = \sqrt{\frac{117}{68}}.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(0.25 cm);<br /> <br /> pair[] O;<br /> real[] r;<br /> pair P;<br /> <br /> r = 4.096;<br /> r = 16.6;<br /> O = (r/(2/3*sqrt(17/13)),r);<br /> O = (r/(2/3*sqrt(17/13)),r);<br /> P = reflect(O,O)*(9,6);<br /> <br /> draw(Circle(O,r));<br /> //draw(Circle(O,r));<br /> draw(arc(O,r,130,300));<br /> draw((0,0)--(8,12*sqrt(221)/49*8));<br /> draw((0,0)--(30,0));<br /> draw((0,0)--O--(O.x,0));<br /> draw(O--(O + reflect((0,0),(10,12*sqrt(221)/49*10))*(O))/2);<br /> <br /> label(&quot;$y = mx$&quot;, (8,12*sqrt(221)/49*8), N);<br /> <br /> dot(&quot;$(9,6)$&quot;, (9,6), NE);<br /> dot(&quot;$(kr,r)$&quot;, O, N);<br /> dot(P,red);<br /> &lt;/asy&gt;<br /> <br /> Since the circle is tangent to the line &lt;math&gt;y = mx,&lt;/math&gt; the distance from the center &lt;math&gt;(kr,r)&lt;/math&gt; to the line is &lt;math&gt;r.&lt;/math&gt; We can write &lt;math&gt;y = mx&lt;/math&gt; as &lt;math&gt;y - mx = 0,&lt;/math&gt; so from the distance formula,<br /> &lt;cmath&gt;\frac{|r - krm|}{\sqrt{1 + m^2}} = r.&lt;/cmath&gt;Squaring both sides, we get<br /> &lt;cmath&gt;\frac{(r - krm)^2}{1 + m^2} = r^2,&lt;/cmath&gt;so &lt;math&gt;(r - krm)^2 = r^2 (1 + m^2).&lt;/math&gt; Since &lt;math&gt;r \neq 0,&lt;/math&gt; we can divide both sides by r, to get<br /> &lt;cmath&gt;(1 - km)^2 = 1 + m^2.&lt;/cmath&gt;Then &lt;math&gt;1 - 2km + k^2 m^2 = 1 + m^2,&lt;/math&gt; so &lt;math&gt;m^2 (1 - k^2) + 2km = 0.&lt;/math&gt; Since &lt;math&gt;m \neq 0,&lt;/math&gt;<br /> &lt;cmath&gt;m(1 - k^2) + 2k = 0.&lt;/cmath&gt;Hence,<br /> &lt;cmath&gt;m = \frac{2k}{k^2 - 1} = \frac{2 \sqrt{\frac{117}{68}}}{\frac{117}{68} - 1} = \boxed{\frac{12 \sqrt{221}}{49}}.&lt;/cmath&gt;<br /> <br /> == Sidenote ==<br /> The two circles are centered at &lt;cmath&gt;\left (\frac{1}{13}\left (117 + 4 \sqrt{221} \pm 2 \sqrt{13(18 \sqrt{221} - 49)} \right), \frac{2}{39} \left(68 + 9 \sqrt {221} \pm 2 \sqrt{17(18 \sqrt{221} - 49)} \right) \right)&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2002|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_11&diff=92466 2017 AIME II Problems/Problem 11 2018-02-28T18:13:11Z <p>Claudeaops: Extension</p> <hr /> <div>==Problem==<br /> Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).<br /> <br /> ==Solution==<br /> It is obvious that any configuration of one-way roads which contains a town whose roads all lead into it or lead out of it cannot satisfy the given. We claim that any configuration which does not have a town whose roads all lead into it or lead out of it does satisfy the given conditions. Now we show that a loop of &lt;math&gt;3&lt;/math&gt; or more towns exist. Pick a town, then choose a neighboring town to travel to &lt;math&gt;5&lt;/math&gt; times. Of these &lt;math&gt;6&lt;/math&gt; towns visited, at least two of them must be the same; therefore there must exist a loop of &lt;math&gt;3&lt;/math&gt; or more towns because a loop of &lt;math&gt;2&lt;/math&gt; towns cannot exist. We want to show that the loop can be reached from any town, and any town can be reached from the loop.<br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;. The loop has &lt;math&gt;5&lt;/math&gt; towns.<br /> Clearly every town can be reached by going around the loop.<br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;. The loop has &lt;math&gt;4&lt;/math&gt; towns.<br /> The town not on the loop must have a road leading to it. This road comes from a town on the loop. Therefore this town can be reached from the loop. This town not on the loop must also have a road leading out of it. This road leads to a town on the loop. Therefore the loop can be reached from the town.<br /> <br /> &lt;math&gt;\textbf{Case 3}&lt;/math&gt;. The loop has &lt;math&gt;3&lt;/math&gt; towns.<br /> There are two towns not on the loop; call them Town &lt;math&gt;A&lt;/math&gt; and Town &lt;math&gt;B&lt;/math&gt;. Without loss of generality assume &lt;math&gt;A&lt;/math&gt; leads to &lt;math&gt;B&lt;/math&gt;. Because a road must lead to &lt;math&gt;A&lt;/math&gt;, the town where this road comes from must be on the loop. Therefore &lt;math&gt;A&lt;/math&gt; and therefore &lt;math&gt;B&lt;/math&gt; can be reached from the loop. Because a road must lead out of &lt;math&gt;B&lt;/math&gt;, the town it leads to must be on the loop. Therefore the loop can be reached from &lt;math&gt;B&lt;/math&gt; and also &lt;math&gt;A&lt;/math&gt;.<br /> <br /> The number of good configurations is the total number of configurations minus the number of bad configurations. There are &lt;math&gt;2^{{5\choose2}}&lt;/math&gt; total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are &lt;math&gt;5&lt;/math&gt; ways to choose this town and &lt;math&gt;2^6&lt;/math&gt; ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are &lt;math&gt;5 \cdot 2^6+5 \cdot 2^6&lt;/math&gt; bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are &lt;math&gt;5&lt;/math&gt; ways to choose the town for which all roads lead to it, &lt;math&gt;4&lt;/math&gt; ways to choose the town for which all roads lead out of it, and &lt;math&gt;2^3&lt;/math&gt; ways to assign the remaining &lt;math&gt;3&lt;/math&gt; roads not connected to either of these towns. Therefore, the answer is &lt;math&gt;2^{{5\choose2}}-(5 \cdot 2^6+5 \cdot 2^6-5\cdot 4 \cdot 2^3)=\boxed{544}&lt;/math&gt;.<br /> <br /> ==Solution 2 (complementary counting)==<br /> The only way a town does not meet the conditions in the question is if the town has either all roads leading towards it or all roads leading away from it. For example, if all roads lead away from Town &lt;math&gt;A&lt;/math&gt;, there is no way to reach the town starting from Towns &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, or &lt;math&gt;D&lt;/math&gt;. If all roads lead towards Town &lt;math&gt;A&lt;/math&gt;, there is no way to reach any other town starting from Town &lt;math&gt;A&lt;/math&gt;. Thus, we will count the ways this occurs. <br /> <br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;. WLOG, Town &lt;math&gt;A&lt;/math&gt; has all roads leading towards it.<br /> <br /> In this case, the four roads leading to Town &lt;math&gt;A&lt;/math&gt; have already been determined. There are &lt;math&gt;6&lt;/math&gt; roads that have not been given directions. Each of these roads have &lt;math&gt;2&lt;/math&gt; options: it can lead towards one town or the other town. Thus, there are &lt;math&gt;2^6&lt;/math&gt; arrangements of roads.<br /> However, we must consider the &lt;math&gt;5&lt;/math&gt; towns in which this scenario can occur: there are &lt;math&gt;5 \cdot 2^6=320&lt;/math&gt; arrangements.<br /> <br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;. WLOG, Town &lt;math&gt;A&lt;/math&gt; has all roads leading away.<br /> <br /> Notice that this case is symmetrical to the first case. Thus, there are &lt;math&gt;320&lt;/math&gt; arrangements.<br /> <br /> <br /> &lt;math&gt;\textbf{Case 3}&lt;/math&gt;. WLOG, Town &lt;math&gt;A&lt;/math&gt; has all roads leading towards it and Town &lt;math&gt;B&lt;/math&gt; has all roads leading away.<br /> <br /> We must check for double-counted cases. Drawing a flow diagram, we see that this case determines &lt;math&gt;7&lt;/math&gt; roads. For the undetermined &lt;math&gt;3&lt;/math&gt; roads, there are &lt;math&gt;2^3=8&lt;/math&gt; arrangements. However, we must again consider the different ways that this case can occur. There are &lt;math&gt;5&lt;/math&gt; choices for towns with roads leading towards it &lt;math&gt;4&lt;/math&gt; choices for roads leading away. Thus, the total double-counted cases are &lt;math&gt;5 \cdot 4 \cdot 8=160&lt;/math&gt; arrangements.<br /> <br /> We must subtract the cases we counted from the total. There are a total of &lt;math&gt;10&lt;/math&gt; roads with &lt;math&gt;2&lt;/math&gt; possible arrangements each.<br /> Therefore the total number of cases is &lt;math&gt;2^{10} =1024&lt;/math&gt;, and the number of cases that meet the conditions is &lt;math&gt;1024 - (320+320-160) = \boxed{544}&lt;/math&gt; arrangements.<br /> <br /> ~ jf<br /> <br /> =See Also=<br /> Another way to state this problem is finding the number of strongly connected labeled tournaments on &lt;math&gt;5&lt;/math&gt; nodes. Finding a strongly connected labeled tournament is the same as finding a way to make all the roads one-way such that there is always a way to get from one town to another. [http://oeis.org/A054946 This link] gives the number of strongly connected labeled tournaments on &lt;math&gt;n&lt;/math&gt; nodes for small &lt;math&gt;n&lt;/math&gt;. <br /> {{AIME box|year=2017|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_12&diff=88967 2001 AIME II Problems/Problem 12 2017-12-19T01:19:58Z <p>Claudeaops: </p> <hr /> <div>== Problem ==<br /> Given a [[triangle]], its [[midpoint]] triangle is obtained by joining the midpoints of its sides. A sequence of [[polyhedra]] &lt;math&gt;P_{i}&lt;/math&gt; is defined recursively as follows: &lt;math&gt;P_{0}&lt;/math&gt; is a regular [[tetrahedron]] whose volume is 1. To obtain &lt;math&gt;P_{i + 1}&lt;/math&gt;, replace the midpoint triangle of every face of &lt;math&gt;P_{i}&lt;/math&gt; by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The [[volume]] of &lt;math&gt;P_{3}&lt;/math&gt; is &lt;math&gt;\frac {m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> On the first construction, &lt;math&gt;P_1&lt;/math&gt;, four new tetrahedra will be constructed with side lengths &lt;math&gt;\frac 12&lt;/math&gt; of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume &lt;math&gt;\left(\frac 12\right)^3 = \frac 18&lt;/math&gt;. The total volume added here is then &lt;math&gt;\Delta P_1 = 4 \cdot \frac 18 = \frac 12&lt;/math&gt;.<br /> <br /> We now note that for each midpoint triangle we construct in step &lt;math&gt;P_{i}&lt;/math&gt;, there are now &lt;math&gt;6&lt;/math&gt; places to construct new midpoint triangles for step &lt;math&gt;P_{i+1}&lt;/math&gt;. The outward tetrahedron for the midpoint triangle provides &lt;math&gt;3&lt;/math&gt; of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other &lt;math&gt;3&lt;/math&gt;. This is because if you read this question carefully, it asks to add new tetrahedra to each face of &lt;math&gt;P_{i}&lt;/math&gt; which also includes the ones that were left over when we did the previous addition of tetrahedra. However, the volume of the tetrahedra being constructed decrease by a factor of &lt;math&gt;\frac 18&lt;/math&gt;. Thus we have the recursion &lt;math&gt;\Delta P_{i+1} = \frac{6}{8} \Delta P_i&lt;/math&gt;, and so &lt;math&gt;\Delta P_i = \frac 12 \cdot \left(\frac{3}{4}\right)^{i-1} P_1&lt;/math&gt;.<br /> <br /> The volume of &lt;math&gt;P_3 = P_0 + \Delta P_1 + \Delta P_2 + \Delta P_3 = 1 + \frac 12 + \frac 38 + \frac 9{32} = \frac{69}{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{101}&lt;/math&gt;. Note that the summation was in fact a [[geometric series]].<br /> <br /> == See also ==<br /> https://www.math.hmc.edu/funfacts/ffiles/30002.2.shtml<br /> http://users.math.yale.edu/public_html/People/frame/Fractals/Labs/KochTetra/KochTetraAns3.html<br /> {{AIME box|year=2001|n=II|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=1998_AHSME_Problems/Problem_27&diff=88966 1998 AHSME Problems/Problem 27 2017-12-19T01:02:33Z <p>Claudeaops: </p> <hr /> <div>== Problem ==<br /> A &lt;math&gt;9 \times 9 \times 9&lt;/math&gt; [[cube]] is composed of twenty-seven &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; cubes. The big cube is ‘tunneled’ as follows: First, the six &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; cubes which make up the center of each [[face]] as well as the center &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; cube are removed. Second, each of the twenty remaining &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; cubes is diminished in the same way. That is, the center facial unit cubes as well as each center cube are removed. The [[surface area]] of the final figure is:<br /> <br /> [[Image:1998 AHSME num. 27.png]]<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 384<br /> \qquad\mathrm{(B)}\ 729<br /> \qquad\mathrm{(C)}\ 864<br /> \qquad\mathrm{(D)}\ 1024<br /> \qquad\mathrm{(E)}\ 1056&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Each &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; cube has eight faces on each side, for a surface area of &lt;math&gt;6 \cdot 8 \cdot (1 \cdot 1) = 48&lt;/math&gt; on the outside. Each face also has to count the surface area in the inside area of the removed cube, for an additional surface area of &lt;math&gt;6 \cdot 4 \cdot (1 \cdot 1) = 24&lt;/math&gt;. Thus the total surface area for each &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; is &lt;math&gt;72&lt;/math&gt;.<br /> <br /> There are &lt;math&gt;20&lt;/math&gt; of these cubes, for an area of &lt;math&gt;72 \cdot 20 = 1440&lt;/math&gt;. However, many of the cubes share a common face; each corner &lt;math&gt;3\times 3\times 3&lt;/math&gt; cube has three hidden faces and each edge cube has two hidden faces, for a total of &lt;math&gt;8\cdot 3 + 12\cdot 2 = 48&lt;/math&gt; hidden faces. Each hidden face has a surface area of &lt;math&gt;8&lt;/math&gt;, so the surface area of the final figure is &lt;math&gt;1440 - 48 \cdot 8 = 1056 \Rightarrow \mathrm{(E)}&lt;/math&gt;. <br /> <br /> === Solution 2 ===<br /> After the first step, twenty &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; cubes remain, with &lt;math&gt;8&lt;/math&gt; corner cubes and &lt;math&gt;12&lt;/math&gt; edge cubes. Each one of these &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; corner cubes contributes &lt;math&gt;27&lt;/math&gt; square units of area, and each edge cube contributes &lt;math&gt;36&lt;/math&gt; square units of area. <br /> <br /> The second stage takes away &lt;math&gt;3&lt;/math&gt; square units of area (&lt;math&gt;1&lt;/math&gt; for each exposed face) from each of the eight &lt;math&gt;3 \times 3 \times 3&lt;/math&gt; corner cubes, and adds an additional &lt;math&gt;24&lt;/math&gt; more units from the center facial cubes removed. Similarly, the twelve &lt;math&gt;3\times 3\times 3&lt;/math&gt; edge cubes each lose &lt;math&gt;4&lt;/math&gt; square nits but gain &lt;math&gt;24&lt;/math&gt; units. Thus, the total surface area is <br /> &lt;cmath&gt;8 \cdot (27 - 3 + 24) + 12 \cdot (36 - 4 + 24) = 1056&lt;/cmath&gt;<br /> <br /> == See also ==<br /> [http://mathworld.wolfram.com/MengerSponge.html Menger Sponge]<br /> {{AHSME box|year=1998|num-b=26|num-a=28}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=File:Cute_Geo_Problem.png&diff=85479 File:Cute Geo Problem.png 2017-04-29T23:40:21Z <p>Claudeaops: </p> <hr /> <div></div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_15&diff=81839 2010 AIME II Problems/Problem 15 2016-12-10T23:01:02Z <p>Claudeaops: Extension</p> <hr /> <div>== Problem 15 ==<br /> <br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AC = 13&lt;/math&gt;, &lt;math&gt;BC = 14&lt;/math&gt;, and &lt;math&gt;AB=15&lt;/math&gt;. Points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; lie on &lt;math&gt;AC&lt;/math&gt; with &lt;math&gt;AM=MC&lt;/math&gt; and &lt;math&gt;\angle ABD = \angle DBC&lt;/math&gt;. Points &lt;math&gt;N&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; lie on &lt;math&gt;A&lt;/math&gt;B with &lt;math&gt;AN=NB&lt;/math&gt; and &lt;math&gt;\angle ACE = \angle ECB&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point, other than &lt;math&gt;A&lt;/math&gt;, of intersection of the circumcircles of &lt;math&gt;\triangle AMN&lt;/math&gt; and &lt;math&gt;\triangle ADE&lt;/math&gt;. Ray &lt;math&gt;AP&lt;/math&gt; meets &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt;. The ratio &lt;math&gt;\frac{BQ}{CQ}&lt;/math&gt; can be written in the form &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m-n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;Y = MN \cap AQ&lt;/math&gt;. &lt;math&gt;\frac {BQ}{QC} = \frac {NY}{MY}&lt;/math&gt; since &lt;math&gt;\triangle AMN \sim \triangle ACB&lt;/math&gt;. Since quadrilateral &lt;math&gt;AMPN&lt;/math&gt; is cyclic, &lt;math&gt;\triangle MYA \sim \triangle PYN&lt;/math&gt; and &lt;math&gt;\triangle MYP \sim \triangle AYN&lt;/math&gt;, yielding &lt;math&gt;\frac {YM}{YA} = \frac {MP}{AN}&lt;/math&gt; and &lt;math&gt;\frac {YA}{YN} = \frac {AM}{PN}&lt;/math&gt;. Multiplying these together yields &lt;math&gt;\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}&lt;/math&gt;. Also, &lt;math&gt;P&lt;/math&gt; is the center of spiral similarity of segments &lt;math&gt;MD&lt;/math&gt; and &lt;math&gt;NE&lt;/math&gt;, so &lt;math&gt;\triangle PMD \sim \triangle PNE&lt;/math&gt;. Therefore, &lt;math&gt;\frac {PN}{PM} = \frac {NE}{MD}&lt;/math&gt;, which can easily be computed by the angle bisector theorem to be &lt;math&gt;\frac {145}{117}&lt;/math&gt;. It follows that &lt;math&gt;\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}&lt;/math&gt;, giving us an answer of &lt;math&gt;725 - 507 = \boxed{218}&lt;/math&gt;.<br /> <br /> '''Note:''' Spiral similarities may sound complex, but they're really not. The fact that &lt;math&gt;\triangle PMD \sim \triangle PNE&lt;/math&gt; is really just a result of simple angle chasing.<br /> <br /> Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero<br /> <br /> == Extension ==<br /> The work done in this problem leads to a nice extension of this problem:<br /> <br /> Given a &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;A_1&lt;/math&gt;, &lt;math&gt;A_2&lt;/math&gt;, &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_2&lt;/math&gt;, &lt;math&gt;C_1&lt;/math&gt;, &lt;math&gt;C_2&lt;/math&gt;, such that &lt;math&gt;A_1&lt;/math&gt;, &lt;math&gt;A_2&lt;/math&gt; &lt;math&gt;\in BC&lt;/math&gt;, &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_2&lt;/math&gt; &lt;math&gt;\in AC&lt;/math&gt;, and &lt;math&gt;C_1&lt;/math&gt;, &lt;math&gt;C_2&lt;/math&gt; &lt;math&gt;\in AB&lt;/math&gt;, then let &lt;math&gt;\omega_1&lt;/math&gt; be the circumcircle of &lt;math&gt;\triangle AB_1C_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; be the circumcircle of &lt;math&gt;\triangle AB_2C_2&lt;/math&gt;. Let &lt;math&gt;A'&lt;/math&gt; be the intersection point of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; distinct from &lt;math&gt;A&lt;/math&gt;. Define &lt;math&gt;B'&lt;/math&gt; and &lt;math&gt;C'&lt;/math&gt; similarly. Then &lt;math&gt;AA'&lt;/math&gt;, &lt;math&gt;BB'&lt;/math&gt;, and &lt;math&gt;CC'&lt;/math&gt; concur. <br /> <br /> This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line &lt;math&gt;AA'&lt;/math&gt; divides the opposite side &lt;math&gt;BC&lt;/math&gt; into and similarly for the other two sides.<br /> <br /> ==See Also==<br /> {{AIME box|year=2010|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2002_USAMO_Problems/Problem_2&diff=77962 2002 USAMO Problems/Problem 2 2016-04-07T02:32:00Z <p>Claudeaops: latex</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;ABC &lt;/math&gt; be a triangle such that<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> where &lt;math&gt;s &lt;/math&gt; and &lt;math&gt;r &lt;/math&gt; denote its [[semiperimeter]] and [[inradius]], respectively. Prove that triangle &lt;math&gt;ABC &lt;/math&gt; is similar to a triangle &lt;math&gt;T &lt;/math&gt; whose side lengths are all positive integers with no common divisor and determine those integers.<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;a,b,c &lt;/math&gt; denote &lt;math&gt;BC, CA, AB &lt;/math&gt;, respectively. Since the line from a triangle's incenter to one of its vertices bisects the angle at the triangle's vertex, the condition of the problem is equivalent to the<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \left( \frac{s-a}{r} \right)^2 + 4\left( \frac{s-b}{r} \right)^2 + 9\left( \frac{s-c}{r} \right)^2 = \left( \frac{6s}{7r} \right)^2<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> or<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} = \frac{s^2}{36 + 9 + 4}<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> But by the [[Cauchy-Schwarz Inequality]], we know<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \begin{matrix}<br /> (36 + 9 + 4) \left[ \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} \right] &amp; \ge &amp;\left[ (s-a) + (s-b) + (s-c) \right]^2\\<br /> &amp; = &amp; s^2 \\<br /> \qquad\qquad \quad \quad \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} &amp; \ge &amp;\frac{s^2}{36 + 9 + 4} \; ,<br /> \end{matrix}<br /> &lt;/math&gt;<br /> &lt;/center&gt;<br /> with equality only when &lt;math&gt; \frac{(s-a)^2}{36}, \frac{(s-b)^2}{9}, \frac{(s-c)^2}{4}&lt;/math&gt; are directly proportional to 36, 9, 4, respectively. Therefore (clearing denominators and taking square roots) our problem requires that &lt;math&gt;(s-a), (s-b), (s-c) &lt;/math&gt; be directly proportional to 36, 9, 4, and since &lt;math&gt;a = (s-b) + (s-c) &lt;/math&gt; etc., this is equivalent to the condition that &lt;math&gt;a,b,c &lt;/math&gt; be in proportion with 13, 40, 45, Q.E.D.<br /> <br /> Sidenote: A 13, 40, 45 obtuse triangle has an integer area of 252 and an inradius of &lt;math&gt;\frac{36}{7}&lt;/math&gt;.<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> {{USAMO newbox|year=2002|num-b=1|num-a=3}}<br /> <br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> [[Category:Olympiad Inequality Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2002_USAMO_Problems/Problem_2&diff=77961 2002 USAMO Problems/Problem 2 2016-04-07T02:31:11Z <p>Claudeaops: Sidenote</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;ABC &lt;/math&gt; be a triangle such that<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> where &lt;math&gt;s &lt;/math&gt; and &lt;math&gt;r &lt;/math&gt; denote its [[semiperimeter]] and [[inradius]], respectively. Prove that triangle &lt;math&gt;ABC &lt;/math&gt; is similar to a triangle &lt;math&gt;T &lt;/math&gt; whose side lengths are all positive integers with no common divisor and determine those integers.<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;a,b,c &lt;/math&gt; denote &lt;math&gt;BC, CA, AB &lt;/math&gt;, respectively. Since the line from a triangle's incenter to one of its vertices bisects the angle at the triangle's vertex, the condition of the problem is equivalent to the<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \left( \frac{s-a}{r} \right)^2 + 4\left( \frac{s-b}{r} \right)^2 + 9\left( \frac{s-c}{r} \right)^2 = \left( \frac{6s}{7r} \right)^2<br /> &lt;/math&gt;,<br /> &lt;/center&gt;<br /> or<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} = \frac{s^2}{36 + 9 + 4}<br /> &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> But by the [[Cauchy-Schwarz Inequality]], we know<br /> &lt;center&gt;<br /> &lt;math&gt;<br /> \begin{matrix}<br /> (36 + 9 + 4) \left[ \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} \right] &amp; \ge &amp;\left[ (s-a) + (s-b) + (s-c) \right]^2\\<br /> &amp; = &amp; s^2 \\<br /> \qquad\qquad \quad \quad \frac{(s-a)^2}{36} + \frac{(s-b)^2}{9} + \frac{(s-c)^2}{4} &amp; \ge &amp;\frac{s^2}{36 + 9 + 4} \; ,<br /> \end{matrix}<br /> &lt;/math&gt;<br /> &lt;/center&gt;<br /> with equality only when &lt;math&gt; \frac{(s-a)^2}{36}, \frac{(s-b)^2}{9}, \frac{(s-c)^2}{4}&lt;/math&gt; are directly proportional to 36, 9, 4, respectively. Therefore (clearing denominators and taking square roots) our problem requires that &lt;math&gt;(s-a), (s-b), (s-c) &lt;/math&gt; be directly proportional to 36, 9, 4, and since &lt;math&gt;a = (s-b) + (s-c) &lt;/math&gt; etc., this is equivalent to the condition that &lt;math&gt;a,b,c &lt;/math&gt; be in proportion with 13, 40, 45, Q.E.D.<br /> <br /> Sidenote: A 13, 40, 45 obtuse triangle has an integer area of 252 and an inradius of 36/7.<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See also ==<br /> {{USAMO newbox|year=2002|num-b=1|num-a=3}}<br /> <br /> <br /> [[Category:Olympiad Geometry Problems]]<br /> [[Category:Olympiad Inequality Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_15&diff=75665 2001 AIME II Problems/Problem 15 2016-02-07T22:38:47Z <p>Claudeaops: wording</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;EFDC&lt;/math&gt;, and &lt;math&gt;EHBC&lt;/math&gt; be three adjacent [[square]] faces of a [[cube]], for which &lt;math&gt;EC = 8&lt;/math&gt;, and let &lt;math&gt;A&lt;/math&gt; be the eighth [[vertex]] of the cube. Let &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;J&lt;/math&gt;, and &lt;math&gt;K&lt;/math&gt;, be the points on &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{EH}&lt;/math&gt;, and &lt;math&gt;\overline{EC}&lt;/math&gt;, respectively, so that &lt;math&gt;EI = EJ = EK = 2&lt;/math&gt;. A solid &lt;math&gt;S&lt;/math&gt; is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to &lt;math&gt;\overline{AE}&lt;/math&gt;, and containing the edges, &lt;math&gt;\overline{IJ}&lt;/math&gt;, &lt;math&gt;\overline{JK}&lt;/math&gt;, and &lt;math&gt;\overline{KI}&lt;/math&gt;. The [[surface area]] of &lt;math&gt;S&lt;/math&gt;, including the walls of the tunnel, is &lt;math&gt;m + n\sqrt {p}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> import three; currentprojection = perspective(5,-40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(&quot;10 2&quot;);<br /> <br /> triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);<br /> draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8));<br /> <br /> draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);<br /> &lt;/asy&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;asy&gt;<br /> import three; currentprojection = perspective(5,40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(&quot;10 2&quot;);<br /> <br /> triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);<br /> draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8));<br /> <br /> draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Set the coordinate system so that vertex &lt;math&gt;E&lt;/math&gt;, where the drilling starts, is at &lt;math&gt;(8,8,8)&lt;/math&gt;. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining &lt;math&gt;(1,0,0)&lt;/math&gt; to &lt;math&gt;(2,2,0)&lt;/math&gt;, and &lt;math&gt;(0,1,0)&lt;/math&gt; to &lt;math&gt;(2,2,0)&lt;/math&gt;, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), &lt;math&gt;S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)&lt;/math&gt;, and the other two faces of the tunnel are congruent to this shape. <br /> <br /> Observe that this shape is made up of two congruent [[trapezoid]]s each with height &lt;math&gt;\sqrt {2}&lt;/math&gt; and bases &lt;math&gt;7\sqrt {3}&lt;/math&gt; and &lt;math&gt;6\sqrt {3}&lt;/math&gt;. Together they make up an area of &lt;math&gt;\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}&lt;/math&gt;. The total area of the tunnel is then &lt;math&gt;3\cdot13\sqrt {6} = 39\sqrt {6}&lt;/math&gt;. Around the corner &lt;math&gt;E&lt;/math&gt; we're missing an area of &lt;math&gt;6&lt;/math&gt;, the same goes for the corner opposite &lt;math&gt;E&lt;/math&gt; . So the outside area is &lt;math&gt;6\cdot 64 - 2\cdot 6 = 372&lt;/math&gt;. Thus the the total surface area is &lt;math&gt;372 + 39\sqrt {6}&lt;/math&gt;, and the answer is &lt;math&gt;372 + 39 + 6 = \boxed{417}&lt;/math&gt;.<br /> <br /> {{stub}}<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_15&diff=75642 2001 AIME II Problems/Problem 15 2016-02-06T19:18:05Z <p>Claudeaops: Typo</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;EFGH&lt;/math&gt;, &lt;math&gt;EFDC&lt;/math&gt;, and &lt;math&gt;EHBC&lt;/math&gt; be three adjacent [[square]] faces of a [[cube]], for which &lt;math&gt;EC = 8&lt;/math&gt;, and let &lt;math&gt;A&lt;/math&gt; be the eighth [[vertex]] of the cube. Let &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;J&lt;/math&gt;, and &lt;math&gt;K&lt;/math&gt;, be the points on &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{EH}&lt;/math&gt;, and &lt;math&gt;\overline{EC}&lt;/math&gt;, respectively, so that &lt;math&gt;EI = EJ = EK = 2&lt;/math&gt;. A solid &lt;math&gt;S&lt;/math&gt; is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to &lt;math&gt;\overline{AE}&lt;/math&gt;, and containing the edges, &lt;math&gt;\overline{IJ}&lt;/math&gt;, &lt;math&gt;\overline{JK}&lt;/math&gt;, and &lt;math&gt;\overline{KI}&lt;/math&gt;. The [[surface area]] of &lt;math&gt;S&lt;/math&gt;, including the walls of the tunnel, is &lt;math&gt;m + n\sqrt {p}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p&lt;/math&gt;.<br /> <br /> == Solution ==<br /> &lt;center&gt;&lt;asy&gt;<br /> import three; currentprojection = perspective(5,-40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(&quot;10 2&quot;);<br /> <br /> triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);<br /> draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8));<br /> <br /> draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);<br /> &lt;/asy&gt; &amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; &lt;asy&gt;<br /> import three; currentprojection = perspective(5,40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(&quot;10 2&quot;);<br /> <br /> triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);<br /> draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8));<br /> <br /> draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Set the coordinate system so that vertex &lt;math&gt;E&lt;/math&gt;, where the drilling starts, is at &lt;math&gt;(8,8,8)&lt;/math&gt;. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining &lt;math&gt;(1,0,0)&lt;/math&gt; to &lt;math&gt;(2,2,0)&lt;/math&gt;, and &lt;math&gt;(0,1,0)&lt;/math&gt; to &lt;math&gt;(2,2,0)&lt;/math&gt;, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), &lt;math&gt;S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)&lt;/math&gt;, and the other two faces of the tunnel are congruent to this shape. <br /> <br /> Observe that this shape is made up of two congruent [[trapezoid]]s each with height &lt;math&gt;\sqrt {2}&lt;/math&gt; and bases &lt;math&gt;7\sqrt {3}&lt;/math&gt; and &lt;math&gt;6\sqrt {3}&lt;/math&gt;. Together they make up an area of &lt;math&gt;\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}&lt;/math&gt;. The total area of the tunnel is then &lt;math&gt;3\cdot13\sqrt {6} = 39\sqrt {6}&lt;/math&gt;. Around the corner &lt;math&gt;E&lt;/math&gt; and the corner opposite &lt;math&gt;E&lt;/math&gt; we're missing an area of &lt;math&gt;6&lt;/math&gt;. So the outside area is &lt;math&gt;6\cdot 64 - 2\cdot 6 = 372&lt;/math&gt;. Thus the the total surface area is &lt;math&gt;372 + 39\sqrt {6}&lt;/math&gt;, and the answer is &lt;math&gt;372 + 39 + 6 = \boxed{417}&lt;/math&gt;.<br /> <br /> {{stub}}<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_16&diff=74750 1970 AHSME Problems/Problem 16 2016-01-22T00:42:42Z <p>Claudeaops: formatting</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;F(n)&lt;/math&gt; is a function such that &lt;math&gt;F(1)=F(2)=F(3)=1&lt;/math&gt;, and such that &lt;math&gt;F(n+1)= \frac{F(n)\cdot F(n-1)+1}{F(n-2)}&lt;/math&gt; for &lt;math&gt;n\ge 3,&lt;/math&gt; <br /> then &lt;math&gt;F(6)=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 2\quad<br /> \text{(B) } 3\quad<br /> \text{(C) } 7\quad<br /> \text{(D) } 11\quad<br /> \text{(E) } 26&lt;/math&gt;<br /> <br /> = Solution =<br /> We can chug through the recursion to find the answer is &lt;math&gt;\fbox{C}&lt;/math&gt;.<br /> <br /> ==Sidenote==<br /> All the numbers in the sequence &lt;math&gt;F(n)&lt;/math&gt; are integers. In fact, the function &lt;math&gt;F&lt;/math&gt; satisfies &lt;math&gt;F(n)=4F(n-2)-F(n-4)&lt;/math&gt;. (Prove it!).<br /> <br /> == See also ==<br /> {{AHSME 35p box|year=1970|num-b=15|num-a=17}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_13&diff=74705 2001 AIME II Problems/Problem 13 2016-01-19T20:05:14Z <p>Claudeaops: extension</p> <hr /> <div>== Problem ==<br /> In [[quadrilateral]] &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;\angle{BAD}\cong\angle{ADC}&lt;/math&gt; and &lt;math&gt;\angle{ABD}\cong\angle{BCD}&lt;/math&gt;, &lt;math&gt;AB = 8&lt;/math&gt;, &lt;math&gt;BD = 10&lt;/math&gt;, and &lt;math&gt;BC = 6&lt;/math&gt;. The length &lt;math&gt;CD&lt;/math&gt; may be written in the form &lt;math&gt;\frac {m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Extend &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; to meet at &lt;math&gt;E&lt;/math&gt;. Then, since &lt;math&gt;\angle BAD = \angle ADC&lt;/math&gt; and &lt;math&gt;\angle ABD = \angle DCE&lt;/math&gt;, we know that &lt;math&gt;\triangle ABD \sim \triangle DCE&lt;/math&gt;. Hence &lt;math&gt;\angle ADB = \angle DEC&lt;/math&gt;, and &lt;math&gt;\triangle BDE&lt;/math&gt; is [[isosceles triangle|isosceles]]. Then &lt;math&gt;BD = BE = 10&lt;/math&gt;. <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> /* We arbitrarily set AD = x */<br /> real x = 60^.5, anglesize = 28;<br /> <br /> pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype(&quot;6 6&quot;)+linewidth(0.7);<br /> pair A=(0,0), D=(x,0), B=IP(CR(A,8),CR(D,10)), E=(-3x/5,0), C=IP(CR(E,16),CR(D,64/5));<br /> D(MP(&quot;A&quot;,A)--MP(&quot;B&quot;,B,NW)--MP(&quot;C&quot;,C,NW)--MP(&quot;D&quot;,D)--cycle); D(B--D); D(A--MP(&quot;E&quot;,E)--B,d);<br /> D(anglemark(D,A,B,anglesize));D(anglemark(C,D,A,anglesize));D(anglemark(A,B,D,anglesize));D(anglemark(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize));<br /> MP(&quot;10&quot;,(B+D)/2,SW);MP(&quot;8&quot;,(A+B)/2,W);MP(&quot;6&quot;,(B+C)/2,NW);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Using the similarity, we have:<br /> <br /> &lt;cmath&gt;\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5&lt;/cmath&gt;<br /> <br /> The answer is &lt;math&gt;m+n = \boxed{069}&lt;/math&gt;.<br /> <br /> <br /> '''Extension''': To Find &lt;math&gt;AD&lt;/math&gt;, use Law of Cosines on &lt;math&gt;\triangle BCD&lt;/math&gt; to get &lt;math&gt;\cos(\angle BCD)=\frac{13}{20}&lt;/math&gt;<br /> Then since &lt;math&gt;\angle BCD=\angle ABD&lt;/math&gt; use Law of Cosines on &lt;math&gt;\triangle ABD&lt;/math&gt; to find &lt;math&gt;AD=2\sqrt{15}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12B_Problems/Problem_21&diff=74513 2011 AMC 12B Problems/Problem 21 2016-01-13T18:17:26Z <p>Claudeaops: sidenote</p> <hr /> <div>==Problem==<br /> <br /> The arithmetic mean of two distinct positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is a two-digit integer. The geometric mean of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; is obtained by reversing the digits of the arithmetic mean. What is &lt;math&gt;|x - y|&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70&lt;/math&gt;<br /> <br /> ==Solution==<br /> Answer: (D)<br /> <br /> &lt;math&gt;\frac{x + y}{2} = 10 a+b&lt;/math&gt; for some &lt;math&gt;1\le a\le 9 &lt;/math&gt;,&lt;math&gt;0\le b\le 9&lt;/math&gt;.<br /> <br /> &lt;math&gt;\sqrt{xy} = 10 b+a&lt;/math&gt;<br /> <br /> &lt;math&gt;100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;xy = 100b^2 + 20ab + a^2&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)&lt;/math&gt;<br /> <br /> <br /> &lt;br /&gt;<br /> &lt;math&gt;|x-y| = 2\sqrt{99(a^2 - b^2)}&lt;/math&gt;<br /> <br /> Note that in order for x-y to be integer, &lt;math&gt;(a^2 - b^2)&lt;/math&gt; has to be &lt;math&gt;11n&lt;/math&gt; for some perfect square &lt;math&gt;n&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; is at most &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;n = 1&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt;<br /> <br /> If &lt;math&gt;n = 1&lt;/math&gt;, &lt;math&gt;|x-y| = 66&lt;/math&gt;, if &lt;math&gt;n = 4&lt;/math&gt;, &lt;math&gt;|x-y| = 132&lt;/math&gt;. In AMC, we are done. Otherwise, we need to show that &lt;math&gt;a^2 -b^2 = 44&lt;/math&gt; is impossible.<br /> <br /> &lt;math&gt;(a-b)(a+b) = 44&lt;/math&gt; -&gt; &lt;math&gt;a-b = 1&lt;/math&gt;, or &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;a+b = 44&lt;/math&gt;, &lt;math&gt;22&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt; respectively. And since &lt;math&gt;a+b \le 18&lt;/math&gt;, &lt;math&gt;a+b = 11&lt;/math&gt;, &lt;math&gt;a-b = 4&lt;/math&gt;, but there is no integer solution for &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;.<br /> <br /> In addition:<br /> Note that &lt;math&gt;11n&lt;/math&gt; with &lt;math&gt;n = 1&lt;/math&gt; may be obtained with &lt;math&gt;a = 6&lt;/math&gt; and &lt;math&gt;b = 5&lt;/math&gt; as &lt;math&gt;a^2 - b^2 = 36 - 25 = 11&lt;/math&gt;.<br /> <br /> ==Sidenote==<br /> It is easy to see that &lt;math&gt;(a,b)=(6,5)&lt;/math&gt; is the only solution. This yields &lt;math&gt;(x,y)=(98,32)&lt;/math&gt;. Their arithmetic mean is &lt;math&gt;65&lt;/math&gt; and their geometric mean is &lt;math&gt;56&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=20|num-a=22|ab=B}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_13&diff=74027 2000 AMC 12 Problems/Problem 13 2015-12-28T20:30:12Z <p>Claudeaops: Sidenote</p> <hr /> <div>{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #13]] and [[2000 AMC 10 Problems|2000 AMC 10 #22]]}}<br /> <br /> == Problem ==<br /> One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?<br /> <br /> &lt;math&gt;\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> '''Solution 1:'''<br /> <br /> Let &lt;math&gt;c&lt;/math&gt; be the total amount of coffee, &lt;math&gt;m&lt;/math&gt; of milk, and &lt;math&gt;p&lt;/math&gt; the number of people in the family. Then each person drinks the same total amount of coffee and milk (8 ounces), so <br /> &lt;cmath&gt;\left(\frac{c}{6} + \frac{m}{4}\right)p = c + m&lt;/cmath&gt;<br /> Regrouping, we get &lt;math&gt;2c(6-p)=3m(p-4)&lt;/math&gt;. Since both &lt;math&gt;c,m&lt;/math&gt; are positive, it follows that &lt;math&gt;6-p&lt;/math&gt; and &lt;math&gt;p-4&lt;/math&gt; are also positive, which is only possible when &lt;math&gt;p = 5\ \mathrm{(C)}&lt;/math&gt;.<br /> <br /> <br /> '''Solution 2 (less rigorous):'''<br /> <br /> One could notice that (since there are only two components to the mixture) Angela must have more than her &quot;fair share&quot; of milk and less then her &quot;fair share&quot; of coffee in order to ensure that everyone has &lt;math&gt;8&lt;/math&gt; ounces. The &quot;fair share&quot; is &lt;math&gt;1/p.&lt;/math&gt; So,<br /> <br /> &lt;cmath&gt;\frac{1}{6} &lt; \frac{1}{p}&lt;\frac{1}{4}&lt;/cmath&gt;<br /> <br /> Which requires that &lt;math&gt;p&lt;/math&gt; be &lt;math&gt;p = 5\ \mathrm{(C)},&lt;/math&gt; since &lt;math&gt;p&lt;/math&gt; is a whole number.<br /> <br /> <br /> '''Solution 3:'''<br /> <br /> Again, let &lt;math&gt;c,&lt;/math&gt; &lt;math&gt;m,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is &lt;math&gt;8p,&lt;/math&gt; and also &lt;math&gt;c+m.&lt;/math&gt; Thus, &lt;math&gt;c+m = 8p,&lt;/math&gt; so &lt;math&gt;m = 8p-c&lt;/math&gt; and &lt;math&gt;c = 8p-m.&lt;/math&gt;<br /> <br /> We also know that the amount Angela drank, which is &lt;math&gt;\frac{c}{6} + \frac{m}{4},&lt;/math&gt; is equal to &lt;math&gt;8&lt;/math&gt; ounces, thus &lt;math&gt;\frac{c}{6} + \frac{m}{4} = 8.&lt;/math&gt; Rearranging gives &lt;cmath&gt;24p - c = 96.&lt;/cmath&gt; Now notice that &lt;math&gt;c &gt; 0&lt;/math&gt; (by the problem statement). In addition, &lt;math&gt;m &gt; 0,&lt;/math&gt; so &lt;math&gt;c = 8p-m &lt; 8p.&lt;/math&gt; Therefore, &lt;math&gt;0 &lt; c &lt; 8p,&lt;/math&gt; and so &lt;math&gt;24p &gt; 24p-c &gt; 16p.&lt;/math&gt; We know that &lt;math&gt;24p-c = 96,&lt;/math&gt; so &lt;cmath&gt;24p &gt; 96 &gt; 16p.&lt;/cmath&gt; From the leftmost inequality, we get &lt;math&gt;p &gt; 4,&lt;/math&gt; and from the rightmost inequality, we get &lt;math&gt;p &lt; 6.&lt;/math&gt; The only possible value of &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;p = 5\ \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==Sidenote==<br /> If we now solve for &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt;, we find that &lt;math&gt;m=16&lt;/math&gt; and &lt;math&gt;c=24&lt;/math&gt;. Thus in total the family drank &lt;math&gt;16&lt;/math&gt; ounces of milk and &lt;math&gt;24&lt;/math&gt; ounces of coffee. Angela drank exactly &lt;math&gt;4&lt;/math&gt; ounces of milk and &lt;math&gt;4&lt;/math&gt; ounces of coffee.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2000|num-b=12|num-a=14}}<br /> {{AMC10 box|year=2000|num-b=21|num-a=23}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_24&diff=73600 2004 AMC 12A Problems/Problem 24 2015-12-10T23:20:48Z <p>Claudeaops: See Math Jam for the 2nd solution</p> <hr /> <div>== Problem 24 ==<br /> A [[plane]] contains points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; with &lt;math&gt;AB = 1&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be the [[union]] of all disks of radius &lt;math&gt;1&lt;/math&gt; in the plane that cover &lt;math&gt;\overline{AB}&lt;/math&gt;. What is the area of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text {(A)} 2\pi + \sqrt3 \qquad \text {(B)} \frac {8\pi}{3} \qquad \text {(C)} 3\pi - \frac {\sqrt3}{2} \qquad \text {(D)} \frac {10\pi}{3} - \sqrt3 \qquad \text {(E)}4\pi - 2\sqrt3&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> &lt;center&gt;&lt;asy&gt;<br /> pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);<br /> <br /> draw(arc(A,2,-60,60),blue);<br /> draw(arc(B,2,120,240),blue);<br /> draw(circle(C,1),red);<br /> <br /> draw(A--(.5,3^.5));<br /> draw(B--(-.5,3^.5));<br /> draw(A--(.5,-3^.5));<br /> draw(B--(-.5,-3^.5));<br /> draw(A--B);<br /> <br /> dot(A);dot(B);dot(C);dot(D);<br /> <br /> label(&quot;$$1$$&quot;,(0,0),N);<br /> label(&quot;$$1$$&quot;,A/2+D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+C/2,E);<br /> label(&quot;$$1$$&quot;,A/2+3D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+3C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+3D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+3C/2,E);<br /> <br /> label(&quot;$$A$$&quot;,A,W);<br /> label(&quot;$$B$$&quot;,B,E);<br /> label(&quot;$$C$$&quot;,C,W);<br /> label(&quot;$$D$$&quot;,D,E);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> As the red circles move about segment &lt;math&gt;AB&lt;/math&gt;, they cover the area we are looking for.<br /> On the left side, the circle must move around pivoted on &lt;math&gt;B&lt;/math&gt;.<br /> On the right side, the circle must move pivoted on &lt;math&gt;A&lt;/math&gt;<br /> However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.<br /> <br /> This egg-like shape is &lt;math&gt;S&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);<br /> <br /> draw(arc(A,2,-60,60),blue);<br /> draw(arc(B,2,120,240),blue);<br /> draw(arc(C,1,60,120),red);<br /> draw(arc(D,1,-120,-60),red);<br /> <br /> draw(A--(.5,3^.5));<br /> draw(B--(-.5,3^.5));<br /> draw(A--(.5,-3^.5));<br /> draw(B--(-.5,-3^.5));<br /> draw(A--B);<br /> <br /> dot(A);dot(B);dot(C);dot(D);<br /> label(&quot;$$A$$&quot;,A,W);<br /> label(&quot;$$B$$&quot;,B,E);<br /> label(&quot;$$C$$&quot;,C,W);<br /> label(&quot;$$D$$&quot;,D,E);<br /> <br /> label(&quot;$$1$$&quot;,(0,0),N);<br /> label(&quot;$$1$$&quot;,A/2+D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+C/2,E);<br /> label(&quot;$$1$$&quot;,A/2+3D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+3C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+3D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+3C/2,E);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The area of the region can be found by dividing it into several sectors, namely<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> A &amp;= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\<br /> A &amp;= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\<br /> A &amp;= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\<br /> A &amp;= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}&lt;/cmath&gt;<br /> <br /> ==See also== <br /> {{AMC12 box|year=2004|ab=A|num-b=23|num-a=25}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_24&diff=73555 2004 AMC 12A Problems/Problem 24 2015-12-07T03:34:26Z <p>Claudeaops: Solution 2</p> <hr /> <div>== Problem 24 ==<br /> A [[plane]] contains points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; with &lt;math&gt;AB = 1&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be the [[union]] of all disks of radius &lt;math&gt;1&lt;/math&gt; in the plane that cover &lt;math&gt;\overline{AB}&lt;/math&gt;. What is the area of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text {(A)} 2\pi + \sqrt3 \qquad \text {(B)} \frac {8\pi}{3} \qquad \text {(C)} 3\pi - \frac {\sqrt3}{2} \qquad \text {(D)} \frac {10\pi}{3} - \sqrt3 \qquad \text {(E)}4\pi - 2\sqrt3&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> &lt;center&gt;&lt;asy&gt;<br /> pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);<br /> <br /> draw(arc(A,2,-60,60),blue);<br /> draw(arc(B,2,120,240),blue);<br /> draw(circle(C,1),red);<br /> <br /> draw(A--(.5,3^.5));<br /> draw(B--(-.5,3^.5));<br /> draw(A--(.5,-3^.5));<br /> draw(B--(-.5,-3^.5));<br /> draw(A--B);<br /> <br /> dot(A);dot(B);dot(C);dot(D);<br /> <br /> label(&quot;$$1$$&quot;,(0,0),N);<br /> label(&quot;$$1$$&quot;,A/2+D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+C/2,E);<br /> label(&quot;$$1$$&quot;,A/2+3D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+3C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+3D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+3C/2,E);<br /> <br /> label(&quot;$$A$$&quot;,A,W);<br /> label(&quot;$$B$$&quot;,B,E);<br /> label(&quot;$$C$$&quot;,C,W);<br /> label(&quot;$$D$$&quot;,D,E);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> As the red circles move about segment &lt;math&gt;AB&lt;/math&gt;, they cover the area we are looking for.<br /> On the left side, the circle must move around pivoted on &lt;math&gt;B&lt;/math&gt;.<br /> On the right side, the circle must move pivoted on &lt;math&gt;A&lt;/math&gt;<br /> However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.<br /> <br /> This egg-like shape is &lt;math&gt;S&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);<br /> <br /> draw(arc(A,2,-60,60),blue);<br /> draw(arc(B,2,120,240),blue);<br /> draw(arc(C,1,60,120),red);<br /> draw(arc(D,1,-120,-60),red);<br /> <br /> draw(A--(.5,3^.5));<br /> draw(B--(-.5,3^.5));<br /> draw(A--(.5,-3^.5));<br /> draw(B--(-.5,-3^.5));<br /> draw(A--B);<br /> <br /> dot(A);dot(B);dot(C);dot(D);<br /> label(&quot;$$A$$&quot;,A,W);<br /> label(&quot;$$B$$&quot;,B,E);<br /> label(&quot;$$C$$&quot;,C,W);<br /> label(&quot;$$D$$&quot;,D,E);<br /> <br /> label(&quot;$$1$$&quot;,(0,0),N);<br /> label(&quot;$$1$$&quot;,A/2+D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+C/2,E);<br /> label(&quot;$$1$$&quot;,A/2+3D/2,W);<br /> label(&quot;$$1$$&quot;,A/2+3C/2,W);<br /> label(&quot;$$1$$&quot;,B/2+3D/2,E);<br /> label(&quot;$$1$$&quot;,B/2+3C/2,E);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The area of the region can be found by dividing it into several sectors, namely<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> A &amp;= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\<br /> A &amp;= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi(1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\<br /> A &amp;= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\<br /> A &amp;= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}&lt;/cmath&gt;<br /> <br /> ==Alternate Solution==<br /> Credit to the Math Jam for the solution. <br /> (Diagram needed)<br /> <br /> Let &lt;math&gt;\triangle ABX&lt;/math&gt; be an equilateral triangle &quot;above&quot; segment &lt;math&gt;AB&lt;/math&gt;. Now imagine having a &quot;moving&quot; circle of radius 1 pegged through point &lt;math&gt;B&lt;/math&gt; and sliding it around as much as possible. The center of the circle must trace an arc of a circle &lt;math&gt;\omega_1&lt;/math&gt; with radius 1 centered at &lt;math&gt;B&lt;/math&gt;. Note that &lt;math&gt;\omega_1&lt;/math&gt; passes through &lt;math&gt;X&lt;/math&gt;. In addition, <br /> the point on the &quot;moving&quot; circle farthest from &lt;math&gt;B&lt;/math&gt; must always be 2 units away from &lt;math&gt;B&lt;/math&gt;, so it traces out an arc of a circle &lt;math&gt;\omega_2&lt;/math&gt; with radius 2 centered at &lt;math&gt;B&lt;/math&gt;. Now as we move around the circle the center will move around arc &lt;math&gt;AX&lt;/math&gt; of &lt;math&gt;\omega_1&lt;/math&gt;. Let &lt;math&gt;W&lt;/math&gt; be the intersection of &lt;math&gt;AB&lt;/math&gt; with &lt;math&gt;\omega_2&lt;/math&gt; such that &lt;math&gt;WA&lt;WB&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;AB&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; be the intersection of &lt;math&gt;BX&lt;/math&gt; with &lt;math&gt;\omega_2&lt;/math&gt; such that &lt;math&gt;ZX&lt;ZB&lt;/math&gt;. <br /> Now the point on the &quot;moving&quot; circle farthest from &lt;math&gt;B&lt;/math&gt; traces out arc &lt;math&gt;WZ&lt;/math&gt; on &lt;math&gt;\omega_2&lt;/math&gt;. Finally, one of the extreme positions of the moving circle is when the circle is centered at &lt;math&gt;X&lt;/math&gt;. Let the circle centered at &lt;math&gt;X&lt;/math&gt; with radius &lt;math&gt;1&lt;/math&gt; be called &lt;math&gt;\omega_3&lt;/math&gt;. Let &lt;math&gt;Y&lt;/math&gt; be the intersection of &lt;math&gt;MX&lt;/math&gt; and &lt;math&gt;\omega_3&lt;/math&gt; such that &lt;math&gt;YX&lt;YM&lt;/math&gt;. Once the center of the moving circle has moved from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;X&lt;/math&gt;, the moving circle will have finished one-fourth of its total motion. Thus, &lt;math&gt;S&lt;/math&gt; is made up of 4 quadrants (consult the diagram in solution 1). Focus on the upper left quadrant of &lt;math&gt;S&lt;/math&gt;. The upper left quadrant is bounded by arc &lt;math&gt;WZ&lt;/math&gt; of &lt;math&gt;\omega_2&lt;/math&gt;, arc &lt;math&gt;ZY&lt;/math&gt; of &lt;math&gt;\omega_3&lt;/math&gt;, segments &lt;math&gt;YM&lt;/math&gt; and &lt;math&gt;WM&lt;/math&gt;. The area of this region is equal to <br /> the area of sector &lt;math&gt;WBZ&lt;/math&gt; of &lt;math&gt;\omega_2&lt;/math&gt; plus the area of sector &lt;math&gt;ZXY&lt;/math&gt; of &lt;math&gt;\omega_3&lt;/math&gt; minus the area of &lt;math&gt;\triangle BMX&lt;/math&gt;. &lt;math&gt;\angle WBZ=60, \angle ZXY=\angle BXM=30&lt;/math&gt;, and &lt;math&gt;\triangle BXM&lt;/math&gt; is a 30-60-90 triangle. Routine calculations yield answer choice &lt;math&gt;\textbf {(C)}&lt;/math&gt;.<br /> <br /> ==See also== <br /> {{AMC12 box|year=2004|ab=A|num-b=23|num-a=25}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_21&diff=73553 2014 AMC 12B Problems/Problem 21 2015-12-07T02:47:30Z <p>Claudeaops: formatting</p> <hr /> <div>==Problem 21==<br /> In the figure, &lt;math&gt; ABCD &lt;/math&gt; is a square of side length &lt;math&gt; 1 &lt;/math&gt;. The rectangles &lt;math&gt; JKHG &lt;/math&gt; and &lt;math&gt; EBCF &lt;/math&gt; are congruent. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2);<br /> draw(A--B--C--D--cycle);<br /> draw(K--H--G--J--cycle);<br /> draw(F--E);<br /> label(&quot;$A$&quot;,A,SE); label(&quot;$B$&quot;,B,SW); label(&quot;$C$&quot;,C,NW); label(&quot;$D$&quot;,D,NE); label(&quot;$E$&quot;,E,S); label(&quot;$F$&quot;,F,N);<br /> label(&quot;$G$&quot;,G,E); label(&quot;$H$&quot;,H,N); label(&quot;$J$&quot;,J,S); label(&quot;$K$&quot;,K,W);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> Draw the attitude from &lt;math&gt;H&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt; and call the foot &lt;math&gt;L&lt;/math&gt;. Then &lt;math&gt;HL=1&lt;/math&gt;. Consider &lt;math&gt;HJ&lt;/math&gt;. It is the hypotenuse of both right triangles &lt;math&gt;\triangle HGJ&lt;/math&gt; and &lt;math&gt;\triangle HLJ&lt;/math&gt;, and we know &lt;math&gt;JG=HL=1&lt;/math&gt;, so we must have &lt;math&gt;\triangle HGJ\cong\triangle JLH&lt;/math&gt; by Hypotenuse-Leg congruence. From this congruence we have &lt;math&gt;LJ=HG=BE&lt;/math&gt;.<br /> <br /> Notice that all four triangles in this picture are similar. Also, we have &lt;math&gt;LA=HD=EJ&lt;/math&gt;. So set &lt;math&gt;x=LJ=HG=BE&lt;/math&gt; and &lt;math&gt;y=LA=HD=EJ&lt;/math&gt;. Now &lt;math&gt;BE+EJ+JL+LA=2(x+y)=1&lt;/math&gt;. This means &lt;math&gt;x+y=\frac{1}{2}=BE+EJ=BJ&lt;/math&gt;, so &lt;math&gt;J&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt;. So &lt;math&gt;\triangle AJG&lt;/math&gt;, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have &lt;math&gt;AG=\sqrt{3} \cdot AJ=\sqrt{3}/2&lt;/math&gt; and subsequently &lt;math&gt;GD=\frac{2-\sqrt{3}}{2}=KE&lt;/math&gt;. This means &lt;math&gt;EJ=\sqrt{3} \cdot KE=\frac{2\sqrt{3}-3}{2}&lt;/math&gt;, which gives &lt;math&gt;BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}&lt;/math&gt;, so the answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;BE = x&lt;/math&gt;. Let &lt;math&gt;JA = y&lt;/math&gt;. Because &lt;math&gt;\angle FKH = \angle EJK = \angle AGJ = \angle DHG&lt;/math&gt; and &lt;math&gt;\angle FHK = \angle EKJ = \angle AJG = \angle DGH&lt;/math&gt;, &lt;math&gt;\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK&lt;/math&gt; are all similar. Using proportions and the pythagorean theorem, we find <br /> &lt;cmath&gt;EK = xy&lt;/cmath&gt;<br /> &lt;cmath&gt;FK = \sqrt{1-y^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;EJ = x\sqrt{1-y^2}&lt;/cmath&gt;<br /> Because we know that &lt;math&gt;BE+EJ+AJ = EK + FK = 1&lt;/math&gt;, we can set up a systems of equations<br /> &lt;cmath&gt;x + x\sqrt{1-y^2} + y = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;xy + \sqrt{1-y^2} = 1&lt;/cmath&gt;<br /> Solving for &lt;math&gt;x&lt;/math&gt; in the second equation, we get <br /> &lt;cmath&gt;x= \frac{1-\sqrt{1-y^2}}{y}&lt;/cmath&gt;<br /> Plugging this into the first equation, we get<br /> &lt;cmath&gt;\frac{1-\sqrt{1-y^2}}{y} + (\sqrt{1-y^2})\frac{1-\sqrt{1-y^2}}{y} + y = 1 \implies \frac{2y^2}{y}=1 \implies y=\frac{1}{2}&lt;/cmath&gt;<br /> Plugging into the previous equation with &lt;math&gt;x&lt;/math&gt;, we get<br /> &lt;cmath&gt;x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;BE = x&lt;/math&gt;, &lt;math&gt;EK = a&lt;/math&gt;, and &lt;math&gt;EJ = b&lt;/math&gt;. Then &lt;math&gt;x^2 = a^2 + b^2&lt;/math&gt; and because &lt;math&gt;\triangle KEJ \cong \triangle GDH&lt;/math&gt; and &lt;math&gt;\triangle KEJ \sim \triangle JAG&lt;/math&gt;, &lt;math&gt;\frac{GA}{1} = 1 - a = \frac{b}{x}&lt;/math&gt;. Furthermore, the area of the four triangles and the two rectangles sums to 1:<br /> <br /> &lt;cmath&gt;1 = 2x + GA\cdot JA + ab&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 = 2x + (1 - a)(1 - (x + b)) + ab&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 = 2x + \frac{b}{x}(1 - x - b) + \left(1 - \frac{b}{x}\right)b&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 = 2x + \frac{b}{x} - b - \frac{b^2}{x} + b - \frac{b^2}{x}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x = 2x^2 + b - 2b^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x - b = 2(x - b)(x + b)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x + b = \frac{1}{2}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;b = \frac{1}{2} - x&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a = 1 - \frac{b}{x} = 2 - \frac{1}{2x}&lt;/cmath&gt;<br /> <br /> By the Pythagorean theorem: &lt;math&gt;x^2 = a^2 + b^2&lt;/math&gt;<br /> <br /> &lt;cmath&gt;x^2 = \left(2 - \frac{1}{2x}\right)^2 + \left(\frac{1}{2} - x\right)^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x^2 = 4 - \frac{2}{x} + \frac{1}{4x^2} + \frac{1}{4} - x + x^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = \frac{1}{4x^2} - \frac{2}{x} + \frac{17}{4} - x&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = 1 - 8x + 17x^2 - 4x^3.&lt;/cmath&gt;<br /> <br /> Then by the rational root theorem, this has roots &lt;math&gt;\frac{1}{4}&lt;/math&gt;, &lt;math&gt;2 - \sqrt{3}&lt;/math&gt;, and &lt;math&gt;2 + \sqrt{3}&lt;/math&gt;. The first and last roots are extraneous because they imply &lt;math&gt;a = 0&lt;/math&gt; and &lt;math&gt;x &gt; 1&lt;/math&gt;, respectively, thus &lt;math&gt;x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12B_Problems/Problem_21&diff=73552 2014 AMC 12B Problems/Problem 21 2015-12-07T02:45:07Z <p>Claudeaops: Clarifying Solution 1</p> <hr /> <div>==Problem 21==<br /> In the figure, &lt;math&gt; ABCD &lt;/math&gt; is a square of side length &lt;math&gt; 1 &lt;/math&gt;. The rectangles &lt;math&gt; JKHG &lt;/math&gt; and &lt;math&gt; EBCF &lt;/math&gt; are congruent. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2);<br /> draw(A--B--C--D--cycle);<br /> draw(K--H--G--J--cycle);<br /> draw(F--E);<br /> label(&quot;$A$&quot;,A,SE); label(&quot;$B$&quot;,B,SW); label(&quot;$C$&quot;,C,NW); label(&quot;$D$&quot;,D,NE); label(&quot;$E$&quot;,E,S); label(&quot;$F$&quot;,F,N);<br /> label(&quot;$G$&quot;,G,E); label(&quot;$H$&quot;,H,N); label(&quot;$J$&quot;,J,S); label(&quot;$K$&quot;,K,W);<br /> &lt;/asy&gt;<br /> &lt;math&gt; \textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> Draw the attitude from &lt;math&gt;H&lt;/math&gt; to &lt;math&gt;AB&lt;/math&gt; and call the foot &lt;math&gt;L&lt;/math&gt;. Then &lt;math&gt;HL=1&lt;/math&gt;. Consider &lt;math&gt;HJ&lt;/math&gt;. It is the hypotenuse of both right triangles &lt;math&gt;\triangle HGJ&lt;/math&gt; and &lt;math&gt;\triangle HLJ&lt;/math&gt;, and we know &lt;math&gt;JG=HL=1&lt;/math&gt;, so we must have &lt;math&gt;\triangle HGJ\cong\triangle JLH&lt;/math&gt; by Hypotenuse-Leg congruence. From this congruence we have &lt;math&gt;LJ=HG=BE&lt;/math&gt;.<br /> <br /> Notice that all four triangles in this picture are similar. Also, we have &lt;math&gt;LA=HD=EJ&lt;/math&gt;. So set &lt;math&gt;x=LJ=HG=BE&lt;/math&gt; and &lt;math&gt;y=LA=HD=EJ&lt;/math&gt;. Now &lt;math&gt;BE+EJ+JL+LA=1=2(x+y)&lt;/math&gt;. This means &lt;math&gt;x+y=\frac{1}{2}=BE+EJ=BJ&lt;/math&gt;, so &lt;math&gt;J&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt;. So &lt;math&gt;\triangle AJG&lt;/math&gt;, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have &lt;math&gt;AG=\sqrt{3} \cdot AJ=\sqrt{3}/2&lt;/math&gt; and subsequently &lt;math&gt;GD=\frac{2-\sqrt{3}}{2}=KE&lt;/math&gt;. This means &lt;math&gt;EJ=\sqrt{3} \cdot KE=\frac{2\sqrt{3}-3}{2}&lt;/math&gt;, which gives &lt;math&gt;BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}&lt;/math&gt;, so the answer is &lt;math&gt;\textbf{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;BE = x&lt;/math&gt;. Let &lt;math&gt;JA = y&lt;/math&gt;. Because &lt;math&gt;\angle FKH = \angle EJK = \angle AGJ = \angle DHG&lt;/math&gt; and &lt;math&gt;\angle FHK = \angle EKJ = \angle AJG = \angle DGH&lt;/math&gt;, &lt;math&gt;\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK&lt;/math&gt; are all similar. Using proportions and the pythagorean theorem, we find <br /> &lt;cmath&gt;EK = xy&lt;/cmath&gt;<br /> &lt;cmath&gt;FK = \sqrt{1-y^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;EJ = x\sqrt{1-y^2}&lt;/cmath&gt;<br /> Because we know that &lt;math&gt;BE+EJ+AJ = EK + FK = 1&lt;/math&gt;, we can set up a systems of equations<br /> &lt;cmath&gt;x + x\sqrt{1-y^2} + y = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;xy + \sqrt{1-y^2} = 1&lt;/cmath&gt;<br /> Solving for &lt;math&gt;x&lt;/math&gt; in the second equation, we get <br /> &lt;cmath&gt;x= \frac{1-\sqrt{1-y^2}}{y}&lt;/cmath&gt;<br /> Plugging this into the first equation, we get<br /> &lt;cmath&gt;\frac{1-\sqrt{1-y^2}}{y} + (\sqrt{1-y^2})\frac{1-\sqrt{1-y^2}}{y} + y = 1 \implies \frac{2y^2}{y}=1 \implies y=\frac{1}{2}&lt;/cmath&gt;<br /> Plugging into the previous equation with &lt;math&gt;x&lt;/math&gt;, we get<br /> &lt;cmath&gt;x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}&lt;/cmath&gt;<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;BE = x&lt;/math&gt;, &lt;math&gt;EK = a&lt;/math&gt;, and &lt;math&gt;EJ = b&lt;/math&gt;. Then &lt;math&gt;x^2 = a^2 + b^2&lt;/math&gt; and because &lt;math&gt;\triangle KEJ \cong \triangle GDH&lt;/math&gt; and &lt;math&gt;\triangle KEJ \sim \triangle JAG&lt;/math&gt;, &lt;math&gt;\frac{GA}{1} = 1 - a = \frac{b}{x}&lt;/math&gt;. Furthermore, the area of the four triangles and the two rectangles sums to 1:<br /> <br /> &lt;cmath&gt;1 = 2x + GA\cdot JA + ab&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 = 2x + (1 - a)(1 - (x + b)) + ab&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 = 2x + \frac{b}{x}(1 - x - b) + \left(1 - \frac{b}{x}\right)b&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1 = 2x + \frac{b}{x} - b - \frac{b^2}{x} + b - \frac{b^2}{x}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x = 2x^2 + b - 2b^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x - b = 2(x - b)(x + b)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x + b = \frac{1}{2}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;b = \frac{1}{2} - x&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a = 1 - \frac{b}{x} = 2 - \frac{1}{2x}&lt;/cmath&gt;<br /> <br /> By the Pythagorean theorem: &lt;math&gt;x^2 = a^2 + b^2&lt;/math&gt;<br /> <br /> &lt;cmath&gt;x^2 = \left(2 - \frac{1}{2x}\right)^2 + \left(\frac{1}{2} - x\right)^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;x^2 = 4 - \frac{2}{x} + \frac{1}{4x^2} + \frac{1}{4} - x + x^2&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = \frac{1}{4x^2} - \frac{2}{x} + \frac{17}{4} - x&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;0 = 1 - 8x + 17x^2 - 4x^3.&lt;/cmath&gt;<br /> <br /> Then by the rational root theorem, this has roots &lt;math&gt;\frac{1}{4}&lt;/math&gt;, &lt;math&gt;2 - \sqrt{3}&lt;/math&gt;, and &lt;math&gt;2 + \sqrt{3}&lt;/math&gt;. The first and last roots are extraneous because they imply &lt;math&gt;a = 0&lt;/math&gt; and &lt;math&gt;x &gt; 1&lt;/math&gt;, respectively, thus &lt;math&gt;x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_11&diff=73534 2001 AIME I Problems/Problem 11 2015-12-05T01:32:50Z <p>Claudeaops: Sidenote</p> <hr /> <div>== Problem ==<br /> In a [[rectangle|rectangular]] array of points, with 5 rows and &lt;math&gt;N&lt;/math&gt; columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through &lt;math&gt;N,&lt;/math&gt; the second row is numbered &lt;math&gt;N + 1&lt;/math&gt; through &lt;math&gt;2N,&lt;/math&gt; and so forth. Five points, &lt;math&gt;P_1, P_2, P_3, P_4,&lt;/math&gt; and &lt;math&gt;P_5,&lt;/math&gt; are selected so that each &lt;math&gt;P_i&lt;/math&gt; is in row &lt;math&gt;i.&lt;/math&gt; Let &lt;math&gt;x_i&lt;/math&gt; be the number associated with &lt;math&gt;P_i.&lt;/math&gt; Now renumber the array consecutively from top to bottom, beginning with the first column. Let &lt;math&gt;y_i&lt;/math&gt; be the number associated with &lt;math&gt;P_i&lt;/math&gt; after the renumbering. It is found that &lt;math&gt;x_1 = y_2,&lt;/math&gt; &lt;math&gt;x_2 = y_1,&lt;/math&gt; &lt;math&gt;x_3 = y_4,&lt;/math&gt; &lt;math&gt;x_4 = y_5,&lt;/math&gt; and &lt;math&gt;x_5 = y_3.&lt;/math&gt; Find the smallest possible value of &lt;math&gt;N.&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Let each point &lt;math&gt;P_i&lt;/math&gt; be in column &lt;math&gt;c_i&lt;/math&gt;. The numberings for &lt;math&gt;P_i&lt;/math&gt; can now be defined as follows.<br /> &lt;cmath&gt;\begin{align*}x_i &amp;= (i - 1)N + c_i\\<br /> y_i &amp;= (c_i - 1)5 + i<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> We can now convert the five given equalities.<br /> &lt;cmath&gt;\begin{align}x_1&amp;=y_2 &amp; \Longrightarrow &amp; &amp; c_1 &amp;= 5 c_2-3\\<br /> x_2&amp;=y_1 &amp; \Longrightarrow &amp; &amp; N+c_2 &amp;= 5 c_1-4\\<br /> x_3&amp;=y_4 &amp; \Longrightarrow &amp; &amp; 2 N+c_3 &amp;= 5 c_4-1\\<br /> x_4&amp;=y_5 &amp; \Longrightarrow &amp; &amp; 3 N+c_4 &amp;= 5 c_5\\<br /> x_5&amp;=y_3 &amp; \Longrightarrow &amp; &amp; 4 N+c_5 &amp;= 5 c_3-2<br /> \end{align}&lt;/cmath&gt;<br /> Equations &lt;math&gt;(1)&lt;/math&gt; and &lt;math&gt;(2)&lt;/math&gt; combine to form<br /> &lt;cmath&gt;N = 24c_2 - 19&lt;/cmath&gt;<br /> Similarly equations &lt;math&gt;(3)&lt;/math&gt;, &lt;math&gt;(4)&lt;/math&gt;, and &lt;math&gt;(5)&lt;/math&gt; combine to form<br /> &lt;cmath&gt;117N +51 = 124c_3&lt;/cmath&gt;<br /> Take this equation modulo 31<br /> &lt;cmath&gt;24N+20\equiv 0 \pmod{31}&lt;/cmath&gt;<br /> And substitute for N<br /> &lt;cmath&gt;24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}&lt;/cmath&gt;<br /> &lt;cmath&gt;18 c_2 \equiv 2 \pmod{31}&lt;/cmath&gt;<br /> <br /> Thus the smallest &lt;math&gt;c_2&lt;/math&gt; might be is &lt;math&gt;7&lt;/math&gt; and by substitution &lt;math&gt;N = 24 \cdot 7 - 19 = 149&lt;/math&gt;<br /> <br /> The column values can also easily be found by substitution<br /> &lt;cmath&gt;\begin{align*}c_1&amp;=32\\<br /> c_2&amp;=7\\<br /> c_3&amp;=141\\<br /> c_4&amp;=88\\<br /> c_5&amp;=107<br /> \end{align*}&lt;/cmath&gt;<br /> As these are all positive and less than &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;\boxed{149}&lt;/math&gt; is the solution.<br /> <br /> == Sidenote ==<br /> If we express all the &lt;math&gt;c_i&lt;/math&gt; in terms of &lt;math&gt;N&lt;/math&gt;, we have<br /> &lt;cmath&gt;24c_1=5N+23&lt;/cmath&gt;<br /> &lt;cmath&gt;24c_2=N+19&lt;/cmath&gt;<br /> &lt;cmath&gt;124c_3=117N+51&lt;/cmath&gt;<br /> &lt;cmath&gt;124c_4=73N+35&lt;/cmath&gt;<br /> &lt;cmath&gt;124c_5=89N+7&lt;/cmath&gt;<br /> <br /> It turns out that there exists such an array satisfying the problem conditions if and only if<br /> &lt;cmath&gt;N\equiv 149 \pmod{744}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|num-b=10|num-a=12|t=384200}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_16&diff=70413 1970 AHSME Problems/Problem 16 2015-05-25T05:01:59Z <p>Claudeaops: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;F(n)&lt;/math&gt; is a function such that &lt;math&gt;F(1)=F(2)=F(3)=1&lt;/math&gt;, and such that &lt;math&gt;F(n+1)= \frac{F(n)\cdot F(n-1)+1}{F(n-2)}&lt;/math&gt; for &lt;math&gt;n\ge 3,&lt;/math&gt; <br /> then &lt;math&gt;F(6)=&lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A) } 2\quad<br /> \text{(B) } 3\quad<br /> \text{(C) } 7\quad<br /> \text{(D) } 11\quad<br /> \text{(E) } 26&lt;/math&gt;<br /> <br /> == Solution =<br /> We can chug through the recursion to find the answer is &lt;math&gt;\fbox{C}&lt;/math&gt;.<br /> <br /> ==Sidenote==<br /> All the numbers in the sequence &lt;math&gt;F(n)&lt;/math&gt; are integers. In fact, the function &lt;math&gt;F&lt;/math&gt; satisfies &lt;math&gt;F(n)=4F(n-2)-F(n-4)&lt;/math&gt;. (Prove it!).<br /> <br /> == See also ==<br /> {{AHSME 35p box|year=1970|num-b=15|num-a=17}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_40&diff=70412 1958 AHSME Problems/Problem 40 2015-05-25T04:23:45Z <p>Claudeaops: /* Sidenote */</p> <hr /> <div>== Problem ==<br /> Given &lt;math&gt; a_0 = 1&lt;/math&gt;, &lt;math&gt; a_1 = 3&lt;/math&gt;, and the general relation &lt;math&gt; a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n&lt;/math&gt; for &lt;math&gt; n \ge 1&lt;/math&gt;. Then &lt;math&gt; a_3&lt;/math&gt; equals:<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{13}{27}\qquad <br /> \textbf{(B)}\ 33\qquad <br /> \textbf{(C)}\ 21\qquad <br /> \textbf{(D)}\ 10\qquad <br /> \textbf{(E)}\ -17&lt;/math&gt;<br /> <br /> == Solution ==<br /> Using the recursive definition, we find that &lt;math&gt;a_3=33&lt;/math&gt;.<br /> <br /> ==Sidenote==<br /> All the terms in the sequence &lt;math&gt;a_n&lt;/math&gt; are integers. In fact, the sequence &lt;math&gt;a_n&lt;/math&gt; satisfies the recursion &lt;math&gt;a_n=3a_{n-1}+a_{n-2}&lt;/math&gt; (Prove it!).<br /> <br /> == See Also ==<br /> <br /> {{AHSME 50p box|year=1958|num-b=39|num-a=41}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_40&diff=70411 1958 AHSME Problems/Problem 40 2015-05-25T04:22:35Z <p>Claudeaops: Sidenote</p> <hr /> <div>== Problem ==<br /> Given &lt;math&gt; a_0 = 1&lt;/math&gt;, &lt;math&gt; a_1 = 3&lt;/math&gt;, and the general relation &lt;math&gt; a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n&lt;/math&gt; for &lt;math&gt; n \ge 1&lt;/math&gt;. Then &lt;math&gt; a_3&lt;/math&gt; equals:<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{13}{27}\qquad <br /> \textbf{(B)}\ 33\qquad <br /> \textbf{(C)}\ 21\qquad <br /> \textbf{(D)}\ 10\qquad <br /> \textbf{(E)}\ -17&lt;/math&gt;<br /> <br /> == Solution ==<br /> Using the recursive definition, we find that &lt;math&gt;a_3=33&lt;/math&gt;.<br /> <br /> ==Sidenote==<br /> All the terms in the sequence &lt;math&gt;a_n&lt;/math&gt; are integers. In fact, the sequence &lt;math&gt;a_n&lt;/math&gt; satisfies the recursion &lt;math&gt;a_n=3a_(n-1)+a_(n-2)&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AHSME 50p box|year=1958|num-b=39|num-a=41}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_40&diff=70410 1958 AHSME Problems/Problem 40 2015-05-25T04:21:20Z <p>Claudeaops: /* Solution */</p> <hr /> <div>== Problem ==<br /> Given &lt;math&gt; a_0 = 1&lt;/math&gt;, &lt;math&gt; a_1 = 3&lt;/math&gt;, and the general relation &lt;math&gt; a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n&lt;/math&gt; for &lt;math&gt; n \ge 1&lt;/math&gt;. Then &lt;math&gt; a_3&lt;/math&gt; equals:<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{13}{27}\qquad <br /> \textbf{(B)}\ 33\qquad <br /> \textbf{(C)}\ 21\qquad <br /> \textbf{(D)}\ 10\qquad <br /> \textbf{(E)}\ -17&lt;/math&gt;<br /> <br /> == Solution ==<br /> Using the recursive definition, we find that &lt;math&gt;a_3=33&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AHSME 50p box|year=1958|num-b=39|num-a=41}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63169 2013 AMC 10A Problems/Problem 20 2014-08-18T23:17:43Z <p>Claudeaops: Fixing Latex</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Image:AMC 10A 2013 20.jpg]]<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the square and &lt;math&gt;C&lt;/math&gt; be the intersection of &lt;math&gt;OB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. The desired area consists of the unit square, plus 4 regions congruent to the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt;, plus 4 triangular regions congruent to right triangle &lt;math&gt;BCD&lt;/math&gt;. The area of the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt; is (Area of Circle-Area of Square)/8. Since the circle has radius &lt;math&gt;\dfrac{1}{\sqrt {2}}&lt;/math&gt;, the area of the region is &lt;math&gt;\dfrac{\dfrac{\pi}{2}-1}{8}&lt;/math&gt;, so 4 times the area of that region is &lt;math&gt;\dfrac{\pi}{4}-\dfrac{1}{2}&lt;/math&gt;. Now we find the area of &lt;math&gt;\triangle BCD&lt;/math&gt;. &lt;math&gt;BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}&lt;/math&gt;. Since &lt;math&gt;\triangle BCD&lt;/math&gt; is a 45-45-90 right triangle, the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}&lt;/math&gt;, so 4 times the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\dfrac{3}{2}-\sqrt {2}&lt;/math&gt;. Finally, the area of the whole region is &lt;math&gt;1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63168 2013 AMC 10A Problems/Problem 20 2014-08-18T23:02:30Z <p>Claudeaops: Fixing Latex</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Image:AMC 10A 2013 20.jpg]]<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the square and &lt;math&gt;C&lt;/math&gt; be the intersection of &lt;math&gt;OB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. The desired area consists of the unit square, plus 4 regions congruent to the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt;, plus 4 triangular regions congruent to right triangle &lt;math&gt;BCD&lt;/math&gt;. The area of the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt; is (Area of Circle-Area of Square)/8. Since the circle has radius &lt;math&gt;\dfrac{1}{\sqrt {2}}&lt;/math&gt;, the area of the region is &lt;math&gt;\dfrac{\dfrac{\pi}{2}-1}{8}&lt;/math&gt;, so 4 times the area of that region is &lt;math&gt;\dfrac{\pi}{4}-\dfrac{1}{2}&lt;/math&gt;. Now we find the area of &lt;math&gt;\triangle BCD&lt;/math&gt;. &lt;math&gt;BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}&lt;/math&gt;. Since &lt;math&gt;\triangle BCD&lt;/math&gt; is a 45-45-90 right triangle, the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\dfrac{BC^2}{2}=\dfrac{({\dfrac{\sqrt {2}}{2}-\dfrac{1}{2})^2}{2}&lt;/math&gt;, so 4 times the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\dfrac{3}{2}-\sqrt {2}&lt;/math&gt;. Finally, the area of the whole region is &lt;math&gt;1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=&lt;/math&gt;\dfrac{\pi}{4}+2-\sqrt {2}\$.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63167 2013 AMC 10A Problems/Problem 20 2014-08-18T23:01:55Z <p>Claudeaops: Solution 2</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Image:AMC 10A 2013 20.jpg]]<br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the square and &lt;math&gt;C&lt;/math&gt; be the intersection of &lt;math&gt;OB&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;. The desired area consists of the unit square, plus 4 regions congruent to the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt;, plus 4 triangular regions congruent to right triangle &lt;math&gt;BCD&lt;/math&gt;. The area of the region bounded by arc &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\overline{AC}&lt;/math&gt;, and &lt;math&gt;\overline{BC}&lt;/math&gt; is (Area of Circle-Area of Square)/8. Since the circle has radius &lt;math&gt;\dfrac{1}{\sqrt {2}}&lt;/math&gt;, the area of the region is &lt;math&gt;\dfrac{\dfrac{\pi}{2}-1}{8}&lt;/math&gt;, so 4 times the area of that region is &lt;math&gt;\dfrac{\pi}{4}-\dfrac{1}{2}&lt;/math&gt;. Now we find the area of &lt;math&gt;\triangle BCD&lt;/math&gt;. &lt;math&gt;BC=BO-OC=&lt;/math&gt;\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}&lt;math&gt;. Since &lt;/math&gt;\triangle BCD&lt;math&gt; is a 45-45-90 right triangle, the area of &lt;/math&gt;\triangle BCD&lt;math&gt; is &lt;/math&gt;\dfrac{BC^2}{2}=\dfrac{({\dfrac{\sqrt {2}}{2}-\dfrac{1}{2})^2}{2}&lt;math&gt;, so 4 times the area of &lt;/math&gt;\triangle BCD&lt;math&gt; is &lt;/math&gt;\dfrac{3}{2}-\sqrt {2}&lt;math&gt;. Finally, the area of the whole region is &lt;/math&gt;1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=&lt;math&gt;\dfrac{\pi}{4}+2-\sqrt {2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63166 2013 AMC 10A Problems/Problem 20 2014-08-18T22:43:30Z <p>Claudeaops: Centering the Image</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Image:AMC 10A 2013 20.jpg|thumb|center]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63165 2013 AMC 10A Problems/Problem 20 2014-08-18T22:41:44Z <p>Claudeaops: Solution 2</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Image:AMC 10A 2013 20.jpg]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=File:AMC_10A_2013_20.jpg&diff=63164 File:AMC 10A 2013 20.jpg 2014-08-18T22:40:40Z <p>Claudeaops: uploaded a new version of &amp;quot;File:AMC 10A 2013 20.jpg&amp;quot;: Resized Version of Image</p> <hr /> <div>Diagram for Solution</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63161 2013 AMC 10A Problems/Problem 20 2014-08-18T05:47:14Z <p>Claudeaops: </p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63160 2013 AMC 10A Problems/Problem 20 2014-08-18T05:46:53Z <p>Claudeaops: </p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> [[Image:AMC 10A 2013 20.jpg|200px|thumb|left|alt text]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63159 2013 AMC 10A Problems/Problem 20 2014-08-18T05:46:08Z <p>Claudeaops: </p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> [[Image:AMC 10A 2013 20.jpg]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=File:AMC_10A_2013_20.jpg&diff=63158 File:AMC 10A 2013 20.jpg 2014-08-18T05:44:49Z <p>Claudeaops: Diagram for Solution</p> <hr /> <div>Diagram for Solution</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63157 2013 AMC 10A Problems/Problem 20 2014-08-18T05:35:54Z <p>Claudeaops: </p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63156 2013 AMC 10A Problems/Problem 20 2014-08-18T05:34:59Z <p>Claudeaops: Image</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> [[Image:AMC 10A 2013 -20.png]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63155 2013 AMC 10A Problems/Problem 20 2014-08-18T05:32:23Z <p>Claudeaops: Image</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63154 2013 AMC 10A Problems/Problem 20 2014-08-18T05:31:58Z <p>Claudeaops: Image</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> [[Image:AMC 10A 2013 -20.png|thumb|left|100px|]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=File:AMC_10A_2013_-20.png&diff=63153 File:AMC 10A 2013 -20.png 2014-08-18T05:25:56Z <p>Claudeaops: uploaded a new version of &amp;quot;File:AMC 10A 2013 -20.png&amp;quot;</p> <hr /> <div>Diagram for the Solution</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63152 2013 AMC 10A Problems/Problem 20 2014-08-18T04:48:32Z <p>Claudeaops: </p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63151 2013 AMC 10A Problems/Problem 20 2014-08-18T04:47:22Z <p>Claudeaops: Image Formatting</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Image:AMC_10A_2013_-20.png|200px|thumb|left|alt text]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=File:AMC_10A_2013_-20.png&diff=63150 File:AMC 10A 2013 -20.png 2014-08-18T04:44:50Z <p>Claudeaops: Diagram for the Solution</p> <hr /> <div>Diagram for the Solution</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63149 2013 AMC 10A Problems/Problem 20 2014-08-18T04:44:04Z <p>Claudeaops: Image Upload</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[Image:AMC_10A_2013_-20.png]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63148 2013 AMC 10A Problems/Problem 20 2014-08-18T04:43:34Z <p>Claudeaops: Image Upload</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[File:AMC_10A_2013_-20.png]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63147 2013 AMC 10A Problems/Problem 20 2014-08-18T04:30:37Z <p>Claudeaops: </p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_20&diff=63146 2013 AMC 10A Problems/Problem 20 2014-08-18T04:28:20Z <p>Claudeaops: Solution 2</p> <hr /> <div>==Problem==<br /> <br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> ==Solution 1==<br /> <br /> First, we need to see what this looks like. Below is a diagram.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;<br /> fill(square^^square2,grey);<br /> for(int i=0;i&lt;=3;i=i+1)<br /> {<br /> path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));<br /> draw(arcrot);<br /> fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);<br /> draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);<br /> }<br /> draw(square^^square2);&lt;/asy&gt;<br /> <br /> For this square with side length 1, the distance from center to vertex is &lt;math&gt;r = \frac{1}{\sqrt{2}}&lt;/math&gt;, hence the area is composed of a semicircle of radius &lt;math&gt;r&lt;/math&gt;, plus &lt;math&gt;4&lt;/math&gt; times a parallelogram with height &lt;math&gt;\frac{1}{2}&lt;/math&gt; and base &lt;math&gt;\frac{\sqrt{2}}{2(1+\sqrt{2})}&lt;/math&gt;. That is to say, the total area is &lt;math&gt;\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> [[File: AMC_10A_2013_-20.png&quot;]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_20&diff=60346 2014 AMC 10B Problems/Problem 20 2014-02-21T20:31:49Z <p>Claudeaops: /* Solution */</p> <hr /> <div>==Problem==<br /> For how many integers &lt;math&gt;x&lt;/math&gt; is the number &lt;math&gt;x^4-51x^2+50&lt;/math&gt; negative?<br /> <br /> &lt;math&gt; \textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, note that &lt;math&gt;50+1=51&lt;/math&gt;, which motivates us to factor the polynomial as &lt;math&gt;(x^2-50)(x^2-1)&lt;/math&gt;. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so &lt;math&gt;x^2-50&lt;0&lt;x^2-1&lt;/math&gt;. Solving this inequality, we find &lt;math&gt;1&lt;x^2&lt;50&lt;/math&gt;. There are exactly 12 integers &lt;math&gt;x&lt;/math&gt; that satisfy this inequality, &lt;math&gt;\pm 2,3,4,5,6,7&lt;/math&gt;.<br /> <br /> Thus our answer is &lt;math&gt;\boxed{\textbf {(C) } 12}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Claudeaops