https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathemagician1729&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T18:41:52ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_24&diff=331272008 AMC 12B Problems/Problem 242010-01-01T01:55:38Z<p>Mathemagician1729: /* Solution */</p>
<hr />
<div>==Problem 24==<br />
Let <math>A_0=(0,0)</math>. Distinct points <math>A_1,A_2,\dots</math> lie on the <math>x</math>-axis, and distinct points <math>B_1,B_2,\dots</math> lie on the graph of <math>y=\sqrt{x}</math>. For every positive integer <math>n,\ A_{n-1}B_nA_n</math> is an equilateral triangle. What is the least <math>n</math> for which the length <math>A_0A_n\geq100</math>?<br />
<br />
<math>\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21</math><br />
<br />
==Solution==<br />
Let <math>a_n=|A_{n-1}A_n|</math>. We need to rewrite the recursion into something manageable. The two strange conditions, <math>B</math>'s lie on the graph of <math>y=\sqrt{x}</math> and <math>A_{n-1}B_nA_n</math> is an equilateral triangle, can be compacted as follows: <cmath>\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1</cmath><br />
which uses <math>y^2=x</math>, where <math>x</math> is the height of the equilateral triangle and therefore <math>\frac{\sqrt{3}}{2}</math> times its base.<br />
<br />
The relation above holds for <math>n=k</math> and for <math>n=k-1</math> <math>(k>1)</math>, so <cmath>\left(a_k\frac{\sqrt{3}}{2}\right)^2-\left(a_{k-1}\frac{\sqrt{3}}{2}\right)^2=</cmath><br />
<cmath>=\left(\frac{a_k}{2}+a_{k-1}+a_{k-2}+\cdots+a_1\right)-\left(\frac{a_{k-1}}{2}+a_{k-2}+a_{k-3}+\cdots+a_1\right)</cmath><br />
Or, <cmath>a_k-a_{k-1}=\frac23</cmath> This implies that each segment of a successive triangle is <math>\frac23</math> more than the last triangle. To find <math>a_{1}</math>, we merely have to plug in <math>k=1</math> into the aforementioned recursion and we have <math>a_{1} - a_{0} = \frac23</math>. Knowing that <math>a_{0}</math> is <math>0</math>, we can deduce that <math>a_{1} = 2/3</math>.Thus, <math>a_n=\frac{2n}{3}</math>, so <math>A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{2}{3} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{3}</math>. We want to find <math>n</math> so that <math>n^2<300<(n+1)^2</math>. <math>n=\boxed{17}</math> is our answer.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=B|num-b=23|num-a=25}}</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&diff=326382008 AMC 12A Problems/Problem 222009-08-11T18:12:56Z<p>Mathemagician1729: /* Solution 2 (without trigonometry) */</p>
<hr />
<div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #22]] and [[2008 AMC 10A Problems/Problem 25|2004 AMC 10A #25]]}}<br />
==Problem==<br />
A round table has radius <math>4</math>. Six rectangular place mats are placed on the table. Each place mat has width <math>1</math> and length <math>x</math> as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length <math>x</math>. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is <math>x</math>?<br />
<br />
<asy>unitsize(4mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(1\)",(-0.5,3.8),S);</asy><br />
<br />
<math>\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}</math><br />
<br />
==Solution==<br />
=== Solution 1 (trigonometry) ===<br />
Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown: <br />
<br />
<asy>unitsize(8mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(x\)",(0.03,1.5),E);<br />
label("\(A\)",(-3.6,2.5513),E);<br />
label("\(B\)",(-3.15,1.35),E);<br />
label("\(C\)",(0.05,3.20),E);<br />
label("\(D\)",(-0.75,4.15),E);<br />
label("\(O\)",(0.00,-0.10),E);<br />
label("\(1\)",(-0.1,3.8),S);<br />
label("\(4\)",(-0.4,2.2),S);<br />
draw(Line(0,0)--(0,3.103));<br />
draw(Line(0,0)--(-2.687,1.5513));<br />
draw(Line(0,0)--(-0.5,3.9686));</asy><br />
<br />
Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]]. So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>. <br />
<br />
By the [[Law of Cosines]]: <math>4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}</math>. <br />
<br />
Since <math>x</math> must be positive, <math>x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C</math>.<br />
<br />
=== Solution 2 (without trigonometry) ===<br />
<br />
[http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=218 See Math Jam]<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}}<br />
{{AMC10 box|year=2008|ab=A|num-b=24|after=Last Question}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Introductory Trigonometry Problems]]</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_23&diff=326372008 AMC 10A Problems/Problem 232009-08-11T18:07:38Z<p>Mathemagician1729: /* Solution */</p>
<hr />
<div>==Problem==<br />
Two subsets of the set <math>S=\lbrace a,b,c,d,e\rbrace</math> are to be chosen so that their union is <math>S</math> and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?<br />
<br />
<math>\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320</math><br />
<br />
==Solution==<br />
<br />
<br />
<br />
First choose the two letters to be repeated in each set. <math>\dbinom{5}{2}=10</math>. Now we have three remaining elements that we wish to place into two separate subsets. There are <math>2^3 = 8 </math> ways to do so (Do you see why?). Unfortunately, we have over-counted (Take for example <math>S_{1} = \{a,b,c,d \}</math> and <math>S_{2} = \{a,b,e \}</math>). Notice how <math>S_{1}</math> and <math>S_{2}</math> are interchangeable. A simple division by two will fix this problem. Thus we have:<br />
<br />
<math> \dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{D}} </math><br />
<br />
This problem was discussed in an AoPS Math Jam a while back. The transcript should be located here: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=218<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=22|num-a=24}}</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_23&diff=326362008 AMC 10A Problems/Problem 232009-08-11T18:06:48Z<p>Mathemagician1729: /* Solution */</p>
<hr />
<div>==Problem==<br />
Two subsets of the set <math>S=\lbrace a,b,c,d,e\rbrace</math> are to be chosen so that their union is <math>S</math> and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?<br />
<br />
<math>\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320</math><br />
<br />
==Solution==<br />
<br />
<br />
<br />
First choose the two letters to be repeated in each set. <math>\dbinom{5}{2}=10</math>. Now we have three remaining elements that we wish to place into two separate subsets. There are <math>2^3 = 8 </math> ways to do so (Do you see why?). Unfortunately, we have over-counted (Take for example <math>S_{1} = \{a,b,c,d \}</math> and <math>S_{1} = \{a,b,e \}</math>). Notice how <math>S_{1}</math> and <math>S_{2}</math> are interchangeable. A simple division by two will fix this problem. Thus we have:<br />
<br />
<math> \dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{D}} </math><br />
<br />
This problem was discussed in an AoPS Math Jam a while back. The transcript should be located here: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=218<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=22|num-a=24}}</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10B_Problems/Problem_23&diff=322612006 AMC 10B Problems/Problem 232009-07-07T18:57:56Z<p>Mathemagician1729: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?<br />
<br />
<asy><br />
unitsize(1.5cm);<br />
defaultpen(.8);<br />
<br />
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br />
pair F = intersectionpoint( A--D, B--Ep );<br />
<br />
draw( A -- B -- C -- cycle );<br />
draw( A -- D );<br />
draw( B -- Ep );<br />
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br />
<br />
label("$7$",(1.25,0.2));<br />
label("$7$",(2.2,0.45));<br />
label("$3$",(0.45,0.35));<br />
</asy><br />
<br />
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math><br />
<br />
== Solution ==<br />
<br />
Label the points in the figure as shown below, and draw the segment <math>CF</math>. This segment divides the quadrilateral into two triangles, let their areas be <math>x</math> and <math>y</math>.<br />
<br />
<asy><br />
unitsize(2cm);<br />
defaultpen(.8);<br />
<br />
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);<br />
pair F = intersectionpoint( A--D, B--Ep );<br />
<br />
draw( A -- B -- C -- cycle );<br />
draw( A -- D );<br />
draw( B -- Ep );<br />
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );<br />
<br />
label("$7$",(1.45,0.15));<br />
label("$7$",(2.2,0.45));<br />
label("$3$",(0.45,0.35));<br />
<br />
draw( C -- F, dashed );<br />
<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$D$",D,NE);<br />
label("$E$",Ep,NW);<br />
label("$F$",F,S);<br />
<br />
label("$x$",(1,1));<br />
label("$y$",(1.6,1));<br />
</asy><br />
<br />
Since [[triangle]]s <math>AFB</math> and <math>DFB</math> share an [[altitude]] from <math>B</math> and have equal area, their bases must be equal, hence <math>AF=DF</math>.<br />
<br />
Since triangles <math>AFC</math> and <math>DFC</math> share an altitude from <math>C</math> and their respective bases are equal, their areas must be equal, hence <math>x+3=y</math>. <br />
<br />
Since triangles <math>EFA</math> and <math>BFA</math> share an altitude from <math>A</math> and their respective areas are in the ratio <math>3:7</math>, their bases must be in the same ratio, hence <math>EF:FB = 3:7</math>.<br />
<br />
Since triangles <math>EFC</math> and <math>BFC</math> share an altitude from <math>C</math> and their respective bases are in the ratio <math>3:7</math>, their areas must be in the same ratio, hence <math>x:(y+7) = 3:7</math>, which gives us <math>7x = 3(y+7)</math>.<br />
<br />
Substituting <math>y=x+3</math> into the second equation we get <math>7x = 3(x+10)</math>, which solves to <math>x=\frac{15}{2}</math>. Then <math>y=x+3 = \frac{15}{2}+3 = \frac{21}{2}</math>, and the total area of the quadrilateral is <math>x+y = \frac{15}{2}+\frac{21}{2} = \boxed{18}</math>.<br />
<br />
== See Also ==<br />
<br />
*[[2006 AMC 10B Problems]]<br />
<br />
*[[2006 AMC 10B Problems/Problem 22|Previous Problem]]<br />
<br />
*[[2006 AMC 10B Problems/Problem 24|Next Problem]]<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_22&diff=322372007 AMC 12A Problems/Problem 222009-07-04T02:57:36Z<p>Mathemagician1729: /* Solution 1 */</p>
<hr />
<div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #22]] and [[2007 AMC 10A Problems/Problem 25|2007 AMC 10A #25]]}}<br />
== Problem ==<br />
For each positive integer <math>n</math>, let <math>S(n)</math> denote the sum of the digits of <math>n.</math> For how many values of <math>n</math> is <math>n + S(n) + S(S(n)) = 2007?</math> <br />
<br />
<math>\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5</math><br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
For the sake of notation let <math>T(n) = n + S(n) + S(S(n))</math>. Obviously <math>n<2007</math>. Then the maximum value of <math>S(n) + S(S(n))</math> is when <math>n = 1999</math>, and the sum becomes <math>28 + 10 = 38</math>. So the minimum bound is <math>1969</math>. We do [[casework]] upon the tens digit:<br />
<br />
Case 1: <math>196u \Longrightarrow u = 9</math>. Easy to directly disprove.<br />
<br />
Case 2: <math>197u</math>. <math>S(n) = 1 + 9 + 7 + u = 17 + u</math>, and <math>S(S(n)) = 8+u</math> if <math>u \le 2</math> and <math>S(S(n)) = 2 + (u-3) = u-1</math> otherwise. <br />
<br />
:Subcase a: <math>T(n) = 1970 + u + 17 + u + 8 + u = 1995 + 3u = 2007 \Longrightarrow u = 4</math>. This exceeds our bounds, so no solution here.<br />
:Subcase b: <math>T(n) = 1970 + u + 17 + u + u - 1 = 1986 + 3u = 2007 \Longrightarrow u = 7</math>. First solution.<br />
<br />
Case 3: <math>198u</math>. <math>S(n) = 18 + u</math>, and <math>S(S(n)) = 9 + u</math> if <math>u \le 1</math> and <math>2 + (u-2) = u</math> otherwise.<br />
<br />
:Subcase a: <math>T(n) = 1980 + u + 18 + u + 9 + u = 2007 + 3u = 2007 \Longrightarrow u = 0</math>. Second solution.<br />
:Subcase b: <math>T(n) = 1980 + u + 18 + u + u = 1998 + 3u = 2007 \Longrightarrow u = 3</math>. Third solution. <br />
<br />
Case 4: <math>199u</math>. But <math>S(n) > 19</math>, and the these clearly sum to <math>> 2007</math>.<br />
<br />
Case 5: <math>200u</math>. So <math>S(n) = 2 + u</math> and <math>S(S(n)) = 2 + u</math> (recall that <math>n < 2007 </math>), and <math>2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1</math>. Fourth solution.<br />
<br />
In total we have <math>4 \mathrm{(D)}</math> solutions, which are <math>1977, 1980, 1983, </math> and <math>2001</math>.<br />
<br />
=== Solution 2 ===<br />
Clearly, <math>n > 1900</math>. We can break this up into three cases:<br />
<br />
Case 1: <math>n \geq 2000</math><br />
<br />
:Inspection gives <math>n = 2001</math>.<br />
<br />
Case 2: <math>n < 2000</math>, <math>n = 19xy</math>, <math>x + y < 10 </math><br />
<br />
:If you set up an equation, it reduces to<br />
<br />
<math>4x + y = 32</math><br />
<br />
:which has as its only solution satisfying the constraints <math>x = 8</math>, <math>y = 0</math>.<br />
<br />
Case 3: <math>n < 2000</math>, <math>n = 19xy</math>, <math>x + y \geq 10</math><br />
<br />
:This reduces to<br />
<br />
:<math>4x + y = 35</math>. The only two solutions satisfying the constraints for this equation are <math>x = 7</math>, <math>y = 7</math> and <math>x = 8</math>, <math>y = 3</math>. <br />
<br />
The solutions are thus <math>1977, 1980, 1983, 2001</math> and the answer is <math>\mathrm{(D)}\ 4</math>.<br />
<br />
=== Solution 3 ===<br />
As in Solution 1, we note that <math>S(n)\leq 28</math> and <math>S(S(n))\leq 10</math>. <br/><br />
Obviously, <math>n\equiv S(n)\equiv S(S(n)) \pmod 9</math>. <br/><br />
As <math>2007\equiv 0 \pmod 9</math>, this means that <math>n\bmod 9 \in\{0,3,6\}</math>, or equivalently that <math>n\equiv S(n)\equiv S(S(n))\equiv 0 \pmod 3</math>.<br />
<br />
Thus <math>S(S(n))\in\{3,6,9\}</math>. For each possible <math>S(S(n))</math> we get three possible <math>S(n)</math>. <br/><br />
(E. g., if <math>S(S(n))=6</math>, then <math>S(n)=x</math> is a number such that <math>x\leq 28</math> and <math>S(x)=6</math>, therefore <math>S(n)\in\{6,15,24\}</math>.)<br />
<br />
For each of these nine possibilities we compute <math>n_{?}</math> as <math>2007-S(n)-S(S(n))</math> and check whether <math>S(n_{?})=S(n)</math>. <br/><br />
We'll find out that out of the 9 cases, in 4 the value <math>n_{?}</math> has the correct sum of digits. <br/><br />
This happens for <math>n_{?}\in \{ 1977, 1980, 1983, 2001 \}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2007|ab=A|num-b=21|num-a=23}}<br />
{{AMC10 box|year=2007|ab=A|num-b=24|after=Last question}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_22&diff=321052009 AMC 10B Problems/Problem 222009-06-10T14:45:26Z<p>Mathemagician1729: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
A cubical cake with edge length <math>2</math> inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where <math>M</math> is the midpoint of a top edge. The piece whose top is triangle <math>B</math> contains <math>c</math> cubic inches of cake and <math>s</math> square inches of icing. What is <math>c+s</math>?<br />
<br />
<asy><br />
unitsize(1cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);<br />
draw((1,1)--(-1,0));<br />
pair P=foot((1,-1),(1,1),(-1,0));<br />
draw((1,-1)--P);<br />
draw(rightanglemark((-1,0),P,(1,-1),4));<br />
<br />
label("$M$",(-1,0),W);<br />
label("$C$",(-0.1,-0.3));<br />
label("$A$",(-0.4,0.7));<br />
label("$B$",(0.7,0.4));<br />
</asy><br />
<br />
<math><br />
\text{(A) } \frac{24}{5}<br />
\qquad<br />
\text{(B) } \frac{32}{5}<br />
\qquad<br />
\text{(C) } 8+\sqrt5<br />
\qquad<br />
\text{(D) } 5+\frac{16\sqrt5}{5}<br />
\qquad<br />
\text{(E) } 10+5\sqrt5<br />
</math><br />
<br />
== Solution ==<br />
<br />
<br />
<asy><br />
unitsize(2cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);<br />
draw((1,1)--(-1,0));<br />
pair P=foot((1,-1),(1,1),(-1,0));<br />
draw((1,-1)--P);<br />
draw(rightanglemark((-1,0),P,(1,-1),4));<br />
<br />
label("$M$",(-1,0),W);<br />
label("$C$",(-0.1,-0.3));<br />
label("$A$",(-0.4,0.7));<br />
label("$B$",(0.7,0.4));<br />
label("$P$",(-1,1),NW);<br />
label("$Q$",(1,1),NE);<br />
label("$R$",(1,-1),SE);<br />
label("$S$",(-1,-1),SW);<br />
label("$N$",P,NW);<br />
</asy><br />
<br />
Let's label the points as in the picture above. Let <math>[RNQ]</math> be the area of <math>\triangle RNQ</math>. Then the volume of the corresponding piece is <math>c=2[RNQ]</math>. This cake piece has icing on the top and on the vertical side that contains the edge <math>QR</math>. Hence the total area with icing is <math>[RNQ]+2^2 = [RNQ]+4</math>. Thus the answer to our problem is <math>3[RNQ]+4</math>, and all we have to do now is to determine <math>[RNQ]</math>.<br />
<br />
=== Solution 1 ===<br />
<br />
Introduce a coordinate system where <math>Q=(0,0)</math>, <math>P=(2,0)</math> and <math>R=(0,2)</math>. <br />
<br />
In this coordinate system we have <math>M=(2,1)</math>, and the line <math>QM</math> has the equation <math>2y-x=0</math>. <br />
<br />
As the line <math>RN</math> is orthogonal to <math>QM</math>, it must have the equation <math>y+2x+q=0</math> for some suitable constant <math>q</math>. As this line contains the point <math>R=(0,2)</math>, we have <math>q=-2</math>.<br />
<br />
Substituting <math>x=2y</math> into <math>y+2x-2=0</math>, we get <math>y=\frac 25</math>, and then <math>x=\frac 45</math>. <br />
<br />
We can note that in <math>\triangle RNQ</math> <math>x</math> is the height from <math>N</math> onto <math>RQ</math>, hence its area is <math>[RNQ] = \frac{x \cdot RQ} 2 = \frac{2x}2 = x = \frac 45</math>, and therefore the answer is <math>3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
Extend <math>RN</math> to intersect <math>PQ</math> at <math>O</math>:<br />
<br />
<asy><br />
unitsize(2cm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
<br />
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);<br />
draw((1,1)--(-1,0));<br />
pair P=foot((1,-1),(1,1),(-1,0));<br />
draw((1,-1)--P);<br />
draw(rightanglemark((-1,0),P,(1,-1),4));<br />
draw(P -- (0,1));<br />
<br />
label("$M$",(-1,0),W);<br />
label("$C$",(-0.1,-0.3));<br />
label("$A$",(-0.4,0.7));<br />
label("$B$",(0.7,0.4));<br />
label("$P$",(-1,1),NW);<br />
label("$Q$",(1,1),NE);<br />
label("$R$",(1,-1),SE);<br />
label("$S$",(-1,-1),SW);<br />
label("$N$",P,1.5*WNW);<br />
label("$O$",(0,1),N);<br />
</asy><br />
<br />
It is now obvious that <math>O</math> is the midpoint of <math>PQ</math>. (Imagine rotating the square <math>PQRS</math> by <math>90^\circ</math> clockwise around its center. This rotation will map the segment <math>MQ</math> to a segment that is orthogonal to <math>MQ</math>, contains <math>R</math> and contains the midpoint of <math>PQ</math>.)<br />
<br />
From <math>\triangle PQM</math> we can compute that <math>QM = \sqrt{1^2 + 2^2} = \sqrt 5</math>.<br />
<br />
Observe that <math>\triangle PQM</math> and <math>\triangle NQO</math> have the same angles and therefore they are similar. The ratio of their sides is <math>\frac{QM}{OQ} = \frac{\sqrt 5}1 = \sqrt 5</math>.<br />
<br />
Hence we have <math>ON = \frac{PM}{\sqrt 5} = \frac 1{\sqrt 5}</math>, and <math>NQ = \frac{PQ}{\sqrt 5} = \frac 2{\sqrt 5}</math>.<br />
<br />
Knowing this, we can compute the area of <math>\triangle NQO</math> as <math>[NQO] = \frac{ON \cdot NQ}2 = \frac 15</math>.<br />
<br />
Finally, we compute <math>[RNQ] = [ROQ] - [NQO] = 1 - \frac 15 = \frac 45</math>, and conclude that the answer is <math>3[RNQ]+4 = 3\cdot \frac 45 + 4 = \boxed{\frac{32}5}</math>.<br />
<br />
*You could also notice that the two triangles in the original figure are similar.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2009|ab=B|num-b=21|num-a=23}}</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283192007 AMC 8 Problems2008-11-02T19:14:52Z<p>Mathemagician1729: </p>
<hr />
<div>1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13<br />
<br />
<br />
2. 650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?<br />
<br />
[[Image:Graph.jpg]]</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283142007 AMC 8 Problems2008-11-02T19:07:27Z<p>Mathemagician1729: </p>
<hr />
<div>1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13<br />
<br />
<br />
2. 650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283132007 AMC 8 Problems2008-11-02T19:06:56Z<p>Mathemagician1729: </p>
<hr />
<div>1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13<br />
<br />
<br />
2. 650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?<br />
<br />
[[Image:Graph.jpg]]</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283112007 AMC 8 Problems2008-11-02T19:05:07Z<p>Mathemagician1729: </p>
<hr />
<div>1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13<br />
<br />
<br />
2. 650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283102007 AMC 8 Problems2008-11-02T19:04:10Z<p>Mathemagician1729: </p>
<hr />
<div>1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13<br />
<br />
<br />
2. 650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?<br />
<br />
[[Image:Graph.jpg]]</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283092007 AMC 8 Problems2008-11-02T19:03:33Z<p>Mathemagician1729: </p>
<hr />
<div>1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13<br />
<br />
<br />
[http://www.artofproblemsolving.com/Wiki/images/f/fb/Graph.JPG]<br />
<br />
2. 650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=File:Graph.JPG&diff=28308File:Graph.JPG2008-11-02T19:02:27Z<p>Mathemagician1729: problem 2 2007 AMC 8</p>
<hr />
<div>problem 2 2007 AMC 8</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283072007 AMC 8 Problems2008-11-02T18:55:33Z<p>Mathemagician1729: </p>
<hr />
<div>1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13<br />
<br />
2. 650 students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_8_Problems&diff=283062007 AMC 8 Problems2008-11-02T18:41:40Z<p>Mathemagician1729: </p>
<hr />
<div><br />
1. Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work for the final week to earn the tickets?<br />
<br />
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13</div>Mathemagician1729https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_2&diff=282381997 AIME Problems/Problem 22008-10-24T02:54:27Z<p>Mathemagician1729: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The nine horizontal and nine vertical lines on an <math>8\times8</math> checkerboard form <math>r</math> [[rectangles]], of which <math>s</math> are [[square]]s. The number <math>s/r</math> can be written in the form <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math><br />
<br />
== Solution ==<br />
To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or <math>{9\choose 2} = 36</math>. Similarily, there are <math>{9\choose 2}</math> ways to pick the vertical sides, giving us <math>r = 1296</math> rectangles.<br />
<br />
For <math>s</math>, there are <math>8^2</math> [[unit square]]s, <math>7^2</math> of the <math>2\times2</math> squares, and so on until <math>1^2</math> of the <math>8\times 8</math> squares. Using the sum of squares formula, that gives us <math>s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204</math>.<br />
<br />
Thus <math>\frac rs = \dfrac{204}{1296}=\dfrac{17}{108}</math>, and <math>m+n=\boxed{125}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Mathemagician1729