https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Quantomaticguy&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T23:46:52ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_13&diff=1765552000 AIME II Problems/Problem 132022-08-02T13:57:42Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The [[equation]] <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n}}r</math>, where <math>m</math>, <math>n</math> and <math>r</math> are integers, <math>m</math> and <math>r</math> are relatively prime, and <math>r>0</math>. Find <math>m+n+r</math>.<br />
<br />
== Solution ==<br />
We may factor the equation as:{{ref|1}}<br />
<br />
<cmath><br />
\begin{align*}<br />
2000x^6+100x^5+10x^3+x-2&=0\\<br />
2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\<br />
2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\<br />
2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\<br />
(20x^2+x-2)(100x^4+10x^2+1)&=0\\<br />
\end{align*}<br />
</cmath><br />
<br />
Now <math>100x^4+10x^2+1\ge 1>0</math> for real <math>x</math>. Thus the real roots must be the roots of the equation <math>20x^2+x-2=0</math>. By the [[quadratic formula]] the roots of this are:<br />
<br />
<cmath>x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.</cmath><br />
<br />
Thus <math>r=\frac{-1+\sqrt{161}}{40}</math>, and so the final answer is <math>-1+161+40 = \boxed{200}</math>.<br />
<br />
<br /><br />
{{note|1}} A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of <math>x</math> with half of the polynomial's degree (in this case, divide through by <math>x^3</math>), and then to use one of the substitutions <math>t = x \pm \frac{1}{x}</math>. In this case, the substitution <math>t = x\sqrt{10} - \frac{1}{x\sqrt{10}}</math> gives <math>t^2 + 2 = 10x^2 + \frac 1{10x^2}</math> and <math>2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}</math>, which reduces the polynomial to just <math>(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0</math>. Then one can backwards solve for <math>x</math>.<br />
<br />
<br />
A slightly different approach using symmetry:<br />
let y = 10x -1/x<br />
notice that the equation can be rewritten( after dividing across by x^3) as 2( (10x)^3 - 1/x^3 ) + (10x)^2 +1/x^2 +10 =0<br />
now it is easy to see that the equation reduces to 2(y^3+30y)+ (y^2+20) + 10 = 0<br />
OR 2y^3 +y^2 +60y+30 = 0<br />
or y^2(2y+1) + 30(2y+1) = 0<br />
which is (2y+1)(y^2+30)= 0<br />
so for real solutions we have y= -1/2<br />
and solve the quadratic in x to get final answer as \boxed{200}$.<br />
<br />
== Solution 2 (Complex Bash)==<br />
It would be really nice if the coefficients were symmetrical. What if we make the substitution, <math>x = -\frac{i}{\sqrt{10}}y</math>. The the polynomial becomes <br />
<br />
<math>-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2</math><br />
<br />
It's symmetric! Dividing by <math>y^3</math> and rearranging, we get <br />
<br />
<math>-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})</math><br />
<br />
Now, if we let <math>z = y + \frac{1}{y}</math>, we can get the equations <br />
<br />
<math>z = y + \frac{1}{y}</math> <br />
<br />
<math>z^2 - 2 = y^2 + \frac{1}{y^2}</math><br />
<br />
and<br />
<br />
<math>z^3 - 3z = y^3 + \frac{1}{y^3}</math><br />
<br />
(These come from squaring <math>z</math> and subtracting <math>2</math>, then multiplying that result by <math>z</math> and subtracting <math>z</math>) <br />
Plugging this into our polynomial, expanding, and rearranging, we get <br />
<br />
<math>-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})</math><br />
<br />
Now, we see that the two <math>i</math> terms must cancel in order to get this polynomial equal to <math>0</math>, so what squared equals 3? Plugging in <math>z = \sqrt{3}</math> into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying <math>z = -\sqrt{3}</math>, we see that it also works! Great, we use long division on the polynomial by <br />
<br />
<math>(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)</math> and we get <br />
<br />
<math>2z -(\frac{i}{\sqrt{10}}) = 0</math>. <br />
<br />
We know that the other two solutions for z wouldn't result in real solutions for <math>x</math> since we have to solve a quadratic with a negative discriminant, then multiply by <math>-(\frac{i}{\sqrt{10}})</math>. We get that <math>z = (\frac{i}{-2\sqrt{10}})</math>. Solving for <math>y</math> (using <math>y + \frac{1}{y} = z</math>) we get that <math>y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}</math>, and multiplying this by <math>-(\frac{i}{\sqrt{10}})</math> (because <math>x = -(\frac{i}{\sqrt{10}})y</math>) we get that <math>x = \frac{-1 \pm \sqrt{161}}{40}</math> for a final answer of <math>-1 + 161 + 40 = \boxed{200}</math><br />
<br />
-Grizzy<br />
<br />
==Solution 3==<br />
Notice the original expression can be written as <math>2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2</math>.<br />
<br />
Which equals to <math>(20x^2+x-2)(100x^4+10x^2+1)=0</math><br />
<br />
So our solution is to find what is the root for <math>20x^2+x-2=0</math> since the determinant of <math>100t^2+10t+1<0</math>(Let <math>x^2=t</math>)<br />
<br />
By solving the equation, we can get that <math>x = \frac{-1 \pm \sqrt{161}}{40}</math> for a final answer of <math>-1 + 161 + 40 = \boxed{200}</math><br />
<br />
~bluesoul<br />
<br />
==Solution 4 (Geometric Series)==<br />
Observe that the given equation may be rearranged as<br />
<math>2000x^6-2+(100x^5+10x^3+x)=0</math>.<br />
The expression in parentheses is a geometric series with common factor <math>10x^2</math>. Using the geometric sum formula, we rewrite as<br />
<math>2000x^6-2+\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\neq0</math>.<br />
Factoring a bit, we get<br />
<math>2(1000x^6-1)+(1000x^6-1)\frac{x}{10x^2-1}=0, 10x^2-1\neq0 \implies</math><br />
<math>(1000x^6-1)(2+\frac{x}{10x^2-1})=0, 10x^2-1\neq0</math>.<br />
Note that setting <math>1000x^6-1=0</math> gives <math>10x^2-1=0</math>, which is clearly extraneous.<br />
Hence, we set <math>2+\frac{x}{10x^2-1}=0</math> and use the quadratic formula to get the desired root<br />
<math>x=\frac{-1+\sqrt{161}}{40} \implies -1+161+40=\boxed{200}</math><br />
<br />
~keeper1098<br />
<br />
==Video solution==<br />
<br />
https://www.youtube.com/watch?v=mAXDdKX52TM<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=II|num-b=12|num-a=14}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_1&diff=1754071990 AIME Problems/Problem 12022-06-28T02:29:25Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neither the [[perfect square | square]] nor the [[perfect cube | cube]] of a positive integer. Find the 500th term of this sequence.<br />
<br />
== Solution ==<br />
Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than <math>500</math>. This happens to be <math>23^2=529</math>. Notice that there are <math>23</math> squares and <math>8</math> cubes less than or equal to <math>529</math>, but <math>1</math> and <math>2^6</math> are both squares and cubes. Thus, there are <math>529-23-8+2=500</math> numbers in our sequence less than <math>529</math>. Magically, we want the <math>500th</math> term, so our answer is the smallest non-square and non-cube less than <math>529</math>, which is <math>\boxed{528}</math>.<br />
<br />
== Solution 2==<br />
similar as above, but to get the intuition why we chose to consider 23^2 = 529 , consider this<br />
<br />
we need n - T = 500, where n = #integers in the list 1,2,..,n and T is the set of numbers which are either k^2 or k^3 and <=n<br />
firstly, we clearly need n > 500<br />
so we think of taking the smallest square greater than 500 and let that be equal to n( u could try letting n = 512 = 8^3 to with similiar logic, but quickly realise that it fails). This is done so that set T is easy to calculate<br />
so n = 529, set T = 23+8-2 by PIE hence n-T = 500<br />
so our answer is 529-1 = 528<br />
<br />
== See also ==<br />
{{AIME box|year=1990|before=First Question|num-a=2}}<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_8&diff=1754061989 AIME Problems/Problem 82022-06-28T02:15:13Z<p>Quantomaticguy: /* Solution 5 (Very Cheap: Not Recommended) */</p>
<hr />
<div>== Problem ==<br />
Assume that <math>x_1,x_2,\ldots,x_7</math> are real numbers such that<br />
<cmath>\begin{align*}<br />
x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\<br />
4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\<br />
9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123.<br />
\end{align*}</cmath><br />
Find the value of <math>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7</math>.<br />
<br />
== Solution 1 (Quadratic Function) ==<br />
Note that each given equation is of the form <cmath>f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7</cmath> for some <math>k\in\{1,2,3\}.</math><br />
<br />
When we expand <math>f(k)</math> and combine like terms, we obtain a quadratic function of <math>k:</math> <cmath>f(k)=ak^2+bk+c,</cmath> where <math>a,b,</math> and <math>c</math> are linear combinations of <math>x_1,x_2,x_3,x_4,x_5,x_6,</math> and <math>x_7.</math> <br />
<br />
We are given that<br />
<cmath>\begin{alignat*}{10}<br />
f(1)&=\phantom{42}a+b+c&&=1, \\<br />
f(2)&=4a+2b+c&&=12, \\<br />
f(3)&=9a+3b+c&&=123,<br />
\end{alignat*}</cmath><br />
and we wish to find <math>f(4).</math><br />
<br />
We eliminate <math>c</math> by subtracting the first equation from the second, then subtracting the second equation from the third:<br />
<cmath>\begin{align*}<br />
3a+b&=11, \\<br />
5a+b&=111.<br />
\end{align*}</cmath><br />
By either substitution or elimination, we get <math>a=50</math> and <math>b=-139.</math> Substituting these back produces <math>c=90.</math><br />
<br />
Finally, the answer is <cmath>f(4)=16a+4b+c=\boxed{334}.</cmath><br />
<br />
~Azjps (Fundamental Logic)<br />
<br />
~MRENTHUSIASM (Reconstruction)<br />
<br />
== Solution 2 (Linear Combination) ==<br />
For simplicity purposes, we number the given equations <math>(1),(2),</math> and <math>(3),</math> in that order. Let <cmath>16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=S. \hspace{29.5mm}(4)</cmath><br />
Subtracting <math>(1)</math> from <math>(2),</math> subtracting <math>(2)</math> from <math>(3),</math> and subtracting <math>(3)</math> from <math>(4),</math> we obtain the following equations, respectively:<br />
<cmath>\begin{align*}<br />
3x_1 + 5x_2 + 7x_3 + 9x_4 + 11x_5 + 13x_6 + 15x_7 &=11, \hspace{20mm}&(5) \\<br />
5x_1 + 7x_2 + 9x_3 + 11x_4 + 13x_5 + 15x_6 + 17x_7 &=111, &(6) \\<br />
7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 + 19x_7 &=S-123. &(7) \\<br />
\end{align*}</cmath><br />
Subtracting <math>(5)</math> from <math>(6)</math> and subtracting <math>(6)</math> from <math>(7),</math> we obtain the following equations, respectively:<br />
<cmath>\begin{align*}<br />
2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \\<br />
2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \hspace{20mm}&(9)<br />
\end{align*}</cmath><br />
Finally, applying the Transitive Property to <math>(8)</math> and <math>(9)</math> gives <math>S-234=100,</math> from which <math>S=\boxed{334}.</math><br />
<br />
~Duohead (Fundamental Logic)<br />
<br />
~MRENTHUSIASM (Reconstruction)<br />
<br />
== Solution 3 (Finite Differences by Arithmetic) ==<br />
Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:<br />
<br />
<asy><br />
/* Made by MRENTHUSIASM */<br />
size(20cm);<br />
<br />
for (real i=1; i<=10; ++i) {<br />
label("\boldmath{$"+string(i^2)+"$}",(i-1,0));<br />
}<br />
<br />
for (real i=1; i<=9; ++i) {<br />
label("$"+string(1+2*i)+"$",(i-0.5,-0.75));<br />
}<br />
<br />
for (real i=1; i<=8; ++i) {<br />
label("$2$",(i,-1.5));<br />
}<br />
<br />
for (real i=1; i<=9; ++i) {<br />
draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red);<br />
}<br />
<br />
for (real i=1; i<=8; ++i) {<br />
draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red);<br />
}<br />
<br />
for (real i=1; i<=9; ++i) {<br />
draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red);<br />
}<br />
<br />
for (real i=1; i<=8; ++i) {<br />
draw((0.1+i,-1.35)--(0.4+i,-0.9),red);<br />
}<br />
<br />
label("\textbf{First Differences}",(-0.75,-0.75),align=W);<br />
label("\textbf{Second Differences}",(-0.75,-1.5),align=W);<br />
</asy><br />
<br />
Label equations <math>(1),(2),(3),</math> and <math>(4)</math> as Solution 2 does. Since the coefficients of <math>x_1,x_2,x_3,x_4,x_5,x_6,x_7,</math> or <math>(1,4,9,16),(4,9,16,25),(9,16,25,36),(16,25,36,49),(25,36,49,64),(36,49,64,81),(49,64,81,100),</math> respectively, all form quadratic sequences with second differences <math>2,</math> we conclude that the second differences of equations <math>(1),(2),(3),(4)</math> must be constant.<br />
<br />
It follows that the second differences of <math>(1,12,123,S)</math> must be constant, as shown below:<br />
<br />
<asy><br />
/* Made by MRENTHUSIASM */<br />
size(10cm);<br />
<br />
label("\boldmath{$1$}",(0,0));<br />
label("\boldmath{$12$}",(1,0));<br />
label("\boldmath{$123$}",(2,0));<br />
label("\boldmath{$S$}",(3,0));<br />
<br />
label("$11$",(0.5,-0.75));<br />
label("$111$",(1.5,-0.75));<br />
label("$d_1$",(2.5,-0.75));<br />
<br />
label("$100$",(1,-1.5));<br />
label("$d_2$",(2,-1.5));<br />
<br />
for (real i=1; i<=3; ++i) {<br />
draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red);<br />
}<br />
<br />
for (real i=1; i<=2; ++i) {<br />
draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red);<br />
}<br />
<br />
for (real i=1; i<=3; ++i) {<br />
draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red);<br />
}<br />
<br />
for (real i=1; i<=2; ++i) {<br />
draw((0.1+i,-1.35)--(0.4+i,-0.9),red);<br />
}<br />
<br />
label("\textbf{First Differences}",(-0.75,-0.75),align=W);<br />
label("\textbf{Second Differences}",(-0.75,-1.5),align=W);<br />
</asy><br />
<br />
Finally, we have <math>d_2=100,</math> from which<br />
<cmath>\begin{align*}<br />
S&=123+d_1 \\<br />
&=123+(111+d_2) \\<br />
&=\boxed{334}.<br />
\end{align*}</cmath><br />
~MRENTHUSIASM<br />
<br />
== Solution 4 (Finite Differences by Algebra) ==<br />
Notice that we may rewrite the equations in the more compact form as:<br />
<cmath>\begin{align*}<br />
\sum_{i=1}^{7}i^2x_i&=c_1, \\<br />
\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\<br />
\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\<br />
\sum_{i=1}^{7}(i+3)^2x_i&=c_4,<br />
\end{align*}</cmath><br />
where <math>c_1=1, c_2=12, c_3=123,</math> and <math>c_4</math> is what we are trying to find.<br />
<br />
Now consider the polynomial given by <math> f(z) = \sum_{i=1}^7 (i+z)^2x_i </math> (we are only treating the <math>x_i</math> as coefficients).<br />
<br />
Notice that <math>f</math> is in fact a quadratic. We are given <math>f(0)=c_1, f(1)=c_2, f(2)=c_3</math> and are asked to find <math>f(3)=c_4</math>. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find <math>c_4=\boxed{334}</math>. <br />
<br />
Alternatively, applying finite differences, one obtains <cmath>c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334.</cmath><br />
<br />
==Solution 5 (Very Cheap: Not Recommended)==<br />
We let <math>(x_4,x_5,x_6,x_7)=(0,0,0,0)</math>. Thus, we have <br />
<cmath>\begin{align*} <br />
x_1+4x_2+9x_3&=1,\\<br />
4x_1+9x_2+16x_3&=12,\\<br />
9x_1+16x_2+25x_3&=123.\\ <br />
\end{align*}</cmath><br />
Grinding this out, we have <math>(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)</math> which gives <math>\boxed{334}</math> as our final answer. <br />
<br />
~Pleaseletmewin<br />
<br />
<br />
== Solution 6 ==<br />
the idea is to multiply the first,second and third equation by a,b,c respectively.<br />
then, we get 3 equations: a+4b+9c = 16 ; 4a+9b+16c =25 ; 9a+16b+25c = 36<br />
We try to find a clever way to solve this system now<br />
observe the following:<br />
3a+5b+7c = 9 ( subtract 1st equation from 2nd) - this is equation 4<br />
now we manipulate to get 4b+16c = 36 or b+4c = 9( this is equation 5)<br />
and we get 7c-a = 20( equation 6)<br />
so (a,b,c) = (7c-20,9-4c,c)<br />
use this triplet in equation 2 to get c = 3 and (a,b,c) = (1,-3,3)<br />
so the answer is just 1 +12*(-3) + 123(3) = <math>\boxed{334}</math><br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=7|num-a=9}}<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_24&diff=1403562016 AMC 12A Problems/Problem 242020-12-23T06:50:44Z<p>Quantomaticguy: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
There is a smallest positive real number <math>a</math> such that there exists a positive real number <math>b</math> such that all the roots of the polynomial <math>x^3-ax^2+bx-a</math> are real. In fact, for this value of <math>a</math> the value of <math>b</math> is unique. What is this value of <math>b</math>?<br />
<br />
<math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math><br />
<br />
==Solution==<br />
<br />
===Solution 1 (calculus) ===<br />
The acceleration must be zero at the <math>x</math>-intercept; this intercept must be an inflection point for the minimum <math>a</math> value.<br />
Derive <math>f(x)</math> so that the acceleration <math>f''(x)=0</math>:<br />
<math>x^3-ax^2+bx-a\rightarrow 3x^2-2ax+b\rightarrow 6x-2a\rightarrow x=\frac{a}{3}</math> for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at <math>x=a/3</math> (if the slope is greater than zero, there will be two complex roots and we do not want that).<br />
<br />
The function with the minimum <math>a</math>:<br />
<br />
<cmath>f(x)=\left(x-\frac{a}{3}\right)^3</cmath><br />
<cmath>x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}</cmath><br />
Since this is equal to the original equation <math>x^3-ax^2+bx-a</math>,<br />
<br />
<cmath>\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}</cmath><br />
<cmath>b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}</cmath><br />
<br />
The actual function:<br />
<math>f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}</math><br />
<br />
<math>f(x)=0\rightarrow x=\sqrt{3}</math> triple root. "Complete the cube."<br />
<br />
===Solution 2===<br />
<br />
Note that since both <math>a</math> and <math>b</math> are positive, all 3 roots of the polynomial are positive as well.<br />
<br />
<br />
Let the roots of the polynomial be <math>r, s, t</math>. By Vieta's <math>a=r+s+t</math> and <math>a=rst</math>.<br />
<br />
<br />
Since <math>r, s, t</math> are positive we can apply AM-GM to get <math>\frac{r+s+t}{3} \ge \sqrt[3]{rst} \rightarrow \frac{a}{3} \ge \sqrt[3]{a}</math>. Cubing both sides and then dividing by <math>a</math> (since <math>a</math> is positive we can divide by <math>a</math> and not change the sign of the inequality) yields <math>\frac{a^2}{27} \ge 1 \rightarrow a \ge 3\sqrt{3}</math>.<br />
<br />
<br />
Thus, the smallest possible value of <math>a</math> is <math>3\sqrt{3}</math> which is achieved when all the roots are equal to <math>\sqrt{3}</math>. For this value of <math>a</math>, we can use Vieta's to get <math>b=\boxed{\textbf{(B) }9}</math>.<br />
<br />
===Solution 3===<br />
All three roots are identical. Therefore, comparing coefficients, the root of this cubic function is <math>\sqrt{3}</math>.<br />
Using Vieta's, the coefficient we desire is the sum of the pairwise products of the roots. Because our root is unique, the answer is simply <math>b=\boxed{\textbf{(B) }9}</math>.<br />
(note that this is only true since for the min value of a, applying AM-GM to the sums and products of roots, equality condition produced min value of a )<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_23&diff=1403552016 AMC 10A Problems/Problem 232020-12-23T06:40:55Z<p>Quantomaticguy: /* Solutions */</p>
<hr />
<div>== Problem ==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
== Solutions ==<br />
== Solution ==<br />
note that (2016 T 6 )x = 100 where T = the diamond symbol <br />
hence we realies that 2016 T 6 = 2010 T 6 + 6 T 6 = 2004 T 6 + 2*(6 T 6 ) = ... = 336 * ( 6 T 6 ) = 336 <br />
=== Solution 1 ===<br />
<br />
We see that <math>a \diamond a = 1</math>, and think of division. Testing, we see that the first condition <math>a \diamond (b \diamond c) = (a \diamond b) \cdot c</math> is satisfied, because <math>\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c</math>. Therefore, division can be the operation <math>\diamond</math>. Solving the equation, <cmath>\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},</cmath> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109}</math>.<br />
<br />
=== Solution 2 ===<br />
We can manipulate the given identities to arrive at a conclusion about the binary operator <math>\diamondsuit</math>. Substituting <math>b = c</math> into the first identity yields <cmath>( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\ b) = a\ \diamondsuit\ 1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a.</cmath> Hence, <math>( a\ \diamondsuit\ b) \cdot b = a,</math> or, dividing both sides of the equation by <math>b,</math> <math>( a\ \diamondsuit\ b) = \frac{a}{b}.</math><br />
<br />
Hence, the given equation becomes <math>\frac{2016}{\frac{6}{x}} = 100</math>. Solving yields <math>x=\frac{100}{336} = \frac{25}{84},</math> so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math><br />
<br />
=== Solution 3 ===<br />
One way to eliminate the <math>\diamondsuit</math> in this equation is to make <math>a = b</math> so that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = c</math>. In this case, we can make <math>b = 2016</math>.<br />
<br />
<cmath>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\implies <br />
(2016\, \diamondsuit\, 6) \cdot x = 100</cmath><br />
<br />
By multiplying both sides by <math>\frac{6}{x}</math>, we get:<br />
<br />
<cmath>(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\implies <br />
2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}</cmath><br />
<br />
Because <math>6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1:</math><br />
<br />
<cmath>2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\implies <br />
(2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\implies <br />
2016 = \frac{600}{x}</cmath><br />
<br />
Therefore, <math>x = \frac{600}{2016} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math><br />
<br />
== Video Solution ==<br />
https://www.youtube.com/watch?v=8GULAMwu5oE<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2016|ab=A|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2016|ab=A|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_11&diff=1400802008 AIME II Problems/Problem 112020-12-21T11:54:58Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In triangle <math>ABC</math>, <math>AB = AC = 100</math>, and <math>BC = 56</math>. [[Circle]] <math>P</math> has [[radius]] <math>16</math> and is tangent to <math>\overline{AC}</math> and <math>\overline{BC}</math>. Circle <math>Q</math> is externally [[tangent]] to <math>P</math> and is tangent to <math>\overline{AB}</math> and <math>\overline{BC}</math>. No point of circle <math>Q</math> lies outside of <math>\triangle ABC</math>. The radius of circle <math>Q</math> can be expressed in the form <math>m - n\sqrt {k}</math>, where <math>m</math>, <math>n</math>, and <math>k</math> are positive integers and <math>k</math> is the product of distinct primes. Find <math>m + nk</math>.<br />
<br />
== Solution ==<br />
<center><asy><br />
size(200);<br />
pathpen=black;pointpen=black;pen f=fontsize(9);<br />
real r=44-6*35^.5;<br />
pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P);<br />
path PC=CR(P,16),QC=CR(Q,r);<br />
D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed);<br />
D(PC); D(QC); <br />
MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f));<br />
</asy></center><br />
<br />
Let <math>X</math> and <math>Y</math> be the feet of the [[perpendicular]]s from <math>P</math> and <math>Q</math> to <math>BC</math>, respectively. Let the radius of <math>\odot Q</math> be <math>r</math>. We know that <math>PQ = r + 16</math>. From <math>Q</math> draw segment <math>\overline{QM} \parallel \overline{BC}</math> such that <math>M</math> is on <math>PX</math>. Clearly, <math>QM = XY</math> and <math>PM = 16-r</math>. Also, we know <math>QPM</math> is a right triangle.<br />
<br />
To find <math>XC</math>, consider the right triangle <math>PCX</math>. Since <math>\odot P</math> is tangent to <math>\overline{AC},\overline{BC}</math>, then <math>PC</math> [[bisect]]s <math>\angle ACB</math>. Let <math>\angle ACB = 2\theta</math>; then <math>\angle PCX = \angle QBX = \theta</math>. Dropping the altitude from <math>A</math> to <math>BC</math>, we recognize the <math>7 - 24 - 25</math> [[right triangle]], except scaled by <math>4</math>.<br />
<br />
So we get that <math>\tan(2\theta) = 24/7</math>. From the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{3}</math>.<br />
<br />
We conclude that <math>XY = 56 - \frac {4r + 64}{3} = \frac {104 - 4r}{3}</math>.<br />
<br />
So our right triangle <math>QPM</math> has sides <math>r + 16</math>, <math>r - 16</math>, and <math>\frac {104 - 4r}{3}</math>.<br />
<br />
By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>.<br />
<br />
<br />
== Solution ==<br />
first let A = <PCB ; now connect the points as shown in the first solution's diagram ; realise that tanA = r/x = 16/y = r + 16/(x+y) where x = BY and y = CX ( the 2 tangents) ; then we have that QM = 64r = 56 - x - y => (x+y) = 56 - 64r<br />
hence r/x = 16+r/(56-64r) ; now drop altitude AY to solve for tan2A ; now since we know tan2A we know tan A = r/x in terms of r hence solve the resulting equation in r<br />
<br />
== Solution 2(pure synthetic) ==<br />
Refer to the above diagram. Let the larger circle have center <math>O_1</math>, the smaller have center <math>O_2</math>, and the incenter be <math>I</math>. We can easily calculate that the area of <math>\triangle ABC = 2688</math>, and <math>s = 128</math> and <math>R = 21</math>, where <math>R</math> is the inradius. <br />
<br />
Now, Line <math>\overline{AI}</math> is the perpendicular bisector of <math>\overline{BC}</math>, as <math>\triangle ABC</math> is isosceles. Letting the point of intersection be <math>X</math>, we get that <math>BX = 28</math> and <math>IX = 21</math>, and <math>B, O_2, I</math> are collinear as <math>O_2</math> is equidistant from <math>\overline{AB}</math> and <math>\overline{BC}</math>. By Pythagoras, <math>BI = 35</math>, and we notice that <math>\triangle BIX</math> is a 3-4-5 right triangle. <br />
<br />
Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam<br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=II|num-b=10|num-a=12}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2006_iTest_Problems/Problem_32&diff=1400792006 iTest Problems/Problem 322020-12-21T10:48:56Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Triangle <math>ABC</math> is scalene. Points <math>P</math> and <math>Q</math> are on segment <math>BC</math> with <math>P</math> between <math>B</math> and <math>Q</math> such that <math>BP=21</math>, <math>PQ=35</math>, and <math>QC=100</math>. If <math>AP</math> and <math>AQ</math> trisect <math>\angle A</math>, then <math>\tfrac{AB}{AC}</math> can be written uniquely as <math>\tfrac{p\sqrt q}r</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is a positive integer not divisible by the square of any prime. Determine <math>p+q+r</math>.<br />
<br />
==Solution==<br />
<br />
Let <math>a = AB</math> and <math>b = AC</math>. Since <math>\angle BAP = \angle PAQ = \angle QAC</math>, by the [[Angle Bisector Theorem]], we have <math>AP = \tfrac{7}{20}b</math> and <math>AQ = \tfrac{5}{3}a</math>.<br />
<br />
<br><br />
By using the [[Law of Cosines]] on <math>\triangle BAP</math> and <math>\triangle PAQ</math>, we have<br />
<cmath>\begin{align*}<br />
\frac{a^2 + \frac{49}{400}b^2 - 21^2}{2ab \cdot \frac{7}{20}} &= \frac{\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2}{2ab \cdot \frac{7}{20} \cdot \frac{5}{3}} \\<br />
\frac{5}{3} \cdot \left( a^2 + \frac{49}{400}b^2 - 21^2 \right) &= \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 \\<br />
\frac53 a^2 + \frac{49}{240}b^2 - 21 \cdot 35 &= \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 \\<br />
\frac{98}{1200} b^2 + 35 \cdot 14 &= \frac{10}{9} a^2 \\<br />
\frac{49}{600} b^2 + 35 \cdot 14 &= \frac{40}{36} a^2.<br />
\end{align*}</cmath><br />
By using the Law of Cosines on <math>\triangle PAQ</math> and <math>\triangle QAC</math>, we have <br />
<cmath>\begin{align*}<br />
\frac{\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2}{2ab \cdot \frac{7}{20} \cdot \frac{5}{3}} &= \frac{b^2 + \frac{25}{9}a^2 - 100^2}{2ab \cdot \frac{5}{3}} \\<br />
\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 &= \frac{7}{20} \cdot \left( b^2 + \frac{25}{9}a^2 - 100^2 \right) \\<br />
\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 &= \frac{7}{20} b^2 + \frac{35}{36} a^2 - 100 \cdot 35 \\<br />
\frac{65}{36} a^2 + 65 \cdot 35 &= \frac{91}{400}b^2 \\<br />
\frac{5}{36} a^2 + 5 \cdot 35 &= \frac{7}{400}b^2.<br />
\end{align*}</cmath><br />
Multiplying the second equation by <math>-8</math> and adding the two equations results in<br />
<cmath>\begin{align*}<br />
-\frac{40}{36} a^2 - 40 \cdot 35 &= -\frac{7}{50} b^2 \\<br />
\frac{40}{36} a^2 &= \frac{49}{600} b^2 + 35 \cdot 14 \\<br />
-40 \cdot 35 &= -\frac{35}{600} b^2 + 35 \cdot 14 \\<br />
-40 &= -\frac{1}{600} b^2 + 14 \\<br />
\frac{b^2}{600} &= 54 \\<br />
b^2 &= 600 \cdot 9 \cdot 6 \\<br />
b &= 6 \cdot 10 \cdot 3 \\<br />
&= 180.<br />
\end{align*}</cmath><br />
After substituting <math>b</math> back, solve for <math>a</math> to get<br />
<cmath>\begin{align*}<br />
\frac{49}{600} \cdot 180 \cdot 180 + 35 \cdot 14 &= \frac{40}{36} a^2 \\<br />
49 \cdot 54 + 35 \cdot 14 &= \frac{40}{36} a^2 \\<br />
49 \cdot 54 + 49 \cdot 10 &= \frac{40}{36} a^2 \\<br />
49 \cdot 64 &= \frac{10}{9} a^2 \\<br />
a^2 &= \frac{49 \cdot 9 \cdot 64}{10} \\<br />
a &= \frac{168}{\sqrt{10}} \\<br />
&= \frac{84\sqrt{10}}{5}<br />
\end{align*}</cmath><br />
Thus, <math>\frac{AB}{AC} = \frac{84\sqrt{10}}{5} \cdot \frac{1}{180} = \frac{7\sqrt{10}}{75}</math>, so <math>p+q+r = \boxed{92}</math>.<br />
<br />
NOTE: SIMPLY USE STEWARTS THEOREM<br />
<br />
==See Also==<br />
{{iTest box|year=2006|num-b=31|num-a=33|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1997_JBMO_Problems/Problem_4&diff=1400751997 JBMO Problems/Problem 42020-12-21T06:24:08Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Determine the triangle with sides <math>a,b,c</math> and circumradius <math>R</math> for which <math>R(b+c) = a\sqrt{bc}</math>.<br />
<br />
== Solution ==<br />
NOTE(not by author): we can conclude b = c by noticing that b + c <=2sqrt(bc) but by AM-GM b = c must hold<br />
Solving for <math>R</math> yields <math>R = \tfrac{a\sqrt{bc}}{b+c}</math>. We can substitute <math>R</math> into the area formula <math>A = \tfrac{abc}{4R}</math> to get<br />
<cmath>\begin{align*}<br />
A &= \frac{abc}{4 \cdot \tfrac{a\sqrt{bc}}{b+c} } \\<br />
&= \frac{abc}{4a\sqrt{bc}} \cdot (b+c) \\<br />
&= \frac{(b+c)\sqrt{bc}}{4}.<br />
\end{align*}</cmath><br />
We also know that <math>A = \tfrac{1}{2}bc \sin(\theta)</math>, where <math>\theta</math> is the angle between sides <math>b</math> and <math>c.</math> Substituting this yields<br />
<cmath>\begin{align*}<br />
\tfrac{1}{2}bc \sin(\theta) &= \frac{(b+c)\sqrt{bc}}{4} \\<br />
2\sqrt{bc} \cdot \sin(\theta) &= b+c \\<br />
\sin(\theta) &= \frac{b+c}{2\sqrt{bc}}<br />
\end{align*}</cmath><br />
Since <math>\theta</math> is inside a triangle, <math>0 < \sin{\theta} \le 1</math>. Substitution yields<br />
<cmath>0 < \frac{b+c}{2\sqrt{bc}} \le 1.</cmath><br />
Note that <math>2\sqrt{bc}</math>, so multiplying both sides by that value would not change the inequality sign. This means<br />
<cmath>0 < b+c \le 2\sqrt{bc}.</cmath><br />
Since all values in the inequality are positive, squaring both sides would not change the inequality sign, so<br />
<cmath>0 < b^2 + 2bc + c^2 \le 4bc</cmath><br />
<cmath>-4bc < b^2 - 2bc + c^2 \le 0</cmath><br />
<cmath>-4bc < (b-c)^2 \le 0</cmath><br />
By the [[Trivial Inequality]], <math>(b-c)^2 \ge 0</math> for all <math>b</math> and <math>c,</math> so the only values of <math>b</math> and <math>c</math> that satisfies is when <math>(b-c)^2 = 0</math>. Thus, <math>b = c</math>. Since <math>-4bc < 0</math> for positive <math>b</math> and <math>c</math>, the value <math>b=c</math> truly satisfies all conditions.<br />
<br />
<br><br />
That means <math>\sin(\theta) = \frac{2b}{2\sqrt{b^2}} = 1,</math> so <math>\theta = 90^\circ.</math> That means the only truangle that satisfies all the conditions is a 45-45-90 triangle where <math>a</math> is the longest side. In other words, <math>(a,b,c) \rightarrow \boxed{(n\sqrt{2},n,n)}</math> for all positive <math>n.</math><br />
<br />
== See Also ==<br />
<br />
{{JBMO box|year=1997|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1998_AHSME_Problems/Problem_28&diff=1400641998 AHSME Problems/Problem 282020-12-21T02:53:26Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In triangle <math>ABC</math>, angle <math>C</math> is a [[right angle]] and <math>CB > CA</math>. Point <math>D</math> is located on <math>\overline{BC}</math> so that angle <math>CAD</math> is twice angle <math>DAB</math>. If <math>AC/AD = 2/3</math>, then <math>CD/BD = m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
<math> \mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26</math><br />
<br />
== Solution == <br />
Let <math>\theta = \angle DAB</math>, so <math>2\theta = \angle CAD</math> and <math>3 \theta = \angle CAB</math>. Then, it is given that <math>\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}</math> and<br />
<br />
<br />
<center><math>\frac{BD}{CD} = \frac{AC(\tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.</math></center><br />
<br />
<br />
Now, through the use of trigonometric identities, <math>\cos 2\theta = 2\cos^2 \theta - 1 = \frac{2}{\sec ^2 \theta} - 1 = \frac{1 - \tan^2 \theta}{1 + \tan ^2 \theta} = \frac{2}{3}</math>. Solving yields that <math>\tan^2 \theta = \frac 15</math>. Using the tangent addition identity, we find that <math>\tan 2\theta = \frac{2\tan \theta}{1 - \tan ^2 \theta},\ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}</math>, and<br />
<br />
<br />
<center><math>\frac{BD}{CD} = \frac{\tan 3\theta}{\tan 2\theta} - 1 = \frac{(3 - \tan^2 \theta)(1-\tan ^2 \theta)}{2(1 - 3\tan^2 \theta)} - 1 = \frac{(1 + \tan^2 \theta)^2}{2(1 - 3\tan^2 \theta)} = \frac{9}{5}</math></center><br />
<br />
<br />
and <math>\frac{CD}{BD} = \frac{5}{9} \Longrightarrow m+n = 14 \Longrightarrow \mathbf{(B)}</math>. (This also may have been done on a calculator by finding <math>\theta</math> directly)<br />
== Solution 3 == <br />
By the application of ratio lemma for CD/BD, we get CD/BD = 2cos3AcosA,where we let A = angleDAB. we already know cos2A hence the rest is easy<br />
<br />
==Solution 2==<br />
<br />
Let <math>AC=2</math> and <math>AD=3</math>. By the Pythagorean Theorem, <math>CD=\sqrt{5}</math>. Let point <math>P</math> be on segment <math>CD</math> such that <math>AP</math> bisects <math>\angle CAD</math>. Thus, angles <math>CAP</math>, <math>PAD</math>, and <math>DAB</math> are congruent. Applying the angle bisector theorem on <math>ACD</math>, we get that <math>CP=\frac{2\sqrt{5}}{5}</math> and <math>PD=\frac{3\sqrt{5}}{5}</math>. Pythagorean Theorem gives <math>AP=\frac{\sqrt{5}\sqrt{24}}{5}</math>.<br />
<br />
Let <math>DB=x</math>. By the Pythagorean Theorem, <math>AB=\sqrt{(x+\sqrt{5})^{2}+2^2}</math>. Applying the angle bisector theorem again on triangle <math>APB</math>, we have <cmath>\frac{\sqrt{(x+\sqrt{5})^{2}+2^2}}{x}=\frac{\frac{\sqrt{5}\sqrt{24}}{5}}{\frac{3\sqrt{5}}{5}}</cmath><br />
The right side simplifies to<math>\frac{\sqrt{24}}{3}</math>. Cross multiplying, squaring, and simplifying, we get a quadratic: <cmath>5x^2-6\sqrt{5}x-27=0</cmath> Solving this quadratic and taking the positive root gives <cmath>x=\frac{9\sqrt{5}}{5}</cmath> Finally, taking the desired ratio and canceling the roots gives <math>\frac{CD}{BD}=\frac{5}{9}</math>. The answer is <math>\fbox{(B) 14}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1998|num-b=27|num-a=29}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_28&diff=1399681970 AHSME Problems/Problem 282020-12-19T06:36:11Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
In triangle <math>ABC</math>, the median from vertex <math>A</math> is perpendicular to the median from vertex <math>B</math>. If the lengths of sides <math>AC</math> and <math>BC</math> are <math>6</math> and <math>7</math> respectively, then the length of side <math>AB</math> is<br />
<br />
<math>\text{(A) } \sqrt{17}\quad<br />
\text{(B) } 4\quad<br />
\text{(C) } 4\tfrac{1}{2}\quad<br />
\text{(D) } 2\sqrt{5}\quad<br />
\text{(E) } 4\tfrac{1}{4}</math><br />
<br />
== Solution ==<br />
<math>\fbox{A}</math><br />
let the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN<br />
<br />
== See also ==<br />
{{AHSME 35p box|year=1970|num-b=27|num-a=29}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1961_AHSME_Problems/Problem_38&diff=1399651961 AHSME Problems/Problem 382020-12-19T05:45:17Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
<math>\triangle ABC</math> is inscribed in a semicircle of radius <math>r</math> so that its base <math>AB</math> coincides with diameter <math>AB</math>. <br />
Point <math>C</math> does not coincide with either <math>A</math> or <math>B</math>. Let <math>s=AC+BC</math>. Then, for all permissible positions of <math>C</math>:<br />
<br />
<math>\textbf{(A)}\ s^2\le8r^2\qquad<br />
\textbf{(B)}\ s^2=8r^2 \qquad<br />
\textbf{(C)}\ s^2 \ge 8r^2 \qquad\\<br />
\textbf{(D)}\ s^2\le4r^2 \qquad<br />
\textbf{(E)}\ s^2=4r^2 </math><br />
<br />
==Solution==<br />
<asy><br />
draw((-50,0)--(-30,40)--(50,0)--(-50,0));<br />
draw(Arc((0,0),50,0,180));<br />
draw(rightanglemark((-50,0),(-30,40),(50,0),200));<br />
dot((-50,0));<br />
label("A",(-50,0),SW);<br />
dot((-30,40));<br />
label("C",(-30,40),NW);<br />
dot((50,0));<br />
label("B",(50,0),SE);<br />
</asy><br />
Since <math>s=AC+BC</math>, <math>s^2 = AC^2 + 2 \cdot AC \cdot BC + BC^2</math>. Since <math>\triangle ABC</math> is inscribed and <math>AB</math> is the diameter, <math>\triangle ABC</math> is a right triangle, and by the [[Pythagorean Theorem]], <math>AC^2 + BC^2 = AC^2 = (2r)^2</math>. Thus, <math>s^2 = 4r^2 + 2 \cdot AC \cdot BC</math>.<br />
<br />
<br><br />
<br />
The area of <math>\triangle ABC</math> is <math>\frac{AC \cdot BC}{2}</math>, so <math>2 \cdot [ABC] = AC \cdot BC</math>. That means <math>s^2 = 4r^2 + 4 \cdot [ABC]</math>. The area of <math>\triangle ABC</math> can also be calculated by using base <math>AB</math> and the altitude from <math>C</math>. The maximum possible value of the altitude is <math>r</math>, so the maximum area of <math>\triangle ABC</math> is <math>r^2</math>.<br />
<br />
<br><br />
<br />
Therefore, <math>s^2 \le 8r^2</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>.<br />
<br />
Solution using Triangle inequality:<br />
s = AC ^2 + BC^2 - 2AC*BC = 4r^2 - 2AC*BC <= 4r^2 - (-4r^2) using triangle inequality in triangle AOC and BOC where O is the center<br />
<br />
==See Also==<br />
{{AHSME 40p box|year=1961|num-b=37|num-a=39}}<br />
<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2020_CIME_II_Problems/Problem_4&diff=1399632020 CIME II Problems/Problem 42020-12-19T04:49:46Z<p>Quantomaticguy: /* Similar Problems */</p>
<hr />
<div>==Problem==<br />
The probability a randomly chosen positive integer <math>N<1000</math> has more digits when written in base <math>7</math> than when written in base <math>8</math> can be expressed in the form <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
If a positive integer <math>N</math> has more digits in base <math>7</math> than base <math>8</math>, then <math>7^k \le N < 8^k</math> for some positive integer <math>k</math>. There are <math>8^k-7^k</math> positive integers <math>N</math> that satisfy this condition for every positive integer <math>k</math>. If <math>k \geq 4</math>, <math>N</math> will be greater than <math>1000</math>, so we only need to consider <math>k \le 3</math>. The number of possible values of <math>N</math> is <cmath> \sum^{3}_{k=1}[8^k-7^k]=8-7+64-49+512-343=185.</cmath> The requested probability is <cmath>\frac{185}{999}=\frac{5}{27},</cmath> and so the answer is <math>5+27=\boxed{032}</math>.<br />
<br />
==Similar Problems==<br />
[[2009 AIME I Problems/Problem 6|2009 AIME I Problem 6]]<br />
<br />
The intuition behind the above solution:<br />
consider 7^3 = 343 <= N < 8^3 = 512 ; notice that any N in this range will have 4 digits in base 7 ( since 1*7^3 <= N) and since 1*8^3 > N , we have almost 3 digits in base 8 representation of N<br />
<br />
its easy to see 3 can be replaced by any positive integer k , where k = 1,2,3<br />
<br />
==See also==<br />
{{CIME box|year=2020|n=II|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAC Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2020_CIME_I_Problems/Problem_7&diff=1399622020 CIME I Problems/Problem 72020-12-19T04:37:51Z<p>Quantomaticguy: </p>
<hr />
<div>==Problem 7==<br />
For every positive integer <math>n</math>, define <cmath>f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.</cmath> Suppose that the sum <math>f(1)+f(2)+\cdots+f(2020)</math> can be expressed as <math>\frac{p}{q}</math> for relatively prime integers <math>p</math> and <math>q</math>. Find the remainder when <math>p</math> is divided by <math>1000</math>.<br />
<br />
==Solution==<br />
{{solution}}<br />
Notice that the product of first 2n+1 odd numbers = (2n+2)! / ((2^n+2)*(n+1)!) ; substituting this for f(n) gives the answer <br />
==See also==<br />
{{CIME box|year=2020|n=I|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAC Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1997_PMWC_Problems/Problem_T9&diff=1399011997 PMWC Problems/Problem T92020-12-18T02:01:43Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Find the two <math>10</math>-digit numbers which become nine times as large if the order of the digits is reversed.<br />
<br />
==Solution==<br />
The pair of numbers are <math>1089001089</math> and is <math>1098910989</math>. <br />
<br />
Notice that the sum of two numbers is 10 times of the smaller one. Let the smaller one be <math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0</math>, the large one becomes <math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9</math>. Then we have<br />
<math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_010</math> = <math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0</math>+<math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9</math>.<br />
It's obvious that <math>a_9=1</math> and <math>a_0=9</math>. Comparing the digits, we have <math>(a_8=0, a_1=8)</math>, <math>(a_7=8, a_2=0)</math>, <math>(a_6=9, a_3=1)</math>, and <math>(a_5=0, a_4=0)</math>.<br />
<br />
<br />
Side note:<br />
if its not clear how to obtain the digits<br />
essentially we want to write an equation first and then compare the units,tens,hundreds digits etc.. on the left and right side. its easy to notice the digits must be one of {0,1,8,9} hence there aren't many cases to consider was well<br />
<br />
==Mistake Above Fix==<br />
<br />
The actual two numbers are <math>1089001089</math>, as mentioned above, but the second number is <math>1098910989</math>, not <math>9801009801</math>.<br />
Someone please fix.<br />
<br />
==See Also==<br />
{{PMWC box|year=1997|num-b=T8|num-a=T10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1997_PMWC_Problems/Problem_T9&diff=1399001997 PMWC Problems/Problem T92020-12-18T01:58:46Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Find the two <math>10</math>-digit numbers which become nine times as large if the order of the digits is reversed.<br />
<br />
==Solution==<br />
The pair of numbers are <math>1089001089</math> and is <math>1098910989</math>. <br />
<br />
Notice that the sum of two numbers is 10 times of the smaller one. Let the smaller one be <math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0</math>, the large one becomes <math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9</math>. Then we have<br />
<math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_010</math> = <math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0</math>+<math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9</math>.<br />
It's obvious that <math>a_9=1</math> and <math>a_0=9</math>. Comparing the digits, we have <math>(a_8=0, a_1=8)</math>, <math>(a_7=8, a_2=0)</math>, <math>(a_6=9, a_3=1)</math>, and <math>(a_5=0, a_4=0)</math>.<br />
<br />
==Mistake Above Fix==<br />
<br />
The actual two numbers are <math>1089001089</math>, as mentioned above, but the second number is <math>1098910989</math>, not <math>9801009801</math>.<br />
Someone please fix.<br />
<br />
==See Also==<br />
{{PMWC box|year=1997|num-b=T8|num-a=T10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=1997_PMWC_Problems/Problem_T9&diff=1398991997 PMWC Problems/Problem T92020-12-18T01:58:31Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>==Problem==<br />
Find the two <math>10</math>-digit numbers which become nine times as large if the order of the digits is reversed.<br />
<br />
==Solution==<br />
The pair of numbers are <math>1089001089</math> and <math>is </math>1098910989<math>. <br />
<br />
Notice that the sum of two numbers is 10 times of the smaller one. Let the smaller one be </math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0<math>, the large one becomes </math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9<math>. Then we have<br />
</math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_010<math> = </math>a_9a_8a_7a_6a_5a_4a_3a_2a_1a_0<math>+</math>a_0a_1a_2a_3a_4a_5a_6a_7a_8a_9<math>.<br />
It's obvious that </math>a_9=1<math> and </math>a_0=9<math>. Comparing the digits, we have </math>(a_8=0, a_1=8)<math>, </math>(a_7=8, a_2=0)<math>, </math>(a_6=9, a_3=1)<math>, and </math>(a_5=0, a_4=0)$.<br />
<br />
==Mistake Above Fix==<br />
<br />
The actual two numbers are <math>1089001089</math>, as mentioned above, but the second number is <math>1098910989</math>, not <math>9801009801</math>.<br />
Someone please fix.<br />
<br />
==See Also==<br />
{{PMWC box|year=1997|num-b=T8|num-a=T10}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_12&diff=1395272005 AIME II Problems/Problem 122020-12-13T06:10:02Z<p>Quantomaticguy: /* Solutions */</p>
<hr />
<div>== Problem ==<br />
[[Square]] <math>ABCD </math> has [[center]] <math> O,\ AB=900,\ E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are [[positive]] [[integer]]s and <math> r </math> is not divisible by the [[square]] of any [[prime]], find <math> p+q+r. </math><br />
__TOC__<br />
<br />
== Solutions ==<br />
=== Solution 1 (trigonometry) ===<br />
<center><asy><br />
size(200);<br />
defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(-1,1));label("\(O\)",O,(1,-1)); label("\(x\)",E/2+G/2,(0,1));label("\(y\)",G/2+F/2,(0,1)); label("\(450\)",(O+G)/2,(-1,1)); <br />
</asy></center> <!-- Asymptote replacement for Image:AIME_2005II_Solution_12_1.png by Minsoens --><br />
<br />
Let <math>G</math> be the foot of the [[perpendicular]] from <math>O</math> to <math>AB</math>. Denote <math>x = EG</math> and <math>y = FG</math>, and <math>x > y</math> (since <math>AE < BF</math> and <math>AG = BG</math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>.<br />
<br />
By the [[trigonometric identity|tangent addition rule]] <math>\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)</math>, we see that <cmath>\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since <math>\tan 45 = 1</math>, this simplifies to <math>1 - \frac{xy}{450^2} = \frac{x + y}{450}</math>. We know that <math>x + y = 400</math>, so we can substitute this to find that <math>1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2</math>.<br />
<br />
Substituting <math>x = 400 - y</math> again, we know have <math>xy = (400 - y)y = 150^2</math>. This is a quadratic with roots <math>200 \pm 50\sqrt{7}</math>. Since <math>y < x</math>, use the smaller root, <math>200 - 50\sqrt{7}</math>.<br />
<br />
Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = \boxed{307}</math>.<br />
<br />
=== Solution 2 (synthetic) ===<br />
<center><asy><br />
size(200);<br />
defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(1,0));label("\(J\)",J,(1,0));label("\(O\)",O,(1,-1)); label("\(x\)",(B+F)/2,(0,1)); label("\(400\)",(E+F)/2,(0,1)); label("\(900\)",(C+D)/2,(0,-1));<br />
</asy></center><br />
<br />
Label <math>BF=x</math>, so <math>EA =</math> <math>500 - x</math>. Rotate <math>\triangle{OEF}</math> about <math>O</math> until <math>EF</math> lies on <math>BC</math>. Now we know that <math>\angle{EOF}=45^\circ</math> therefore <math>\angle BOF+\angle AOE=45^\circ</math> also since <math>O</math> is the center of the square. Label the new triangle that we created <math>\triangle OGJ</math>. Now we know that rotation preserves angles and side lengths, so <math>BG=500-x</math> and <math>JC=x</math>. Draw <math>GF</math> and <math>OB</math>. Notice that <math>\angle BOG =\angle OAE</math> since rotations preserve the same angles so <br />
<math>\angle{FOG}=45^\circ</math> too. By SAS we know that <math>\triangle FOE\cong \triangle FOG,</math> so <math>FG=400</math>. Now we have a right <math>\triangle BFG</math> with legs <math>x</math> and <math>500-x</math> and hypotenuse <math>400</math>. By the [[Pythagorean Theorem]], <br />
<br />
<cmath>\begin{align*}<br />
(500-x)^2+x^2&=400^2 \\<br />
250000-1000x+2x^2&=16000 \\<br />
90000-1000x+2x^2&=0 \end{align*}</cmath><br />
<br />
and applying the [[quadratic formula]] we get that <br />
<math>x=250\pm 50\sqrt{7}</math>. Since <math>BF > AE,</math> we take the positive root, and our answer is <math>p+q+r = 250 + 50 + 7 = 307</math>.<br />
<br />
=== Solution 3 (similar triangles)===<br />
<asy><br />
size(3inch);<br />
pair A, B, C, D, M, O, X, Y;<br />
A = (0,900); B = (900,900); C = (900,0); D = (0,0);<br />
M = (450,900); O = (450,450); X = (250 - 50*sqrt(7),900); Y = (650 - 50*sqrt(7),900);<br />
draw(A--B--C--D--cycle);<br />
draw(X--O--Y);<br />
draw(M--O--A);<br />
label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",X,N); label("$F$",Y,NNE); label("$O$",O,S); label("$M$",M,N);<br />
</asy><br />
Let the midpoint of <math>\overline{AB}</math> be <math>M</math> and let <math>FB = x</math>, so then <math>MF = 450 - x</math> and <math>AF = 900 - x</math>. Drawing <math>\overline{AO}</math>, we have <math>\triangle OEF\sim\triangle AOF</math>, so<br />
<cmath>\frac{OF}{EF} = \frac{AF}{OF}\Rightarrow (OF)^2 = 400(900 - x).</cmath><br />
By the Pythagorean Theorem on <math>\triangle OMF</math>,<br />
<cmath>(OF)^2 = 450^2 + (450 - x)^2.</cmath><br />
Setting these two expressions for <math>(OF)^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. Since <math>BF > AE</math>, we want the value <math>x = 250 + 50\sqrt{7}</math>, and the answer is <math>250 + 50 + 7 = \boxed{307}</math>.<br />
<br />
=== Solution 4 (Abusing Stewart) ===<br />
Let <math>x = BF</math>, so <math>AE = 500-x</math>. Let <math>a = OE</math>, <math>b = OF</math>. Applying Stewart's Theorem on triangles <math>AOB</math> twice, first using <math>E</math> as the base point and then <math>F</math>, we arrive at the equations <cmath>(450 \sqrt{2})^2 (900) = 900(500-x)(400+x) + a^2 (900)</cmath> and <cmath>(450 \sqrt{2})^2 (900) = 900x(900-x) + b^2 (900)</cmath> Now applying law of sines and law of cosines on <math>\triangle EOF</math> yields <cmath>\frac{1}{2} ab \sin 45^{\circ} = \frac{4}{9} \times \frac{1}{4} \times 900^2 = 202500</cmath> and <cmath>a^2+b^2- 2 ab \cos 45^{\circ} = 160000</cmath> Solving for <math>ab</math> from the sines equation and plugging into the law of cosines equation yields <math>a^2+b^2 = 290000</math>. We now finish by adding the two original stewart equations and obtaining: <cmath>2(450\sqrt{2})^2 = (500-x)(400+x)+x(900-x)+520000</cmath> This is a quadratic which only takes some patience to solve for <math>x = 250 + 50\sqrt{7}</math><br />
<br />
=== Solution 5 (Complex Numbers) ===<br />
Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with <math>o = 0, a = -450 + 450i, b = 450 + 450i</math>, and <math>f = x + 450i</math>. Since <math>EF</math> = 400, <math>e = (x-400) + 450i</math>. From <math>\angle{EOF} = 45^{\circ}</math>, we can deduce that the rotation of point <math>F</math> 45 degrees counterclockwise, <math>E</math>, and the origin are collinear. In other words, <cmath>\dfrac{e^{i \frac{\pi}{4}} \cdot (x + 450i)}{(x - 400) + 450i}</cmath> is a real number. Simplyfying using the fact that <math>e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}</math>, clearing the denominator, and setting the imaginary part equal to <math>0</math>, we eventually get the quadratic <cmath>x^2 - 400x + 22500 = 0</cmath> which has solutions <math>x = 200 \pm 50\sqrt{7}</math>. It is given that <math>AE < BF</math>, so <math>x = 200 - 50\sqrt{7}</math> and <cmath>BF = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7} \Rightarrow \boxed{307}.</cmath><br />
<br />
-MP8148<br />
<br />
=== Solution 6 ===<br />
<asy><br />
size(250);<br />
pair A,B,C,D,O,E,F,G,H,K;<br />
A = (0,0);<br />
B = (900,0);<br />
C = (900,900);<br />
D = (0,900);<br />
O = (450,450);<br />
E = (600,0);<br />
F = (150,0);<br />
G = (-600,0);<br />
H = (450,0);<br />
K = (0,270);<br />
draw(A--B--C--D--cycle);<br />
draw(O--E);<br />
draw(O--F);<br />
draw(O--G);<br />
draw(A--G);<br />
draw(O--H);<br />
label("O",O,N);<br />
label("A",A,S);<br />
label("B",B,SE);<br />
label("C",C,NE);<br />
label("D",D,NW);<br />
label("E",E,SE);<br />
label("F",F,S);<br />
label("H",H,SW);<br />
label("G",G,SW);<br />
label("x",H--E,S);<br />
label("K",K,NW);<br />
</asy><br />
Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.<br />
Since <math>\triangle GOE \sim \triangle OHE</math>, <math>\frac{GO}{OE} = \frac{450}{x}</math>, and by [[Angle Bisector Theorem]], <math>\frac{GF}{FE} = \frac{450}{x}</math>. Thus, <math>GF = \frac{450 \cdot 400}{x}</math>. <math>AF = AH-FH = 50+x</math>, and <math>KA = EB</math> (90 degree rotation), and now we can bash on 2 similar triangles <math>\triangle GAK \sim \triangle GHO</math>.<br />
<br />
<cmath>\frac{GA}{AK} = \frac{GH}{OH}</cmath> <br />
<cmath>\frac{\frac{450 \cdot 400}{x}-50-x}{450-x} = \frac{\frac{450 \cdot 400}{x}+400-x}{450}</cmath><br />
I hope you like expanding<br />
<cmath>x^2 - 850x + \frac{81000000}{x} = -450x - 22500 + \frac{81000000}{x}</cmath><br />
<cmath>x^2 - 400x + 22500 = 0</cmath><br />
Quadratic formula gives us<br />
<cmath>x = 200 \pm 50 \sqrt{7}</cmath><br />
Since AE < BF <br />
<cmath>x = 200 - 50 \sqrt{7}</cmath><br />
Thus, <br />
<cmath>BF = 250 + 50 \sqrt{7}</cmath><br />
So, our answer is <math>\boxed{307}</math>.<br />
<br />
-AlexLikeMath<br />
<br />
=== Solution 7 (Using a Circle) ===<br />
<center><asy><br />
size(200);<br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5-0.5*sqrt(7),7), O=(4.5,4.5), H=(4.5-0.5*sqrt(7),4.5), I=(0,4.5), J=(4.5-0.5*sqrt(7),9);<br />
draw(A--B--C--D--A);<br />
draw(E--O--F);<br />
draw(J--G);<br />
draw(E--G--F);<br />
draw(G--H--O--G);<br />
draw(I--H);<br />
draw(circle(G,2*sqrt(2)));<br />
markscalefactor=0.05;<br />
draw(rightanglemark(E,G,F));<br />
dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^O);<br />
label("\(A\)",A,(-1,1));<br />
label("\(B\)",B,(1,1));<br />
label("\(C\)",C,(1,-1));<br />
label("\(D\)",D,(-1,-1));<br />
label("\(E\)",E,(0,1));<br />
label("\(F\)",F,(1,1));<br />
label("\(G\)",G,(1,0));<br />
label("\(H\)",H,(-1,1));<br />
label("\(I\)",I,(-1,0));<br />
label("\(J\)",J,(0,1));<br />
label("\(O\)",O,(1,-1));<br />
</asy></center><br />
<br />
We know that G is on the perpendicular bisector of <math>EF</math>, which means that <math>EJ=JF=200</math>, <math>EG=GF=200\sqrt{2}</math> and <math>GH=250</math>. Now, let <math>HO</math> be equal to <math>x</math>. We can set up an equation with the Pythagorean Theorem: <br />
<br />
<cmath><br />
\begin{align*}<br />
\sqrt{x^2+250^2}&=(200\sqrt{2})^2 \\<br />
x^2+62500&=80000 \\<br />
x^2&=17500 \\<br />
x&=50\sqrt{7}<br />
\end{align*}<br />
</cmath><br />
<br />
Now, since <math>IO=450</math>, <br />
<br />
<cmath><br />
\begin{align*}<br />
HI&=450-x \\<br />
&=450-50\sqrt{7} \\<br />
\end{align*} \\<br />
</cmath><br />
<br />
Since <math>HI=AJ</math>, we now have:<br />
<br />
<cmath><br />
\begin{align*}<br />
BF&=AB-AJ-JF \\<br />
&=900-(450-50\sqrt{7})-200 \\<br />
&=250+50\sqrt{7} \\<br />
\end{align*}<br />
</cmath> <br />
<br />
This means that our answer would be <math>250+50+7=\boxed{307}</math><br />
<br />
~Jerry_Guo<br />
<br />
=== Solution 8 (More Similar Triangles) ===<br />
<br />
Construct <math>BO, AO.</math> Let <math>\angle{FOB} = \alpha.</math> Also let <math>FB = x</math> then <math>AE = 500-x.</math> We then have from simple angle-chasing:<br />
<cmath><br />
\begin{align*}<br />
\angle{BFO} = 135 - \alpha \\<br />
\angle{OFE} = 45 + \alpha \\<br />
\angle{EOA} = 45 - \alpha \\<br />
\angle{AEO} = 90 + \alpha \\<br />
\angle{OEF} = 90 - \alpha.<br />
\end{align*}<br />
</cmath><br />
From AA similarity we have <cmath>\triangle{EOB} \sim \triangle{EFO}.</cmath> This gives the ratios, <br />
<cmath>\dfrac{400 + x}{EO} = \dfrac{450\sqrt{2}}{FO}.</cmath><br />
Similarly from AA similarity <cmath>\triangle{FOA} \sim \triangle{FEO}.</cmath> So we get the ratios <cmath>\dfrac{EO}{450\sqrt{2}} = \dfrac{FO}{900-x}.</cmath> We can multiply to get <br />
<cmath>\dfrac{400 + x}{450\sqrt{2}} = \dfrac{450\sqrt{2}}{900 - x}.</cmath> Cross-multiplying reveals<br />
<cmath>360000 + 500x - x^2 = 405000.</cmath> Bringing everything to one side we have <cmath>x^2 - 500x + 45000 = 0.</cmath> By the quadratic formula we get <cmath>x = \dfrac{500 + \sqrt{500^2 - 4\cdot45000}}{2} = \dfrac{500 + \sqrt{70000}}{2} = \dfrac{500 + 100\sqrt{7}}{2} = 250 + 50\sqrt{7}.</cmath><br />
Therefore <br />
<cmath>p+q+r = 250 + 50 + 7 = \boxed{307}.</cmath><br />
~aa1024<br />
<br />
<br />
Solution 9:<br />
we use ratio lemma and stewarts theorem<br />
Connect OA,OE,OF,OB and let AE = x while BF = 500 - x<br />
let angle AOE = y, hence BOF = 45 - y<br />
now we apply stewarts theorem in triangles AOF and BOE to get OE and OF in terms of X<br />
finally calculate x/400 and 500-x/400 using ratio lemma to find x and y<br />
<br />
== See also ==<br />
{{AIME box|year=2005|n=II|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_16&diff=1319052011 AMC 10B Problems/Problem 162020-08-16T07:50:44Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>== Problem==<br />
<br />
A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?<br />
<br />
<center><asy><br />
unitsize(10mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2));<br />
<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
draw(A--D);<br />
draw(B--G);<br />
draw(C--F);<br />
draw(E--H);<br />
<br />
</asy><br />
</center><br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{2} - 1}{2} \qquad\textbf{(B)}\ \frac{1}{4} \qquad\textbf{(C)}\ \frac{2 - \sqrt{2}}{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{4} \qquad\textbf{(E)}\ 2 - \sqrt{2}</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
== Solution ==<br />
<br />
<center><asy><br />
unitsize(10mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=1;<br />
<br />
pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2));<br />
pair I=(1,1), J=(1+sqrt(2),1), K=(1+sqrt(2),1+sqrt(2)), L=(1,1+sqrt(2));<br />
<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
draw(A--D);<br />
draw(B--G);<br />
draw(C--F);<br />
draw(E--H);<br />
<br />
pair[] ps={A,B,C,D,E,F,G,H,I,J,K,L};<br />
dot(ps);<br />
label("$A$",A,W);<br />
label("$B$",B,S);<br />
label("$C$",C,S);<br />
label("$D$",D,E);<br />
label("$E$",E,E);<br />
label("$F$",F,N);<br />
label("$G$",G,N);<br />
label("$H$",H,W);<br />
label("$I$",I,NE);<br />
label("$J$",J,NW);<br />
label("$K$",K,SW);<br />
label("$L$",L,SE);<br />
label("$\sqrt{2}$",midpoint(B--C),S);<br />
label("$1$",midpoint(A--I),N);<br />
</asy><br />
</center><br />
<br />
If the side lengths of the dart board and the side lengths of the center square are all <math>\sqrt{2},</math> then the side length of the legs of the triangles are <math>1</math>.<br />
<br />
<cmath>\begin{align*}<br />
\text{area of center square} &: \sqrt{2} \times \sqrt{2} = 2\\<br />
\text{total area} &: (\sqrt{2})^2 + 4(1 \times \sqrt{2}) + 4(1 \times 1 \times \frac{1}{2}) = 2 + 4\sqrt{2} + 2 = 4 + 4\sqrt{2}<br />
\end{align*}</cmath><br />
<br />
Use [[Geometric probability]] by putting the area of the desired region over the area of the entire region.<br />
<br />
<cmath> \frac{2}{4+4\sqrt{2}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}</cmath><br />
<br />
== Solution 2==<br />
area of a regular octagon = 2(1+sqrt{2})a^2 where a is the side hence the answer is obvious now.<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_15&diff=1319042012 AMC 10B Problems/Problem 152020-08-16T07:30:40Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>==Problem==<br />
In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end on the tournament?<br />
<br />
<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6 </math><br />
==Solution==<br />
The total number of games (and wins) in the tournament is <math>\frac{6 \times 5}{2}= 15</math>. A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:<br />
| 1 2 3 4 5 6 |<br />
|1 X W L W L W |<br />
|2 L X W L W W |<br />
|3 W L X W L W |<br />
|4 L W L X W W |<br />
|5 W L W L X W |<br />
|6 L L L L L X |<br />
The "X's" are for when it is where a team is set against itself, which cannot happen. The chart says that Team 6 has lost all of its matches, which means that each of the other teams won against it. Then, alternating Wins and Losses were tried. It shows that it is possible for 5 teams to tie for the same amount of wins, which in this case is 3 wins.<br />
Thus, the answer is <math>\boxed{\textbf{(D)}\ 5}</math>.<br />
<br />
==Solution 2==<br />
just note that total matches are 15, hence and if 4 teams tie for most wins then they can tie for 3 wins each, but then 5th team also has 3 wins hence 4 + 1 = 5 is the answer<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_7&diff=1319032013 AMC 10A Problems/Problem 72020-08-16T07:14:19Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br />
<br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
<br />
Let us split this up into two cases.<br />
<br />
Case <math>1</math>: The student chooses both algebra and geometry.<br />
<br />
This means that <math>3</math> courses have already been chosen. We have <math>3</math> more options for the last course, so there are <math>3</math> possibilities here. <br />
<br />
Case <math>2</math>: The student chooses one or the other. <br />
<br />
Here, we simply count how many ways we can do one, multiply by <math>2</math>, and then add to the previous. <br />
<br />
Assume the mathematics course is algebra. This means that we can choose <math>2</math> of History, Art, and Latin, which is simply <math>\dbinom{3}{2} = 3</math>. If it is geometry, we have another <math>3</math> options, so we have a total of <math>6</math> options if only one mathematics course is chosen.<br />
<br />
Thus, overall, we can choose a program in <math>6 + 3 = \boxed{\textbf{(C) }9}</math> ways<br />
<br />
===Solution 2===<br />
<br />
We can use complementary counting. Since there must be an English class, we will add that to our list of classes for <math>3</math> remaining spots for the classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing <math>3</math> classes out of the <math>5</math> is <math>\binom{5}3</math>. The total number of ways of choosing only non-mathematical classes is <math>\binom{3}3</math>. Therefore the amount of ways you can pick classes with at least one math class is <math>\binom{5}3-\binom{3}3=10-1=\boxed{\textbf{(C) }9}</math> ways.<br />
<br />
===Solution 3===<br />
Similar to solution 1, note that for case 1 of solution the answer is simply 4 choose 2 and for second case it is 3 choose 2 hence \boxed{\textbf{(C) }9}$ ways<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2013|ab=A|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_18&diff=1319022014 AMC 10B Problems/Problem 182020-08-16T07:02:26Z<p>Quantomaticguy: /* See Also */</p>
<hr />
<div>==Problem==<br />
<br />
A list of <math>11</math> positive integers has a mean of <math>10</math>, a median of <math>9</math>, and a unique mode of <math>8</math>. What is the largest possible value of an integer in the list?<br />
<br />
<math> \textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35</math><br />
<br />
==Solution==<br />
We start off with the fact that the median is <math>9</math>, so we must have <math>a, b, c, d, e, 9, f, g, h, i, j</math>, listed in ascending order. Note that the integers do not have to be distinct. <br />
<br />
Since the mode is <math>8</math>, we have to have at least <math>2</math> occurrences of <math>8</math> in the list. If there are <math>2</math> occurrences of <math>8</math> in the list, we will have <math>a, b, c, 8, 8, 9, f, g, h, i, j</math>. In this case, since <math>8</math> is the unique mode, the rest of the integers have to be distinct. So we minimize <math>a,b,c,f,g,h,i</math> in order to maximize <math>j</math>. If we let the list be <math>1,2,3,8,8,9,10,11,12,13,j</math>, then <math>j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33</math>. <br />
<br />
Next, consider the case where there are <math>3</math> occurrences of <math>8</math> in the list. Now, we can have two occurrences of another integer in the list. We try <math>1,1,8,8,8,9,9,10,10,11,j</math>. Following the same process as above, we get <math>j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35</math>. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is <math>\boxed{\textbf{(E) }35}</math><br />
<br />
<br />
Solution 2:<br />
Solution 2: Note that x1 + ... + x11 = 110 let x6 = 9 so x1 + ... + x5 + x7+ ... + x11 = 101 now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7 so x1 + x2 + x3 = 57 now we have to let one of above 3 values = 8 hence x1 + x2 = 49 now let x1 = 35 , x2 =14 hence 35 is the answe<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_18&diff=1319012014 AMC 10B Problems/Problem 182020-08-16T07:02:11Z<p>Quantomaticguy: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A list of <math>11</math> positive integers has a mean of <math>10</math>, a median of <math>9</math>, and a unique mode of <math>8</math>. What is the largest possible value of an integer in the list?<br />
<br />
<math> \textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35</math><br />
<br />
==Solution==<br />
We start off with the fact that the median is <math>9</math>, so we must have <math>a, b, c, d, e, 9, f, g, h, i, j</math>, listed in ascending order. Note that the integers do not have to be distinct. <br />
<br />
Since the mode is <math>8</math>, we have to have at least <math>2</math> occurrences of <math>8</math> in the list. If there are <math>2</math> occurrences of <math>8</math> in the list, we will have <math>a, b, c, 8, 8, 9, f, g, h, i, j</math>. In this case, since <math>8</math> is the unique mode, the rest of the integers have to be distinct. So we minimize <math>a,b,c,f,g,h,i</math> in order to maximize <math>j</math>. If we let the list be <math>1,2,3,8,8,9,10,11,12,13,j</math>, then <math>j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33</math>. <br />
<br />
Next, consider the case where there are <math>3</math> occurrences of <math>8</math> in the list. Now, we can have two occurrences of another integer in the list. We try <math>1,1,8,8,8,9,9,10,10,11,j</math>. Following the same process as above, we get <math>j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35</math>. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is <math>\boxed{\textbf{(E) }35}</math><br />
<br />
<br />
Solution 2:<br />
Solution 2: Note that x1 + ... + x11 = 110 let x6 = 9 so x1 + ... + x5 + x7+ ... + x11 = 101 now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7 so x1 + x2 + x3 = 57 now we have to let one of above 3 values = 8 hence x1 + x2 = 49 now let x1 = 35 , x2 =14 hence 35 is the answe<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}<br />
Solution 2:<br />
Note that x1 + ... + x11 = 110<br />
let x6 = 9 <br />
so x1 + ... + x5 + x7+ ... + x11 = 101<br />
now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7<br />
so x1 + x2 + x3 = 57<br />
now we have to let one of above 3 values = 8 hence x1 + x2 = 49<br />
now let x1 = 35 , x2 =14 hence 35 is the answer</div>Quantomaticguyhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_18&diff=1319002014 AMC 10B Problems/Problem 182020-08-16T07:01:42Z<p>Quantomaticguy: /*Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
A list of <math>11</math> positive integers has a mean of <math>10</math>, a median of <math>9</math>, and a unique mode of <math>8</math>. What is the largest possible value of an integer in the list?<br />
<br />
<math> \textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35</math><br />
<br />
==Solution==<br />
We start off with the fact that the median is <math>9</math>, so we must have <math>a, b, c, d, e, 9, f, g, h, i, j</math>, listed in ascending order. Note that the integers do not have to be distinct. <br />
<br />
Since the mode is <math>8</math>, we have to have at least <math>2</math> occurrences of <math>8</math> in the list. If there are <math>2</math> occurrences of <math>8</math> in the list, we will have <math>a, b, c, 8, 8, 9, f, g, h, i, j</math>. In this case, since <math>8</math> is the unique mode, the rest of the integers have to be distinct. So we minimize <math>a,b,c,f,g,h,i</math> in order to maximize <math>j</math>. If we let the list be <math>1,2,3,8,8,9,10,11,12,13,j</math>, then <math>j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33</math>. <br />
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Next, consider the case where there are <math>3</math> occurrences of <math>8</math> in the list. Now, we can have two occurrences of another integer in the list. We try <math>1,1,8,8,8,9,9,10,10,11,j</math>. Following the same process as above, we get <math>j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35</math>. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is <math>\boxed{\textbf{(E) }35}</math><br />
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==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}}<br />
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Solution 2:<br />
Note that x1 + ... + x11 = 110<br />
let x6 = 9 <br />
so x1 + ... + x5 + x7+ ... + x11 = 101<br />
now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7<br />
so x1 + x2 + x3 = 57<br />
now we have to let one of above 3 values = 8 hence x1 + x2 = 49<br />
now let x1 = 35 , x2 =14 hence 35 is the answer</div>Quantomaticguy