https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=I-+-i&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T04:37:35ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_17&diff=1612632003 AMC 12A Problems/Problem 172021-08-31T16:04:15Z<p>I- -i: </p>
<hr />
<div>== Problem ==<br />
Square <math>ABCD</math> has sides of length <math>4</math>, and <math>M</math> is the midpoint of <math>\overline{CD}</math>. A circle with radius <math>2</math> and center <math>M</math> intersects a circle with radius <math>4</math> and center <math>A</math> at points <math>P</math> and <math>D</math>. What is the distance from <math>P</math> to <math>\overline{AD}</math>?<br />
<br />
<asy><br />
pair A,B,C,D,M,P;<br />
D=(0,0);<br />
C=(10,0);<br />
B=(10,10);<br />
A=(0,10);<br />
M=(5,0);<br />
P=(8,4);<br />
dot(M);<br />
dot(P);<br />
draw(A--B--C--D--cycle,linewidth(0.7));<br />
draw((5,5)..D--C..cycle,linewidth(0.7));<br />
draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7));<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$M$",M,S);<br />
label("$P$",P,N);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math><br />
<br />
== Solutions ==<br />
=== Solution 1===<br />
<br />
Let <math>D</math> be the origin. <math>A</math> is the point <math>(0,4)</math> and <math>M</math> is the point <math>(2,0)</math>. We are given the radius of the quarter circle and semicircle as <math>4</math> and <math>2</math>, respectively, so their equations, respectively, are:<br />
<br />
<math>x^2 + (y-4)^2 = 4^2</math><br />
<br />
<math>(x-2)^2 + y^2 = 2^2</math><br />
<br />
Subtract the second equation from the first:<br />
<br />
<math>x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12</math><br />
<br />
<math>4x - 8y + 12 = 12</math><br />
<br />
<math>x = 2y.</math><br />
<br />
Then substitute:<br />
<br />
<math>(2y)^2 + (y - 4)^2 = 16</math><br />
<br />
<math>4y^2 + y^2 - 8y + 16 = 16</math><br />
<br />
<math>5y^2 - 8y = 0</math><br />
<br />
<math>y(5y - 8) = 0.</math><br />
<br />
Thus <math>y = 0</math> and <math>y = \frac{8}{5}</math> making <math>x = 0</math> and <math>x = \frac{16}{5}</math>.<br />
<br />
The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow \boxed{B}.</math><br />
<br />
=== Solution 2 ===<br />
<math>APMD</math> obviously forms a kite. Let the intersection of the diagonals be <math>E</math>. <math>AE+EM=AM=2\sqrt{5}</math> Let <math>AE=x</math>. Then, <math>EM=2\sqrt{5}-x</math>.<br />
<br />
<br />
By Pythagorean Theorem, <math>DE^2=4^2-AE^2=2^2-EM^2</math>. Thus, <math>16-x^2=4-(2\sqrt{5}-x)^2</math>. Simplifying, <math>x=\frac{8}{\sqrt{5}}</math>. By Pythagoras again, <math>DE=\frac{4}{\sqrt{5}}</math>. Then, the area of <math>ADP</math> is <math>DE\cdot AE=\frac{32}{5}</math>. <br />
<br />
<br />
Using <math>4</math> instead as the base, we can drop a altitude from P. <math>\frac{32}{5}=\frac{bh}{2}</math>. <math>\frac{32}{5}=\frac{4h}{2}</math>. Thus, the horizontal distance is <math>\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}</math><br />
<br />
<br />
<br />
~BJHHar<br />
<br />
=== Solution 3 ===<br />
<br />
Note that <math>P</math> is merely a reflection of <math>D</math> over <math>AM</math>. Call the intersection of <math>AM</math> and <math>DP</math> <math>X</math>. Drop perpendiculars from <math>X</math> and <math>P</math> to <math>AD</math>, and denote their respective points of intersection by <math>J</math> and <math>K</math>. We then have <math>\triangle DXJ\sim\triangle DPK</math>, with a scale factor of 2. Thus, we can find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16}{5}</math>.<br />
<br />
=== Solution 4 ===<br />
As in Solution 2, draw in <math>DP</math> and <math>AM</math> and denote their intersection point <math>X</math>. Next, drop a perpendicular from <math>P</math> to <math>AD</math> and denote the foot as <math>Z</math>. <math>AP \cong AD</math> as they are both radii and similarly <math>DM \cong MP</math> so <math>APMD</math> is a kite and <math>DX \perp XM</math> by a well-known theorem. <br />
<br />
Pythagorean theorem gives us <math>AM=2 \sqrt{5}</math>. Clearly <math>\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP</math> by angle-angle and <math>\triangle XMD \cong \triangle XMP</math> by Hypotenuse Leg.<br />
Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math><br />
<br />
=== Solution 5 ===<br />
Using the double-angle formula for sine, what we need to find is <math>AP\cdot \sin(DAP) = AP\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}</math>.<br />
==Solution 6(LoC)==<br />
We use the Law of Cosines:<br />
<br />
<math>32-32 \cos \theta = 8 + 8 \cos \theta </math><br />
<br />
<math>\frac{3}{5} = \cos \theta </math><br />
<br />
<math>2 + 2*\frac{3}{5} = \frac{16}{5}</math><br />
<br />
=== Solution 7 ===<br />
<br />
Let <math>H</math> be the foot of the perpendicular from <math>P</math> to <math>CD</math>, and let <math>HD = x</math>. Then we have <math>HC = 4-x</math>, and <math>PH = 4 - \sqrt{16 - x^2}</math>. Since <math>\triangle DHP \sim \triangle PHC</math>, we have <math>HP^2 = DH \cdot HC</math>, or <math>-x^2 + 4x = 16 - 8\sqrt{16-x^2}</math>. Solving gives <math>x = \frac{16}{5}</math>.<br />
<br />
== See Also ==<br />
{{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=File:Awesome_Face.png&diff=150328File:Awesome Face.png2021-03-26T02:30:57Z<p>I- -i: </p>
<hr />
<div></div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=1489622020 AIME I Problems/Problem 142021-03-09T03:03:31Z<p>I- -i: /* Solution 6 */</p>
<hr />
<div><br />
== Problem ==<br />
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math><br />
<br />
== Solution 1 ==<br />
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333<br />
<br />
<b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>.<br />
<br />
== Solution 2 ==<br />
Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now casework: <br />
If <math>P(3)=P(4)=m</math>, then <br />
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath><br />
<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath><br />
so this gives <math>(a+b)^2=7^2=49</math>. <br />
Next, if <math>P(3)=P(a)=m</math>, then <br />
<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath><br />
<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath><br />
Subtracting the first part of the first equation from the first part of the second equation gives <br />
<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath><br />
Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. <br />
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong<br />
<br />
== Solution 3 ==<br />
Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.<br />
<br />
Case 1: We have <math>P(3) \ne P(4)</math>. We must have <br />
<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath><br />
Rearrange and divide through by <math>8</math> to obtain<br />
<cmath>w = \frac{-25-2z}{8}.</cmath><br />
Now, note that<br />
<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath><br />
<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath><br />
Now, rearrange to get<br />
<cmath>\frac{z}{8} = -\frac{21}{16}</cmath><br />
and thus<br />
<cmath>z = -\frac{21}{2}.</cmath><br />
Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.<br />
<br />
Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.<br />
<br />
Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton<br />
<br />
==Solution 4==<br />
Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>.<br />
<br />
In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>.<br />
<br />
Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>.<br />
<br />
The requested sum is <math>36+49=85</math>.~TheUltimate123<br />
<br />
==Solution 5 (Official MAA)==<br />
Note that because <math>P\big(P(3)\big)=P\big(P(4)\big)= 0</math>, <math>P(3)</math> and <math>P(4)</math> are roots of <math>P(x)</math>. There are two cases.<br />
CASE 1: <math>P(3) = P(4)</math>. Then <math>P(x)</math> is symmetric about <math>x=\tfrac72</math>; that is to say, <math>P(r) = P(7-r)</math> for all <math>r</math>. Thus the remaining two roots must sum to <math>7</math>. Indeed, the polynomials <math>P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3</math> satisfy the conditions.<br />
CASE 2: <math>P(3)\neq P(4)</math>. Then <math>P(3)</math> and <math>P(4)</math> are the two distinct roots of <math>P(x)</math>, so<cmath>P(x) = \big(x-P(3)\big)\big(x-P(4)\big)</cmath>for all <math>x</math>. Note that any solution to <math>P\big(P(x)\big) = 0</math> must satisfy either <math>P(x) = P(3)</math> or <math>P(x) = P(4)</math>. Because <math>P(x)</math> is quadratic, the polynomials <math>P(x) - P(3)</math> and <math>P(x) - P(4)</math> each have the same sum of roots as the polynomial <math>P(x)</math>, which is <math>P(3) + P(4)</math>. Thus the answer in this case is <math>2\big(P(3) + P(4)\big)-7</math>, and so it suffices to compute the value of <math>P(3)+P(4)</math>.<br />
<br />
Let <math>P(3)=u</math> and <math>P(4) = v</math>. Substituting <math>x=3</math> and <math>x=4</math> into the above quadratic polynomial yields the system of equations<br />
<cmath>\begin{align*}<br />
u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\<br />
v &= (4-u)(4-v) = 16 - 4u - 4v + uv.<br />
\end{align*}</cmath>Subtracting the first equation from the second gives <math>v - u = 7 - u - v</math>, yielding <math>v = \frac72.</math> Substituting this value into the second equation gives<cmath>\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),</cmath>yielding <math>u = -3.</math> The sum of the two solutions is <math>2\left(\tfrac72-3\right)-7 = -6</math>. In this case, <math>P(x)= (x+3)\left(x-\frac72\right)</math>.<br />
<br />
The requested sum of squares is <math>7^2+(-6)^2 = {85}</math>.<br />
<br />
==Solution 6==<br />
<br />
Let <math>P(x) = (x-c)^2 - d</math> for some <math>c</math>, <math>d</math>.<br />
<br />
Then, we can write <math>P(P(x)) = ((x-c)^2 - d - c)^2 - d</math>. Setting the expression equal to <math>0</math> and solving for <math>x</math> gives:<br />
<br />
<math>x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c</math><br />
<br />
Therefore, we have that <math>x</math> takes on the four values <math>\sqrt{\sqrt{d} + d + c} + c</math>, <math>-\sqrt{\sqrt{d} + d + c} + c</math>, <math>\sqrt{-\sqrt{d} + d + c} + c</math>, and <math>-\sqrt{-\sqrt{d} + d + c} + c</math>. Two of these values are <math>3</math> and <math>4</math>, and the other two are <math>a</math> and <math>b</math>.<br />
<br />
We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.<br />
<br />
<math>\textbf{Case 1}</math>: Both the 3 and 4 values are from the same group.<br />
<br />
In this case, the <math>a</math> and <math>b</math> values are both from the other group. The sum of this is just <math>2c</math> because the radical cancels out. Because of this, we can see that <math>c</math> is just the average of <math>3</math> and <math>4</math>, so we have <math>2c = 3 + 4 = 7</math>, so <math>(a+b)^2 = 7^2 = 49</math>.<br />
<br />
<math>\textbf{Case 2}</math>: The 3 and 4 values come from different groups.<br />
<br />
It is easy to see that all possibilities in this case are basically symmetric and yield the same value for <math>(a+b)^2</math>. Without loss of generality, assume that <math>\sqrt{\sqrt{d} + d + c} + c = 4</math> and <math>\sqrt{-\sqrt{d} + d + c} + c = 3</math>. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.<br />
<br />
We can write <math>\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c</math>.<br />
<br />
Moving most terms to the left side and simplifying gives <math>\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1</math>.<br />
<br />
We can square both sides and simplify:<br />
<br />
<math>\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}</math><br />
<br />
Squaring both sides again gives the following:<br />
<br />
<math>d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}</math><br />
<br />
Nearly all terms cancel out, yielding <math>c = \frac{1}{4}</math>.<br />
<br />
By substituting this back in, we obtain <math>\sqrt{\sqrt{d} + d + c} = \frac{15}{4}</math> and <math>\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}</math>.<br />
<br />
The sum of <math>a</math> and <math>b</math> is equal to <math>-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6</math>, so <math>(a+b)^2 = 36</math>.<br />
<br />
Adding up both values gives <math>49 + 36 = \boxed{085}</math> as our final answer.<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_14&diff=1489612020 AIME I Problems/Problem 142021-03-09T02:56:49Z<p>I- -i: /* Solution 5 (Official MAA) */</p>
<hr />
<div><br />
== Problem ==<br />
Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math><br />
<br />
== Solution 1 ==<br />
Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333<br />
<br />
<b>Remark</b>: We know that <math>c=\frac{8a-1}{2}</math> from <math>P(3)+P(4)=3+a</math>.<br />
<br />
== Solution 2 ==<br />
Let the roots of <math>P(x)</math> be <math>m</math> and <math>n</math>, then we can write <math>P(x)=x^2-(m+n)x+mn</math>. The fact that <math>P(P(x))=0</math> has solutions <math>x=3,4,a,b</math> implies that some combination of <math>2</math> of these are the solution to <math>P(x)=m</math>, and the other <math>2</math> are the solution to <math>P(x)=n</math>. It's fairly easy to see there are only <math>2</math> possible such groupings: <math>P(3)=P(4)=m</math> and <math>P(a)=P(b)=n</math>, or <math>P(3)=P(a)=m</math> and <math>P(4)=P(b)=n</math> (Note that <math>a,b</math> are interchangeable, and so are <math>m</math> and <math>n</math>). We now casework: <br />
If <math>P(3)=P(4)=m</math>, then <br />
<cmath>9-3(m+n)+mn=16-4(m+n)+mn=m \implies m+n=7</cmath><br />
<cmath>a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \implies a+b=m+n=7</cmath><br />
so this gives <math>(a+b)^2=7^2=49</math>. <br />
Next, if <math>P(3)=P(a)=m</math>, then <br />
<cmath>9-3(m+n)+mn=a^2-a(m+n)+mn=m \implies a+3=m+n</cmath><br />
<cmath>16-4(m+n)+mn=b^2-b(m+n)+mn=n \implies b+4=m+n</cmath><br />
Subtracting the first part of the first equation from the first part of the second equation gives <br />
<cmath>7-(m+n)=n-m \implies 2n=7 \implies n=\frac{7}{2} \implies m=-3</cmath><br />
Hence, <math>a+b=2(m+n)-7=2\cdot \frac{1}{2}-7=-6</math>, and so <math>(a+b)^2=(-6)^2=36</math>. <br />
Therefore, the solution is <math>49+36=\boxed{085}</math> ~ktong<br />
<br />
== Solution 3 ==<br />
Write <math>P(x) = x^2+wx+z</math>. Split the problem into two cases: <math>P(3)\ne P(4)</math> and <math>P(3) = P(4)</math>.<br />
<br />
Case 1: We have <math>P(3) \ne P(4)</math>. We must have <br />
<cmath>w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.</cmath><br />
Rearrange and divide through by <math>8</math> to obtain<br />
<cmath>w = \frac{-25-2z}{8}.</cmath><br />
Now, note that<br />
<cmath>z = P(3)P(4) = (9+3w+z)(16+4w+z) = \left(9 + 3\cdot \frac{-25-2z}{8} + z\right)\left(16 + 4 \cdot \frac{-25-2z}{8} + z\right) =</cmath><br />
<cmath>\left(-\frac{3}{8} + \frac{z}{4}\right)\left(\frac{7}{2}\right) = -\frac{21}{16} + \frac{7z}{8}.</cmath><br />
Now, rearrange to get<br />
<cmath>\frac{z}{8} = -\frac{21}{16}</cmath><br />
and thus<br />
<cmath>z = -\frac{21}{2}.</cmath><br />
Substituting this into our equation for <math>w</math> yields <math>w = -\frac{1}{2}</math>. Then, it is clear that <math>P</math> does not have a double root at <math>P(3)</math>, so we must have <math>P(a) = P(3)</math> and <math>P(b) = P(4)</math> or vice versa. This gives <math>3+a = \frac{1}{2}</math> and <math>4+b = \frac{1}{2}</math> or vice versa, implying that <math>a+b = 1-3-4 = -6</math> and <math>(a+b)^2 = 6</math>.<br />
<br />
Case 2: We have <math>P(3) = P(4)</math>. Then, we must have <math>w = -7</math>. It is clear that <math>P(a) = P(b)</math> (we would otherwise get <math>P(a)=P(3)=P(4)</math> implying <math>a \in \{3,4\}</math> or vice versa), so <math>a+b=-w=7</math> and <math>(a+b)^2 = 49</math>.<br />
<br />
Thus, our final answer is <math>49+36=\boxed{085}</math>. ~GeronimoStilton<br />
<br />
==Solution 4==<br />
Let <math>P(x)=(x-r)(x-s)</math>. There are two cases: in the first case, <math>(3-r)(3-s)=(4-r)(4-s)</math> equals <math>r</math> (without loss of generality), and thus <math>(a-r)(a-s)=(b-r)(b-s)=s</math>. By Vieta's formulas <math>a+b=r+s=3+4=7</math>.<br />
<br />
In the second case, say without loss of generality <math>(3-r)(3-s)=r</math> and <math>(4-r)(4-s)=s</math>. Subtracting gives <math>-7+r+s=r-s</math>, so <math>s=7/2</math>. From this, we have <math>r=-3</math>.<br />
<br />
Note <math>r+s=1/2</math>, so by Vieta's, we have <math>\{a,b\}=\{1/2-3,1/2-4\}=\{-5/2,-7/2\}</math>. In this case, <math>a+b=-6</math>.<br />
<br />
The requested sum is <math>36+49=85</math>.~TheUltimate123<br />
<br />
==Solution 5 (Official MAA)==<br />
Note that because <math>P\big(P(3)\big)=P\big(P(4)\big)= 0</math>, <math>P(3)</math> and <math>P(4)</math> are roots of <math>P(x)</math>. There are two cases.<br />
CASE 1: <math>P(3) = P(4)</math>. Then <math>P(x)</math> is symmetric about <math>x=\tfrac72</math>; that is to say, <math>P(r) = P(7-r)</math> for all <math>r</math>. Thus the remaining two roots must sum to <math>7</math>. Indeed, the polynomials <math>P(x) = \left(x-\frac72\right)^2 + \frac{11}4 \pm i\sqrt3</math> satisfy the conditions.<br />
CASE 2: <math>P(3)\neq P(4)</math>. Then <math>P(3)</math> and <math>P(4)</math> are the two distinct roots of <math>P(x)</math>, so<cmath>P(x) = \big(x-P(3)\big)\big(x-P(4)\big)</cmath>for all <math>x</math>. Note that any solution to <math>P\big(P(x)\big) = 0</math> must satisfy either <math>P(x) = P(3)</math> or <math>P(x) = P(4)</math>. Because <math>P(x)</math> is quadratic, the polynomials <math>P(x) - P(3)</math> and <math>P(x) - P(4)</math> each have the same sum of roots as the polynomial <math>P(x)</math>, which is <math>P(3) + P(4)</math>. Thus the answer in this case is <math>2\big(P(3) + P(4)\big)-7</math>, and so it suffices to compute the value of <math>P(3)+P(4)</math>.<br />
<br />
Let <math>P(3)=u</math> and <math>P(4) = v</math>. Substituting <math>x=3</math> and <math>x=4</math> into the above quadratic polynomial yields the system of equations<br />
<cmath>\begin{align*}<br />
u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\<br />
v &= (4-u)(4-v) = 16 - 4u - 4v + uv.<br />
\end{align*}</cmath>Subtracting the first equation from the second gives <math>v - u = 7 - u - v</math>, yielding <math>v = \frac72.</math> Substituting this value into the second equation gives<cmath>\dfrac72 = \left(4 - u\right)\left(4 - \dfrac72\right),</cmath>yielding <math>u = -3.</math> The sum of the two solutions is <math>2\left(\tfrac72-3\right)-7 = -6</math>. In this case, <math>P(x)= (x+3)\left(x-\frac72\right)</math>.<br />
<br />
The requested sum of squares is <math>7^2+(-6)^2 = {85}</math>.<br />
<br />
==Solution 6==<br />
<br />
Let <math>P(x) = (x-c)^2 - d</math> for some <math>c</math>, <math>d</math>.<br />
<br />
Then, we can write <math>P(P(x)) = ((x-c)^2 - d - c)^2 - d</math>. Setting the expression equal to <math>0</math> and solving for <math>x</math> gives:<br />
<br />
<math>x = \pm \sqrt{ \pm \sqrt{d} + d + c} + c</math><br />
<br />
Therefore, we have that <math>x</math> takes on the four values <math>\sqrt{\sqrt{d} + d + c} + c</math>, <math>-\sqrt{\sqrt{d} + d + c} + c</math>, <math>\sqrt{-\sqrt{d} + d + c} + c</math>, and <math>-\sqrt{-\sqrt{d} + d + c} + c</math>. Two of these values are <math>3</math> and <math>4</math>, and the other two are <math>a</math> and <math>b</math>.<br />
<br />
We can split these four values into two "groups" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values.<br />
<br />
<math>\textbf{Case 1}</math>: Both the 3 and 4 values are from the same group.<br />
<br />
In this case, the <math>a</math> and <math>b</math> values are both from the other group. The sum of this is just <math>2c</math> because the radical cancels out. Because of this, we can see that <math>c</math> is just the average of <math>3</math> and <math>4</math>, so we have <math>2c = 3 + 4 = 7</math>, so <math>(a+b)^2 = 7^2 = 49</math>.<br />
<br />
<math>\textbf{Case 2}</math>: The 3 and 4 values come from different groups.<br />
<br />
It is easy to see that all possibilities in this case are basically symmetric and yield the same value for <math>(a+b)^2</math>. Without loss of generality, assume that <math>\sqrt{\sqrt{d} + d + c} + c = 4</math> and <math>\sqrt{-\sqrt{d} + d + c} + c = 3</math>. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger.<br />
<br />
We can write <math>\sqrt{\sqrt{d} + d + c} + c = 1 + \sqrt{-\sqrt{d} + d + c} + c</math>.<br />
<br />
Moving most terms to the left side and simplifying gives <math>\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} = 1</math>.<br />
<br />
We can square both sides and simplify:<br />
<br />
<math>\sqrt{d} + d + c - \sqrt{d} + d + c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>2d + 2c - 2\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = 1</math><br />
<br />
<math>\sqrt{(d + c + \sqrt{d})(d + c - \sqrt{d})} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{(d+c)^2 - (\sqrt{d})^2} = (d+c) - \frac{1}{2}</math><br />
<br />
<math>\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \frac{1}{2}</math><br />
<br />
Squaring both sides again gives the following:<br />
<br />
<math>d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \frac{1}{4}</math><br />
<br />
Nearly all terms cancel out, yielding <math>c = \frac{1}{4}</math>.<br />
<br />
By substituting this back in, we obtain <math>\sqrt{\sqrt{d} + d + c} = \frac{15}{4}</math> and <math>\sqrt{-\sqrt{d} + d + c} = \frac{11}{4}</math>.<br />
<br />
The sum of <math>a</math> and <math>b</math> is equal to <math>-\sqrt{\sqrt{d} + d + c} - \sqrt{-\sqrt{d} + d + c} + 2c = -\frac{15}{4} - \frac{11}{4} + \frac{1}{2} = -6</math>, so <math>(a+b)^2 = 36</math>.<br />
<br />
Adding up both values gives <math>49 + 36 = \boxed{085}</math> as our final answer. ~ I-_-I<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_9&diff=1131822013 AMC 10A Problems/Problem 92019-12-22T20:48:13Z<p>I- -i: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on <math>20\%</math> of her three-point shots and <math>30\%</math> of her two-point shots. Shenille attempted <math>30</math> shots. How many points did she score?<br />
<br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math><br />
<br />
==Solution==<br />
<br />
Let the number of attempted three-point shots made be <math>x</math> and the number of attempted two-point shots be <math>y</math>. We know that <math>x+y=30</math>, and we need to evaluate <math>(0.2\cdot3)x +(0.3\cdot2)y</math>, as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.<br />
<br />
Simplifying, we see that this is equal to <math>0.6x + 0.6y = 0.6(x+y)</math>. Plugging in <math>x+y=30</math>, we get <math>0.6(30) = \boxed{\textbf{(B) }18}</math><br />
<br />
==Solution 2 (cheap)== <br />
<br />
The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }18}</math> points.<br />
<br />
-- I-_-I<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}<br />
{{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_9&diff=1131812013 AMC 10A Problems/Problem 92019-12-22T20:47:13Z<p>I- -i: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on <math>20\%</math> of her three-point shots and <math>30\%</math> of her two-point shots. Shenille attempted <math>30</math> shots. How many points did she score?<br />
<br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math><br />
<br />
==Solution==<br />
<br />
Let the number of attempted three-point shots made be <math>x</math> and the number of attempted two-point shots be <math>y</math>. We know that <math>x+y=30</math>, and we need to evaluate <math>(0.2\cdot3)x +(0.3\cdot2)y</math>, as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.<br />
<br />
Simplifying, we see that this is equal to <math>0.6x + 0.6y = 0.6(x+y)</math>. Plugging in <math>x+y=30</math>, we get <math>0.6(30) = \boxed{\textbf{(B) }18}</math><br />
<br />
==Solution 2== <br />
<br />
The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }18}</math> points.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}<br />
{{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_9&diff=1131802013 AMC 10A Problems/Problem 92019-12-22T20:46:53Z<p>I- -i: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on <math>20\%</math> of her three-point shots and <math>30\%</math> of her two-point shots. Shenille attempted <math>30</math> shots. How many points did she score?<br />
<br />
<br />
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math><br />
<br />
==Solution==<br />
<br />
Let the number of attempted three-point shots made be <math>x</math> and the number of attempted two-point shots be <math>y</math>. We know that <math>x+y=30</math>, and we need to evaluate <math>(0.2\cdot3)x +(0.3\cdot2)y</math>, as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.<br />
<br />
Simplifying, we see that this is equal to <math>0.6x + 0.6y = 0.6(x+y)</math>. Plugging in <math>x+y=30</math>, we get <math>0.6(30) = \boxed{\textbf{(B) }18}</math><br />
<br />
==Solution 2== <br />
<br />
The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }18</math> points.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}<br />
{{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=1104292006 AIME II Problems/Problem 112019-10-19T17:44:34Z<p>I- -i: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A [[sequence]] is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the [[remainder]] when <math>\sum^{28}_{k=1} a_k </math> is divided by 1000.<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Define the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be:<br />
<center><math>s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br />
s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br />
s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br />
s = -s + a_{28} + a_{30}<br />
</math></center><br />
<br />
Thus <math>s = \frac{a_{28} + a_{30}}{2}</math>, and <math>a_{28},\,a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>\boxed{834}</math>.<br />
<br />
=== Solution 2 (bash) ===<br />
<br />
Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br />
<br />
<math>a_{1}\equiv 1 \pmod {1000}<br />
\newline<br />
a_{2}\equiv 1 \pmod {1000}<br />
\newline<br />
a_{3}\equiv 1 \pmod {1000}<br />
\newline<br />
a_{4}\equiv 3 \pmod {1000}<br />
\newline<br />
a_{5}\equiv 5 \pmod {1000}<br />
\newline<br />
\cdots <br />
\newline<br />
a_{25} \equiv 793 \pmod {1000}<br />
\newline<br />
a_{26} \equiv 281 \pmod {1000}<br />
\newline<br />
a_{27} \equiv 233 \pmod {1000}<br />
\newline<br />
a_{28} \equiv 307 \pmod {1000}</math><br />
<br />
Adding all the residues shows the sum is congruent to <math>\boxed{834}</math> mod 1000.<br />
<br />
~ I-_-I<br />
<br />
=== Solution 3 (some guessing involved) ===<br />
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given <math>a_{28}, a_{29}, </math> and <math>a_{30}</math>, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some <math>p, q, r</math> such that <math>\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}</math>. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that <math>(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})</math>, at least for the first few terms. From this, we have that <math>\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)</math>.<br />
<br />
Solution by zeroman<br />
== See also ==<br />
{{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=1104282006 AIME II Problems/Problem 112019-10-19T17:44:19Z<p>I- -i: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
A [[sequence]] is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the [[remainder]] when <math>\sum^{28}_{k=1} a_k </math> is divided by 1000.<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Define the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be:<br />
<center><math>s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br />
s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br />
s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br />
s = -s + a_{28} + a_{30}<br />
</math></center><br />
<br />
Thus <math>s = \frac{a_{28} + a_{30}}{2}</math>, and <math>a_{28},\,a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>\boxed{834}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br />
<br />
<math>a_{1}\equiv 1 \pmod {1000}<br />
\newline<br />
a_{2}\equiv 1 \pmod {1000}<br />
\newline<br />
a_{3}\equiv 1 \pmod {1000}<br />
\newline<br />
a_{4}\equiv 3 \pmod {1000}<br />
\newline<br />
a_{5}\equiv 5 \pmod {1000}<br />
\newline<br />
\cdots <br />
\newline<br />
a_{25} \equiv 793 \pmod {1000}<br />
\newline<br />
a_{26} \equiv 281 \pmod {1000}<br />
\newline<br />
a_{27} \equiv 233 \pmod {1000}<br />
\newline<br />
a_{28} \equiv 307 \pmod {1000}</math><br />
<br />
Adding all the residues shows the sum is congruent to <math>\boxed{834}</math> mod 1000.<br />
<br />
~ I-_-I<br />
<br />
=== Solution 3 (some guessing involved) ===<br />
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given <math>a_{28}, a_{29}, </math> and <math>a_{30}</math>, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some <math>p, q, r</math> such that <math>\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}</math>. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that <math>(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})</math>, at least for the first few terms. From this, we have that <math>\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)</math>.<br />
<br />
Solution by zeroman<br />
== See also ==<br />
{{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>I- -ihttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_11&diff=1104232006 AIME II Problems/Problem 112019-10-19T17:40:34Z<p>I- -i: /* Solution 2 */</p>
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<div>== Problem ==<br />
A [[sequence]] is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the [[remainder]] when <math>\sum^{28}_{k=1} a_k </math> is divided by 1000.<br />
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== Solution ==<br />
=== Solution 1 ===<br />
Define the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be:<br />
<center><math>s = a_{28} + \sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\<br />
s = a_{28} + \left(\sum^{30}_{k=4} a_{k} - \sum^{29}_{k=3} a_{k}\right) - \left(\sum^{28}_{k=2} a_{k}\right)\\<br />
s = a_{28} + (a_{30} - a_{3}) - \left(\sum^{28}_{k=2} a_{k}\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\<br />
s = -s + a_{28} + a_{30}<br />
</math></center><br />
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Thus <math>s = \frac{a_{28} + a_{30}}{2}</math>, and <math>a_{28},\,a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>\boxed{834}</math>.<br />
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=== Solution 2 ===<br />
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Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:<br />
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<math>a_{1}\equiv 1 \pmod 1000</math><br />
<math>a_{2}\equiv 1 \pmod 1000</math><br />
<math>a_{3}\equiv 1 \pmod 1000</math><br />
<math>a_{4}\equiv 3 \pmod 1000</math><br />
<math>a_{5}\equiv 5 \pmod 1000</math><br />
<math>\cdots</math><br />
<math>a_{25} \equiv 793 \pmod 1000</math><br />
<math>a_{26} \equiv 281 \pmod 1000</math><br />
<math>a_{27} \equiv 233 \pmod 1000</math><br />
<math>a_{28} \equiv 307 \pmod 1000</math><br />
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Adding all the residues shows the sum is congruent to <math>\boxed{834}</math> mod 1000.<br />
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~ I-_-I<br />
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=== Solution 3 (some guessing involved) ===<br />
All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given <math>a_{28}, a_{29}, </math> and <math>a_{30}</math>, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some <math>p, q, r</math> such that <math>\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}</math>. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that <math>(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})</math>, at least for the first few terms. From this, we have that <math>\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}}(\mod 1000)</math>.<br />
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Solution by zeroman<br />
== See also ==<br />
{{AIME box|year=2006|n=II|num-b=10|num-a=12}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>I- -i