https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Tomg418&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:26:58ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=1236602012 AIME II Problems/Problem 152020-06-04T17:45:51Z<p>Tomg418: Removed incorrect solution</p>
<hr />
<div>== Problem 15 ==<br />
Triangle <math>ABC</math> is inscribed in circle <math>\omega</math> with <math>AB=5</math>, <math>BC=7</math>, and <math>AC=3</math>. The bisector of angle <math>A</math> meets side <math>\overline{BC}</math> at <math>D</math> and circle <math>\omega</math> at a second point <math>E</math>. Let <math>\gamma</math> be the circle with diameter <math>\overline{DE}</math>. Circles <math>\omega</math> and <math>\gamma</math> meet at <math>E</math> and a second point <math>F</math>. Then <math>AF^2 = \frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
<br />
==Quick Solution using Olympiad Terms ==<br />
<br />
Take a force-overlaid inversion about <math>A</math> and note <math>D</math> and <math>E</math> map to each other. As <math>DE</math> was originally the diameter of <math>\gamma</math>, <math>DE</math> is still the diameter of <math>\gamma</math>. Thus <math>\gamma</math> is preserved. Note that the midpoint <math>M</math> of <math>BC</math> lies on <math>\gamma</math>, and <math>BC</math> and <math>\omega</math> are swapped. Thus points <math>F</math> and <math>M</math> map to each other, and are isogonal. It follows that <math>AF</math> is a symmedian of <math>\triangle{ABC}</math>, or that <math>ABFC</math> is harmonic. Then <math>(AB)(FC)=(BF)(CA)</math>, and thus we can let <math>BF=5x, CF=3x</math> for some <math>x</math>. By the LoC, it is easy to see <math>\angle{BAC}=120^\circ</math> so <math>(5x)^2+(3x)^2-2\cos{60^\circ}(5x)(3x)=49</math>. Solving gives <math>x^2=\frac{49}{19}</math>, from which by Ptolemy's we see <math>AF=\frac{30}{\sqrt{19}}</math>. We conclude the answer is <math>900+19=\boxed{919}</math>, as desired.<br />
<br />
'''- Emathmaster'''<br />
<br />
Quick Side Note: You might be wondering what the motivation for this solution is. Most of the people who've done EGMO Chapter 8 should recognize this as problem 8.32 (2009 Russian Olympiad) with the computational finish afterwards.<br />
Now if you haven't done this, but still know what inversion is, here's the motivation. We'd see that it's kinda hard to angle chase, and if we could, it would still be a bit hard to apply (you could use trig, but it won't be so clean most likely). If you give up after realizing that angle chasing won't work, you'd likely go in a similar approach to Solution 1 (below) or maybe be a bit more insightful and go with the elementary solution above.<br />
<br />
Finally, we notice there's circles! Classic setup for inversion! Since we're involving an angle-bisector, the first thing that comes to mind is a force overlaid inversion described in Lemma 8.16 of EGMO (where we invert with radius <math>\sqrt{AB \cdot AC}</math> and center <math>A</math>, then reflect over the <math>A</math>-angle bisector, which fixes <math>B, C</math>). We try applying this to the problem, and it's fruitful - we end up with this solution.<br />
-MSC<br />
<br />
== Solution 1==<br />
Use the angle bisector theorem to find <math>CD=\frac{21}{8}</math>, <math>BD=\frac{35}{8}</math>, and use Stewart's Theorem to find <math>AD=\frac{15}{8}</math>. Use Power of the Point to find <math>DE=\frac{49}{8}</math>, and so <math>AE=8</math>. Use law of cosines to find <math>\angle CAD = \frac{\pi} {3}</math>, hence <math>\angle BAD = \frac{\pi}{3}</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>.<br />
<br />
I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br />
<br />
<math>AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.</math> (1)<br />
<br />
<math>AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.</math> Adding these two and simplifying we get:<br />
<br />
<math>EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF</math> (2). Ah, but <math>\angle AFE = \angle ACE</math> (since <math>F</math> lies on <math>\omega</math>), and we can find <math>cos \angle ACE</math> using the law of cosines:<br />
<br />
<math>AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE</math>, and plugging in <math>AE = 8, AC = 3, BE = BC = 7,</math> we get <math>\cos \angle ACE = -1/7 = \cos \angle AFE</math>.<br />
<br />
Also, <math>\angle AEF = \angle DEF</math>, and <math>\angle DFE = \pi/2</math> (since <math>F</math> is on the circle <math>\gamma</math> with diameter <math>DE</math>), so <math>\cos \angle AEF = EF/DE = 8 \cdot EF/49</math>. <br />
<br />
Plugging in all our values into equation (2), we get:<br />
<br />
<math>EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}</math>, or <math>EF = \frac{7}{15} \cdot AF</math>.<br />
<br />
Finally, we plug this into equation (1), yielding:<br />
<br />
<math>8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}</math>. Thus,<br />
<br />
<math>64 = \frac{AF^2}{225} \cdot (225+49+30),</math> or <math>AF^2 = \frac{900}{19}.</math> The answer is <math>\boxed{919}</math>.<br />
<br />
== Solution 2==<br />
<br />
Let <math>a = BC</math>, <math>b = CA</math>, <math>c = AB</math> for convenience. We claim that <math>AF</math> is a symmedian. Indeed, let <math>M</math> be the midpoint of segment <math>BC</math>. Since <math>\angle EAB=\angle EAC</math>, it follows that <math>EB = EC</math> and consequently <math>EM\perp BC</math>. Therefore, <math>M\in \gamma</math>. Now let <math>G = FD\cap \omega</math>. Since <math>EG</math> is a diameter, <math>G</math> lies on the perpendicular bisector of <math>BC</math>; hence <math>E</math>, <math>M</math>, <math>G</math> are collinear. From <math>\angle DAG = \angle DMG = 90</math>, it immediately follows that quadrilateral <math>ADMG</math> is cyclic. Therefore, <math>\angle MAD = \angle MGD=\angle EAF</math>, implying that <math>AF</math> is a symmedian, as claimed.<br />
<br />
The rest is standard; here's a quick way to finish. From above, quadrilateral <math>ABFC</math> is harmonic, so <math>\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}</math>. In conjunction with <math>\triangle ABF\sim\triangle AMC</math>, it follows that <math>AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}</math>. (Notice that this holds for all triangles <math>ABC</math>.) To finish, substitute <math>a = 7</math>, <math>b=3</math>, <math>c=5</math> to obtain <math>AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}</math> as before.<br />
<br />
'''-Solution by thecmd999'''<br />
<br />
==Solution 3==<br />
<asy><br />
size(6cm);<br />
pair E,X,B,C,A,D,M,F,R,I;<br />
real z=sqrt(3)*14/3;<br />
real y=2*sqrt(3)/21;<br />
real x=224*sqrt(3)/57;<br />
E=(z,0);<br />
X=(0,0);<br />
D=(sqrt(3)*7/6,-7/8);<br />
M=(sqrt(3)*7/6,0);<br />
B=z/2*dir(60);<br />
C=z/2*dir(300);<br />
A=(y,-8/7);<br />
F=(x,-sqrt(3)*x/4);<br />
R=circumcenter(A,B,C);<br />
I=circumcenter(M,E,F);<br />
draw(E--X);<br />
draw(A--E);<br />
draw(A--B);<br />
draw(A--C);<br />
draw(B--C);<br />
draw(A--F);<br />
draw(X--F);<br />
draw(E--F);<br />
draw(circumcircle(A,B,C));<br />
draw(circumcircle(M,F,E));<br />
dot(D);<br />
dot(F);<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(E);<br />
dot(X);<br />
dot(R);<br />
dot(I);<br />
label("$A$",A,dir(220));<br />
label("$B$",B,dir(110));<br />
label("$C$",C,dir(250));<br />
label("$D$",D,dir(60));<br />
label("$E$",E,dir(0));<br />
label("$F$",F,dir(315));<br />
label("$X$",X,dir(180));<br />
</asy><br />
First of all, use the [[Angle Bisector Theorem]] to find that <math>BD=35/8</math> and <math>CD=21/8</math>, and use [[Stewart's Theorem]] to find that <math>AD=15/8</math>. Then use [[Power of a Point Theorem|Power of a Point]] to find that <math>DE=49/8</math>. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of <math>\triangle ABC</math> is <math>\frac{7\sqrt{3}}{3}</math>.<br />
<br />
Since <math>DE</math> is the diameter of circle <math>\gamma</math>, <math>\angle DFE</math> is <math>90^\circ</math>. Extending <math>DF</math> to intersect circle <math>\omega</math> at <math>X</math>, we find that <math>XE</math> is the diameter of the circumcircle of <math>\triangle ABC</math> (since <math>\angle DFE</math> is <math>90^\circ</math>). Therefore, <math>XE=\frac{14\sqrt{3}}{3}</math>.<br />
<br />
Let <math>EF=x</math>, <math>XD=a</math>, and <math>DF=b</math>. Then, by the [[Pythagorean Theorem]],<br />
<br />
<cmath>x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}</cmath><br />
<br />
and<br />
<br />
<cmath>x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.</cmath><br />
<br />
Subtracting the first equation from the second, the <math>x^2</math> term cancels out and we obtain:<br />
<br />
<cmath>(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}</cmath><br />
<br />
<cmath>a^2+2ab = \frac{5341}{192}.</cmath><br />
<br />
By Power of a Point, <math>ab=BD \cdot DC=735/64=2205/192</math>, so<br />
<br />
<cmath>a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}</cmath><br />
<br />
<cmath>a^2=\frac{931}{192}.</cmath><br />
<br />
Since <math>a=XD</math>, <math>XD=\frac{7\sqrt{19}}{8\sqrt{3}}</math>.<br />
<br />
Because <math>\angle EXF</math> and <math>\angle EAF</math> intercept the same arc in circle <math>\omega</math> and the same goes for <math>\angle XFA</math> and <math>\angle XEA</math>, <math>\angle EXF\cong\angle EAF</math> and <math>\angle XFA\cong\angle XEA</math>. Therefore, <math>\triangle XDE\sim\triangle ADF</math> by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br />
<br />
<cmath>\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}</cmath><br />
<br />
<cmath>\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}</cmath><br />
<br />
<cmath>AF \cdot \sqrt{19} = 30</cmath><br />
<br />
<cmath>AF = \frac{30}{\sqrt{19}}.</cmath><br />
<br />
However, the problem asks for <math>AF^2</math>, so <math>AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}</math>.<br />
<br />
'''-Solution by TheBoomBox77'''<br />
<br />
==Solution 4==<br />
It can be verified with law of cosines that <math>\angle BAC=120^\circ.</math> Also, <math>E</math> is the midpoint of major arc <math>BC</math> so <math>BE=CE,</math> and <math>\angle BEC=60.</math> Thus <math>CBE</math> is equilateral. Notice now that <math>\angle BFC=\angle BFE= 60.</math> But <math>\angle DFE=90</math> so <math>FD</math> bisects <math>\angle BFC.</math> Thus, <math>\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.</math> <br />
<br />
Let <math>BF=5a, CF=3a.</math> By law of cosines on <math>BFC</math> we find <math>a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.</math> But by ptolemy on <math>BFCA</math>, <math>15a+15a=7*AF,</math> so <math>AF= \frac{30}{\sqrt{19}},</math> so <math>AF^2=\frac{900}{19}</math> and the answer is <math>900+19=\boxed{919}</math><br />
<br />
~abacadaea<br />
<br />
== See Also ==<br />
{{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_15&diff=1188702019 AIME I Problems/Problem 152020-03-07T19:27:57Z<p>Tomg418: </p>
<hr />
<div>==Problem 15==<br />
<br />
Let <math>\overline{AB}</math> be a chord of a circle <math>\omega</math>, and let <math>P</math> be a point on the chord <math>\overline{AB}</math>. Circle <math>\omega_1</math> passes through <math>A</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circle <math>\omega_2</math> passes through <math>B</math> and <math>P</math> and is internally tangent to <math>\omega</math>. Circles <math>\omega_1</math> and <math>\omega_2</math> intersect at points <math>P</math> and <math>Q</math>. Line <math>PQ</math> intersects <math>\omega</math> at <math>X</math> and <math>Y</math>. Assume that <math>AP=5</math>, <math>PB=3</math>, <math>XY=11</math>, and <math>PQ^2 = \tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
<br />
<asy><br />
size(8cm);<br />
pair O, A, B, P, O1, O2, Q, X, Y;<br />
O=(0, 0);<br />
A=dir(140); B=dir(40);<br />
P=(3A+5B)/8;<br />
O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O);<br />
O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O);<br />
Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1];<br />
X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1));<br />
Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1));<br />
<br />
draw(circle(O, 1));<br />
draw(circle(O1, length(A-O1)));<br />
draw(circle(O2, length(B-O2)));<br />
draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2);<br />
<br />
dot("$O$", O, S);<br />
dot("$A$", A, A);<br />
dot("$B$", B, B);<br />
dot("$P$", P, dir(70));<br />
dot("$Q$", Q, dir(200));<br />
dot("$O_1$", O1, SW);<br />
dot("$O_2$", O2, SE);<br />
dot("$X$", X, X);<br />
dot("$Y$", Y, Y);<br />
</asy><br />
Let <math>O_1</math> and <math>O_2</math> be the centers of <math>\omega_1</math> and <math>\omega_2</math>, respectively. There is a homothety at <math>A</math> sending <math>\omega</math> to <math>\omega_1</math> that sends <math>B</math> to <math>P</math> and <math>O</math> to <math>O_1</math>, so <math>\overline{OO_2}\parallel\overline{O_1P}</math>. Similarly, <math>\overline{OO_1}\parallel\overline{O_2P}</math>, so <math>OO_1PO_2</math> is a parallelogram. Moreover, <cmath>\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,</cmath>whence <math>OO_1O_2Q</math> is cyclic. However, <cmath>OO_1=O_2P=O_2Q,</cmath>so <math>OO_1O_2Q</math> is an isosceles trapezoid. Since <math>\overline{O_1O_2}\perp\overline{XY}</math>, <math>\overline{OQ}\perp\overline{XY}</math>, so <math>Q</math> is the midpoint of <math>\overline{XY}</math>.<br />
<br />
By Power of a Point, <math>PX\cdot PY=PA\cdot PB=15</math>. Since <math>PX+PY=XY=11</math> and <math>XQ=11/2</math>, <cmath>XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,</cmath><br />
and the requested sum is <math>61+4=\boxed{065}</math>.<br />
<br />
(Solution by TheUltimate123)<br />
<br />
==Solution 2==<br />
<br />
Let the tangents to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at <math>R</math>. Then, since <math>RA^2=RB^2</math>, <math>R</math> lies on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, which is <math>\overline{PQ}</math>. It follows that <cmath>-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).</cmath><br />
Let <math>Q'</math> denote the midpoint of <math>\overline{XY}</math>. By the Midpoint of Harmonic Bundles Lemma, <cmath>RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,</cmath><br />
whence <math>Q=Q'</math>. Like above, <math>XP=\tfrac{11-\sqrt{61}}2</math>. Since <math>XQ=\tfrac{11}2</math>, we establish that <math>PQ=\tfrac{\sqrt{61}}2</math>, from which <math>PQ^2=\tfrac{61}4</math>, and the requested sum is <math>61+4=\boxed{065}</math>.<br />
<br />
(Solution by TheUltimate123)<br />
<br />
==Solution 3==<br />
<br />
Firstly we need to notice that <math>Q</math> is the middle point of <math>XY</math>. Assume the center of circle <math>w, w_1, w_2</math> are <math>O, O_1, O_2</math>, respectively. Then <math>A, O_2, O</math> are collinear and <math>O, O_1, B</math> are collinear. Link <math>O_1P, O_2P, O_1Q, O_2Q</math>. Notice that, <math>\angle B=\angle A=\angle APO_2=\angle BPO_1</math>. As a result, <math>PO_1\parallel O_2O</math> and <math>QO_1\parallel O_2P</math>. So we have parallelogram <math>PO_2O_1O</math>. So <math>\angle O_2PO_1=\angle O</math> Notice that, <math>O_1O_2\bot PQ</math> and <math>O_1O_2</math> divide <math>PQ</math> into two equal length pieces, So we have <math>\angle O_2PO_1=\angle O_2QO_1=\angle O</math>. As a result, <math>O_2, Q, O, O_1,</math> lie on one circle. So <math>\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P</math>. Notice that <math>\angle O_1PQ+\angle O_2O_1P=90^{\circ}</math>, we have <math>\angle OQP=90^{\circ}</math>. As a result, <math>OQ\bot PQ</math>. So <math>Q</math> is the middle point of <math>XY</math>.<br />
<br />
Back to our problem. Assume <math>XP=x</math>, <math>PY=y</math> and <math>x<y</math>. Then we have <math>AP\cdot PB=XP\cdot PY</math>, that is, <math>xy=15</math>. Also, <math>XP+PY=x+y=XY=11</math>. Solve these above, we have <math>x=\frac{11-\sqrt{61}}{2}=XP</math>. As a result, we have <math>PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}</math>. So, we have <math>PQ^2=\frac{61}{4}</math>. As a result, our answer is <math>m+n=61+4=\boxed{065}</math>.<br />
<br />
<br />
<br />
Solution By BladeRunnerAUG (Fanyuchen20020715).<br />
<br />
==Solution 4==<br />
Note that the tangents to the circles at <math>A</math> and <math>B</math> intersect at a point <math>Z</math> on <math>XY</math> by radical center. Then, since <math>\angle ZAB = \angle ZQA</math> and <math>\angle ZBA = \angle ZQB</math>, we have <br />
<cmath>\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},</cmath><br />
so <math>ZAQB</math> is cyclic. But if <math>O</math> is the center of <math>\omega</math>, clearly <math>ZAOB</math> is cyclic with diameter <math>ZO</math>, so <math>\angle ZQO = 90^{\circ} \implies Q</math> is the midpoint of <math>XY</math>. Then, by Power of a Point, <math>PY \cdot PX = PA \cdot PB = 15</math> and it is given that <math>PY+PX = 11</math>. Thus <math>PY, PX = \frac{11 \pm \sqrt{61}}{2}</math> so <math>PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}</math> and the answer is <math>61+4 = \boxed{065}</math>.<br />
<br />
==Solution 5 (Easy)==<br />
First we solve for <math>PX</math> with PoAP, <math>PX = \frac{11}{2} - \frac{\sqrt{61}}{2}</math>. Notice that <math>PQ^2</math> is rational but <math>PX^2</math> is not, also <math>PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}</math>. The most likely explanation for this is that <math>Q</math> is the midpoint of <math>XY</math>, so that <math>XQ = \frac{11}{2}</math> and <math>PQ=\frac{\sqrt{61}}{2}</math>. Then our answer is <math>m+n=61+4=\boxed{065}</math>.<br />
<br />
<br />
<br />
-Albert Einstein<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=14|after=Last Problem}}<br />
{{MAA Notice}}</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_11&diff=933022016 AIME I Problems/Problem 112018-03-17T18:15:00Z<p>Tomg418: </p>
<hr />
<div>==Problem==<br />
Let <math>P(x)</math> be a nonzero polynomial such that <math>(x-1)P(x+1)=(x+2)P(x)</math> for every real <math>x</math>, and <math>\left(P(2)\right)^2 = P(3)</math>. Then <math>P(\tfrac72)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
==Solution 1==<br />
Plug in <math>x=1</math> to get <math>(1-1)P(1+1) = 0 = (1+2)P(1) \Rightarrow P(1) = 0</math>. Plug in <math>x=0</math> to get <math>(0-1)P(0+1) = (0+2)P(0)\Rightarrow P(0) = -\frac{1}{2}P(1) = 0</math>. Plug in <math>x=-1</math> to get <math>(-1-1)P(-1+1) = (-1+2)P(-1)\Rightarrow (-2)P(0)=P(-1)\Rightarrow P(-1) = 0</math>. So <math>P(x) = x(x-1)(x+1)Q(x)</math> for some polynomial <math>Q(x)</math>. Using the initial equation, once again, <cmath>(x-1)P(x+1) = (x+2)P(x)</cmath><br />
<cmath>(x-1)((x+1)(x+1-1)(x+1+1)Q(x+1)) = (x+2)((x)(x-1)(x+1)Q(x))</cmath><br />
<cmath>(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)</cmath><br />
<cmath>Q(x+1) = Q(x)</cmath><br />
From here, we know that <math>Q(x) = C</math> for a constant <math>C</math> (<math>Q(x)</math> cannot be periodic since it is a polynomial), so <math>P(x) = Cx(x-1)(x+1)</math>. We know that <math>\left(P(2)\right)^2 = P(3)</math>. Plugging those into our definition of <math>P(x)</math>: <math>(C \cdot 2 \cdot (2-1) \cdot (2+1))^2 = C \cdot 3 \cdot (3-1) \cdot (3+1) \Rightarrow (6C)^2 = 24C \Rightarrow 36C^2 - 24C = 0 \Rightarrow C = 0</math> or <math>\frac{2}{3}</math>. So we know that <math>P(x) = \frac{2}{3}x(x-1)(x+1)</math>. So <math>P(\frac{7}{2}) = \frac{2}{3} \cdot \frac{7}{2} \cdot (\frac{7}{2} - 1) \cdot (\frac{7}{2} + 1) = \frac{105}{4}</math>. Thus, the answer is <math>105 + 4 = \boxed{109}</math>.<br />
<br />
==Solution 2==<br />
From the equation we see that <math>x-1</math> divides <math>P(x)</math> and <math>(x+2)</math> divides <math>P(x+1)</math> so we can conclude that <math>x-1</math> and <math>x+1</math> divide <math>P(x)</math> (if we shift the function right by 1, we get <math>(x-2)P(x) = (x+1)P(x-1)</math>, and from here we can see that <math>x+1</math> divides <math>P(x)</math>). This means that <math>1</math> and <math>-1</math> are roots of <math>P(x)</math>. Plug in <math>x = 0</math> and we see that <math>P(0) = 0</math> so <math>0</math> is also a root. <br />
<br />
Suppose we had another root that is not one of those <math>3</math>. Notice that the equation above indicates that if <math>r</math> is a root then <math>r+1</math> and <math>r-1</math> is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three. <br />
<br />
That means <math>P(x) = cx(x-1)(x+1)</math>. We can use <math>P(2)^2 = P(3)</math> to get <math>c = \frac{2}{3}</math>. Plugging in <math>\frac{7}{2}</math> is now trivial and we see that it is <math>\frac{105}{4}</math> so our answer is <math>\boxed{109}</math><br />
<br />
==Solution 3==<br />
Although this may not be the most mathematically rigorous answer, we see that <math>\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}</math>. Using a bit of logic, we can make a guess that <math>P(x+1)</math> has a factor of <math>x+2</math>, telling us <math>P(x)</math> has a factor of <math>x+1</math>. Similarly, we guess that <math>P(x)</math> has a factor of <math>x-1</math>, which means <math>P(x+1)</math> has a factor of <math>x</math>. Now, since <math>P(x)</math> and <math>P(x+1)</math> have so many factors that are off by one, we may surmise that when you plug <math>x+1</math> into <math>P(x)</math>, the factors "shift over," i.e. <math>P(x)=(A)(A+1)(A+2)...(A+n)</math>, which goes to <math>P(x+1)=(A+1)(A+2)(A+3)...(A+n+1)</math>. This is useful because these, when divided, result in <math>\frac{P(x+1)}{P(x)}=\frac{A+n+1}{A}</math>. If <math>\frac{A+n+1}{A}=\frac{x+2}{x-1}</math>, then we get <math>A=x-1</math> and <math>A+n+1=x+2</math>, <math>n=2</math>. This gives us <math>P(x)=(x-1)x(x+1)</math> and <math>P(x+1)=x(x+1)(x+2)</math>, and at this point we realize that there has to be some constant <math>a</math> multiplied in front of the factors, which won't affect our fraction <math>\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}</math> but will give us the correct values of <math>P(2)</math> and <math>P(3)</math>. Thus <math>P(x)=a(x-1)x(x+1)</math>, and we utilize <math>P(2)^2=P(3)</math> to find <math>a=\frac{2}{3}</math>. Evaluating <math>P \left ( \frac{7}{2} \right )</math> is then easy, and we see it equals <math>\frac{105}{4}</math>, so the answer is <math>\boxed{109}</math><br />
<br />
==Solution 4==<br />
Substituting <math>x=2</math> into the given equation, we find that <math>P(3)=4P(2)=P(2)^2</math>. Therefore, either <math>P(2)=0</math> or <math>P(2)=4</math>. Now for integers <math>n\ge 2</math>, we know that<br />
<cmath>P(n+1)=\frac{n+2}{n-1}P(n).</cmath><br />
Applying this repeatedly, we find that<br />
<cmath>P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).</cmath><br />
If <math>P(2)=0</math>, this shows that <math>P(x)</math> has infinitely many roots, meaning that <math>P(x)</math> is identically equal to zero. But this contradicts the problem statement. Therefore, <math>P(2)=4</math>, and we find <math>P(n+1)=\frac{2}{3}(n+2)(n+1)n</math> for all positive integers <math>n\ge2</math>. This cubic polynomial matches the values <math>P(n+1)</math> for infinitely many numbers, hence the two polynomials are identically equal. In particular, <math>P\left(\frac72\right)=\frac23\cdot\frac92\cdot\frac72\cdot\frac52=\frac{105}{4}</math>, and the answer is <math>\boxed{109}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2016|n=I|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=AIME_Problems_and_Solutions&diff=92710AIME Problems and Solutions2018-03-04T22:30:18Z<p>Tomg418: </p>
<hr />
<div>This is a list of all [[AIME]] exams in the AoPSWiki. Some of them contain complete questions and solutions, others complete questions, and others are lacking both questions and solutions. Many of these problems and solutions are also available in the [http://www.artofproblemsolving.com/Forum/resources.php?c=182 AoPS Resources] section. If you find problems that are in the Resources section which are not in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Finally, additions to and improvements on the solutions in the AoPSWiki are always welcome.<br />
<br />
{| class="wikitable" style="text-align:center"<br />
|-<br />
!Year<br />
!Test I<br />
!Test II<br />
<br />
|-<br />
| 2018 || [[2018 AIME I | AIME I]] || [[2018 AIME II | AIME II]]<br />
|-<br />
| 2017 || [[2017 AIME I | AIME I]] || [[2017 AIME II | AIME II]]<br />
|-<br />
| 2016 || [[2016 AIME I | AIME I]] || [[2016 AIME II | AIME II]]<br />
|-<br />
| 2015 || [[2015 AIME I | AIME I]] || [[2015 AIME II | AIME II]]<br />
|-<br />
| 2014 || [[2014 AIME I | AIME I]] || [[2014 AIME II | AIME II]]<br />
|-<br />
| 2013 || [[2013 AIME I | AIME I]] || [[2013 AIME II | AIME II]]<br />
|-<br />
| 2012 || [[2012 AIME I | AIME I]] || [[2012 AIME II | AIME II]]<br />
|-<br />
| 2011 || [[2011 AIME I | AIME I]] || [[2011 AIME II | AIME II]]<br />
|-<br />
| 2010 || [[2010 AIME I | AIME I]] || [[2010 AIME II | AIME II]]<br />
|-<br />
| 2009 || [[2009 AIME I | AIME I]] || [[2009 AIME II | AIME II]]<br />
|-<br />
| 2008 || [[2008 AIME I | AIME I]] || [[2008 AIME II | AIME II]]<br />
|-<br />
| 2007 || [[2007 AIME I | AIME I]] || [[2007 AIME II | AIME II]]<br />
|-<br />
| 2006 || [[2006 AIME I | AIME I]] || [[2006 AIME II | AIME II]]<br />
|-<br />
| 2005 || [[2005 AIME I | AIME I]] || [[2005 AIME II | AIME II]]<br />
|-<br />
| 2004 || [[2004 AIME I | AIME I]] || [[2004 AIME II | AIME II]]<br />
|-<br />
| 2003 || [[2003 AIME I | AIME I]] || [[2003 AIME II | AIME II]]<br />
|-<br />
| 2002 || [[2002 AIME I | AIME I]] || [[2002 AIME II | AIME II]]<br />
|-<br />
| 2001 || [[2001 AIME I | AIME I]] || [[2001 AIME II | AIME II]]<br />
|-<br />
| 2000 || [[2000 AIME I | AIME I]] || [[2000 AIME II | AIME II]]<br />
|-<br />
| 1999<br />
| colspan="2"| [[1999 AIME | AIME]]<br />
|-<br />
| 1998<br />
| colspan="2"| [[1998 AIME | AIME]]<br />
|-<br />
| 1997<br />
| colspan="2"| [[1997 AIME | AIME]] <br />
|-<br />
| 1996<br />
| colspan="2"| [[1996 AIME | AIME]]<br />
|-<br />
| 1995<br />
| colspan="2"| [[1995 AIME | AIME]]<br />
|-<br />
| 1994<br />
| colspan="2"| [[1994 AIME | AIME]]<br />
|-<br />
| 1993<br />
| colspan="2"| [[1993 AIME | AIME]]<br />
|-<br />
| 1992<br />
| colspan="2"| [[1992 AIME | AIME]]<br />
|-<br />
| 1991<br />
| colspan="2"| [[1991 AIME | AIME]]<br />
|-<br />
| 1990<br />
| colspan="2"| [[1990 AIME | AIME]]<br />
|-<br />
| 1989<br />
| colspan="2"| [[1989 AIME | AIME]]<br />
|-<br />
| 1988 <br />
| colspan="2"| [[1988 AIME | AIME]]<br />
|-<br />
| 1987<br />
| colspan="2"| [[1987 AIME | AIME]]<br />
|-<br />
| 1986<br />
| colspan="2"| [[1986 AIME | AIME]]<br />
|-<br />
| 1985<br />
| colspan="2"| [[1985 AIME | AIME]] <br />
|-<br />
| 1984 <br />
| colspan="2"| [[1984 AIME | AIME]] <br />
|-<br />
| 1983<br />
| colspan="2"| [[1983 AIME | AIME]]<br />
|}<br />
<br />
== Resources ==<br />
* [[American Mathematics Competitions]]<br />
* [[AMC Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
<br />
<br />
[[Category:Math Contest Problems]]</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=2030_AIME_I&diff=927082030 AIME I2018-03-04T22:28:10Z<p>Tomg418: </p>
<hr />
<div><br />
Work in progress<br />
<br />
==Problem 1==<br />
<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?<br />
<br />
<asy><br />
size(170);<br />
defaultpen(linewidth(0.9)+fontsize(13pt));<br />
draw(unitcircle^^circle((0,1.5),0.5));<br />
path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle;<br />
for(int i=1;i<=12;i=i+1)<br />
{<br />
draw(0.9*dir(90-30*i)--dir(90-30*i));<br />
label("$"+(string) i+"$",0.78*dir(90-30*i));<br />
}<br />
dot(origin);<br />
draw(shift((0,1.87))*arrow);<br />
draw(arc(origin,1.5,68,30),EndArrow(size=12));<br />
</asy><br />
<br />
<math> \textbf{(A) }\mathrm{2 o'clock} \qquad\textbf{(B) }\mathrm{3 o'clock} \qquad\textbf{(C) }\mathrm{4 o'clock} \qquad\textbf{(D) }\mathrm{6 o'clock} \qquad\textbf{(E) }\mathrm{8 o'clock} </math><br />
<br />
[[2015 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
===Problem 2===<br />
<br />
Prove that for any positive integer <math>k,</math><br />
<cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath><br />
is an integer.<br />
<br />
===Problem 3===<br />
(<math>*</math>) Let <math>ABC</math> be a scalene triangle with circumcircle <math>\Omega</math> and incenter <math>I</math>. Ray <math>AI</math> meets <math>\overline{BC}</math> at <math>D</math> and meets <math>\Omega</math> again at <math>M</math>; the circle with diameter <math>\overline{DM}</math> cuts <math>\Omega</math> again at <math>K</math>. Lines <math>MK</math> and <math>BC</math> meet at <math>S</math>, and <math>N</math> is the midpoint of <math>\overline{IS}</math>. The circumcircles of <math>\triangle KID</math> and <math>\triangle MAN</math> intersect at points <math>L_1</math> and <math>L_2</math>. Prove that <math>\Omega</math> passes through the midpoint of either <math>\overline{IL_1}</math> or <math>\overline{IL_2}</math>.<br />
<br />
[[2017 USAMO Problems/Problem 3|Solution]]<br />
<br />
===Problem 4===<br />
Find the minimum possible value of <cmath>\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}</cmath>given that <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> are nonnegative real numbers such that <math>a+b+c+d=4</math>.<br />
<br />
[[2017 USAMO Problems/Problem 4|Solution]]</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=2030_AIME_I&diff=927072030 AIME I2018-03-04T22:27:02Z<p>Tomg418: Created page with " Work in progress ==Problem 1== The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externa..."</p>
<hr />
<div><br />
Work in progress<br />
<br />
==Problem 1==<br />
The diagram below shows the circular face of a clock with radius <math>20</math> cm and a circular disk with radius <math>10</math> cm externally tangent to the clock face at <math>12</math> o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction? [asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("<math>"+(string) i+"</math>",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]<br />
<br />
<math>\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}</math> <br />
<br />
===Problem 2===<br />
<br />
Prove that for any positive integer <math>k,</math><br />
<cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath><br />
is an integer.<br />
<br />
===Problem 3===<br />
(<math>*</math>) Let <math>ABC</math> be a scalene triangle with circumcircle <math>\Omega</math> and incenter <math>I</math>. Ray <math>AI</math> meets <math>\overline{BC}</math> at <math>D</math> and meets <math>\Omega</math> again at <math>M</math>; the circle with diameter <math>\overline{DM}</math> cuts <math>\Omega</math> again at <math>K</math>. Lines <math>MK</math> and <math>BC</math> meet at <math>S</math>, and <math>N</math> is the midpoint of <math>\overline{IS}</math>. The circumcircles of <math>\triangle KID</math> and <math>\triangle MAN</math> intersect at points <math>L_1</math> and <math>L_2</math>. Prove that <math>\Omega</math> passes through the midpoint of either <math>\overline{IL_1}</math> or <math>\overline{IL_2}</math>.<br />
<br />
[[2017 USAMO Problems/Problem 3|Solution]]<br />
<br />
===Problem 4===<br />
Find the minimum possible value of <cmath>\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}</cmath>given that <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> are nonnegative real numbers such that <math>a+b+c+d=4</math>.<br />
<br />
[[2017 USAMO Problems/Problem 4|Solution]]</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=2017_USAMO_Problems/Problem_6&diff=892092017 USAMO Problems/Problem 62017-12-27T00:57:29Z<p>Tomg418: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Find the minimum possible value of <br />
<br />
<cmath>\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4},</cmath><br />
<br />
given that <math>a,b,c,d,</math> are nonnegative real numbers such that <math>a+b+c+d=4</math>.<br />
<br />
==Solution==<br />
See here: https://artofproblemsolving.com/community/c5t211539f5h1434574_looks_like_mount_inequality_erupted_<br />
<br />
or:<br />
<br />
https://www.youtube.com/watch?v=LSYP_KMbBNc</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=2008_USAMO_Problems/Problem_2&diff=892002008 USAMO Problems/Problem 22017-12-26T20:27:22Z<p>Tomg418: /* Solution 1 (synthetic) */</p>
<hr />
<div>== Problem ==<br />
(''Zuming Feng'') Let <math>ABC</math> be an acute, [[scalene]] triangle, and let <math>M</math>, <math>N</math>, and <math>P</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively. Let the [[perpendicular]] [[bisect]]ors of <math>\overline{AB}</math> and <math>\overline{AC}</math> intersect ray <math>AM</math> in points <math>D</math> and <math>E</math> respectively, and let lines <math>BD</math> and <math>CE</math> intersect in point <math>F</math>, inside of triangle <math>ABC</math>. Prove that points <math>A</math>, <math>N</math>, <math>F</math>, and <math>P</math> all lie on one circle.<br />
<br />
== Solutions ==<br />
<br />
=== Solution 1 (synthetic) ===<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(0,1),s)));<br />
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));<br />
D(C--D(MP("F",F,NW,s))); <br />
D(B--O--C,linetype("4 4")+linewidth(0.7));<br />
D(M--N,linetype("4 4")+linewidth(0.7));<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));<br />
D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
<br />
/* D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); */<br />
</asy></center><br />
<br />
[[Without loss of generality]] <math>AB < AC</math>. The intersection of <math>NE</math> and <math>PD</math> is <math>O</math>, the circumcenter of <math>\triangle ABC</math>. <br />
<br />
Let <math>\angle BAM = y</math> and <math>\angle CAM = z</math>. Note <math>D</math> lies on the perpendicular bisector of <math>AB</math>, so <math>AD = BD</math>. So <math>\angle FBC = \angle B - \angle ABD = B - y</math>. Similarly, <math>\angle FCB = C - z</math>, so <math>\angle BFC = 180 - (B + C) + (y + z) = 2A</math>. Notice that <math>\angle BOC</math> intercepts the minor arc <math>BC</math> in the [[circumcircle]] of <math>\triangle ABC</math>, which is double <math>\angle A</math>. Hence <math>\angle BFC = \angle BOC</math>, so <math>BFOC</math> is cyclic. <br />
<br />
'''Lemma.''' <math>\triangle FEO</math> is directly similar to <math>\triangle NEM</math><br />
<br />
''Proof.''<br />
<cmath>\angle OFE = \angle OFC = \angle OBC = \frac {1}{2}\cdot (180 - 2A) = 90 - A</cmath><br />
since <math>F</math>, <math>E</math>, <math>C</math> are collinear, <math>BFOC</math> is cyclic, and <math>OB = OC</math>. Also<br />
<cmath>\angle ENM = 90 - \angle MNC = 90 - A</cmath><br />
because <math>NE\perp AC</math>, and <math>MNP</math> is the medial triangle of <math>\triangle ABC</math> so <math>AB \parallel MN</math>. Hence <math>\angle OFE = \angle ENM</math>.<br />
<br />
Notice that <math>\angle AEN = 90 - z = \angle CEN</math> since <math>NE\perp AC</math>. <math>\angle FED = \angle MEC = 2z</math>. Then<br />
<cmath>\angle FEO = \angle FED + \angle AEN = \angle CEM + \angle CEN = \angle NEM</cmath><br />
Hence <math>\angle FEO = \angle NEM</math>. <br />
<br />
Hence <math>\triangle FEO</math> is similar to <math>\triangle NEM</math> by AA similarity. It is easy to see that they are oriented such that they are directly similar.<br />
<br />
'''End Lemma'''<br />
<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(1,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(B--D(MP("D",D,NE,s))--MP("P",P,(-1,0),s)--D(MP("O",O,(1,0),s)));<br />
D(D(MP("E",E,SW,s))--MP("N",N,(1,0),s));<br />
D(C--D(MP("F",F,NW,s))); <br />
D(B--O--C,linetype("4 4")+linewidth(0.7));<br />
D(F--N); D(O--M);<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
<br />
/* commented in above asy<br />
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7)); <br />
D(anglemark(B,A,C)); MP("y",A,(0,-6));MP("z",A,(4,-6));<br />
D(anglemark(B,F,C,4),linewidth(0.6));D(anglemark(B,O,C,4),linewidth(0.6));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
*/<br />
</asy></center><br />
<br />
By the similarity in the Lemma, <math>FE: EO = NE: EM\implies FE: EN = OE: EM</math>. <math>\angle FEN = \angle OEM</math> so <math>\triangle FEN\sim\triangle OEM</math> by SAS similarity. Hence<br />
<cmath>\angle EMO = \angle ENF = \angle ONF</cmath><br />
Using essentially the same angle chasing, we can show that <math>\triangle PDM</math> is directly similar to <math>\triangle FDO</math>. It follows that <math>\triangle PDF</math> is directly similar to <math>\triangle MDO</math>. So<br />
<cmath>\angle EMO = \angle DMO = \angle DPF = \angle OPF</cmath><br />
Hence <math>\angle OPF = \angle ONF</math>, so <math>FONP</math> is cyclic. In other words, <math>F</math> lies on the circumcircle of <math>\triangle PON</math>. Note that <math>\angle ONA = \angle OPA = 90</math>, so <math>APON</math> is cyclic. In other words, <math>A</math> lies on the circumcircle of <math>\triangle PON</math>. <math>A</math>, <math>P</math>, <math>N</math>, <math>O</math>, and <math>F</math> all lie on the circumcircle of <math>\triangle PON</math>. Hence <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle, as desired.<br />
<br />
=== Solution 2 (synthetic) ===<br />
Without Loss of Generality, assume <math>AB >AC</math>. It is sufficient to prove that <math>\angle OFA = 90^{\circ}</math>, as this would immediately prove that <math>A,P,O,F,N</math> are concyclic.<br />
By applying the Menelaus' Theorem in the Triangle <math>\triangle BFC</math> for the transversal <math>E,M,D</math>, we have (in magnitude)<br />
<cmath> \frac{FE}{EC} \cdot \frac{CM}{MB} \cdot \frac{BD}{DF} = 1 \iff \frac{FE}{EC} = \frac{DF}{BD} <br />
</cmath> <br />
Here, we used that <math>BM=MC</math>, as <math>M</math> is the midpoint of <math>BC</math>. Now, since <math>EC =EA</math> and <math>BD=DA</math>, we have<br />
<cmath> \frac{FE}{EA} = \frac{DF}{DA} \iff \frac{DA}{AE} = \frac{DF}{FE} \iff AF \text{ bisects exterior } \angle EFD <br />
</cmath><br />
Now, note that <math>OE</math> bisects the exterior <math>\angle FED</math> and <math>OD</math> bisects exterior <math>\angle FDE</math>, making <math>O</math> the <math>F</math>-excentre of <math>\triangle FED</math>. This implies that <math>OF</math> bisects interior <math>\angle EFD</math>, making <math>OF \perp AF</math>, as was required.<br />
<br />
=== Solution 3 (synthetic) ===<br />
Hint: consider <math>CF</math> intersection with <math>PM</math>; show that the resulting intersection lies on the desired circle.<br />
<br />
{{stub}}<br />
<br />
=== Solution 4 (synthetic) ===<br />
This solution utilizes the ''phantom point method.'' Clearly, APON are cyclic because <math>\angle OPA = \angle ONA = 90</math>. Let the circumcircles of triangles <math>APN</math> and <math>BOC</math> intersect at <math>F'</math> and <math>O</math>. <br />
<br />
'''Lemma.''' If <math>A,B,C</math> are points on circle <math>\omega</math> with center <math>O</math>, and the tangents to <math>\omega</math> at <math>B,C</math> intersect at <math>Q</math>, then <math>AP</math> is the symmedian from <math>A</math> to <math>BC</math>.<br />
<br />
''Proof.'' This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon.<br />
<br />
'''End Lemma'''<br />
<br />
It is easy to see <math>Q</math> (the intersection of ray <math>OM</math> and the circumcircle of <math>\triangle BOC</math>) is colinear with <math>A</math> and <math>F'</math>, and because line <math>OM</math> is the diameter of that circle, <math>\angle QBO = \angle QCO = 90</math>, so <math>Q</math> is the point <math>Q</math> in the lemma; hence, we may apply the lemma. From here, it is simple angle-chasing to show that <math>F'</math> satisfies the original construction for <math>F</math>, showing <math>F=F'</math>; we are done.<br />
<br />
{{stub}}<br />
<br />
=== Solution 5 (trigonometric) ===<br />
By the [[Law of Sines]], <math>\frac {\sin\angle BAM}{\sin\angle CAM} = \frac {\sin B}{\sin C} = \frac bc = \frac {b/AF}{c/AF} = \frac {\sin\angle AFC\cdot\sin\angle ABF}{\sin\angle ACF\cdot\sin\angle AFB}</math>. Since <math>\angle ABF = \angle ABD = \angle BAD = \angle BAM</math> and similarly <math>\angle ACF = \angle CAM</math>, we cancel to get <math>\sin\angle AFC = \sin\angle AFB</math>. Obviously, <math>\angle AFB + \angle AFC > 180^\circ</math> so <math>\angle AFC = \angle AFB</math>.<br />
<br />
Then <math>\angle FAB + \angle ABF = 180^\circ - \angle AFB = 180^\circ - \angle AFC = \angle FAC + \angle ACF</math> and <math>\angle ABF + \angle ACF = \angle A = \angle FAB + \angle FAC</math>. Subtracting these two equations, <math>\angle FAB - \angle FCA = \angle FCA - \angle FAB</math> so <math>\angle BAF = \angle ACF</math>. Therefore, <math>\triangle ABF\sim\triangle CAF</math> (by AA similarity), so a spiral similarity centered at <math>F</math> takes <math>B</math> to <math>A</math> and <math>A</math> to <math>C</math>. Therefore, it takes the midpoint of <math>\overline{BA}</math> to the midpoint of <math>\overline{AC}</math>, or <math>P</math> to <math>N</math>. So <math>\angle APF = \angle CNF = 180^\circ - \angle ANF</math> and <math>APFN</math> is cyclic.<br />
<br />
=== Solution 6 (isogonal conjugates) ===<br />
<center><asy><br />
/* setup and variables */<br />
size(280);<br />
pathpen = black + linewidth(0.7); pointpen = black; pen s = fontsize(8);<br />
pair B=(0,0),C=(5,0),A=(4,4); /* A.x > C.x/2 */<br />
/* construction and drawing */<br />
pair P=(A+B)/2,M=(B+C)/2,N=(A+C)/2,D=IP(A--M,P--P+5*(P-bisectorpoint(A,B))),E=IP(A--M,N--N+5*(bisectorpoint(A,C)-N)),F=IP(B--B+5*(D-B),C--C+5*(E-C)),O=circumcenter(A,B,C);<br />
D(MP("A",A,(0,1),s)--MP("B",B,SW,s)--MP("C",C,SE,s)--A--MP("M",M,s));<br />
D(C--D(MP("E",E,NW,s))--MP("N",N,(1,0),s)--D(MP("O",O,SW,s)));<br />
D(D(MP("D",D,SE,s))--MP("P",P,W,s));<br />
D(B--D(MP("F",F,s))); D(O--A--F,linetype("4 4")+linewidth(0.7));<br />
D(MP("O'",circumcenter(A,P,N),NW,s));<br />
D(circumcircle(A,P,N),linetype("4 4")+linewidth(0.7));<br />
D(rightanglemark(A,P,D,3.5));D(rightanglemark(A,N,E,3.5));<br />
picture p = new picture;<br />
draw(p,circumcircle(B,O,C),linetype("1 4")+linewidth(0.7));<br />
draw(p,circumcircle(A,B,C),linetype("1 4")+linewidth(0.7));<br />
clip(p,B+(-5,0)--B+(-5,A.y+2)--C+(5,A.y+2)--C+(5,0)--cycle); add(p);<br />
</asy></center><br />
<br />
Construct <math>T</math> on <math>AM</math> such that <math>\angle BCT = \angle ACF</math>. Then <math>\angle BCT = \angle CAM</math>. Then <math>\triangle AMC\sim\triangle CMT</math>, so <math>\frac {AM}{CM} = \frac {CM}{TM}</math>, or <math>\frac {AM}{BM} = \frac {BM}{TM}</math>. Then <math>\triangle AMB\sim\triangle BMT</math>, so <math>\angle CBT = \angle BAM = \angle FBA</math>. Then we have<br />
<br />
<math>\angle CBT = \angle ABF</math> and <math>\angle BCT = \angle ACF</math>. So <math>T</math> and <math>F</math> are [[isogonal conjugate|isogonally conjugate]]. Thus <math>\angle BAF = \angle CAM</math>. Then<br />
<br />
<math>\angle AFB = 180 - \angle ABF - \angle BAF = 180 - \angle BAM - \angle CAM = 180 - \angle BAC</math>.<br />
<br />
If <math>O</math> is the [[circumcenter]] of <math>\triangle ABC</math> then <math>\angle BFC = 2\angle BAC = \angle BOC</math> so <math>BFOC</math> is cyclic. Then <math>\angle BFO = 180 - \angle BOC = 180 - (90 - \angle BAC) = 90 + \angle BAC</math>.<br />
<br />
Then <math>\angle AFO = 360 - \angle AFB - \angle BFO = 360 - (180 - \angle BAC) - (90 + \angle BAC) = 90</math>. Then <math>\triangle AFO</math> is a right triangle.<br />
<br />
Now by the [[homothety]] centered at <math>A</math> with ratio <math>\frac {1}{2}</math>, <math>B</math> is taken to <math>P</math> and <math>C</math> is taken to <math>N</math>. Thus <math>O</math> is taken to the circumcenter of <math>\triangle APN</math> and is the midpoint of <math>AO</math>, which is also the circumcenter of <math>\triangle AFO</math>, so <math>A,P,N,F,O</math> all lie on a circle.<br />
<br />
=== Solution 7 (symmedians) ===<br />
Median <math>AM</math> of a triangle <math>ABC</math> implies <math>\frac {\sin{BAM}}{\sin{CAM}} = \frac {\sin{B}}{\sin{C}}</math>.<br />
Trig ceva for <math>F</math> shows that <math>AF</math> is a symmedian.<br />
Then <math>FP</math> is a median, use the lemma again to show that <math>AFP = C</math>, and similarly <math>AFN = B</math>, so you're done.<br />
<br />
{{stub}}<br />
<br />
=== Solution 8 (inversion) ===<br />
Invert the figure about a circle centered at <math>A</math>, and let <math>X'</math> denote the image of the point <math>X</math> under this inversion. Find point <math>F_1'</math> so that <math>AB'F_1'C'</math> is a parallelogram and let <math>Z'</math> denote the center of this parallelogram. Note that <math>\triangle BAC\sim\triangle C'AB'</math> and <math>\triangle BAD\sim\triangle D'AB'</math>. Because <math>M</math> is the midpoint of <math>BC</math> and <math>Z'</math> is the midpoint of <math>B'C'</math>, we also have <math>\triangle BAM\sim\triangle C'AZ'</math>. Thus<br />
<cmath>\angle AF_1'B' = \angle F_1'AC' = \angle Z'AC' = \angle MAB = \angle DAB = \angle DBA = \angle AD'B'.</cmath><br />
Hence quadrilateral <math>AB'D'F_1'</math> is cyclic and, by a similar argument, quadrilateral <math>AC'E'F_1'</math> is also cyclic. Because the images under the inversion of lines <math>BDF</math> and <math>CFE</math> are circles that intersect in <math>A</math> and <math>F'</math>, it follows that <math>F_1' = F'</math>.<br />
<br />
Next note that <math>B'</math>, <math>Z'</math>, and <math>C'</math> are collinear and are the images of <math>P'</math>, <math>F'</math>, and <math>N'</math>, respectively, under a homothety centered at <math>A</math> and with ratio <math>1/2</math>. It follows that <math>P'</math>, <math>F'</math>, and <math>N'</math> are collinear, and then that the points <math>A</math>, <math>P</math>, <math>F</math>, and <math>N</math> lie on a circle.<br />
<br />
<center>[[File:2008usamo2-sol8.png]]</center><br />
<br />
=== Solution 9 ===<br />
Let <math>O</math> be the circumcenter of triangle <math>ABC</math>. We prove that<br />
<cmath>\angle APO = \angle ANO = \angle AFO = 90^\circ.\qquad\qquad (1)</cmath><br />
It will then follow that <math>A, P, O, F, N</math> lie on the circle with diameter <math>\overline{AO}</math>. Indeed, the fact that the first two angles in <math>(1)</math> are right is immediate because <math>\overline{OP}</math> and <math>\overline{ON}</math> are the perpendicular bisectors of <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively. Thus we need only prove that <math>\angle AFO = 90^\circ</math>.<br />
<br />
<center>[[File:2008usamo2-sol9.png]]</center><br />
<br />
We may assume, without loss of generality, that <math>AB > AC</math>. This leads to configurations similar to the ones shown above. The proof can be adapted to other configurations. Because <math>\overline{PO}</math> is the perpendicular bisector of <math>\overline{AB}</math>, it follows that triangle <math>ADB</math> is an isosceles triangle with <math>AD = BD</math>. Likewise, triangle <math>AEC</math> is isosceles with <math>AE = CE</math>. Let <math>x = \angle ABD = \angle BAD</math> and <math>y = \angle CAE = \angle ACE</math>, so <math>x + y = \angle BAC</math>.<br />
<br />
Applying the Law of Sines to triangles <math>ABM</math> and <math>ACM</math> gives<br />
<cmath>\frac{BM}{\sin x} = \frac{AB}{\sin\angle BMA}\quad\text{and}\quad\frac{CM}{\sin y} = \frac{AC}{\sin\angle CMA}.</cmath><br />
Taking the quotient of the two equations and noting that <math>\sin\angle BMA = \sin\angle CMA</math>, we find<br />
<cmath>\frac{BM}{CM}\frac{\sin y}{\sin x} = \frac{AB}{AC}\frac{\sin\angle CMA}{\sin\angle BMA} = \frac{AB}{AC}.</cmath><br />
Because <math>BM = MC</math>, we have<br />
<cmath>\frac{\sin x}{\sin y} = \frac{AC}{AB}.\qquad\qquad (2)</cmath><br />
Applying the Law of Sines to triangles <math>ABF</math> and <math>ACF</math>, we find<br />
<cmath>\frac{AF}{\sin x} = \frac{AB}{\sin\angle AFB}\quad\text{and}\quad\frac{AF}{\sin y} = \frac{AC}{\sin\angle AFC}.</cmath><br />
Taking the quotient of the two equations yields<br />
<cmath>\frac{\sin x}{\sin y} = \frac{AC}{AB}\frac{\sin\angle AFB}{\sin\angle AFC},</cmath><br />
so by <math>(2)</math>,<br />
<cmath>\sin\angle AFB = \sin\angle AFC.\qquad\qquad (3)</cmath><br />
Because <math>\angle ADF</math> is an exterior angle to triangle <math>ADB</math>, we have <math>\angle EDF = 2x</math>. Similarly, <math>\angle DEF = 2y</math>. Hence<br />
<cmath>\angle EFD = 180^\circ - 2x - 2y = 180^\circ - 2\angle BAC.</cmath><br />
Thus <math>\angle BFC = 2\angle BAC = \angle BOC</math>, so <math>BOFC</math> is cyclic. In addition,<br />
<cmath>\angle AFB + \angle AFC = 360^\circ - 2\angle BAC > 180^\circ,</cmath><br />
and hence, from <math>(3)</math>, <math>\angle AFB = \angle AFC = 180^\circ - \angle BAC</math>. Because <math>BOFC</math> is cyclic and <math>\triangle BOC</math> is isosceles with vertex angle <math>\angle BOC = 2\angle BAC</math>, we have <math>\angle OFB = \angle OCB = 90^\circ - \angle BAC</math>. Therefore,<br />
<cmath>\angle AFO = \angle AFB - \angle OFB = (180^\circ - \angle BAC) - (90^\circ - \angle BAC) = 90^\circ.</cmath><br />
This completes the proof.<br />
<br />
<br />
<br />
{{alternate solutions}}<br />
<br />
== See Also ==<br />
* <url>viewtopic.php?t=202907 Discussion on AoPS/MathLinks</url><br />
<br />
{{USAMO newbox|year=2008|num-b=1|num-a=3}}<br />
<br />
[[Category:Olympiad Geometry Problems]]<br />
{{MAA Notice}}</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=863871984 AIME Problems/Problem 152017-07-15T02:40:25Z<p>Tomg418: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Determine <math>w^2+x^2+y^2+z^2</math> if<br />
<br />
<div style="text-align:center;"><math> \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 </math><br /><math> \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 </math><br /><math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div><br />
<br />
== Solution 1 ==<br />
Rewrite the system of equations as <math> \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. </math> This equation is satisfied when <math>t = 4,16,36,64</math>, as then the equation is equivalent to the given equations.<br />
After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the [[equation]] <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)</math>. We can move the expression <math>(t-1)(t-9)(t-25)(t-49)</math> to the left hand side to obtain the difference of the polynomials: <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> and <math>(t-1)(t-9)(t-25)(t-49)</math><br />
<br />
Since the polynomials are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials with a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so<br />
<br />
<div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = (t-4)(t-16)(t-36)(t-64) </math><br />
</div><br />
<br />
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with<br />
<br />
<cmath>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</cmath><br />
<br />
so that<br />
<br />
<cmath>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</cmath><br />
<br />
Similarly, we can plug in <math>t=9,25,49</math> and get<br />
<br />
<cmath>\begin{align*}<br />
y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br />
z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br />
w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}</cmath><br />
<br />
Now adding them up,<br />
<br />
<cmath>\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br />
x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}</cmath><br />
<br />
with a sum of<br />
<br />
<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath><br />
<br />
/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:<br />
Let the two polynomials be <math>A(t)</math> and <math>B(t)</math> and let them be equal at <math>t=a,b,c,d</math>. Thus we have <math>A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0</math>. Also the polynomial <math>A(t) - B(t)</math> is cubic, but it equals 0 at 4 values of <math>t</math>. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into <math>(t-a)(t-b)(t-c)(t-d)(</math>some nonzero polynomial<math>)</math> which would have a degree greater than or equal to 4, contradicting the statement that <math>A(t) - B(t)</math> is cubic. Because <math>A(t) - B(t) = 0, A(t)</math> and <math>B(t)</math> are equivalent and must be equal for all <math>t</math>.<br />
<br />
'''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.<br />
<br />
== Solution 2 ==<br />
As in Solution 1, we have <br />
<div style="text-align:center;"><math>(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)</math> <math>-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)</math><br />
<math>=(t-4)(t-16)(t-36)(t-64)</math><br />
</div><br />
Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1984|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=863861984 AIME Problems/Problem 152017-07-15T02:39:27Z<p>Tomg418: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Determine <math>w^2+x^2+y^2+z^2</math> if<br />
<br />
<div style="text-align:center;"><math> \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 </math><br /><math> \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 </math><br /><math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div><br />
<br />
== Solution 1 ==<br />
Rewrite the system of equations as <math> \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. </math> This equation is satisfied when <math>t = 4,16,36,64</math>, as then the equation is equivalent to the given equations.<br />
After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the [[equation]] <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)</math>. We can move the expression <math>(t-1)(t-9)(t-25)(t-49)</math> to the left hand side to obtain the difference of the polynomials <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> and <math>(t-1)(t-9)(t-25)(t-49)</math><br />
<br />
Since the polynomials are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials with a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so<br />
<br />
<div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = (t-4)(t-16)(t-36)(t-64) </math><br />
</div><br />
<br />
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with<br />
<br />
<cmath>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</cmath><br />
<br />
so that<br />
<br />
<cmath>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</cmath><br />
<br />
Similarly, we can plug in <math>t=9,25,49</math> and get<br />
<br />
<cmath>\begin{align*}<br />
y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br />
z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br />
w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}</cmath><br />
<br />
Now adding them up,<br />
<br />
<cmath>\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br />
x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}</cmath><br />
<br />
with a sum of<br />
<br />
<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath><br />
<br />
/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:<br />
Let the two polynomials be <math>A(t)</math> and <math>B(t)</math> and let them be equal at <math>t=a,b,c,d</math>. Thus we have <math>A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0</math>. Also the polynomial <math>A(t) - B(t)</math> is cubic, but it equals 0 at 4 values of <math>t</math>. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into <math>(t-a)(t-b)(t-c)(t-d)(</math>some nonzero polynomial<math>)</math> which would have a degree greater than or equal to 4, contradicting the statement that <math>A(t) - B(t)</math> is cubic. Because <math>A(t) - B(t) = 0, A(t)</math> and <math>B(t)</math> are equivalent and must be equal for all <math>t</math>.<br />
<br />
'''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.<br />
<br />
== Solution 2 ==<br />
As in Solution 1, we have <br />
<div style="text-align:center;"><math>(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)</math> <math>-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)</math><br />
<math>=(t-4)(t-16)(t-36)(t-64)</math><br />
</div><br />
Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1984|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=863851984 AIME Problems/Problem 152017-07-15T02:38:14Z<p>Tomg418: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Determine <math>w^2+x^2+y^2+z^2</math> if<br />
<br />
<div style="text-align:center;"><math> \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 </math><br /><math> \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 </math><br /><math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div><br />
<br />
== Solution 1 ==<br />
Rewrite the system of equations as <math> \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. </math> This equation is satisfied when <math>t = 4,16,36,64</math>, as then the equation is equivalent to the given equations.<br />
After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the [[equation]] <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)</math>. We can move the expression <math>(t-1)(t-9)(t-25)(t-49)</math> to the left hand side to obtain the difference of the polynomials.<br />
<br />
<div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = 0 </math><br />
</div><br />
<br />
Since the polynomials are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials with a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so<br />
<br />
<div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = (t-4)(t-16)(t-36)(t-64) </math><br />
</div><br />
<br />
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with<br />
<br />
<cmath>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</cmath><br />
<br />
so that<br />
<br />
<cmath>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</cmath><br />
<br />
Similarly, we can plug in <math>t=9,25,49</math> and get<br />
<br />
<cmath>\begin{align*}<br />
y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br />
z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br />
w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}</cmath><br />
<br />
Now adding them up,<br />
<br />
<cmath>\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br />
x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}</cmath><br />
<br />
with a sum of<br />
<br />
<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath><br />
<br />
/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:<br />
Let the two polynomials be <math>A(t)</math> and <math>B(t)</math> and let them be equal at <math>t=a,b,c,d</math>. Thus we have <math>A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0</math>. Also the polynomial <math>A(t) - B(t)</math> is cubic, but it equals 0 at 4 values of <math>t</math>. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into <math>(t-a)(t-b)(t-c)(t-d)(</math>some nonzero polynomial<math>)</math> which would have a degree greater than or equal to 4, contradicting the statement that <math>A(t) - B(t)</math> is cubic. Because <math>A(t) - B(t) = 0, A(t)</math> and <math>B(t)</math> are equivalent and must be equal for all <math>t</math>.<br />
<br />
'''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.<br />
<br />
== Solution 2 ==<br />
As in Solution 1, we have <br />
<div style="text-align:center;"><math>(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)</math> <math>-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)</math><br />
<math>=(t-4)(t-16)(t-36)(t-64)</math><br />
</div><br />
Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1984|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=863841984 AIME Problems/Problem 152017-07-15T02:35:50Z<p>Tomg418: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Determine <math>w^2+x^2+y^2+z^2</math> if<br />
<br />
<div style="text-align:center;"><math> \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 </math><br /><math> \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 </math><br /><math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div><br />
<br />
== Solution 1 ==<br />
Rewrite the system of equations as <math> \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. </math> This equation is satisfied when <math>t = 4,16,36,64</math>, as then the equation is equivalent to the given equations.<br />
After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the [[equation]] <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)</math>. We can move the expression <math>(t-1)(t-9)(t-25)(t-49)</math> to the left hand side to obtain the difference of the polynomials <br />
<br />
<math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> <br />
<br />
and <math>(t-1)(t-9)(t-25)(t-49)</math> <br />
<br />
Since the polynomials are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials with a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so<br />
<br />
<div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = (t-4)(t-16)(t-36)(t-64) </math><br />
</div><br />
<br />
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with<br />
<br />
<cmath>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</cmath><br />
<br />
so that<br />
<br />
<cmath>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</cmath><br />
<br />
Similarly, we can plug in <math>t=9,25,49</math> and get<br />
<br />
<cmath>\begin{align*}<br />
y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br />
z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br />
w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}</cmath><br />
<br />
Now adding them up,<br />
<br />
<cmath>\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br />
x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}</cmath><br />
<br />
with a sum of<br />
<br />
<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath><br />
<br />
/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:<br />
Let the two polynomials be <math>A(t)</math> and <math>B(t)</math> and let them be equal at <math>t=a,b,c,d</math>. Thus we have <math>A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0</math>. Also the polynomial <math>A(t) - B(t)</math> is cubic, but it equals 0 at 4 values of <math>t</math>. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into <math>(t-a)(t-b)(t-c)(t-d)(</math>some nonzero polynomial<math>)</math> which would have a degree greater than or equal to 4, contradicting the statement that <math>A(t) - B(t)</math> is cubic. Because <math>A(t) - B(t) = 0, A(t)</math> and <math>B(t)</math> are equivalent and must be equal for all <math>t</math>.<br />
<br />
'''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.<br />
<br />
== Solution 2 ==<br />
As in Solution 1, we have <br />
<div style="text-align:center;"><math>(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)</math> <math>-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)</math><br />
<math>=(t-4)(t-16)(t-36)(t-64)</math><br />
</div><br />
Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1984|num-b=14|after=Last Question}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Tomg418https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=863831984 AIME Problems/Problem 152017-07-15T02:34:16Z<p>Tomg418: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Determine <math>w^2+x^2+y^2+z^2</math> if<br />
<br />
<div style="text-align:center;"><math> \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 </math><br /><math> \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 </math><br /><math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div><br />
<br />
== Solution 1 ==<br />
Rewrite the system of equations as <math> \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. </math> This equation is satisfied when <math>t = 4,16,36,64</math>, as then the equation is equivalent to the given equations.<br />
After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the [[equation]] <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)</math>. We can move the expression <math>(t-1)(t-9)(t-25)(t-49)</math> to the left hand side to obtain the difference of the polynomials <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> and <math>(t-1)(t-9)(t-25)(t-49)</math> . <br />
Since the polynomials are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials with a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so<br />
<br />
<div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = (t-4)(t-16)(t-36)(t-64) </math><br />
</div><br />
<br />
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with<br />
<br />
<cmath>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</cmath><br />
<br />
so that<br />
<br />
<cmath>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</cmath><br />
<br />
Similarly, we can plug in <math>t=9,25,49</math> and get<br />
<br />
<cmath>\begin{align*}<br />
y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br />
z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br />
w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}</cmath><br />
<br />
Now adding them up,<br />
<br />
<cmath>\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br />
x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}</cmath><br />
<br />
with a sum of<br />
<br />
<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath><br />
<br />
/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:<br />
Let the two polynomials be <math>A(t)</math> and <math>B(t)</math> and let them be equal at <math>t=a,b,c,d</math>. Thus we have <math>A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0</math>. Also the polynomial <math>A(t) - B(t)</math> is cubic, but it equals 0 at 4 values of <math>t</math>. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into <math>(t-a)(t-b)(t-c)(t-d)(</math>some nonzero polynomial<math>)</math> which would have a degree greater than or equal to 4, contradicting the statement that <math>A(t) - B(t)</math> is cubic. Because <math>A(t) - B(t) = 0, A(t)</math> and <math>B(t)</math> are equivalent and must be equal for all <math>t</math>.<br />
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'''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.<br />
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== Solution 2 ==<br />
As in Solution 1, we have <br />
<div style="text-align:center;"><math>(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)</math> <math>-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)</math><br />
<math>=(t-4)(t-16)(t-36)(t-64)</math><br />
</div><br />
Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>.<br />
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== See also ==<br />
{{AIME box|year=1984|num-b=14|after=Last Question}}<br />
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[[Category:Intermediate Algebra Problems]]</div>Tomg418