https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mthjjs&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T02:21:07ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_11&diff=1289552020 AIME II Problems/Problem 112020-07-23T17:57:48Z<p>Mthjjs: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>P(x) = x^2 - 3x - 7</math>, and let <math>Q(x)</math> and <math>R(x)</math> be two quadratic polynomials also with the coefficient of <math>x^2</math> equal to <math>1</math>. David computes each of the three sums <math>P + Q</math>, <math>P + R</math>, and <math>Q + R</math> and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If <math>Q(0) = 2</math>, then <math>R(0) = \frac{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 />
Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following:<br />
<cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath><br />
<cmath>P + R = 2x^2 + (b - 3)x + (c - 7)</cmath><br />
<cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath><br />
Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>. We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>.<br />
<br />
By Vieta's, we have:<br />
<cmath> r + t = \dfrac{3 - a}{2}\tag{1}</cmath><br />
<cmath>r + s = \dfrac{3 - b}{2}\tag{2}</cmath><br />
<cmath>s + t = \dfrac{-a - b}{2}\tag{3}</cmath><br />
<cmath>rt = \dfrac{-5}{2}\tag{4}</cmath><br />
<cmath>rs = \dfrac{c - 7}{2}\tag{5}</cmath><br />
<cmath>st = \dfrac{c + 2}{2}\tag{6}</cmath><br />
<br />
Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath<br />
<br />
==Solution 2==<br />
<br />
We know that <math>P(x)=x^2-3x-7</math>. <br />
<br />
Since <math>Q(0)=2</math>, the constant term in <math>Q(x)</math> is <math>2</math>. Let <math>Q(x)=x^2+ax+2</math>. <br />
<br />
Finally, let <math>R(x)=x^2+bx+c</math>.<br />
<br />
<math>P(x)+Q(x)=2x^2+(a-3)x-5</math>. Let its roots be <math>p</math> and <math>q</math>.<br />
<br />
<math>P(x)+R(x)=2x^2+(b-3)x+(c-7)</math> Let its roots be <math>p</math> and <math>r</math>.<br />
<br />
<math>Q(x)+R(x)=2x^2+(a+b)x+(c+2)</math>. Let its roots be <math>q</math> and <math>r</math>.<br />
<br />
By vietas, <math>p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}</math><br />
<br />
We could work out the system of equations, but it's pretty easy to see that <math>p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}</math>.<br />
<br />
<math>\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}\text{, } \text{multiplying everything together a}\text{nd taking the sqrt of both sides,}</math><br />
<cmath>(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)</cmath><br />
<cmath>pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} </cmath><br />
<math>\text{Dividing this }\text{equation by }qr=\frac{c+2}{2} </math><br />
<cmath>\frac{pqr}{qr}=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}} </cmath><br />
<cmath>p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} </cmath><br />
<math>\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}</math><br />
<cmath>\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} </cmath><br />
<math>\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}</math><br />
<br />
~quacker88<br />
<br />
==Solution 3 (Official MAA)==<br />
Let the common root of <math>P+Q</math> and <math>P+R</math> be <math>p</math>, the common root of <math>P+Q</math> and <math>Q+R</math> be <math>q</math>, and the common root of <math>Q+R</math> and <math>P+R</math> be <math>r</math>. Because <math>p</math> and <math>q</math> are both roots of <math>P+Q</math> and <math>P+Q</math> has leading coefficient <math>2</math>, it follows that <math>P(x) + Q(x) = 2(x-p)(x-q).</math> Similarly, <math>P(x) + R(x) = 2(x-p)(x-r)</math> and <math>Q(x) + R(x) = 2(x-q)(x-r)</math>. Adding these three equations together and dividing by <math>2</math> yields<cmath>P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),</cmath>so<br />
<cmath>P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))</cmath><br />
<cmath>= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r) </cmath><br />
<cmath>= x^2 - 2px + (pq + pr - qr).</cmath><br />
Similarly,<br />
<cmath>Q(x) = x^2 - 2qx + (pq + qr - pr) \text{~ and}</cmath><br />
<cmath>R(x) = x^2 - 2rx + (pr + qr - pq).</cmath><br />
Comparing the <math>x</math> coefficients yields <math>p = \tfrac32</math>, and comparing the constant coefficients yields <math>-7 = pq + pr - qr = \tfrac32(q+r) - qr</math>. The fact that <math>Q(0) = 2</math> implies that <math>\tfrac32(q-r) + qr = 2</math>. Adding these two equations yields <math>q = -\tfrac53</math>, and so substituting back in to solve for <math>r</math> gives <math>r=-\tfrac{27}{19}</math>. Finally,<cmath>R(0) = pr + qr - pq = \left(-\frac{27}{19}\right)\left(\frac32-\frac53\right) + \frac52 = \frac{9}{38} + \frac52 = \frac{52}{19}.</cmath>The requested sum is <math>52 + 19 = 71</math>. Note that <math>Q(x) = x^2 + \frac{10}3x + 2</math> and <math>R(x) = x^2 + \frac{54}{19}x + \frac{52}{19}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/BQlab3vjjxw ~ CNCM<br />
<br />
Another one:<br />
<br />
https://www.youtube.com/watch?v=AXN9x51KzNI<br />
<br />
==See Also==<br />
{{AIME box|year=2020|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mthjjshttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_11&diff=1289542020 AIME II Problems/Problem 112020-07-23T17:57:34Z<p>Mthjjs: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>P(x) = x^2 - 3x - 7</math>, and let <math>Q(x)</math> and <math>R(x)</math> be two quadratic polynomials also with the coefficient of <math>x^2</math> equal to <math>1</math>. David computes each of the three sums <math>P + Q</math>, <math>P + R</math>, and <math>Q + R</math> and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If <math>Q(0) = 2</math>, then <math>R(0) = \frac{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 />
Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following:<br />
<cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath><br />
<cmath>P + R = 2x^2 + (b - 3)x + (c - 7)</cmath><br />
<cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath><br />
Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>. We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>.<br />
<br />
By Vieta's, we have:<br />
<cmath> r + t = \dfrac{3 - a}{2}\tag{1}</cmath><br />
<cmath>r + s = \dfrac{3 - b}{2}\tag{2}</cmath><br />
<cmath>s + t = \dfrac{-a - b}{2}\tag{3}</cmath><br />
<cmath>rt = \dfrac{-5}{2}\tag{4}</cmath><br />
<cmath>rs = \dfrac{c - 7}{2}\tag{5}</cmath><br />
<cmath>st = \dfrac{c + 2}{2}\tag{6}</cmath><br />
<br />
Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath<br />
<br />
==Solution 2==<br />
<br />
We know that <math>P(x)=x^2-3x-7</math>. <br />
<br />
Since <math>Q(0)=2</math>, the constant term in <math>Q(x)</math> is <math>2</math>. Let <math>Q(x)=x^2+ax+2</math>. <br />
<br />
Finally, let <math>R(x)=x^2+bx+c</math>.<br />
<br />
<math>P(x)+Q(x)=2x^2+(a-3)x-5</math>. Let its roots be <math>p</math> and <math>q</math>.<br />
<br />
<math>P(x)+R(x)=2x^2+(b-3)x+(c-7)</math> Let its roots be <math>p</math> and <math>r</math>.<br />
<br />
<math>Q(x)+R(x)=2x^2+(a+b)x+(c+2)</math>. Let its roots be <math>q</math> and <math>r</math>.<br />
<br />
By vietas, <math>p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}</math><br />
<br />
We could work out the system of equations, but it's pretty easy to see that <math>p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}</math>.<br />
<br />
<math>\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}\text{, } \text{multiplying everything together a}\text{nd taking the sqrt of both sides,}</math><br />
<cmath>(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)</cmath><br />
<cmath>pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} </cmath><br />
<math>\text{Dividing this }\text{equation by }qr=\frac{c+2}{2} </math><br />
<cmath>\frac{pqr}{qr}=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}} </cmath><br />
<cmath>p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} </cmath><br />
<math>\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}</math><br />
<cmath>\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} </cmath><br />
<math>\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}</math><br />
<br />
~quacker88<br />
<br />
==Solution 4 (Official MAA)==<br />
Let the common root of <math>P+Q</math> and <math>P+R</math> be <math>p</math>, the common root of <math>P+Q</math> and <math>Q+R</math> be <math>q</math>, and the common root of <math>Q+R</math> and <math>P+R</math> be <math>r</math>. Because <math>p</math> and <math>q</math> are both roots of <math>P+Q</math> and <math>P+Q</math> has leading coefficient <math>2</math>, it follows that <math>P(x) + Q(x) = 2(x-p)(x-q).</math> Similarly, <math>P(x) + R(x) = 2(x-p)(x-r)</math> and <math>Q(x) + R(x) = 2(x-q)(x-r)</math>. Adding these three equations together and dividing by <math>2</math> yields<cmath>P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),</cmath>so<br />
<cmath>P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))</cmath><br />
<cmath>= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r) </cmath><br />
<cmath>= x^2 - 2px + (pq + pr - qr).</cmath><br />
Similarly,<br />
<cmath>Q(x) = x^2 - 2qx + (pq + qr - pr) \text{~ and}</cmath><br />
<cmath>R(x) = x^2 - 2rx + (pr + qr - pq).</cmath><br />
Comparing the <math>x</math> coefficients yields <math>p = \tfrac32</math>, and comparing the constant coefficients yields <math>-7 = pq + pr - qr = \tfrac32(q+r) - qr</math>. The fact that <math>Q(0) = 2</math> implies that <math>\tfrac32(q-r) + qr = 2</math>. Adding these two equations yields <math>q = -\tfrac53</math>, and so substituting back in to solve for <math>r</math> gives <math>r=-\tfrac{27}{19}</math>. Finally,<cmath>R(0) = pr + qr - pq = \left(-\frac{27}{19}\right)\left(\frac32-\frac53\right) + \frac52 = \frac{9}{38} + \frac52 = \frac{52}{19}.</cmath>The requested sum is <math>52 + 19 = 71</math>. Note that <math>Q(x) = x^2 + \frac{10}3x + 2</math> and <math>R(x) = x^2 + \frac{54}{19}x + \frac{52}{19}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/BQlab3vjjxw ~ CNCM<br />
<br />
Another one:<br />
<br />
https://www.youtube.com/watch?v=AXN9x51KzNI<br />
<br />
==See Also==<br />
{{AIME box|year=2020|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mthjjshttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_11&diff=1289532020 AIME II Problems/Problem 112020-07-23T17:57:22Z<p>Mthjjs: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>P(x) = x^2 - 3x - 7</math>, and let <math>Q(x)</math> and <math>R(x)</math> be two quadratic polynomials also with the coefficient of <math>x^2</math> equal to <math>1</math>. David computes each of the three sums <math>P + Q</math>, <math>P + R</math>, and <math>Q + R</math> and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If <math>Q(0) = 2</math>, then <math>R(0) = \frac{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 />
Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following:<br />
<cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath><br />
<cmath>P + R = 2x^2 + (b - 3)x + (c - 7)</cmath><br />
<cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath><br />
Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>. We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>.<br />
<br />
By Vieta's, we have:<br />
<cmath> r + t = \dfrac{3 - a}{2}\tag{1}</cmath><br />
<cmath>r + s = \dfrac{3 - b}{2}\tag{2}</cmath><br />
<cmath>s + t = \dfrac{-a - b}{2}\tag{3}</cmath><br />
<cmath>rt = \dfrac{-5}{2}\tag{4}</cmath><br />
<cmath>rs = \dfrac{c - 7}{2}\tag{5}</cmath><br />
<cmath>st = \dfrac{c + 2}{2}\tag{6}</cmath><br />
<br />
Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath<br />
<br />
==Solution 3==<br />
<br />
We know that <math>P(x)=x^2-3x-7</math>. <br />
<br />
Since <math>Q(0)=2</math>, the constant term in <math>Q(x)</math> is <math>2</math>. Let <math>Q(x)=x^2+ax+2</math>. <br />
<br />
Finally, let <math>R(x)=x^2+bx+c</math>.<br />
<br />
<math>P(x)+Q(x)=2x^2+(a-3)x-5</math>. Let its roots be <math>p</math> and <math>q</math>.<br />
<br />
<math>P(x)+R(x)=2x^2+(b-3)x+(c-7)</math> Let its roots be <math>p</math> and <math>r</math>.<br />
<br />
<math>Q(x)+R(x)=2x^2+(a+b)x+(c+2)</math>. Let its roots be <math>q</math> and <math>r</math>.<br />
<br />
By vietas, <math>p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}</math><br />
<br />
We could work out the system of equations, but it's pretty easy to see that <math>p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}</math>.<br />
<br />
<math>\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}\text{, } \text{multiplying everything together a}\text{nd taking the sqrt of both sides,}</math><br />
<cmath>(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)</cmath><br />
<cmath>pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} </cmath><br />
<math>\text{Dividing this }\text{equation by }qr=\frac{c+2}{2} </math><br />
<cmath>\frac{pqr}{qr}=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}} </cmath><br />
<cmath>p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} </cmath><br />
<math>\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}</math><br />
<cmath>\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} </cmath><br />
<math>\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}</math><br />
<br />
~quacker88<br />
<br />
==Solution 4 (Official MAA)==<br />
Let the common root of <math>P+Q</math> and <math>P+R</math> be <math>p</math>, the common root of <math>P+Q</math> and <math>Q+R</math> be <math>q</math>, and the common root of <math>Q+R</math> and <math>P+R</math> be <math>r</math>. Because <math>p</math> and <math>q</math> are both roots of <math>P+Q</math> and <math>P+Q</math> has leading coefficient <math>2</math>, it follows that <math>P(x) + Q(x) = 2(x-p)(x-q).</math> Similarly, <math>P(x) + R(x) = 2(x-p)(x-r)</math> and <math>Q(x) + R(x) = 2(x-q)(x-r)</math>. Adding these three equations together and dividing by <math>2</math> yields<cmath>P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),</cmath>so<br />
<cmath>P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))</cmath><br />
<cmath>= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r) </cmath><br />
<cmath>= x^2 - 2px + (pq + pr - qr).</cmath><br />
Similarly,<br />
<cmath>Q(x) = x^2 - 2qx + (pq + qr - pr) \text{~ and}</cmath><br />
<cmath>R(x) = x^2 - 2rx + (pr + qr - pq).</cmath><br />
Comparing the <math>x</math> coefficients yields <math>p = \tfrac32</math>, and comparing the constant coefficients yields <math>-7 = pq + pr - qr = \tfrac32(q+r) - qr</math>. The fact that <math>Q(0) = 2</math> implies that <math>\tfrac32(q-r) + qr = 2</math>. Adding these two equations yields <math>q = -\tfrac53</math>, and so substituting back in to solve for <math>r</math> gives <math>r=-\tfrac{27}{19}</math>. Finally,<cmath>R(0) = pr + qr - pq = \left(-\frac{27}{19}\right)\left(\frac32-\frac53\right) + \frac52 = \frac{9}{38} + \frac52 = \frac{52}{19}.</cmath>The requested sum is <math>52 + 19 = 71</math>. Note that <math>Q(x) = x^2 + \frac{10}3x + 2</math> and <math>R(x) = x^2 + \frac{54}{19}x + \frac{52}{19}</math>.<br />
<br />
==Video Solution==<br />
https://youtu.be/BQlab3vjjxw ~ CNCM<br />
<br />
Another one:<br />
<br />
https://www.youtube.com/watch?v=AXN9x51KzNI<br />
<br />
==See Also==<br />
{{AIME box|year=2020|n=II|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mthjjshttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_7&diff=1289522020 AIME II Problems/Problem 72020-07-23T17:55:04Z<p>Mthjjs: /* Solution 2 (Official MAA) */</p>
<hr />
<div>==Problem==<br />
Two congruent right circular cones each with base radius <math>3</math> and height <math>8</math> have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance <math>3</math> from the base of each cone. A sphere with radius <math>r</math> lies withing both cones. The maximum possible value of <math>r^2</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution (Official MAA)==<br />
Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let <math>A</math> be the point in the cross section where the bases of the cones meet, and let <math>C</math> be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, <math>B</math>, to the center of its base, <math>D</math>. Let the sphere be tangent to <math>\overline{AB}</math> at <math>E</math>. The right triangles <math>\triangle ABD</math> and <math>\triangle CBE</math> are similar, implying that the radius of the sphere is<cmath>CE = AD \cdot\frac{BC}{AB} = AD \cdot\frac{BD-CD}{AB} =3\cdot\frac5{\sqrt{8^2+3^2}} = \frac{15}{\sqrt{73}}=\sqrt{\frac{225}{73}}.</cmath>The requested sum is <math>225+73=298</math>.<br />
<asy><br />
unitsize(0.6cm);<br />
pair A = (0,0); <br />
pair TriangleOneLeft = (-6,0); <br />
pair TriangleOneDown = (-3,-8); <br />
pair TriangleOneMid = (-3,0);<br />
<br />
pair D = (0,-3); <br />
pair TriangleTwoDown = (0,-6); <br />
pair B = (-8,-3);<br />
<br />
pair C = IP(TriangleOneMid -- TriangleOneDown, B--D);<br />
pair EE = foot(C, A, B); <br />
real radius = arclength(C--EE); <br />
path circ = Circle(C, radius);<br />
<br />
<br />
<br />
draw(A--B--TriangleTwoDown--cycle);<br />
draw(B--D); <br />
draw(A--TriangleOneLeft--TriangleOneDown--cycle); <br />
draw(circ); <br />
draw(C--EE); <br />
draw(TriangleOneMid -- TriangleOneDown, gray);<br />
<br />
dot("$B$", B, W); <br />
dot("$E$", EE, NW); <br />
dot("$A$", A, NE); <br />
dot("$D$", D, E); <br />
dot("$C$", C, SE);<br />
</asy><br />
<br />
==Video Solution==<br />
https://youtu.be/bz5N-jI2e0U?t=44<br />
<br />
==Video Solution 2==<br />
https://www.youtube.com/watch?v=0XJddG43pIk ~ MathEx<br />
<br />
==Video Solution 3==<br />
https://youtu.be/dHGXtB0FxXs<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
{{AIME box|year=2020|n=II|num-b=6|num-a=8}}<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:3D Geometry Problems]]<br />
{{MAA Notice}}</div>Mthjjshttps://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_7&diff=1289512020 AIME II Problems/Problem 72020-07-23T17:54:53Z<p>Mthjjs: </p>
<hr />
<div>==Problem==<br />
Two congruent right circular cones each with base radius <math>3</math> and height <math>8</math> have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance <math>3</math> from the base of each cone. A sphere with radius <math>r</math> lies withing both cones. The maximum possible value of <math>r^2</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution 2 (Official MAA)==<br />
Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let <math>A</math> be the point in the cross section where the bases of the cones meet, and let <math>C</math> be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, <math>B</math>, to the center of its base, <math>D</math>. Let the sphere be tangent to <math>\overline{AB}</math> at <math>E</math>. The right triangles <math>\triangle ABD</math> and <math>\triangle CBE</math> are similar, implying that the radius of the sphere is<cmath>CE = AD \cdot\frac{BC}{AB} = AD \cdot\frac{BD-CD}{AB} =3\cdot\frac5{\sqrt{8^2+3^2}} = \frac{15}{\sqrt{73}}=\sqrt{\frac{225}{73}}.</cmath>The requested sum is <math>225+73=298</math>.<br />
<asy><br />
unitsize(0.6cm);<br />
pair A = (0,0); <br />
pair TriangleOneLeft = (-6,0); <br />
pair TriangleOneDown = (-3,-8); <br />
pair TriangleOneMid = (-3,0);<br />
<br />
pair D = (0,-3); <br />
pair TriangleTwoDown = (0,-6); <br />
pair B = (-8,-3);<br />
<br />
pair C = IP(TriangleOneMid -- TriangleOneDown, B--D);<br />
pair EE = foot(C, A, B); <br />
real radius = arclength(C--EE); <br />
path circ = Circle(C, radius);<br />
<br />
<br />
<br />
draw(A--B--TriangleTwoDown--cycle);<br />
draw(B--D); <br />
draw(A--TriangleOneLeft--TriangleOneDown--cycle); <br />
draw(circ); <br />
draw(C--EE); <br />
draw(TriangleOneMid -- TriangleOneDown, gray);<br />
<br />
dot("$B$", B, W); <br />
dot("$E$", EE, NW); <br />
dot("$A$", A, NE); <br />
dot("$D$", D, E); <br />
dot("$C$", C, SE);<br />
</asy><br />
<br />
==Video Solution==<br />
https://youtu.be/bz5N-jI2e0U?t=44<br />
<br />
==Video Solution 2==<br />
https://www.youtube.com/watch?v=0XJddG43pIk ~ MathEx<br />
<br />
==Video Solution 3==<br />
https://youtu.be/dHGXtB0FxXs<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
{{AIME box|year=2020|n=II|num-b=6|num-a=8}}<br />
[[Category:Intermediate Geometry Problems]]<br />
[[Category:3D Geometry Problems]]<br />
{{MAA Notice}}</div>Mthjjs