https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Brian6liu&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T19:12:21ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Lagrange_Multipliers&diff=89295Lagrange Multipliers2017-12-31T20:16:59Z<p>Brian6liu: /* Problems */</p>
<hr />
<div>This article discusses Lagrange multipliers, a topic of multivariable calculus.<br />
__TOC__<br />
<br />
== Definition ==<br />
Useful in optimization, Lagrange multipliers, based on a calculus approach, can be used to find local minimums and maximums of a function given a constraint. Suppose there is a continuous function <math>f(x,y)</math> and there exists a continuous constraint function on the values of the function <math>c = g(x,y)</math>. The method of Lagrange multipliers states that, to find the minimum or maximum satisfying both requirements (<math>\lambda</math> is a constant):<br />
<br />
<cmath>\dfrac{\partial f}{\partial x} = \lambda \dfrac{\partial g}{\partial x}</cmath><br />
<cmath>\dfrac{\partial f}{\partial y} = \lambda \dfrac{\partial g}{\partial y}</cmath><br />
<br />
The method can be extended to multiple variables, as well as multiple constraints. If we had a continuous function <math>f(x, y, z)</math> and the constraint <math>c = g(x, y, z)</math>, we would just have another equation relating the partial derivatives with respect to <math>z</math> through a factor of <math>\lambda</math>. Similarly, if we have another constraint <math>d = h(x, y, z)</math>, then we would add the partial with respect to each variable to their respective equation with another factor <math>\mu</math>. Thus, we have the general form for Lagrange multipliers for the function <math>f(x_1, x_2,...,x_n)</math> bounded by a certain number of constraints <math>c_1 = g_1(x_1, x_2,..., x_n)</math>, <math>c_2 = g_2(x_1, x_2,..., x_n)</math>,...,<math>c_m = g_m(x_1, x_2,..., x_n)</math>:<br />
<br />
<cmath>\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 + ... + \lambda_m \nabla g_m</cmath><br />
<br />
Where <math>\nabla</math> is the del operator, which when applied to a scalar function <math>f</math> results in a vector with each component the partial in that component's direction, representing total change.<br />
<br />
It is important to remember that sometimes there maybe a function that is bounded on some interval. In such cases, Lagrange multipliers may give a result, but the answer may not be the one this method results. After using this method, it is imperative to check the endpoints and also plug back in the numbers resulted from this method to verify the final answer. Also, sometimes the function may be over-constrained and there maybe no point that satisfies all requirements.<br />
<br />
== Example ==<br />
This method can become extremely difficult to use when there are several variables and constraints, but it is very effective and useful in some situations, including some contest math problems. Consider the following question:<br />
<br />
A rectangular prism lies on the <math>xy</math> plane with one vertex at the origin. The vertex that does not share a face with this vertex at the origin lies on the plane bounded by the points <math>(0, 0, 4)</math>, <math>(0, 8, 0)</math>, <math>(16, 0, 0)</math>. Find the maximum volume of the box.<br />
<br />
<br />
Solution:<br />
The volume of the box is given by the equation <math>V(x, y, z) = xyz</math>. Because the way the box is described, the point that lies on the plane given determines the volume of the box. Thus we need only need to consider that point. The constraint we are given for that point is given by the plane <math>x + 2y + 4z = 16</math>, since that is the plane that goes through the three points mentioned. Let us call that function <math>16 = g(x, y, z)</math>. Now we can begin to find our partial derivatives:<br />
<br />
<cmath>\dfrac{\partial V}{\partial x} = yz</cmath> <cmath>\dfrac{\partial V}{\partial y} = xz</cmath> <cmath>\dfrac{\partial V}{\partial z} = xy</cmath><br />
<cmath>\dfrac{\partial g}{\partial x} = 1</cmath> <cmath>\dfrac{\partial g}{\partial y} = 2</cmath> <cmath>\dfrac{\partial g}{\partial z} = 4</cmath><br />
<br />
We equate the corresponding partials through a constant factor, and we also use our original constraint:<br />
<cmath>x + 2y + 4z = 16</cmath><br />
<cmath>yz = \left(\lambda \right)1</cmath> <cmath>xz = \left(\lambda \right)2</cmath> <cmath>xy = \left(\lambda \right)4</cmath><br />
<br />
Multiplying each side of the last three equations by the variable not in the equation, we have:<br />
<cmath>xyz = \left(\lambda \right)1x</cmath> <cmath>xyz = \left(\lambda \right)2y</cmath> <cmath>xyz = \left(\lambda \right)4z</cmath><br />
<br />
We can equate these to result:<br />
<math>x = 2y = 4z</math><br />
<br />
Plugging back into the constraint, we have:<br />
<cmath>x + x + x = 16</cmath><br />
<cmath>x = \dfrac{16}{3}</cmath> <cmath>y = \dfrac{8}{3}</cmath> <cmath>z = \dfrac{4}{3}</cmath><br />
<br />
Thus, the maximum volume, given by <math>xyz</math>, is:<br />
<math>\framebox{\tfrac{512}{9}}</math><br />
<br />
==Problems==<br />
'''Introductory''' <br />
<br />
Find the maximum area of a rectangular prism with a surface area of <math>624</math>.<br />
<br />
'''Olympiad'''<br />
<br />
Prove that <cmath>0\le yz+zx+xy-2xyz\le \frac{7}{27}</cmath>, where <math>x,y</math> and <math>z</math> are non-negative real numbers satisfying <cmath>x+y+z=1</cmath>.</div>Brian6liuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_9&diff=694582015 AIME I Problems/Problem 92015-03-24T22:49:16Z<p>Brian6liu: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.<br />
<br />
==Solution==<br />
Let <math>a_1=x, a_2=y, a_3=z</math>. First note that if any absolute value equals 0, then <math>a_n</math>=0.<br />
Also note that if at any position, <math>a_n=a_{n-1}</math>, then <math>a_{n+2}=0</math>.<br />
Then, if any absolute value equals 1, then <math>a_n</math>=0.<br />
Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> is less than or equal to 1, then that ordered triple meets the criteria.<br />
Assume that to be the only way the criteria is met.<br />
To prove, let <math>|y-x|>1</math>, and <math>|z-y|>1</math>. Then, <math>a_4 \ge 2z</math>, <math>a_5 \ge 4z</math>, and <math>a_6 \ge 4z</math>.<br />
However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be <math>z=1</math>, <math>|y-x|=2</math>. Again assume that any other scenario will not meet criteria.<br />
To prove, divide the other scenarios into two cases: <math>z>1</math>, <math>|y-x|>1</math>, and <math>|z-y|>1</math>; and <math>z=1</math>, <math>|y-x|>2</math>, and <math>|z-y|>1</math>.<br />
For the first one, <math>a_4</math>>=2z, <math>a_5</math>>=4z, <math>a_6</math>>=8z, and <math>a_7</math>>=16z, by which point we see that this function diverges.<br />
For the second one, <math>a_4 \ge 3</math>, <math>a_5 \ge 6</math>, <math>a_6 \ge 18</math>, and <math>a_7 \ge 54</math>, by which point we see that this function diverges.<br />
Therefore, the only scenarios where <math>a_n</math>=0 is when any of the following are met:<br />
<math>|y-x|</math><2 (280 options)<br />
<math>|z-y|</math><2 (280 options, 80 of which coincide with option 1)<br />
z=1, <math>|y-x|</math>=2. (14 options, none of which coincide with either option 1 or option 2)<br />
Adding the total number of such ordered triples yields <math>280+280-80+14=494</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Brian6liuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_9&diff=694572015 AIME I Problems/Problem 92015-03-24T22:48:21Z<p>Brian6liu: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.<br />
<br />
==Solution==<br />
Let <math>a_1=x, a_2=y, a_3=z</math>. First note that if any absolute value equals 0, then <math>a_n</math>=0.<br />
Also note that if at any position, <math>a_n=a_{n-1}</math>, then <math>a_{n+2}=0</math>.<br />
Then, if any absolute value equals 1, then <math>a_n</math>=0.<br />
Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> is less than or equal to 1, then that ordered triple meets the criteria.<br />
Assume that to be the only way the criteria is met.<br />
To prove, let <math>|y-x|>1</math>, and <math>|z-y|>1</math>. Then, <math>a_4 \ge 2z</math>, <math>a_5 \ge 4z</math>, and <math>a_6 \ge 4z</math>.<br />
However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be <math>z=1</math>, <math>|y-x|=2</math>. Again assume that any other scenario will not meet criteria.<br />
To prove, divide the other scenarios into two cases: <math>z>1</math>, <math>|y-x|>1</math>, and <math>|z-y|>1</math>; and <math>z=1</math>, <math>|y-x|>2</math>, and <math>|z-y|>1</math>.<br />
For the first one, <math>a_4</math>>=2z, <math>a_5</math>>=4z, <math>a_6</math>>=8z, and <math>a_7</math>>=16z, by which point we see that this function diverges.<br />
For the second one, <math>a_4</math>>=3, <math>a_5</math>>=6, <math>a_6</math>>=18, and <math>a_7</math>>=54, by which point we see that this function diverges.<br />
Therefore, the only scenarios where <math>a_n</math>=0 is when any of the following are met:<br />
<math>|y-x|</math><2 (280 options)<br />
<math>|z-y|</math><2 (280 options, 80 of which coincide with option 1)<br />
z=1, <math>|y-x|</math>=2. (14 options, none of which coincide with either option 1 or option 2)<br />
Adding the total number of such ordered triples yields <math>280+280-80+14=494</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Brian6liuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_13&diff=694072015 AIME I Problems/Problem 132015-03-22T22:33:25Z<p>Brian6liu: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
===Solution 1===<br />
Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity<br />
<cmath>\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath><br />
we deduce that (taking absolute values and noticing <math>|x| = 1</math>)<br />
<cmath>|2\sin 1| = |x^2 - 1|.</cmath><br />
But because <math>\csc</math> is the reciprocal of <math>\sin</math> and because <math>\sin z = \sin (180^\circ - z)</math>, if we let our product be <math>M</math> then<br />
<cmath>\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ</cmath><br />
<cmath> = \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|</cmath><br />
because <math>\sin</math> is positive in the first and second quadrants. Now, notice that <math>x^2, x^6, x^{10}, \dots, x^{358}</math> are the roots of <math>z^{90} + 1 = 0.</math> Hence, we can write <math>(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1</math>, and so<br />
<cmath>\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.</cmath><br />
It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>.<br />
<br />
===Solution 2===<br />
Let <math>p=\sin1\sin3\sin5...\sin89</math><br />
<br />
<cmath>p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}</cmath><br />
<br />
<cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}</cmath><br />
<br />
<cmath>=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89)}}</cmath><br />
<br />
<cmath>=\sqrt{\frac{1}{2^{89}}\frac{\sin90\sin91\sin92\sin93...\sin177\sin178\sin179}{\cos1\cos2\cos3\cos4...\cos89}}</cmath><br />
<br />
<math>=\sqrt{\frac{1}{2^{89}}}</math> because of the identity <math>\sin(90+x)=\cos(x)</math><br />
<br />
we want <math>\frac{1}{p^2}=2^{89}</math><br />
<br />
Thus the answer is <math>2+89=091</math><br />
<br />
=== Solution 3 ===<br />
Similar to Solution <math>2</math>, so we use <math>\sin{2\theta}=2\sin\theta\cos\theta</math> and we find that:<br />
<cmath>\begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\<br />
&=(2\sin(2)\sin(88))(2\sin(4))\sin(86))(2\sin(6)\sin(84))(2\sin(8)\sin(82))\cdots(2\sin(42)\sin(48))(2\sin(44)\sin(46))\\<br />
&=2^{22}(\sin(2)\sin(88)\sin(4)\sin(86)\sin(6)\sin(84)\sin(8)\sin(82)\cdots\sin(42)\sin(48)\sin(44)\sin(46))\\<br />
&=2^{22}(\sin(2)\sin(4)\sin(6)\sin(8)\cdots\sin(82)\sin(84)\sin(86)\sin(88))\end{align*}</cmath><br />
Now we can cancel the sines of the multiples of <math>4</math>:<br />
<cmath>1=2^{22}(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86))</cmath><br />
So <math>\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)=2^{-22}</math> and we can apply the double-angle formula again:<br />
<cmath>\begin{align*}2^{-22}&=(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)\\<br />
&=(2\sin(1)\cos(1))(2\sin(3)\cos(3))(2\sin(5)\cos(5))(2\sin(7)\cos(7))\cdots(2\sin(41)\cos(41))(2\sin(43)\cos(43))\\<br />
&=(2\sin(1)\sin(89))(2\sin(3)\sin(87))(2\sin(5)\sin(85))(2\sin(7)\sin(87))\cdots(2\sin(41)\sin(49))(2\sin(43)\sin(47))\\<br />
&=2^{22}(\sin(1)\sin(89)\sin(3)\sin(87)\sin(5)\sin(85)\sin(7)\sin(83)\cdots\sin(41)\sin(49)\sin(43)\sin(47))\\<br />
&=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\end{align*}</cmath><br />
Of course, <math>\sin(45)=2^{-\frac{1}{2}}</math> is missing, so we multiply it to both sides:<br />
<cmath>2^{-22}\sin(45)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(45))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))</cmath><br />
<cmath>\left(2^{-22}\right)\left(2^{-\frac{1}{2}}\right)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))</cmath><br />
<cmath>2^{-\frac{45}{2}}=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))</cmath><br />
Now isolate the product of the sines:<br />
<cmath>\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89)=2^{-\frac{89}{2}}</cmath><br />
And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number:<br />
<cmath>\csc^2(1)\csc^2(3)\csc^2(5)\csc^2(7)\cdots\csc^2(83)\csc^2(85)\csc^2(87)\csc^2(89)=\left(\frac{1}{2^{-\frac{89}{2}}}\right)^2=\left(2^{\frac{89}{2}}\right)^2=2^{89}</cmath><br />
The answer is therefore <math>m+n=(2)+(89)=\boxed{091}</math>.<br />
<br />
===Solution 4===<br />
Let <math>p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ</math>.<br />
<br />
Then, <math>\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(2k-1)^\circ</math>.<br />
<br />
Since <math>\sin\theta=\cos(180^{\circ}-\theta)</math>, we can multiply both sides by <math>\frac{\sqrt{2}}{2}</math> to get <math>\sqrt{\frac{1}{2p}}=\prod_{k=1}^{23} \sin(2k-1)^\circ\cos(2k-1)^\circ</math>.<br />
<br />
Using the double-angle identity <math>\sin2\theta=2\sin\theta\cos\theta</math>, we get <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{23}}\prod_{k=1}^{23} \sin(4k-2)^\circ</math>.<br />
<br />
Note that the right-hand side is equal to <math>\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \sin(4k)^\circ</math>, which is equal to <math>\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} 2\sin(2k)^\circ\cos(2k)^\circ</math>, again, from using our double-angle identity.<br />
<br />
Putting this back into our equation and simplifying gives us <math>\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ</math>.<br />
<br />
Using the fact that <math>\sin\theta=\cos(180^{\circ}-\theta)</math> again, our equation simplifies to <math>\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}</math>, and since <math>\sin90^\circ=1</math>, it follows that <math>2p = 2^{90}</math>, which implies <math>p=2^{89}</math>. Thus, <math>m+n=2+89=\boxed{091}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Brian6liuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_9&diff=694052015 AIME I Problems/Problem 92015-03-22T22:22:12Z<p>Brian6liu: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.<br />
<br />
==Solution==<br />
Let <math>a_1=x, a_2=y, a_3=z</math>. First note that if any absolute value equals 0, then <math>a_n</math>=0.<br />
Also note that if at any position, <math>a_n=a_{n-1}</math>, then <math>a_{n+2}=0</math>.<br />
Then, if any absolute value equals 1, then <math>a_n</math>=0.<br />
Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> is less than or equal to 1, then that ordered triple meets the criteria.<br />
Assume that to be the only way the criteria is met.<br />
To prove, let <math>|y-x|>1</math>, and <math>|z-y|>1</math>. Then, <math>a_4 \ge 2z</math>, <math>a_5 \ge 4z</math>, and <math>a_6 \ge 4z</math>.<br />
However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be z=1, <math>|y-x|</math>=2. Again assume that any other scenario will not meet criteria.<br />
To prove, divide the other scenarios into two cases: z>1, <math>|y-x|</math>>1, and <math>|z-y|</math>>1; and z=1, <math>|y-x|</math>>2, and <math>|z-y|</math>>1.<br />
For the first one, <math>a_4</math>>=2z, <math>a_5</math>>=4z, <math>a_6</math>>=8z, and <math>a_7</math>>=16z, by which point we see that this function diverges.<br />
For the second one, <math>a_4</math>>=3, <math>a_5</math>>=6, <math>a_6</math>>=18, and <math>a_7</math>>=54, by which point we see that this function diverges.<br />
Therefore, the only scenarios where <math>a_n</math>=0 is when any of the following are met:<br />
<math>|y-x|</math><2 (280 options)<br />
<math>|z-y|</math><2 (280 options, 80 of which coincide with option 1)<br />
z=1, <math>|y-x|</math>=2. (14 options, none of which coincide with either option 1 or option 2)<br />
Adding the total number of such ordered triples yields 280+280-80+14=494.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Brian6liuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_9&diff=694042015 AIME I Problems/Problem 92015-03-22T21:57:20Z<p>Brian6liu: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.<br />
<br />
==Solution==<br />
Let <math>a_1=x, a_2=y, a_3=z</math>. First note that if any absolute value equals 0, then <math>a_n</math>=0.<br />
Also note that if at any position, <math>a_n=a_{n-1}</math>, then <math>a_{n+2}=0</math>.<br />
Then, if any absolute value equals 1, then <math>a_n</math>=0.<br />
Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> is less than or equal to 1, then that ordered triple meets the criteria.<br />
Assume that to be the only way the criteria is met.<br />
To prove, let <math>|y-x|</math>>1, and <math>|z-y|</math>>1. Then, <math>a_4</math>>=2z, <math>a_5</math>>=4z, and <math>a_6</math>>=4z.<br />
However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be z=1, <math>|y-x|</math>=2. Again assume that any other scenario will not meet criteria.<br />
To prove, divide the other scenarios into two cases: z>1, <math>|y-x|</math>>1, and <math>|z-y|</math>>1; and z=1, <math>|y-x|</math>>2, and <math>|z-y|</math>>1.<br />
For the first one, <math>a_4</math>>=2z, <math>a_5</math>>=4z, <math>a_6</math>>=8z, and <math>a_7</math>>=16z, by which point we see that this function diverges.<br />
For the second one, <math>a_4</math>>=3, <math>a_5</math>>=6, <math>a_6</math>>=18, and <math>a_7</math>>=54, by which point we see that this function diverges.<br />
Therefore, the only scenarios where <math>a_n</math>=0 is when any of the following are met:<br />
<math>|y-x|</math><2 (280 options)<br />
<math>|z-y|</math><2 (280 options, 80 of which coincide with option 1)<br />
z=1, <math>|y-x|</math>=2. (14 options, none of which coincide with either option 1 or option 2)<br />
Adding the total number of such ordered triples yields 280+280-80+14=494.<br />
<br />
==See Also==<br />
{{AIME box|year=2015|n=I|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Brian6liuhttps://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems&diff=684941966 AHSME Problems2015-03-05T22:55:13Z<p>Brian6liu: /* Problem 8 */</p>
<hr />
<div>== Problem 1 ==<br />
Given that the ratio of <math>3x - 4</math> to <math>y + 15</math> is constant, and <math>y = 3</math> when <math>x = 2</math>, then, when <math>y = 12</math>, <math>x</math> equals:<br />
<br />
<math>\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 8</math><br />
<br />
[[1966 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
When the base of a triangle is increased 10% and the altitude to this base is decreased 10%, the change in area is <br />
<br />
<math> \text{(A)}\ 1\%~\text{increase}\qquad\text{(B)}\ \frac{1}2\%~\text{increase}\qquad\text{(C)}\ 0\%\qquad\text{(D)}\ \frac{1}2\% ~\text{decrease}\qquad\text{(E)}\ 1\% ~\text{decrease} </math><br />
<br />
[[1966 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
If the arithmetic mean of two numbers is <math>6</math> and their geometric mean is <math>10</math>, then an equation with the given two numbers as roots is:<br />
<br />
<math>\text{(A)} \ x^2 + 12x + 100 = 0 ~~ \text{(B)} \ x^2 + 6x + 100 = 0 ~~ \text{(C)} \ x^2 - 12x - 10 = 0</math><br />
<math>\text{(D)} \ x^2 - 12x + 100 = 0 \qquad \text{(E)} \ x^2 - 6x + 100 = 0</math><br />
<br />
[[1966 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Circle I is circumscribed about a given square and circle II is inscribed in the given square. If <math>r</math> is the ratio of the area of circle <math>I</math> to that of circle <math>II</math>, then <math>r</math> equals:<br />
<br />
<math>\text{(A)} \sqrt 2 \qquad \text{(B)} 2 \qquad \text{(C)} \sqrt 3 \qquad \text{(D)} 2\sqrt 2 \qquad \text{(E)} 2\sqrt 3</math><br />
<br />
[[1966 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
The number of values of <math>x</math> satisfying the equation<br />
<br />
<math>\frac {2x^2 - 10x}{x^2 - 5x} = x - 3</math><br />
<br />
is:<br />
<br />
<math>\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qquad \text{(E)} \ \text{an integer greater than 3}</math><br />
<br />
[[1966 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<math>AB</math> is the diameter of a circle centered at <math>O</math>. <math>C</math> is a point on the circle such that angle <math>BOC</math> is <math>60^\circ</math>. If the diameter of the circle is <math>5</math> inches, the length of chord <math>AC</math>, expressed in inches, is:<br />
<br />
<math>\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}</math><br />
<br />
[[1966 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Let <math>\frac {35x - 29}{x^2 - 3x + 2} = \frac {N_1}{x - 1} + \frac {N_2}{x - 2}</math> be an identity in <math>x</math>. The numerical value of <math>N_1N_2</math> is:<br />
<br />
<math>\text{(A)} \ - 246 \qquad \text{(B)} \ - 210 \qquad \text{(C)} \ - 29 \qquad \text{(D)} \ 210 \qquad \text{(E)} \ 246</math><br />
<br />
[[1966 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
The length of the common chord of two intersecting circles is <math>16</math> feet. If the radii are <math>10</math> feet and <math>17</math> feet, a possible value for the distance between the centers of the circles, expressed in feet, is:<br />
<br />
<math>\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}</math><br />
<br />
[[1966 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
If <math>x = (\log_82)^{(\log_28)}</math>, then <math>\log_3x</math> equals:<br />
<br />
<math>\text{(A)} \ - 3 \qquad \text{(B)} \ - \frac13 \qquad \text{(C)} \ \frac13 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 9</math><br />
<br />
[[1966 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:<br />
<br />
<math>\text{(A)} \ 2 \qquad \text{(B)} \ - 2 - \frac {3i\sqrt {3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ - \frac {3i\sqrt {3}}{4} \qquad \text{(E)} \ - 2</math><br />
<br />
[[1966 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
The sides of triangle <math>BAC</math> are in the ratio <math>2: 3: 4</math>. <math>BD</math> is the angle-bisector drawn to the shortest side <math>AC</math>, dividing it into segments <math>AD</math> and <math>CD</math>. If the length of <math>AC</math> is <math>10</math>, then the length of the longer segment of <math>AC</math> is:<br />
<br />
<math>\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12</math><br />
<br />
[[1966 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
The number of real values of <math>x</math> that satisfy the equation <cmath>(2^{6x+3})(4^{3x+6})=8^{4x+5}</cmath> is:<br />
<br />
<math>\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}</math><br />
<br />
[[1966 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
The number of points with positive rational coordinates selected from the set of points in the <math>xy</math>-plane such that <math>x+y \le 5</math>, is:<br />
<br />
<math>\text{(A)} \ 9 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E) infinite}</math><br />
<br />
[[1966 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
The length of rectangle <math>ABCD</math> is 5 inches and its width is 3 inches. Diagonal <math>AC</math> is divided into three equal segments by points <math>E</math> and <math>F</math>. The area of triangle <math>BEF</math>, expressed in square inches, is:<br />
<br />
<math>\text{(A)} \frac{3}{2} \qquad \text{(B)} \frac {5}{3} \qquad \text{(C)} \frac{5}{2} \qquad \text{(D)} \frac{1}{3}\sqrt{34} \qquad \text{(E)} \frac{1}{3}\sqrt{68}</math><br />
<br />
[[1966 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
If <math>x-y>x</math> and <math>x+y<y</math>, then<br />
<br />
<math>\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0</math><br />
<br />
[[1966 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
If <math>\frac{4^x}{2^{x+y}}=8</math> and <math>\frac{9^{x+y}}{3^{5y}}=243</math>, <math>x</math> and <math>y</math> real numbers, then <math>xy</math> equals:<br />
<br />
<math>\text{(A) } \frac{12}{5} \quad \text{(B) } 4 \quad \text{(C) } 6 \quad \text{(D)} 12 \quad \text{(E) } -4</math><br />
<br />
[[1966 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
The number of distinct points common to the curves <math>x^2+4y^2=1</math> and <math>4x^2+y^2=4</math> is:<br />
<br />
<math>\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4</math><br />
<br />
[[1966 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
In a given arithmetic sequence the first term is <math>2</math>, the last term is <math>29</math>, and the sum of all the terms is <math>155</math>. The common difference is:<br />
<br />
<math>\text{(A) } 3 \qquad \text{(B) } 2 \qquad \text{(C) } \frac{27}{19} \qquad \text{(D) } \frac{13}{9} \qquad \text{(E) } \frac{23}{38}</math><br />
<br />
[[1966 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
Let <math>s_1</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>8,12,\cdots</math> and let <math>s_2</math> be the sum of the first <math>n</math> terms of the arithmetic sequence <math>17,19,\cdots</math>. Assume <math>n \ne 0</math>. Then <math>s_1=s_2</math> for:<br />
<br />
<math>\text{(A) no value of } n \quad \text{(B) one value of } n \quad \text{(C) two values of } n \quad \text{(D) four values of } n \quad \text{(E) more than four values of } n</math><br />
<br />
[[1966 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
The negation of the proposition "For all pairs of real numbers <math>a,b</math>, if <math>a=0</math>, then <math>ab=0</math>" is: There are real numbers <math>a,b</math> such that<br />
<br />
<math>\text{(A) } a\ne 0 \text{ and } ab\ne 0 \qquad \text{(B) } a\ne 0 \text{ and } ab=0 \qquad \text{(C) } a=0 \text{ and } ab\ne 0</math><br />
<br />
<math>\text{(D) } ab\ne 0 \text{ and } a\ne 0 \qquad \text{(E) } ab=0 \text{ and } a\ne 0</math><br />
<br />
[[1966 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
<asy><br />
draw((0,-5)--(-6,10),black+dashed+linewidth(1));<br />
draw((-6,10)--(10,0),black+dashed+linewidth(1));<br />
draw((10,0)--(-10,0),black+dashed+linewidth(1));<br />
draw((-10,-0)--(10,10),black+dashed+linewidth(1));<br />
draw((10,10)--(0,-5),black+dashed+linewidth(1));<br />
draw((-2,0)--(-10/3,10/3),black+linewidth(2));<br />
draw((-10/3,10/3)--(10/9,50/9),black+linewidth(2));<br />
draw((10/9,50/9)--(90/17,50/17),black+linewidth(2));<br />
draw((90/17,50/17)--(10/3,0),black+linewidth(2));<br />
draw((10/3,0)--(-2,0),black+linewidth(2));<br />
MP("1", (1,0), N); MP("2", (-2.5,2), E); MP("3", (-.5,4.6), S); MP("4",(3.5,3.6), W);MP("5",(3.5,1), N);<br />
</asy><br />
<br />
An "<math>n</math>-pointed star" is formed as follows: the sides of a convex polygon are numbered consecutively <math>1,2,\cdots ,k,\cdots,n,\text{ }n\ge 5</math>; for all <math>n</math> values of <math>k</math>, sides <math>k</math> and <math>k+2</math> are non-parallel, sides <math>n+1</math> and <math>n+2</math> being respectively identical with sides <math>1</math> and <math>2</math>; prolong the <math>n</math> pairs of sides numbered <math>k</math> and <math>k+2</math> until they meet. (A figure is shown for the case <math>n=5</math>).<br />
<br />
Let <math>S</math> be the degree-sum of the interior angles at the <math>n</math> points of the star; then <math>S</math> equals:<br />
<br />
<math>\text{(A) } 180 \quad \text{(B) } 360 \quad \text{(C) } 180(n+2) \quad \text{(D) } 180(n-2) \quad \text{(E) } 180(n-4)</math><br />
<br />
[[1966 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
Consider the statements: (I)<math>\sqrt{a^2+b^2}=0</math>, (II) <math>\sqrt{a^2+b^2}=ab</math>, (III) <math>\sqrt{a^2+b^2}=a+b</math>, (IV) <math>\sqrt{a^2+b^2}=a\cdot b</math>, where we allow <math>a</math> and <math>b</math> to be real or complex numbers. Those statements for which there exist solutions other than <math>a=0</math> and <math>b=0</math>, are:<br />
<br />
<math>\text{(A) (I),(II),(III),(IV)}\quad \text{(B) (II),(III),(IV) only} \quad \text{(C) (I),(III),(IV) only}\quad \text{(D) (III),(IV) only} \quad \text{(E) (I) only}</math><br />
<br />
[[1966 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
If <math>x</math> is real and <math>4y^2+4xy+x+6=0</math>, then the complete set of values of <math>x</math> for which <math>y</math> is real, is:<br />
<br />
<math>\text{(A) } x\le-2 \text{ or } x\ge3 \quad \text{(B) } x\le2 \text{ or } x\ge3 \quad \text{(C) } x\le-3 \text{ or } x\ge2 \quad \\ \text{(D) } -3\le x\le2 \quad \text{(E) } -2\le x\le3</math><br />
<br />
[[1966 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
If <math>Log_M{N}=Log_N{M},M \ne N,MN>0,M \ne 1, N \ne 1</math>, then <math>MN</math> equals:<br />
<br />
<math>\text{(A) } \frac{1}{2} \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 10 \\ \text{(E) a number greater than 2 and less than 10}</math><br />
<br />
[[1966 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
If <math>F(n+1)=\frac{2F(n)+1}{2}</math> for <math>n=1,2,\cdots</math> and <math>F(1)=2</math>, then <math>F(101)</math> equals:<br />
<br />
<math>\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53</math><br />
<br />
[[1966 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26 ==<br />
Let <math>m</math> be a positive integer and let the lines <math>13x+11y=700</math> and <math>y=mx-1</math> intersect in a point whose coordinates are integers. Then m can be:<br />
<br />
<math>\text{(A) 4 only} \quad \text{(B) 5 only} \quad \text{(C) 6 only} \quad \text{(D) 7 only} \\ \text{(E) one of the integers 4,5,6,7 and one other positive integer}</math><br />
<br />
[[1966 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27 ==<br />
At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:<br />
<br />
<math>\text{(A) } 2 \quad \text{(B) } \frac{5}{2} \quad \text{(C) } 3 \quad \text{(D) } \frac{7}{2} \quad \text{(E) } 4</math><br />
<br />
[[1966 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
Five points <math>O,A,B,C,D</math> are taken in order on a straight line with distances <math>OA = a</math>, <math>OB = b</math>, <math>OC = c</math>, and <math>OD = d</math>. <math>P</math> is a point on the line between <math>B</math> and <math>C</math> and such that <math>AP: PD = BP: PC</math>. Then <math>OP</math> equals:<br />
<br />
<math>\textbf{(A)} \frac {b^2 - bc}{a - b + c - d} \qquad \textbf{(B)} \frac {ac - bd}{a - b + c - d} \\<br />
\textbf{(C)} - \frac {bd + ac}{a - b + c - d} \qquad \textbf{(D)} \frac {bc + ad}{a + b + c + d} \qquad \textbf{(E)} \frac {ac - bd}{a + b + c + d}</math><br />
<br />
[[1966 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
The number of positive integers less than <math>1000</math> divisible by neither <math>5</math> nor <math>7</math> is:<br />
<br />
<math>\text{(A) } 688 \quad \text{(B) } 686 \quad \text{(C) } 684 \quad \text{(D) } 658 \quad \text{(E) } 630</math><br />
<br />
[[1966 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
If three of the roots of <math>x^4+ax^2+bx+c=0</math> are <math>1</math>, <math>2</math>, and <math>3</math> then the value of <math>a+c</math> is:<br />
<br />
<math>\text{(A) } 35 \quad \text{(B) } 24 \quad \text{(C) } -12 \quad \text{(D) } -61 \quad \text{(E) } -63</math><br />
<br />
[[1966 AHSME Problems/Problem 30|Solution]]<br />
<br />
== Problem 31 ==<br />
<asy><br />
draw(circle((0,0),10),black+linewidth(1));<br />
draw(circle((-1.25,2.5),4.5),black+linewidth(1));<br />
dot((0,0));<br />
dot((-1.25,2.5));<br />
draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5));<br />
draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5));<br />
draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5));<br />
draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5));<br />
MP("O'", (0,0), W);<br />
MP("O", (-2,2), W);<br />
MP("A", (-10,-2), W);<br />
MP("B", (10,-2), E);<br />
MP("C", (-2,sqrt(96)), N);<br />
MP("D", (sqrt(96)-2.5,7), NE);<br />
</asy><br />
Triangle <math>ABC</math> is inscribed in a circle with center <math>O'</math>. A circle with center <math>O</math> is inscribed in triangle <math>ABC</math>. <math>AO</math> is drawn, and extended to intersect the larger circle in <math>D</math>. Then we must have:<br />
<br />
<math>\text{(A) } CD=BD=O'D \quad \text{(B) } AO=CO=OD \quad \text{(C) } CD=CO=BD \\ \text{(D) } CD=OD=BD \quad \text{(E) } O'B=O'C=OD</math><br />
<br />
[[1966 AHSME Problems/Problem 31|Solution]]<br />
<br />
== Problem 32 ==<br />
Let <math>M</math> be the midpoint of side <math>AB</math> of triangle <math>ABC</math>. Let <math>P</math> be a point on <math>AB</math> between <math>A</math> and <math>M</math>, and let <math>MD</math> be drawn parallel to <math>PC</math> and intersecting <math>BC</math> at <math>D</math>. If the ratio of the area of triangle <math>BPD</math> to that of triangle <math>ABC</math> is denoted by <math>r</math>, then<br />
<br />
<math>\text{(A) } \frac{1}{2}<r<1 \text{, depending upon the position of P} \\ \text{(B) } r=\frac{1}{2} \text{, independent of the position of P} \\ \text{(C) } \frac{1}{2} \le r <1 \text{, depending upon the position of P} \\ \text{(D) } \frac{1}{3}<r<\frac{2}{3} \text{, depending upon the position of P}\\ \text{(E) } r=\frac{1}{3} \text{, independent of the position of P}</math><br />
<br />
[[1966 AHSME Problems/Problem 32|Solution]]<br />
== Problem 33 ==<br />
If <math>ab \ne 0</math> and <math>|a| \ne |b|</math>, the number of distinct values of <math>x</math> satisfying the equation<br />
<br />
<cmath>\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},</cmath><br />
<br />
is:<br />
<br />
<math>\text{(A) zero} \quad \text{(B) one} \quad \text{(C) two} \quad \text{(D) three} \quad \text{(E) four} </math><br />
<br />
[[1966 AHSME Problems/Problem 33|Solution]]<br />
== Problem 34 ==<br />
Let <math>r</math> be the speed in miles per hour at which a wheel, <math>11</math> feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by <math>\frac{1}{4}</math> of a second, the speed <math>r</math> is increased by <math>5</math> miles per hour. Then <math>r</math> is:<br />
<br />
<math>\text{(A) } 9 \quad \text{(B) } 10 \quad \text{(C) } 10\frac{1}{2} \quad \text{(D) } 11 \quad \text{(E) } 12</math><br />
<br />
[[1966 AHSME Problems/Problem 34|Solution]]<br />
== Problem 35 ==<br />
Let <math>O</math> be an interior point of triangle <math>ABC</math>, and let <math>s_1=OA+OB+OC</math>. If <math>s_2=AB+BC+CA</math>, then<br />
<br />
<math>\text{(A) for every triangle } s_2>2s_1,s_1 \le s_2 \\ <br />
\text{(B) for every triangle } s_2>2s_1,s_1 < s_2 \\ <br />
\text{(C) for every triangle } s_1> \tfrac{1}{2}s_2,s_1 < s_2 \\ <br />
\text{(D) for every triangle } s_2\ge 2s_1,s_1 \le s_2 \\ <br />
\text{(E) neither (A) nor (B) nor (C) nor (D) applies to every triangle}</math><br />
<br />
[[1966 AHSME Problems/Problem 35|Solution]]<br />
== Problem 36 ==<br />
Let <math>(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}</math> be an identity in <math>x</math>. If we let <math>s=a_0+a_2+a_4+\cdots +a_{2n}</math>, then <math>s</math> equals:<br />
<br />
<math>\text{(A) } 2^n \quad \text{(B) } 2^n+1 \quad \text{(C) } \frac{3^n-1}{2} \quad \text{(D) } \frac{3^n}{2} \quad \text{(E) } \frac{3^n+1}{2}</math><br />
<br />
[[1966 AHSME Problems/Problem 36|Solution]]<br />
== Problem 37 ==<br />
Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let <math>h</math> be the number of hours needed by Alpha and Beta, working together, to do the job. Then <math>h</math> equals:<br />
<br />
<math>\text{(A) } \frac{5}{2} \quad \text{(B) } \frac{3}{2} \quad \text{(C) } \frac{4}{3} \quad \text{(D) } \frac{5}{4} \quad \text{(E) } \frac{3}{4}</math><br />
<br />
[[1966 AHSME Problems/Problem 37|Solution]]<br />
== Problem 38 ==<br />
In triangle <math>ABC</math> the medians <math>AM</math> and <math>CN</math> to sides <math>BC</math> and <math>AB</math>, respectively, intersect in point <math>O</math>. <math>P</math> is the midpoint of side <math>AC</math>, and <math>MP</math> intersects <math>CN</math> in <math>Q</math>. If the area of triangle <math>OMQ</math> is <math>n</math>, then the area of triangle <math>ABC</math> is:<br />
<br />
<math>\text{(A) } 16n \quad \text{(B) } 18n \quad \text{(C) } 21n \quad \text{(D) } 24n \quad \text{(E) } 27n</math><br />
<br />
[[1966 AHSME Problems/Problem 38|Solution]]<br />
== Problem 39 ==<br />
In base <math>R_1</math> the expanded fraction <math>F_1</math> becomes <math>.373737\cdots</math>, and the expanded fraction <math>F_2</math> becomes <math>.737373\cdots</math>. In base <math>R_2</math> fraction <math>F_1</math>, when expanded, becomes <math>.252525\cdots</math>, while the fraction <math>F_2</math> becomes <math>.525252\cdots</math>. The sum of <math>R_1</math> and <math>R_2</math>, each written in the base ten, is:<br />
<br />
<math>\text{(A) } 24 \quad \text{(B) } 22 \quad \text{(C) } 21 \quad \text{(D) } 20 \quad \text{(E) } 19</math><br />
<br />
[[1966 AHSME Problems/Problem 39|Solution]]<br />
== Problem 40 ==<br />
<asy><br />
draw(circle((0,0),10),black+linewidth(1));<br />
MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N);<br />
MP("a", (-5,0), S);MP("E", (-6,3), N);<br />
dot((0,0));dot((-6,2));<br />
draw((-10,0)--(10,0),black+linewidth(1));<br />
draw((-10,0)--(10,10),black+linewidth(1));<br />
draw((-10,-12)--(-10,12),black+linewidth(1));<br />
draw((10,-12)--(10,12),black+linewidth(1));<br />
</asy><br />
In this figure <math>AB</math> is a diameter of a circle, centered at <math>O</math>, with radius <math>a</math>. A chord <math>AD</math> is drawn and extended to meet the tangent to the circle at <math>B</math> in point <math>C</math>. Point <math>E</math> is taken on <math>AC</math> so the <math>AE=DC</math>. Denoting the distances of <math>E</math> from the tangent through <math>A</math> and from the diameter <math>AB</math> by <math>x</math> and <math>y</math>, respectively, we can deduce the relation:<br />
<br />
<math>\text{(A) } y^2=\frac{x^3}{2a-x} \quad \text{(B) } y^2=\frac{x^3}{2a+x} \quad \text{(C) } y^4=\frac{x^2}{2a-x} \\ \text{(D) } x^2=\frac{y^2}{2a-x} \quad \text{(E) } x^2=\frac{y^2}{2a+x}</math><br />
<br />
[[1966 AHSME Problems/Problem 40|Solution]]<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1966|before=[[1965 AHSME]]|after=[[1967 AHSME]]}} <br />
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{{MAA Notice}}</div>Brian6liu