https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=4everwise&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T07:05:48ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=FMO_Problems_and_Solutions&diff=8868FMO Problems and Solutions2006-07-30T19:57:22Z<p>4everwise: </p>
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<div>[[Flanders Mathematical Olympiad]] problems and solutions by test:<br />
<br />
* [[1986 FMO]]<br />
* [[1987 FMO]]<br />
* [[1988 FMO]]<br />
* [[1989 FMO]]<br />
* [[1990 FMO]]<br />
* [[1991 FMO]]<br />
* [[1992 FMO]]<br />
* [[1993 FMO]]<br />
* [[1994 FMO]]<br />
* [[1995 FMO]]<br />
* [[1996 FMO]]<br />
* [[1997 FMO]]<br />
* [[1998 FMO]]<br />
* [[1999 FMO]]<br />
* [[2000 FMO]]<br />
* [[2001 FMO]]<br />
* [[2002 FMO]]<br />
* [[2003 FMO]]<br />
* [[2004 FMO]]<br />
* [[2005 FMO]]<br />
* [[2006 FMO]]<br />
<br />
<br />
== Resources ==<br />
<br />
<br />
[[Category:Math Contest Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1970_Canadian_MO_Problems/Problem_5&diff=88531970 Canadian MO Problems/Problem 52006-07-30T17:51:55Z<p>4everwise: </p>
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<div>== Problem ==<br />
A quadrilateral has one vertex on each side of a square of side-length 1. Show that the lengths $a$, $b$, $c$ and $d$ of the sides of the quadrilateral satisfy the inequalities 2\le a^2+b^2+c^2+d^2\le 4.$<br />
<br />
<br />
== Solution ==<br />
<br />
<br />
<br />
----<br />
* [[1970 Canadian MO Problems/Problem 4|Previous Problem]]<br />
* [[1970 Canadian MO Problems/Problem 6|Next Problem]]<br />
* [[1970 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1970_Canadian_MO&diff=88521970 Canadian MO2006-07-30T17:47:07Z<p>4everwise: </p>
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<div>'''1971 Canadian MO''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br />
<br />
* [[1970 Canadian MO Problems]]<br />
* [[1970 Canadian MO Problems/Problem 1]] <br />
* [[1970 Canadian MO Problems/Problem 2]] <br />
* [[1970 Canadian MO Problems/Problem 3]] <br />
* [[1970 Canadian MO Problems/Problem 4]] <br />
* [[1970 Canadian MO Problems/Problem 5]] <br />
* [[1970 Canadian MO Problems/Problem 6]] <br />
* [[1970 Canadian MO Problems/Problem 7]] <br />
* [[1970 Canadian MO Problems/Problem 8]] <br />
* [[1970 Canadian MO Problems/Problem 9]] <br />
* [[1970 Canadian MO Problems/Problem 10]] <br />
<br />
== See also ==</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Canadian_MO_Problems_and_Solutions&diff=8850Canadian MO Problems and Solutions2006-07-30T17:46:30Z<p>4everwise: </p>
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<div>[[Canadian Mathematical Olympiad]] (CMO) problems and solutions by test:<br />
<br />
* [[1969 Canadian MO]]<br />
* [[1970 Canadian MO]]<br />
* [[1971 Canadian MO]]<br />
* [[1977 Canadian MO]]<br />
* [[2005 Canadian MO]]<br />
<br />
== Resources ==<br />
<br />
<br />
[[Category:Math Contest Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_8&diff=86761969 Canadian MO Problems/Problem 82006-07-28T22:53:12Z<p>4everwise: </p>
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<div>== Problem ==<br />
Let <math>\displaystyle f</math> be a function with the following properties:<br />
<br />
1) <math>\displaystyle f(n)</math> is defined for every positive integer <math>\displaystyle n</math>;<br />
<br />
2) <math>\displaystyle f(n)</math> is an integer;<br />
<br />
3) <math>\displaystyle f(2)=2</math>;<br />
<br />
4) <math>\displaystyle f(mn)=f(m)f(n)</math> for all <math>\displaystyle m</math> and <math>\displaystyle n</math>;<br />
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5) <math>\displaystyle f(m)>f(n)</math> whenever <math>m>n</math>.<br />
<br />
Prove that <math>\displaystyle f(n)=n</math>.<br />
<br />
== Solution ==<br />
It's easily shown that <math>\displaystyle f(1)=1</math> and <math>\displaystyle f(4)=4</math>. Since <math>\displaystyle f(2)<f(3)<f(4),</math> <math>\displaystyle f(3) = 3.</math><br />
<br />
Now, assume that <math>\displaystyle f(2n+2)=f(2(n+1))=f(2)f(n+1)=2n+2</math> is true for all <math>\displaystyle f(k)</math> where <math>\displaystyle k\leq 2n.</math> <br />
<br />
It follows that <math>\displaystyle 2n<f(2n+1)<2n+2.</math> Hence, <math>\displaystyle f(2n+1)=2n+1</math>, and by induction <math>\displaystyle f(n) = n.</math><br />
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----<br />
* [[1969 Canadian MO Problems/Problem 7|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 9|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_6&diff=86671969 Canadian MO Problems/Problem 62006-07-28T22:20:52Z<p>4everwise: </p>
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<div>== Problem ==<br />
Find the sum of <math>\displaystyle 1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!</math>, where <math>\displaystyle n!=n(n-1)(n-2)\cdots2\cdot1</math>.<br />
<br />
== Solution ==<br />
Note that for any positive integer <math>\displaystyle n,</math> <math>\displaystyle n\cdot n!+(n-1)\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!.</math><br />
Hence, pairing terms in the series will telescope most of the terms.<br />
<br />
If <math>\displaystyle n</math> is odd, <math>\displaystyle (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -2!+2!-0!.</math><br />
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If <math>\displaystyle n</math> is even, <math>\displaystyle (n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots -3!+3!-1!.</math><br />
In both cases, the expression telescopes into <math>\displaystyle (n+1)!-1.</math><br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 5|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 7|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_7&diff=86441969 Canadian MO Problems/Problem 72006-07-28T17:50:39Z<p>4everwise: </p>
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<div>== Problem ==<br />
Show that there are no integers <math>\displaystyle a,b,c</math> for which <math>\displaystyle a^2+b^2-8c=6</math>.<br />
<br />
== Solution ==<br />
Note that all perfect squares are equivilant to <math>\displaystyle 0,1,4\pmod8.</math> Hence, we have <math>\displaystyle a^2+b^2\equiv 6\pmod8.</math> It's impossible to obtain a sum of <math>\displaystyle 6</math> with two of <math>\displaystyle 0,1,4,</math> so our proof is complete.<br />
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----<br />
* [[1969 Canadian MO Problems/Problem 6|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 8|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_9&diff=86431969 Canadian MO Problems/Problem 92006-07-28T17:10:59Z<p>4everwise: </p>
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<div>== Problem ==<br />
Show that for any quadrilateral inscribed in a circle of radius <math>\displaystyle 1,</math> the length of the shortest side is less than or equal to <math>\displaystyle \sqrt{2}</math>.<br />
<br />
== Solution ==<br />
Let <math>\displaystyle a,b,c,d</math> be the sides and <math>\displaystyle e,f</math> be the diagonals. By Ptolemy's theorem, <math>\displaystyle ab+cd = ef</math>. However, the diameter is the longest possible diagonal, so <math>\displaystyle e,f \le 2</math> and <math>\displaystyle ab+cd \le 4</math>. <br />
<br />
If <math>\displaystyle a,b,c,d > \sqrt{2}</math>, then <math>\displaystyle ab+cd > 4,</math> which is impossible. Proof by contradiction.<br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 8|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 10|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_10&diff=86401969 Canadian MO Problems/Problem 102006-07-28T16:36:45Z<p>4everwise: </p>
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<div>== Problem ==<br />
Let <math>\displaystyle ABC</math> be the right-angled isosceles triangle whose equal sides have length 1. <math>\displaystyle P</math> is a point on the hypotenuse, and the feet of the perpendiculars from <math>\displaystyle P</math> to the other sides are <math>\displaystyle Q</math> and <math>\displaystyle R</math>. Consider the areas of the triangles <math>\displaystyle APQ</math> and <math>\displaystyle PBR</math>, and the area of the rectangle <math>\displaystyle QCRP</math>. Prove that regardless of how <math>\displaystyle P</math> is chosen, the largest of these three areas is at least <math>\displaystyle 2/9</math>.<br />
<br />
== Solution ==<br />
Let <math>\displaystyle AQ=x.</math> Because triangles <math>\displaystyle APQ</math> and <math>\displaystyle BPR</math> both contain a right and a <math>\displaystyle 45^\circ</math> angle, they are isosceles-right. Hence, <math>\displaystyle PQ=RC=x</math> and <math>\displaystyle QC=PR=BR=1-x.</math><br />
<br />
Now let's consider when <math>\displaystyle \frac13 <x<\frac23,</math> or else one of triangles <math>\displaystyle APQ</math> and <math>\displaystyle PBR</math> will automatically have area greater than <math>\displaystyle \frac29.</math> In this case, <math>\displaystyle [QCRP]>[ABC]-[APQ]-[PBR]>\frac29.</math> Therefore, one of these three figures will always have area greater than <math>\displaystyle \frac29,</math> regardless of where <math>\displaystyle P</math> is chosen.<br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 9|Previous Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_5&diff=86041969 Canadian MO Problems/Problem 52006-07-28T06:04:08Z<p>4everwise: </p>
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<div>== Problem ==<br />
Let <math>\displaystyle ABC</math> be a triangle with sides of length <math>\displaystyle a</math>, <math>\displaystyle b</math> and <math>\displaystyle c</math>. Let the bisector of the <math>\displaystyle \angle C</math> cut <math>\displaystyle AB</math> at <math>\displaystyle D</math>. Prove that the length of <math>\displaystyle CD</math> is <math>\displaystyle \frac{2ab\cos \frac{C}{2}}{a+b}.</math><br />
<br />
<br />
== Solution ==<br />
Let <math>\displaystyle CD=d.</math> Note that <math>\displaystyle [\triangle ABC]=[\triangle ACD]+[\triangle BCD].</math> This can be rewritten as <math>\displaystyle \frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .</math><br />
<br />
Because <math>\displaystyle \sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math>\displaystyle2ab\cos \frac C2=d(a+b).</math> Dividing by <math>\displaystyle a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.<br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 4|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 6|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_4&diff=86031969 Canadian MO Problems/Problem 42006-07-28T05:58:40Z<p>4everwise: </p>
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<div>== Problem ==<br />
Let <math>\displaystyle ABC</math> be an equilateral triangle, and <math>\displaystyle P</math> be an arbitrary point within the triangle. Perpendiculars <math>\displaystyle PD,PE,PF</math> are drawn to the three sides of the triangle. Show that, no matter where <math>\displaystyle P</math> is chosen, <math>\displaystyle \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}</math>.<br />
<br />
== Solution ==<br />
Let a side of the triangle be <math>\displaystyle s</math> and let <math>\displaystyle [ABC]</math> denote the area of <math>\displaystyle ABC.</math> Note that because <math>\displaystyle 2[\triangle ABC]=2[\triangle APB]+2[\triangle APC]+2[\triangle BPC],</math> <math>\displaystyle\frac{\sqrt3}{2}s^2=s(PD+PE+PF).</math> Dividing both sides by <math>\displaystyle s</math>, the sum of the perpendiculars from <math>\displaystyle P</math> equals <math>PD+PE+PF=\frac{\sqrt3}{2}s.</math> (It is independant of point <math>\displaystyle P</math>) Because the sum of the sides is <math>\displaystyle 3s</math>, the ratio is always <math>\displaystyle\cfrac{s\frac{\sqrt3}{2}}{3s}=\frac{1}{2\sqrt3}.</math><br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 3|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 5|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_3&diff=86021969 Canadian MO Problems/Problem 32006-07-28T05:54:21Z<p>4everwise: </p>
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<div>== Problem ==<br />
Let <math>\displaystyle c</math> be the length of the hypotenuse of a right angle triangle whose two other sides have lengths <math>\displaystyle a</math> and <math>\displaystyle b</math>. Prove that <math>\displaystyle a+b\le c\sqrt{2}</math>. When does the equality hold?<br />
<br />
== Solution ==<br />
Since <math>\displaystyle a,b,c</math> are all positive, squaring preserves the inequality; <math>\displaystyle 2c^2\ge (a+b)^2.</math><br />
<br />
By the Pythagorean Theorem, <math>\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0,</math> since the square of a real number is always positive.<br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 2|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 4|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_2&diff=86011969 Canadian MO Problems/Problem 22006-07-28T05:44:20Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
Determine which of the two numbers <math>\displaystyle \sqrt{c+1}-\sqrt{c}</math>, <math>\displaystyle\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>\displaystyle c\ge 1</math>.<br />
<br />
== Solution ==<br />
Multiplying and dividing <math>\displaystyle \sqrt{c+1}-\sqrt c</math> by its conjugate,<br />
<br />
<math>\displaystyle \sqrt{c+1}-\sqrt c=\frac{(\sqrt{c+1})^2-(\sqrt c)^2}{\sqrt{c+1}+\sqrt{c}}=\frac1{\sqrt{c+1}+\sqrt{c}}.</math><br />
<br />
Similarly, <math>\displaystyle \sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c-\sqrt{c-1}}</math>. We know that <math>\displaystyle \frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c-\sqrt{c-1}}</math> for all positive <math>\displaystyle c</math>, so <math>\displaystyle \sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}</math>.<br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 1|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 3|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_1&diff=86001969 Canadian MO Problems/Problem 12006-07-28T05:35:53Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
Show that if <math>\displaystyle a_1/b_1=a_2/b_2=a_3/b_3</math> and <math>\displaystyle p_1,p_2,p_3</math> are not all zero, then <math>\displaystyle\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}</math> for every positive integer <math>\displaystyle n.</math><br />
<br />
== Solution ==<br />
Instead of proving the two expressions equal, we prove that their difference equals zero.<br />
<br />
Subtracting the LHS from the RHS,<br />
<math>0=\displaystyle \frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}-\frac{a_1^n}{b_1^n}.</math><br />
<br />
Finding a common denominator, the numerator becomes<br />
<math>\displaystyle b_1^n(p_1a_1^n+p_2a_2^n+p_3a_3^n)-a_1^n(p_1b_1^n+p_2b_2^n+p_3b_3^n)=p_2(a_2^nb_1^n-a_1^nb_2^n)+p_3(a_3^nb_1^n-a_1^nb_3^n)=0.</math><br />
(The denominator is irrelevant since it never equals zero)<br />
<br />
From <math>\displaystyle a_1/b_1=a_2b_2,</math> <math>\displaystyle a_1^nb_2^n=a_2^nb_1^n.</math> Similarly, <math>\displaystyle a_1^nb_3^n=a_3^nb_1^n</math> from <math>\displaystyle a_1/b_1=a_3/b_3.</math> <br />
<br />
Hence, <math>\displaystyle a_2^nb_1^n-a_1^nb_2^n=a_3^nb_1^n-a_1^nb_3^n=0</math> and our proof is complete.<br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 2|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_3&diff=85161992 AIME Problems/Problem 32006-07-26T23:21:10Z<p>4everwise: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly <math>\displaystyle.500</math>. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than <math>\displaystyle .503</math>. What's the largest number of matches she could've won before the weekend began?<br />
<br />
== Solution ==<br />
Let <math>\displaystyle n</math> be the number of matches won, so that <math>\displaystyle \frac{n}{2n}=\frac{1}{2}</math>, and <math>\displaystyle \frac{n+3}{2n+4}>\frac{503}{1000}</math>. Cross multiplying, <math>\displaystyle 1000n+3000>1006n+2012</math>, and <math>\displaystyle n<\frac{988}{6}</math>. Thus, the answer is <math>\displaystyle 164</math>.<br />
<br />
== See also ==<br />
* [[1992 AIME Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_3&diff=85151992 AIME Problems/Problem 32006-07-26T23:14:06Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly <math>\displaystyle.500</math>. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than <math>\displaystyle .503</math>. What's the largest number of matches she could've won before the weekend began?<br />
<br />
== Solution ==<br />
<br />
== See also ==<br />
* [[1992 AIME Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_2&diff=85141992 AIME Problems/Problem 22006-07-26T23:12:48Z<p>4everwise: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?<br />
<br />
== Solution ==<br />
First, we line up the nine digits that may be used: <math>\displaystyle 1,2,3,4,5,6,7,8,9.</math> Note that each digit may be present or may not be present. Hence, there are <math>\displaystyle 2^9=512</math> working numbers.<br />
<br />
Subtracting off the one-digit numbers and the null set, our answer is <math>\displaystyle 512-10=502.</math><br />
<br />
== See also ==<br />
* [[1992 AIME Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_2&diff=85131992 AIME Problems/Problem 22006-07-26T23:10:05Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?<br />
<br />
== Solution ==<br />
<br />
== See also ==<br />
* [[1992 AIME Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_1&diff=85121992 AIME Problems/Problem 12006-07-26T23:08:40Z<p>4everwise: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms.<br />
<br />
== Solution ==<br />
There are 8 fractions which fit the conditions from 0 to 1: <math>\displaystyle \frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}</math><br />
<br />
Their sum is 4. Note that there are also 8 terms from 1 to 2, and we can obtain them by adding 1 to each of our first 8 terms. For example, <math>\displaystyle 1+\frac{19}{30}=\frac{49}{30}.</math> Following this pattern, our answer is <math>4(10)+8(1+2+3+\cdots+9)=400.</math><br />
<br />
== See also ==<br />
* [[1992 AIME Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_1&diff=85111992 AIME Problems/Problem 12006-07-26T23:05:31Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms.<br />
<br />
== Solution ==<br />
<br />
== See also ==<br />
* [[1992 AIME Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO_Problems/Problem_4&diff=85081971 Canadian MO Problems/Problem 42006-07-26T20:21:57Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Determine all real numbers <math>\displaystyle a</math> such that the two polynomials <math>\displaystyle x^2+ax+1</math> and <math>\displaystyle x^2+x+a</math> have at least one root in common.<br />
<br />
== Solution == <br />
<br />
----<br />
* [[1971 Canadian MO Problems/Problem 3|Previous Problem]]<br />
* [[1971 Canadian MO Problems/Problem 5|Next Problem]]<br />
* [[1971 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO_Problems/Problem_4&diff=85071971 Canadian MO Problems/Problem 42006-07-26T20:21:41Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
Determine all real numbers <math>a</math> such that the two polynomials <math>\displaystyle x^2+ax+1</math> and <math>\displaystyle x^2+x+a</math> have at least one root in common. <br />
<br />
== Solution == <br />
<br />
----<br />
* [[1971 Canadian MO Problems/Problem 3|Previous Problem]]<br />
* [[1971 Canadian MO Problems/Problem 5|Next Problem]]<br />
* [[1971 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO_Problems/Problem_1&diff=84991971 Canadian MO Problems/Problem 12006-07-26T19:43:48Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
<math>\displaystyle DEB</math> is a chord of a circle such that <math>\displaystyle DE=3</math> and <math>\displaystyle EB=5 .</math> Let <math>\displaystyle O</math> be the center of the circle. Join <math>\displaystyle OE</math> and extend <math>\displaystyle OE</math> to cut the circle at <math>\displaystyle C.</math> Given <math>\displaystyle EC=1,</math> find the radius of the circle<br />
<br />
[[Image:CanadianMO_1971-1.jpg]]<br />
<br />
== Solution ==<br />
First, extend <math>\displaystyle CO</math> to meet the circle at <math>\displaystyle P.</math> Let the radius be <math>\displaystyle r.</math> Applying Power of a Point,<br />
<math>\displaystyle (EP)(CE)=(BE)(ED)</math> and <math>\displaystyle 2r-1=15.</math> Hence, <math>\displaystyle r=8.</math> <br />
<br />
----<br />
* [[1971 Canadian MO Problems/Problem 2|Next Problem]]<br />
* [[1971 Canadian MO Problems|Back to Exam]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=File:CanadianMO_1971-1.jpg&diff=8462File:CanadianMO 1971-1.jpg2006-07-25T22:21:58Z<p>4everwise: </p>
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<div></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=1971_Canadian_MO&diff=84611971 Canadian MO2006-07-25T22:16:01Z<p>4everwise: </p>
<hr />
<div>'''1971 Canadian MO''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.<br />
<br />
* [[1971 Canadian MO Problems]]<br />
* [[1971 Canadian MO Problems/Problem 1]] <br />
* [[1971 Canadian MO Problems/Problem 2]] <br />
* [[1971 Canadian MO Problems/Problem 3]] <br />
* [[1971 Canadian MO Problems/Problem 4]] <br />
* [[1971 Canadian MO Problems/Problem 5]] <br />
* [[1971 Canadian MO Problems/Problem 6]] <br />
* [[1971 Canadian MO Problems/Problem 7]] <br />
* [[1971 Canadian MO Problems/Problem 8]] <br />
* [[1971 Canadian MO Problems/Problem 9]] <br />
* [[1971 Canadian MO Problems/Problem 10]] <br />
<br />
== See also ==</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Canadian_MO_Problems_and_Solutions&diff=8460Canadian MO Problems and Solutions2006-07-25T22:15:14Z<p>4everwise: </p>
<hr />
<div>[[Canadian Mathematical Olympiad]] (CMO) problems and solutions by test:<br />
<br />
* [[1969 Canadian MO]]<br />
* [[1971 Canadian MO]]<br />
* [[1977 Canadian MO]]<br />
<br />
<br />
== Resources ==<br />
<br />
<br />
[[Category:Math Contest Problems]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=File:Mock_AIME_2_2007_Problem8.jpg&diff=8439File:Mock AIME 2 2007 Problem8.jpg2006-07-25T18:10:32Z<p>4everwise: </p>
<hr />
<div></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_7&diff=8438Mock AIME 2 2006-2007 Problems/Problem 72006-07-25T18:09:55Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
<br />
A right circular cone of base radius <math>\displaystyle 17</math>cm and slant height <math>\displaystyle 34</math>cm is given. <math>\displaystyle P</math> is a point on the circumference of the base and the shortest path from <math>\displaystyle P</math> around the cone and back is drawn (see diagram). If the length of this path is <math>\displaystyle m\sqrt{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
[[Image:Mock_AIME_2_2007_Problem8.jpg]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems&diff=8437Mock AIME 2 2006-2007 Problems2006-07-25T18:09:22Z<p>4everwise: /* Problem 7 */</p>
<hr />
<div>== Problem 1 ==<br />
A positive integer is called a dragon if it can be [[partitioned]] into four positive integers <math>\displaystyle a,b,c,</math> and <math>\displaystyle d</math> such that <math>\displaystyle a+4=b-4=4c=d/4.</math> Find the smallest dragon.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_1|Solution]]<br />
<br />
== Problem 2 ==<br />
The set <math>\displaystyle S</math> consists of all integers from <math>\displaystyle 1</math> to <math>\displaystyle 2007,</math> inclusive. For how many elements <math>\displaystyle n</math> in <math>\displaystyle S</math> is <math>\displaystyle f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_2|Solution]]<br />
<br />
== Problem 3 ==<br />
Let <math>\displaystyle S</math> be the sum of all positive integers <math>\displaystyle n</math> such that <math>\displaystyle n^2+12n-2007</math> is a perfect square. Find the remainder when <math>\displaystyle S</math> is divided by <math>\displaystyle 1000.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_3|Solution]]<br />
<br />
== Problem 4 ==<br />
Let <math>\displaystyle n</math> be the smallest positive integer for which there exist positive real numbers <math>\displaystyle a</math> and <math>\displaystyle b</math> such that <math>\displaystyle (a+bi)^n=(a-bi)^n</math>. Compute <math>\displaystyle \frac{b^2}{a^2}</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_4|Solution]]<br />
<br />
== Problem 5 ==<br />
Given that <math>\displaystyle iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>\displaystyle z=n\pm \sqrt{-i},</math> find <math>\displaystyle \lfloor 100n \rfloor.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_5|Solution]]<br />
<br />
== Problem 6 ==<br />
If <math>\displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>\displaystyle 0^\circ \le \theta \le 180^\circ, </math> find <math>\displaystyle \theta.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_6|Solution]]<br />
<br />
== Problem 7 ==<br />
A right circular cone of base radius <math>\displaystyle 17</math>cm and slant height <math>\displaystyle 34</math>cm is given. <math>\displaystyle P</math> is a point on the circumference of the base and the shortest path from <math>\displaystyle P</math> around the cone and back is drawn (see diagram). If the length of this path is <math>\displaystyle m\sqrt{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
[[Image:Mock_AIME_2_2007_Problem8.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_7|Solution]]<br />
<br />
== Problem 8 ==<br />
The positive integers <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three digits of <math>\displaystyle x_7</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_8|Solution]]<br />
<br />
== Problem 9 ==<br />
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> intersect at <math>\displaystyle P</math> and are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_9|Solution]]<br />
<br />
== Problem 10 ==<br />
Find the number of solutions, in degrees, to the equation <math>\displaystyle \sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>\displaystyle 0^\circ \le x^\circ \le 2007^\circ.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_10|Solution]]<br />
<br />
== Problem 11 ==<br />
Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations<br />
<br />
<math>\displaystyle x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_11|Solution]]<br />
<br />
== Problem 12 ==<br />
In quadrilateral <math>\displaystyle ABCD,</math> <math>\displaystyle m \angle DAC= m\angle DBC </math> and <math>\displaystyle \frac{[ADB]}{[ABC]}=\frac12.</math> If <math>\displaystyle AD=4,</math> <math>\displaystyle BC=6</math>, <math>\displaystyle BO=1,</math> and the area of <math>\displaystyle ABCD</math> is <math>\displaystyle \frac{a\sqrt{b}}{c},</math> where <math>\displaystyle a,b,c</math> are relatively prime positive integers, find <math>\displaystyle a+b+c.</math><br />
<br />
<br />
Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of triangles <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_12|Solution]]<br />
<br />
== Problem 13 ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle m/n,</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
In triangle ABC, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35.</math> Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> intersect at <math>\displaystyle P</math> and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of <math>\displaystyle BC.</math><br />
<br />
[[Image:Mock AIME 2 2007 Problem14.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is <math>\mathfrak{Intriguing}</math> if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of <math>\mathfrak{Intriguing}</math> colorings.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_15|Solution]]<br />
<br />
[[Image:CubeArt.jpg]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=File:Mock_AIME_2_2007_Problem7.jpg&diff=8436File:Mock AIME 2 2007 Problem7.jpg2006-07-25T18:07:11Z<p>4everwise: </p>
<hr />
<div></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_9&diff=8435Mock AIME 2 2006-2007 Problems/Problem 92006-07-25T18:05:37Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> intersect at <math>\displaystyle P</math> and are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems&diff=8434Mock AIME 2 2006-2007 Problems2006-07-25T18:04:41Z<p>4everwise: /* Problem 9 */</p>
<hr />
<div>== Problem 1 ==<br />
A positive integer is called a dragon if it can be [[partitioned]] into four positive integers <math>\displaystyle a,b,c,</math> and <math>\displaystyle d</math> such that <math>\displaystyle a+4=b-4=4c=d/4.</math> Find the smallest dragon.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_1|Solution]]<br />
<br />
== Problem 2 ==<br />
The set <math>\displaystyle S</math> consists of all integers from <math>\displaystyle 1</math> to <math>\displaystyle 2007,</math> inclusive. For how many elements <math>\displaystyle n</math> in <math>\displaystyle S</math> is <math>\displaystyle f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_2|Solution]]<br />
<br />
== Problem 3 ==<br />
Let <math>\displaystyle S</math> be the sum of all positive integers <math>\displaystyle n</math> such that <math>\displaystyle n^2+12n-2007</math> is a perfect square. Find the remainder when <math>\displaystyle S</math> is divided by <math>\displaystyle 1000.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_3|Solution]]<br />
<br />
== Problem 4 ==<br />
Let <math>\displaystyle n</math> be the smallest positive integer for which there exist positive real numbers <math>\displaystyle a</math> and <math>\displaystyle b</math> such that <math>\displaystyle (a+bi)^n=(a-bi)^n</math>. Compute <math>\displaystyle \frac{b^2}{a^2}</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_4|Solution]]<br />
<br />
== Problem 5 ==<br />
Given that <math>\displaystyle iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>\displaystyle z=n\pm \sqrt{-i},</math> find <math>\displaystyle \lfloor 100n \rfloor.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_5|Solution]]<br />
<br />
== Problem 6 ==<br />
If <math>\displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>\displaystyle 0^\circ \le \theta \le 180^\circ, </math> find <math>\displaystyle \theta.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_6|Solution]]<br />
<br />
== Problem 7 ==<br />
A right circular cone of base radius <math>\displaystyle 17</math>cm and slant height <math>\displaystyle 34</math>cm is given. <math>\displaystyle P</math> is a point on the circumference of the base and the shortest path from <math>\displaystyle P</math> around the cone and back is drawn (see diagram). If the minimum distance from the vertex <math>\displaystyle V</math> to this path is <math>\displaystyle m\sqrt{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math> <br />
<br />
[[Image:Mock_AIME_2_2007_Problem7.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_7|Solution]]<br />
<br />
== Problem 8 ==<br />
The positive integers <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three digits of <math>\displaystyle x_7</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_8|Solution]]<br />
<br />
== Problem 9 ==<br />
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> intersect at <math>\displaystyle P</math> and are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_9|Solution]]<br />
<br />
== Problem 10 ==<br />
Find the number of solutions, in degrees, to the equation <math>\displaystyle \sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>\displaystyle 0^\circ \le x^\circ \le 2007^\circ.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_10|Solution]]<br />
<br />
== Problem 11 ==<br />
Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations<br />
<br />
<math>\displaystyle x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_11|Solution]]<br />
<br />
== Problem 12 ==<br />
In quadrilateral <math>\displaystyle ABCD,</math> <math>\displaystyle m \angle DAC= m\angle DBC </math> and <math>\displaystyle \frac{[ADB]}{[ABC]}=\frac12.</math> If <math>\displaystyle AD=4,</math> <math>\displaystyle BC=6</math>, <math>\displaystyle BO=1,</math> and the area of <math>\displaystyle ABCD</math> is <math>\displaystyle \frac{a\sqrt{b}}{c},</math> where <math>\displaystyle a,b,c</math> are relatively prime positive integers, find <math>\displaystyle a+b+c.</math><br />
<br />
<br />
Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of triangles <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_12|Solution]]<br />
<br />
== Problem 13 ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle m/n,</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
In triangle ABC, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35.</math> Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> intersect at <math>\displaystyle P</math> and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of <math>\displaystyle BC.</math><br />
<br />
[[Image:Mock AIME 2 2007 Problem14.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is <math>\mathfrak{Intriguing}</math> if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of <math>\mathfrak{Intriguing}</math> colorings.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_15|Solution]]<br />
<br />
[[Image:CubeArt.jpg]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_6&diff=8424Mock AIME 2 2006-2007 Problems/Problem 62006-07-25T17:05:28Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
If <math>\displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>\displaystyle 0^\circ \le \theta \le 180^\circ, </math> find <math>\displaystyle \theta.</math></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_5&diff=8423Mock AIME 2 2006-2007 Problems/Problem 52006-07-25T17:05:11Z<p>4everwise: </p>
<hr />
<div>==Problem==<br />
<br />
Given that <math>\displaystyle iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>\displaystyle z=n\pm \sqrt{-i},</math> find <math>\displaystyle \lfloor 100n \rfloor.</math></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems&diff=8422Mock AIME 2 2006-2007 Problems2006-07-25T17:03:56Z<p>4everwise: /* Problem 5 */</p>
<hr />
<div>== Problem 1 ==<br />
A positive integer is called a dragon if it can be [[partitioned]] into four positive integers <math>\displaystyle a,b,c,</math> and <math>\displaystyle d</math> such that <math>\displaystyle a+4=b-4=4c=d/4.</math> Find the smallest dragon.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_1|Solution]]<br />
<br />
== Problem 2 ==<br />
The set <math>\displaystyle S</math> consists of all integers from <math>\displaystyle 1</math> to <math>\displaystyle 2007,</math> inclusive. For how many elements <math>\displaystyle n</math> in <math>\displaystyle S</math> is <math>\displaystyle f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_2|Solution]]<br />
<br />
== Problem 3 ==<br />
Let <math>\displaystyle S</math> be the sum of all positive integers <math>\displaystyle n</math> such that <math>\displaystyle n^2+12n-2007</math> is a perfect square. Find the remainder when <math>\displaystyle S</math> is divided by <math>\displaystyle 1000.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_3|Solution]]<br />
<br />
== Problem 4 ==<br />
Let <math>\displaystyle n</math> be the smallest positive integer for which there exist positive real numbers <math>\displaystyle a</math> and <math>\displaystyle b</math> such that <math>\displaystyle (a+bi)^n=(a-bi)^n</math>. Compute <math>\displaystyle \frac{b^2}{a^2}</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_4|Solution]]<br />
<br />
== Problem 5 ==<br />
Given that <math>\displaystyle iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>\displaystyle z=n\pm \sqrt{-i},</math> find <math>\displaystyle \lfloor 100n \rfloor.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_5|Solution]]<br />
<br />
== Problem 6 ==<br />
If <math>\displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>\displaystyle 0^\circ \le \theta \le 180^\circ, </math> find <math>\displaystyle \theta.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_6|Solution]]<br />
<br />
== Problem 7 ==<br />
A right circular cone of base radius <math>\displaystyle 17</math>cm and slant height <math>\displaystyle 34</math>cm is given. <math>\displaystyle P</math> is a point on the circumference of the base and the shortest path from <math>\displaystyle P</math> around the cone and back is drawn (see diagram). If the minimum distance from the vertex <math>\displaystyle V</math> to this path is <math>\displaystyle m\sqrt{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math> <br />
<br />
[[Image:Mock_AIME_2_2007_Problem7.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_7|Solution]]<br />
<br />
== Problem 8 ==<br />
The positive integers <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three digits of <math>\displaystyle x_7</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_8|Solution]]<br />
<br />
== Problem 9 ==<br />
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_9|Solution]]<br />
<br />
== Problem 10 ==<br />
Find the number of solutions, in degrees, to the equation <math>\displaystyle \sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>\displaystyle 0^\circ \le x^\circ \le 2007^\circ.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_10|Solution]]<br />
<br />
== Problem 11 ==<br />
Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations<br />
<br />
<math>\displaystyle x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_11|Solution]]<br />
<br />
== Problem 12 ==<br />
In quadrilateral <math>\displaystyle ABCD,</math> <math>\displaystyle m \angle DAC= m\angle DBC </math> and <math>\displaystyle \frac{[ADB]}{[ABC]}=\frac12.</math> If <math>\displaystyle AD=4,</math> <math>\displaystyle BC=6</math>, <math>\displaystyle BO=1,</math> and the area of <math>\displaystyle ABCD</math> is <math>\displaystyle \frac{a\sqrt{b}}{c},</math> where <math>\displaystyle a,b,c</math> are relatively prime positive integers, find <math>\displaystyle a+b+c.</math><br />
<br />
<br />
Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of triangles <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_12|Solution]]<br />
<br />
== Problem 13 ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle m/n,</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
In triangle ABC, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35.</math> Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> intersect at <math>\displaystyle P</math> and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of <math>\displaystyle BC.</math><br />
<br />
[[Image:Mock AIME 2 2007 Problem14.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is <math>\mathfrak{Intriguing}</math> if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of <math>\mathfrak{Intriguing}</math> colorings.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_15|Solution]]<br />
<br />
[[Image:CubeArt.jpg]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems&diff=8421Mock AIME 2 2006-2007 Problems2006-07-25T16:52:22Z<p>4everwise: /* Problem 6 */</p>
<hr />
<div>== Problem 1 ==<br />
A positive integer is called a dragon if it can be [[partitioned]] into four positive integers <math>\displaystyle a,b,c,</math> and <math>\displaystyle d</math> such that <math>\displaystyle a+4=b-4=4c=d/4.</math> Find the smallest dragon.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_1|Solution]]<br />
<br />
== Problem 2 ==<br />
The set <math>\displaystyle S</math> consists of all integers from <math>\displaystyle 1</math> to <math>\displaystyle 2007,</math> inclusive. For how many elements <math>\displaystyle n</math> in <math>\displaystyle S</math> is <math>\displaystyle f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_2|Solution]]<br />
<br />
== Problem 3 ==<br />
Let <math>\displaystyle S</math> be the sum of all positive integers <math>\displaystyle n</math> such that <math>\displaystyle n^2+12n-2007</math> is a perfect square. Find the remainder when <math>\displaystyle S</math> is divided by <math>\displaystyle 1000.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_3|Solution]]<br />
<br />
== Problem 4 ==<br />
Let <math>\displaystyle n</math> be the smallest positive integer for which there exist positive real numbers <math>\displaystyle a</math> and <math>\displaystyle b</math> such that <math>\displaystyle (a+bi)^n=(a-bi)^n</math>. Compute <math>\displaystyle \frac{b^2}{a^2}</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_4|Solution]]<br />
<br />
== Problem 5 ==<br />
Given that <math>\displaystyle iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>\displaystyle z=a\pm \sqrt{b+i},</math> find <math>\displaystyle \lfloor 100ab \rfloor.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_5|Solution]]<br />
<br />
== Problem 6 ==<br />
If <math>\displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>\displaystyle 0^\circ \le \theta \le 180^\circ, </math> find <math>\displaystyle \theta.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_6|Solution]]<br />
<br />
== Problem 7 ==<br />
A right circular cone of base radius <math>\displaystyle 17</math>cm and slant height <math>\displaystyle 34</math>cm is given. <math>\displaystyle P</math> is a point on the circumference of the base and the shortest path from <math>\displaystyle P</math> around the cone and back is drawn (see diagram). If the minimum distance from the vertex <math>\displaystyle V</math> to this path is <math>\displaystyle m\sqrt{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math> <br />
<br />
[[Image:Mock_AIME_2_2007_Problem7.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_7|Solution]]<br />
<br />
== Problem 8 ==<br />
The positive integers <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three digits of <math>\displaystyle x_7</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_8|Solution]]<br />
<br />
== Problem 9 ==<br />
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_9|Solution]]<br />
<br />
== Problem 10 ==<br />
Find the number of solutions, in degrees, to the equation <math>\displaystyle \sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>\displaystyle 0^\circ \le x^\circ \le 2007^\circ.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_10|Solution]]<br />
<br />
== Problem 11 ==<br />
Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations<br />
<br />
<math>\displaystyle x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_11|Solution]]<br />
<br />
== Problem 12 ==<br />
In quadrilateral <math>\displaystyle ABCD,</math> <math>\displaystyle m \angle DAC= m\angle DBC </math> and <math>\displaystyle \frac{[ADB]}{[ABC]}=\frac12.</math> If <math>\displaystyle AD=4,</math> <math>\displaystyle BC=6</math>, <math>\displaystyle BO=1,</math> and the area of <math>\displaystyle ABCD</math> is <math>\displaystyle \frac{a\sqrt{b}}{c},</math> where <math>\displaystyle a,b,c</math> are relatively prime positive integers, find <math>\displaystyle a+b+c.</math><br />
<br />
<br />
Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of triangles <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_12|Solution]]<br />
<br />
== Problem 13 ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle m/n,</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
In triangle ABC, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35.</math> Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> intersect at <math>\displaystyle P</math> and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of <math>\displaystyle BC.</math><br />
<br />
[[Image:Mock AIME 2 2007 Problem14.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is <math>\mathfrak{Intriguing}</math> if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of <math>\mathfrak{Intriguing}</math> colorings.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_15|Solution]]<br />
<br />
[[Image:CubeArt.jpg]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems&diff=8420Mock AIME 2 2006-2007 Problems2006-07-25T16:08:07Z<p>4everwise: </p>
<hr />
<div>== Problem 1 ==<br />
A positive integer is called a dragon if it can be [[partitioned]] into four positive integers <math>\displaystyle a,b,c,</math> and <math>\displaystyle d</math> such that <math>\displaystyle a+4=b-4=4c=d/4.</math> Find the smallest dragon.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_1|Solution]]<br />
<br />
== Problem 2 ==<br />
The set <math>\displaystyle S</math> consists of all integers from <math>\displaystyle 1</math> to <math>\displaystyle 2007,</math> inclusive. For how many elements <math>\displaystyle n</math> in <math>\displaystyle S</math> is <math>\displaystyle f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_2|Solution]]<br />
<br />
== Problem 3 ==<br />
Let <math>\displaystyle S</math> be the sum of all positive integers <math>\displaystyle n</math> such that <math>\displaystyle n^2+12n-2007</math> is a perfect square. Find the remainder when <math>\displaystyle S</math> is divided by <math>\displaystyle 1000.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_3|Solution]]<br />
<br />
== Problem 4 ==<br />
Let <math>\displaystyle n</math> be the smallest positive integer for which there exist positive real numbers <math>\displaystyle a</math> and <math>\displaystyle b</math> such that <math>\displaystyle (a+bi)^n=(a-bi)^n</math>. Compute <math>\displaystyle \frac{b^2}{a^2}</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_4|Solution]]<br />
<br />
== Problem 5 ==<br />
Given that <math>\displaystyle iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>\displaystyle z=a\pm \sqrt{b+i},</math> find <math>\displaystyle \lfloor 100ab \rfloor.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_5|Solution]]<br />
<br />
== Problem 6 ==<br />
If <math>\displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ =\tan \theta</math> and <math>\displaystyle 0^\circ \le \theta \le 360^\circ, </math> find <math>\displaystyle \theta.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_6|Solution]]<br />
<br />
== Problem 7 ==<br />
A right circular cone of base radius <math>\displaystyle 17</math>cm and slant height <math>\displaystyle 34</math>cm is given. <math>\displaystyle P</math> is a point on the circumference of the base and the shortest path from <math>\displaystyle P</math> around the cone and back is drawn (see diagram). If the minimum distance from the vertex <math>\displaystyle V</math> to this path is <math>\displaystyle m\sqrt{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math> <br />
<br />
[[Image:Mock_AIME_2_2007_Problem7.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_7|Solution]]<br />
<br />
== Problem 8 ==<br />
The positive integers <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three digits of <math>\displaystyle x_7</math>.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_8|Solution]]<br />
<br />
== Problem 9 ==<br />
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_9|Solution]]<br />
<br />
== Problem 10 ==<br />
Find the number of solutions, in degrees, to the equation <math>\displaystyle \sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>\displaystyle 0^\circ \le x^\circ \le 2007^\circ.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_10|Solution]]<br />
<br />
== Problem 11 ==<br />
Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations<br />
<br />
<math>\displaystyle x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_11|Solution]]<br />
<br />
== Problem 12 ==<br />
In quadrilateral <math>\displaystyle ABCD,</math> <math>\displaystyle m \angle DAC= m\angle DBC </math> and <math>\displaystyle \frac{[ADB]}{[ABC]}=\frac12.</math> If <math>\displaystyle AD=4,</math> <math>\displaystyle BC=6</math>, <math>\displaystyle BO=1,</math> and the area of <math>\displaystyle ABCD</math> is <math>\displaystyle \frac{a\sqrt{b}}{c},</math> where <math>\displaystyle a,b,c</math> are relatively prime positive integers, find <math>\displaystyle a+b+c.</math><br />
<br />
<br />
Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of triangles <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_12|Solution]]<br />
<br />
== Problem 13 ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle m/n,</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
[[Mock_AIME_2_2006-2007/Problem_13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
In triangle ABC, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35.</math> Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> intersect at <math>\displaystyle P</math> and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of <math>\displaystyle BC.</math><br />
<br />
[[Image:Mock AIME 2 2007 Problem14.jpg]]<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is <math>\mathfrak{Intriguing}</math> if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of <math>\mathfrak{Intriguing}</math> colorings.<br />
<br />
[[Mock_AIME_2_2006-2007/Problem_15|Solution]]<br />
<br />
[[Image:CubeArt.jpg]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_15&diff=8419Mock AIME 2 2006-2007 Problems/Problem 152006-07-25T15:54:47Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is <math>\mathfrak{Intriguing}</math> if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of <math>\mathfrak{Intriguing}</math> colorings.<br />
<br />
[[Image:CubeArt.jpg]]<br />
<br />
== Problem Source ==</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=File:CubeArt.jpg&diff=8418File:CubeArt.jpg2006-07-25T15:53:20Z<p>4everwise: </p>
<hr />
<div></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_8&diff=8410Mock AIME 2 2006-2007 Problems/Problem 82006-07-25T05:13:27Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
The positive integers <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three digits of <math>\displaystyle x_7</math>.</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_8&diff=8409Mock AIME 2 2006-2007 Problems/Problem 82006-07-25T05:13:14Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
The positive integers <math>\displaystyle x_1, x_2, ... , x_7</math> satisfy <math>\displaystyle x_6 = 144</math> and <math>\displaystyle x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>\displaystyle n = 1, 2, 3, 4</math>. Find the last three digits of <math>x_7</math>.</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_13&diff=8408Mock AIME 2 2006-2007 Problems/Problem 132006-07-25T04:32:10Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle m/n,</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
== Problem Source ==<br />
4everwise thought of this problem when watching Round 4 of the Professional Poker Tour. (What else can one do during the commercial breaks? [[Image:Razz.gif]]</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_13&diff=8407Mock AIME 2 2006-2007 Problems/Problem 132006-07-25T04:30:27Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of 52 playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle \frac{m}{n},</math> where <math>\displaystyle m</math> and <math>\displaystyle n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
== Problem Source ==<br />
4everwise enjoys playing and watching card games. In fact, he thought of this problem when watching Round 4 of the Professional Poker Tour. (Guy caught an Ace on the river.)</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_13&diff=8406Mock AIME 2 2006-2007 Problems/Problem 132006-07-25T04:28:41Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
In his spare time, Richard Rusczyk shuffles a standard deck of <math>\displaystyle 52</math> playing cards. He then turns the cards up one by one from the top of the deck until the third ace appears. If the expected (average) number of cards Richard will turn up is <math>\displaystyle \frac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>\displaystyle m+n.</math><br />
<br />
== Problem Source ==<br />
4everwise enjoys playing and watching card games. In fact, he thought of this problem when watching Round 4 of the Professional Poker Tour. (Guy caught an Ace on the river.)</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_15&diff=8405Mock AIME 2 2006-2007 Problems/Problem 152006-07-25T04:15:08Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is <math>\mathfrak{Intriguing}</math> if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of <math>\mathfrak{Intriguing}</math> colorings.<br />
<br />
== Problem Source ==<br />
4everwise did not write this problem. The souce cannot be revealed at this moment, as the contest is still running.</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_15&diff=8404Mock AIME 2 2006-2007 Problems/Problem 152006-07-25T04:12:34Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
A <math>\displaystyle 4\times4\times4</math> cube is composed of <math>\displaystyle 64</math> unit cubes. The faces of <math>\displaystyle 16</math> unit cubes are colored red. An arrangement of the cubes is '''''Intriguing''''' if there is exactly <math>\displaystyle 1</math> red unit cube in every <math>\displaystyle 1\times1\times4</math> rectangular box composed of <math>\displaystyle 4</math> unit cubes. Determine the number of '''''Intriguing''''' colorings.<br />
<br />
== Problem Source ==<br />
4everwise did not write this problem. The souce cannot be revealed at this moment, as the contest is still running.</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_14&diff=8403Mock AIME 2 2006-2007 Problems/Problem 142006-07-25T04:08:50Z<p>4everwise: /* Problem */</p>
<hr />
<div>== Problem ==<br />
In triangle ABC, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35.</math> Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> intersect at <math>\displaystyle P</math> and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of <math>\displaystyle BC.</math><br />
<br />
[[Image:Mock AIME 2 2007 Problem14.jpg]]<br />
<br />
== Problem Source ==<br />
4everwise thought of this problem after reading the first chapter of Geometry Revisited.</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=File:Mock_AIME_2_2007_Problem14.jpg&diff=8402File:Mock AIME 2 2007 Problem14.jpg2006-07-25T04:08:19Z<p>4everwise: </p>
<hr />
<div></div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_14&diff=8401Mock AIME 2 2006-2007 Problems/Problem 142006-07-25T04:03:54Z<p>4everwise: </p>
<hr />
<div>== Problem ==<br />
In triangle ABC, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35.</math> Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> intersect at <math>\displaystyle P</math> and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of <math>\displaystyle BC.</math> <br />
<br />
== Problem Source ==<br />
4everwise thought of this problem after reading the first chapter of Geometry Revisited.</div>4everwisehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_11&diff=8400Mock AIME 2 2006-2007 Problems/Problem 112006-07-25T04:01:06Z<p>4everwise: /* Problem */</p>
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<div>== Problem ==<br />
Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations<br />
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<math>\displaystyle x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math><br />
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== Problem Source==<br />
This problem was given to 4everwise by a friend, Henry Tung. Upper classmen bullying freshmen. (Just kidding; it's a nice problem. [[Image:Razz.gif]])</div>4everwise