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https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10

Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. (Students are not penalized for problems left unanswered.)

Given that Mary's score is $30+4c-w$ , two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$ . Since it is clear that $c>1$ , we must have $w<4$ . In order to minimize the score, assume that $w=3$ . The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In order to minimize the score, assume that the number of problems left blank is $4$ , making the number of correct problems $23$ . Substituting, we get that $s=30+23{\,\cdot\,}4-3$ , so $s=\boxed{119}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and "non-birch" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed to those with few branches. Indeed, you could keep dividing until you have them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange the oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.) The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 5 dividers of 8 slots that the birch trees can go in, making ${8\choose5} = 56$ different ways to arrange this. There are ${12 \choose 5} = 792$ total ways to arrange the twelve trees, so the probability is $\frac{56}{792} = \frac{7}{99}$ The answer is $7 + 99 = \boxed{106}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Let $b$ $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$ s to separate the $5$ $b$ s. Specifically, \[b,n,b,n,b,n,b,n,b\] Since we have $7$ $n$ s, we are placing the extra $3$ $n$ s into the $6$ intervals beside the $b$ s. Now doing simple casework. If all $3$ $n$ s are in the same interval, there are $6$ ways. If $2$ of the $3$ $n$ s are in the same interval, there are $6\cdot5=30$ ways. If the $n$ s are in $3$ different intervals, there are ${6 \choose 3} =20$ ways. In total there are $6+30+20=56$ ways. There are ${12\choose5}=792$ ways to distribute the birch trees among all $12$ trees. Thus the probability equals $\frac{56}{792}=\frac{7}{99}\Longrightarrow m+n=7+99=\boxed{106}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion. The number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees. The total number of configurations is given by $\frac{12!}{3! \cdot 4! \cdot 5!}$ . To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE. $\#$ (configurations with at least one pair of adjacent Birch trees) $=$ $\#$ (configurations with one pair) $-$ $\#$ (configurations with two pairs) $+$ $\#$ (configurations with three pairs) $-$ $\#$ (configurations with four pairs). To compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\frac{11!}{3! \cdot 3! \cdot 4!}$ configurations. For the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\frac{10!}{2! \cdot 3! \cdot 4!}$ cases. So our second term is $\frac{2 \cdot 10!}{2! \cdot 3! \cdot 4!}$ The third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\frac{2 \cdot 9!}{3! \cdot 4!}$ arrangements. The final term can happen in one way (BBBBB). This gives $\frac{8!}{3! \cdot 4!}$ arrangements. Substituting these into our PIE expression, we find that there are $25760$ configurations with at least one pair of adjacent Birch trees. Therefore, there are a total of $\frac{12!}{3! \cdot 4! \cdot 5!} - 25760 = 1960$ configurations with no adjacent Birch trees. Thus, the probability of a given configuration having no two adjacent Birch trees is given by $\frac{1960}{\frac{12!}{3! \cdot 4! \cdot 5!}} = \frac{7}{99}$ Therefore, the desired result is given by $7+99 = \boxed{106}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11

A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$

Here is a solution leaving out nothing. This solution is dedicated to those that are in self study and wish to learn the most they can. I will make it as elementary as possible and intuition based. Arrange first the $3$ maple and $4$ oaks as $MMMOOOO$ . We then notice that for none of the $5$ birch trees to be adjacent, they must be put in between these $M$ 's and $O$ 's. We then see that there are $8$ spots to put these $5$ birch trees in. So we can select $5$ spots for these birch trees in $\binom{8}{5}$ . But then, we can rearrange the $M$ 's and $O$ 's in $7!/(3!4!)=\binom{7}{3}$ ways. So then there are $\binom{8}{5}\binom{7}{3}$ valid arrangements with no given consecutive birch trees. There are then a total of $\frac{12!}{3!4!5!}$ different total arrangements. Therefore the probability is given as $\frac{\binom{8}{5}\binom{7}{3}}{\frac{12!}{3!4!5!}}=\frac{7}{99}$ , so the answer is $7+99=\boxed{106}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

If $f(2+x)=f(2-x)$ , then substituting $t=2+x$ gives $f(t)=f(4-t)$ . Similarly, $f(t)=f(14-t)$ . In particular, \[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\] Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function \[f(x) = \sin \frac{\pi x}{10}\sin \frac{\pi (x-4)}{10}\] satisfies the conditions and has no other roots. In the interval $-1000\leq x\leq 1000$ , there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \pmod{10}$ , therefore the minimum number of roots is $\boxed{401}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

We notice that the function has reflectional symmetry across both $x=2$ and $x=7$ . We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 10$ . There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $\boxed{401}$ $QED \blacksquare$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \pm 5, \pm 10, \pm 15... \pm 1000$ so the answer is 400 + 1 = $\boxed{401}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_12

function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$

Let $z$ be an arbitrary zero. If $z=2-x$ , then $x=2-z$ and $2+x=4-z$ . Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$ . From $0$ , we get $4$ and $14$ . Now note that applying either of these twice will return $z$ , so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$ , respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\cdot200+1=\boxed{401}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_14

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35... For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9(ex. 34, 44 . . .). Hence the largest number that ends with a 4 that satisfies the conditions is 14. If you list out all the numbers(15, 27, 9, 21, 33), you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and any number greater that ends with a 3 is bad(ex. 58, 68. . .), so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $\boxed{38}$

https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_15

Determine $x^2+y^2+z^2+w^2$ if

As in Solution 1, we have \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$ Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as \[(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots\] Rearranging gives us \[t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.\] By Vieta's, we know that the sum of the roots of this equation is \[1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.\] (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$ ). Thus, \[x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.\]

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_1

Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_2

Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$

Let $p$ be equal to $15 - \varepsilon$ , where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$ , the domain of $f(x)$ is basically the set ${15}$ . plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$ , or $15$ , so the answer is $\boxed{15}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_3

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

We are given the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] Squaring both sides yields \[(x^2+18x+30)^2=4(x^2+18x+45)\] \[(x^2+18x+30)^2=4(x^2+18x+30+15)\] \[(x^2+18x+30)^2=4(x^2+18x+30)+60\] \[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\] Substituting $y=x^2+18x+30$ yields \[y^2-4y-60=0\] \[(y+6)(y-10)=0\] Thus $y=x^2+18x+30=-6,10$ . However if $y=-6$ , the left side of the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have \[x^2+18x+30=10\] \[x^2+18x+20=0\] Since the discriminant $\sqrt{18^2-4\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $\boxed{20}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_5

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

$x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$ . Letting $p = x+y$ , we have that $p^3 - 21p + 20 = 0$ . Testing $p = 1$ , we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_8

What is the largest $2$ -digit prime factor of the integer $n = {200\choose 100}$

We know that \[{200\choose100}=\frac{200!}{100!100!}\] Since $p<100$ , there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\frac{200!}{100!100!}$ . (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,) So basically, $p$ is the largest prime number such that \[\left \lfloor\frac{200}{p}\right \rfloor>3\] Since $p<\frac{200}{3}=66.66...$ , the largest prime value for $p$ is $p=\boxed{61}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Suppose that the two identical digits are both $1$ . Since the thousands digit must be $1$ , only one of the other three digits can be $1$ . This means the possible forms for the number are Because the number must have exactly two identical digits, $x\neq y$ $x\neq1$ , and $y\neq1$ . Hence, there are $3\cdot9\cdot8=216$ numbers of this form. Now suppose that the two identical digits are not $1$ . Reasoning similarly to before, we have the following possibilities: Again, $x\neq y$ $x\neq 1$ , and $y\neq 1$ . There are $3\cdot9\cdot8=216$ numbers of this form. Thus the answer is $216+216=\boxed{432}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Consider a sequence of $4$ digits instead of a $4$ -digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\frac{1}{10}$ . This means we can find all possible sequences with one digit repeated twice, and then divide by $10$ If we let the three distinct digits of the sequence be $a, b,$ and $c$ , with $a$ repeated twice, we can make a table with all possible sequences: \[\begin{tabular}{ccc} aabc & abac & abca \\ baac & baca & \\ bcaa && \\ \end{tabular}\] There are $6$ possible sequences. Next, we can see how many ways we can pick $a$ $b$ , and $c$ . This is $10(9)(8) = 720$ , because there are $10$ digits, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $\boxed{432}$ as our answer.

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated. Case 1: There are $9$ choices for the hundreds digit (it cannot be $1$ ), $8$ choices for the tens digit (it cannot be $1$ or what was chosen for the hundreds digit), and $7$ for the units digit. This is a total of $9\cdot8\cdot7=504$ numbers. Case 2: One of the three numbers must be $1$ , and the other two numbers must be the same number, but cannot be $1$ (That will be dealt with in case 4). There are $3$ choices to put the $1$ , and there are $9$ choices (any number except for $1$ ) to pick the other number that is repeated, so a total of $3\cdot9=27$ numbers. Case 3: We will split it into $2$ subcases: one where $1$ is repeated $3$ times, and one where another number is repeated $3$ times. When $1$ is repeated $3$ times, then one of the digits is not $1$ . There are $9$ choices for that number, and $3$ choices for its location,so a total of $9\cdot3=27$ numbers. When a number other than $1$ is repeated $3$ times, then there are $9$ choices for the number, and you don't have any choices on where to put that number. So in Case 3 there are $27+9=36$ numbers Case 4: There is only $1$ number: $1111$ There are a total of $1000$ $4$ -digit numbers that begin with $1$ (from $1000$ to $1999$ ), so by complementary counting you get $1000-(504+27+36+1)=\boxed{432}$ numbers.

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_10

The numbers $1447$ $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Let us proceed by casework. Case 1: We will count the amount of numbers that have two identical digits that are not one. The thousands digit is fixed, and we are choosing two spots to hold two identical digits that are chosen from $0, 2-9$ , which is $9$ options. For the last digit, their are $8$ possibilities since it can be neither $1$ or the other number that is chosen. The final outcome is ${3}\choose{2}$ $* 9 * 8 = 216$ possibilities for this case. Case 2: The last case will be the amount of numbers that have two identical digits thare are $1$ . There are ${3}\choose{1}$ places to pick the $1$ . For the other $2$ digits, there are $9$ and $8$ options respectively. Thus, we have $3 * 9 * 8 = 216$ . Summing the two cases, we get $216 + 216 = \boxed{432}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$ . The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$ , and one of the legs is $3\sqrt{2}$ . We apply the Pythagorean Theorem to deduce that the height is $6$ Next, we complete t he figure into a triangular prism, and find its volume, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$ Now, we subtract off the two extra pyramids that we included, whose combined volume is $2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$ Thus, our answer is $432-144=\boxed{288}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

Extend $EA$ and $FB$ to meet at $G$ , and $ED$ and $FC$ to meet at $H$ . Now, we have a regular tetrahedron $EFGH$ , which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\sqrt{2}$ . Using the formula for the volume of a regular tetrahedron, which is $V = \frac{\sqrt{2}S^3}{12}$ , where S is the side length of the tetrahedron, the volume of our original solid is: $V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$ ; thus, we will integrate with respect to height from $0$ to $6$ , noting that each cross section of height $dh$ is a rectangle. The volume is then $\int_0^h(wl) \ \text{d}h$ , where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$ as $w=6\sqrt{2}-\sqrt{2}h$ since it decreases linearly with respect to $h$ , and $l=6\sqrt{2}+\sqrt{2}h$ since it similarly increases linearly with respect to $h$ . Now we solve: \[\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}\]

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_11

The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid?

Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \sqrt{6}$ , and that little portion that hangs out has a length of $3\sqrt2$ . This is a triangular pyramid with a base of $3\sqrt6, 3\sqrt6, 3\sqrt2$ , and a height of $3\sqrt{2}$ . Since there are two of these, we can compute the sum of the volumes of these two to be $72$ . Now we are left with a triangular prism with a base of dimensions $3\sqrt6, 3\sqrt6, 3\sqrt2$ and a height of $6\sqrt2$ . We can compute the volume of this to be 216, and thus our answer is $\boxed{288}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13

For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$

Let $S$ be a non- empty subset of $\{1,2,3,4,5,6\}$ Then the alternating sum of $S$ , plus the alternating sum of $S \cup \{7\}$ , is $7$ . This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $S\cup \{7\}$ Because there are $2^{6}=64$ of these pairs of sets, the sum of all possible subsets of our given set is $64 \cdot 7$ , giving an answer of $\boxed{448}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13

For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$

Consider a given subset $T$ of $S$ that contains $7$ ; then there is a subset $T'$ which contains all the elements of $T$ except for $7$ , and only those elements . Since each element of $T'$ has one fewer element preceding it than it does in $T$ , their signs are opposite. Thus the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 \cdot 2^6 = \boxed{448}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13

For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$

Denote the desired total of all alternating sums of an $n$ -element set as $S_n$ . We are looking for $S_7$ . Notice that all alternating sums of an $n$ -element set are also alternating sums of an $n+1$ -element set. However, when we go from an $n$ to $n+1$ element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the $n$ -element set. There are $2^n$ subsets of an $n+1$ -element set that includes the new element, giving us the relationship $S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)$ . When $n = 6$ , we therefore get $S_ 7 = 2^6(7) = \boxed{448}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13

For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$

We analyze all the numbers from 1 to 7 separately to see where the number contributes its positive or negative to the sum of the alternating sums. Whenever 7 appears, which it does 64 times, it contributes a positive because it is always first. This gives a net gain of $7 \cdot 64=448$ If we look at when 6 appears, which it also does 64 times, whether it comes as positive or negative depends on the presence of 7. Half of the subsets with 6 have 7 resulting in subtracting 6 each time, while the other half does not have 7 adding 6 each time, so these contributions of sixes cancel each other out giving a net gain of 0. The same thing happens to any positive integer less than 7. This is because the determination of a positive or negative contribution is dependent on the number of larger numbers in front of it(For example, the sign of 3 is dependent on the presence of 4, 5, 6, and 7 in the subset). If the number of larger numbers is even, it gives in a positive copy while odd produces its negative. We know that the frequencies of these two cases occurring are the same because $0=(1-1)^{n}=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...\binom{n}{n}$ via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be $\boxed{448}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13

For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$

Let $\mathbb{N}_n := \{1, 2, 3, \dots n\}$ . Let the alternating sum of a certain subset of $S$ of $\mathbb{N}_n$ be $\xi(S),$ and let \[\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).\] We see that \[\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \mathbb{N}_n} \xi(S) + \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = \mathcal{A}(\mathbb{N}_{n-1}) + \sum_{n \in S, S \subseteq \mathbb{N}_n} \left( n - \xi(S - \{n\}) \right),\] as if $n \in S,$ $n$ is the largest element in $S.$ Now, we know that \[\sum_{n \in S, S \subseteq \mathbb{N}_n} n - \xi(S - \{n\}) = \sum_{S \subseteq \mathbb{N}_{n-1}} n - \xi(S) = n \cdot 2^{n-1} - \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = n \cdot 2^{n-1} - \mathcal{A}(\mathbb{N}_{n-1}),\] so \[\mathcal{A}(\mathbb{N}_{n}) = n \cdot 2^{n-1}.\] Thus, our answer (which is the $n = 7$ case) is $\mathcal{A}(\mathbb{N}_{7}) = 7 \cdot 2^6 = \boxed{448}.$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

Firstly, notice that if we reflect $R$ over $P$ , we get $Q$ . Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$ , we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ , and with radius $6$ ) that intersects circle $A$ at $Q$ . The rest is just finding lengths, as follows. Since $P$ is the midpoint of segment $BC$ $AP$ is a median of $\triangle ABC$ . Because we know $AB=12$ $BP=PC=6$ , and $AP=8$ , we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \sqrt{56}$ . Now we have a kite $AQCP$ with $AQ=AP=8$ $CQ=CP=6$ , and $AC=\sqrt{56}$ , and all we need is the length of the other diagonal $PQ$ . The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$ . Then Solving this equation, we find that $x^2=\frac{65}{2}$ , so $PQ^2 = 4x^2 = \boxed{130}.$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label("$A$",a,S); label("$B$",b,S); label("$M$",m,NE); label("$N$",n,NE); label("$P$",p,N); label("$Q$",q,NW); label("$R$",r,E); label("$12$",(14,0),SW); label("$6$",(23,0),S); [/asy] Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$ Since $\frac{AR}{MR}=\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$ , we have $AM=3BN=3y$ Applying the Pythagorean Theorem on triangle $BNR$ , we have $x^2+y^2=36$ . Similarly, for triangle $QMA$ , we have $x^2+9y^2=64$ Subtracting, $8y^2=28\Rightarrow y^2=\frac72\Rightarrow x^2=\frac{65}2\Rightarrow QP^2=4x^2=\boxed{130}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

Let $QP=PR=x$ . Angles $QPA$ $APB$ , and $BPR$ must add up to $180^{\circ}$ . By the Law of Cosines $\angle APB=\cos^{-1}\left(\frac{{-11}}{24}\right)$ . Also, angles $QPA$ and $BPR$ equal $\cos^{-1}\left(\frac{x}{16}\right)$ and $\cos^{-1}\left(\frac{x}{12}\right)$ . So we have Taking the cosine of both sides, and simplifying using the addition formula for $\cos$ as well as the identity $\sin^{2}{x} + \cos^{2}{x} = 1$ , gives $x^2=\boxed{130}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

Let $QP = PR = x$ . Extend the line containing the centers of the two circles to meet $R$ , and to meet the other side of the large circle at a point $S$ The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$ . The length of the diameter of the larger circle is $16$ Thus by Power of a Point in the circle passing through $Q$ $R$ , and $S$ , we have $x \cdot 2x = 10 \cdot (10+16) = 260$ , so $x^2 = \boxed{130}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label("$A$",a,S); label("$B$",b,S); label("$M$",m,NE); label("$N$",n,NE); label("$P$",p,N); label("$Q$",q,NW); label("$R$",r,E); label("$12$",(14,0),SW); label("$T$", t , NW); [/asy] Note that the midpoint of $AB$ is $T.$ Also, since $AM,NB$ bisect $QP$ and $PR,$ respectively, $P$ is the midpoint of $MN.$ Thus, $AM+NB=2PT.$ let $AM=a,BN=b.$ This means that $a+b=2PT.$ From the median formula, $PT=\sqrt{14}.$ Thus, $a+b=2\sqrt{14}.$ Also, since $MP=PN$ , from the Pythagorean Theorem, $8^2-a^2=6^2-b^2\implies a^2-b^2=28.$ Thus, $a-b=\frac{28}{2\sqrt{14}}=\sqrt{14}.$ We conclude that $QP=MN=\sqrt{12^2-(a-b)^2}=\sqrt{130}\implies\boxed{130}.$ ~pinkpig

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

Looking at Drawing 2 (by the way, we don't need point $R$ ), we set $AM=a$ and $BN=b$ , and the desired length $QP=x=PR$ . We know that a radius perpendicular to a chord bisects the chord, so $MP=\frac{x}{2}$ and $PN=\frac{x}{2}$ . Draw line $AP$ and $PB$ , and we see that they are radii of Circles $A$ and $B$ , respectively. We can write the Pythagorean relationships $a^2+(\frac{x}{2})^2=8^2$ for triangle $AMP$ and $b^2+(\frac{x}{2})^2=6^2$ for triangle $BNP$ . We also translate segment $MN$ down so that $N$ coincides with $B$ , and form another right traingle. From that triangle, you can see that the shorter leg is on the left side, having length $a-b$ , the longer leg is the same as $MN=x$ , and the hypotenuse is $AB=12$ . We can write the Pythagorean relationship $(a-b)^2+x^2=12^2$ . Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for $a$ in the first equation and $b$ in the second equation, and substitute into the third equation, get an equation only in terms of $x$ , and solve), you find that $x=\sqrt{130}$ , so $x^2 = \boxed{130}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

The centers are collinear, you can prove it (but that is already given in the later section [Proof that R,A, and B are collinear]). Drop a perpendicular from $P$ to the lines that the centers are on. You then have 2 separate segments, separated by the foot of the altitude of $P$ . Call them $a$ and $b$ respectively. Call the measure of the foot of the altitude of $P$ $h$ . You then have 3 equations: \[(1)a+b=12\] (this is given by the fact that the distance between the centers is 12. \[(2)a^2+h^2=64\] . This is given by the fact that P is on the circle with radius 8. \[(3)b^2+h^2=36\] . This is given by the fact that P is on the circle with radius 6. Subtract (3) from (2) to get that $a^2-b^2=28$ . As per (1), then you have $a-b=\frac{7}{3}$ (4). Add (1) and (4) to get that $2a=\frac{43}{3}$ . Then substitute into (1) to get $b=\frac{29}{6}$ . Substitute either a or b into (2) or (3) to get $h=\sqrt{455}{6}$ . Then to get $PQ=PR$ it is just $\sqrt{(b+6)^2+h^2}=\sqrt{\frac{65^2}{6^2}+\frac{455}{6^2}}=\sqrt{\frac{4680}{36}}=\sqrt{130}$ $PQ^2=\boxed{130}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

We use coordinate geometry to approach this problem. Let the center of larger circle be the origin $O_1$ , the smaller circle be $O_2$ , and the x-axis be $O_1O_2$ . Hence, we can get the two circle equations: $x^2+y^2 = 64$ and $(x-12)^2+y^2=36$ Let point $P$ be $(a, b)$ . Noting that it lies on both circles, we can plug the coordinates into both equations: Substituting $a^2+b^2 = 64$ into equation 2 and solving for $a$ , we get $a = \frac{43}{6}$ . The problem asks us to find $QP^2$ , which is congruent to $PR^2$ . Using the distance formula for $P(a, b)$ and $R(18, 0)$ (by Solution 7's collinear proof), we get $PR^2 = (18-a)^2 +b^2$ . Using $a^2+b^2 = 64$ , we find that $b^2 = \frac{455}{36}$ . Plugging the variables $a$ and $b^2$ in, we get $PR^2 = QP^2 = \boxed{130}$ ~SoilMilk

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14

In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy]

Let the center of the circle with radius $8$ be $A,$ and let the center of the one with radius $6$ be $B.$ Also, let $QP = PR = x.$ Using law of cosines on triangle $APB,$ we have that $\cos {APB} = \left(-\frac{{11}}{24}\right).$ Angle chasing gives that $\angle{QAR} = \angle{APB},$ so its cosines must be the same. Applying law of cosines again on triangle $QAR,$ we have $\left(2x^2\right)=64+324-2(8)(18)\left(-\frac{11}{24}\right),$ which gives that $x^2 = \boxed{130}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ [asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Let $M$ be the midpoint of the chord $BC$ . From right triangle $OMB$ , we have $OM = \sqrt{OB^2 - BM^2} =4$ . This gives $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$ Notice that the distance $OM$ equals $PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)$ , where $r$ is the radius of circle $P$ Hence \[\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5\] (where $R$ represents the radius, $5$ , of the large circle given in the question). Therefore, since $\angle AOM$ is clearly acute, we see that \[\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3\] Next, notice that $\angle AOB = \angle AOM - \angle BOM$ . We can therefore apply the subtraction formula for $\tan$ to obtain \[\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}\] It follows that $\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}$ , such that the answer is $7 \cdot 25=\boxed{175}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ [asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

This solution, while similar to Solution 2, is arguably more motivated and less contrived. Firstly, we note the statement in the problem that " $AD$ is the only chord starting at $A$ and bisected by $BC$ " – what is its significance? What is the criterion for this statement to be true? We consider the locus of midpoints of the chords from $A$ . It is well-known that this is the circle with diameter $AO$ , where $O$ is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor $\frac{1}{2}$ and center $A$ . Thus, the locus is the result of the dilation with scale factor $\frac{1}{2}$ and centre $A$ of circle $O$ . Let the center of this circle be $P$ Now, $AD$ is bisected by $BC$ if they cross at some point $N$ on the circle. Moreover, since $AD$ is the only chord, $BC$ must be tangent to the circle $P$ The rest of this problem is straightforward. Our goal is to find $\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}$ , where $M$ is the midpoint of $BC$ . We have $BM=3$ and $OM=4$ . Let $R$ be the projection of $A$ onto $OM$ , and similarly let $Q$ be the projection of $P$ onto $OM$ . Then it remains to find $AR$ so that we can use the addition formula for $\sin$ As $PN$ is a radius of circle $P$ $PN=2.5$ , and similarly, $PO=2.5$ . Since $OM=4$ , we have $OQ=OM-QM=OM-PN=4-2.5=1.5$ . Thus $PQ=\sqrt{2.5^2-1.5^2}=2$ Further, we see that $\triangle OAR$ is a dilation of $\triangle OPQ$ about center $O$ with scale factor $2$ , so $AR=2PQ=4$ Lastly, we apply the formula: \[\sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}\] Thus the answer is $7\cdot25=\boxed{175}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ [asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Let the center of the circle be $O$ . Fix $B,C,$ and $A$ . Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$ , it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$ Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$ . Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$ . Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$ . Clearly, $KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\frac{5}{2}$ . The same applies for $FO$ , which also equals $\frac{5}{2}$ . By the Pythagorean theorem, we deduce that $FL = 2$ , so $EK = 2$ . This is very important information! Now we know that $BE = 1$ , so by Power of a Point, $AE = ED = \sqrt{5}$ We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 25$ , we deduce that $EO = 2\sqrt{5}$ $EC=OC=5$ , so $\sin (CEO) = \frac{2\sqrt{5}}{5}$ . Furthermore, since $\sin (CEO) = \cos(DEC)$ , we know that $\cos (DEC) = \frac{2\sqrt{5}}{5}$ . By the law of cosines, \[DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10\] Therefore, $DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}$ . Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$ . Then, by the Pythagorean theorem, $OZ = \frac{7\sqrt{2}}{2}$ . Thus, $\sin (BOZ) = \frac{\sqrt{2}}{10}$ and $\cos (BOZ) = \frac{7\sqrt{2}}{10}$ . As a result, $\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}$ $7 \cdot 25 = \boxed{175}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ [asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Let I be the intersection of AD and BC. Lemma: $AI = ID$ if and only if $\angle AIO = 90$ Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\angle AIO = 90$ , We can get $\triangle AIO \cong \triangle OID$ Let be this the circle with diameter AO. Thus, we get $\angle AIO = 90$ , implying I must lie on $\omega$ . I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC. Now, create a coordinate system such that the origin is O and the y-axis is perpendicular to BC. The positive x direction is from B to C. Let Z be (0,5). Let Y be (-5,0). Let X be the center of $\omega$ . Since $\omega$ 's radius is $\frac{5}{2}$ , the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so $sin(XOY) = sin(AOY) = \frac{3}{5}$ $sin(BOZ) = \frac{3}{5}$ . If we let $sin(\theta) = \frac{3}{5}$ , we can find that what we are looking for is $sin(90 - 2\theta)$ , which we can evaluate and get $\frac{7}{25} \implies \boxed{175}$

https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ [asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\overline{AO}$ . Additionally, this circle must be tangent to $\overline{BC}$ . Let the center of this circle be $P$ . Let $M$ be the midpoint of $BC$ $N$ be the foot of the perpendicular from $P$ to $\overline{BM}$ , and $K$ be the foot of the perpendicular from $B$ to $\overline{AP}$ . Let $x=BK$ From right triangle $BKO$ , we get $KO = \sqrt{25-x^2}$ . Thus, $KP = \sqrt{25-x^2}-\frac52$ Since $BO = 5$ $BM = 3$ , and $\angle BMO$ is right, $MO=4$ . From quadrilateral $MNPO$ , we get $MN = \sqrt{PO^2 - (MO - NP)^2} = \sqrt{(5/2)^2 - (4 - 5/2)^2} = \sqrt{(5/2)^2 - (3/2)^2} = 2$ . Thus, $BN = 1$ Since angles $BNP$ and $BKP$ are right, we get \[BK^2+KP^2 = BN^2 + NP^2 \implies x^2 + \left(\sqrt{25-x^2}-\frac52\right)^2 = \left(\frac52\right)^2 + 1\] \[25 - 5\sqrt{25-x^2} = 1\] \[5\sqrt{25-x^2} = 24\] \[25(25-x^2) = 24^2\] \[25x^2 = 25^2 - 24^2 = 49\] \[x = \frac75\] Thus, $\sin \angle AOB = \frac{x}{5} = \frac{7}{25}\implies \boxed{175}$