Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides.  A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle.  What is the volume of this pyramid?

Solution: As shown in the image above, let $D$ $E$ , and $F$ be the midpoints of $\overline{BC}$ $\overline{CA}$ , and $\overline{AB}$ , respectively.  Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$ .  The crux of this problem is the following lemma.
Lemma: The point $O$ is the orthocenter of $\triangle ABC$
Proof. Observe that \[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$ .  Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$ .  Analogously, $O$ lies on the $B$ -altitude and $C$ -altitude of $\triangle ABC$ , and so $O$ is, indeed, the orthocenter of $\triangle ABC$
To find the coordinates of $O$ , we need to find the intersection point of altitudes $BE$ and $AD$ . The equation of $BE$ is simply $x=16$ $AD$ is perpendicular to line $BC$ , so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$ $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$ , therefore $y=\frac{3}{4} x$ . These two lines intersect at $(16,12)$ , so that's the base of the height of the tetrahedron.
Let $S$ be the foot of altitude $BS$ in $\triangle BPQ$ . From the Pythagorean Theorem $h=\sqrt{BS^2-SO^2}$ . However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$
The area of the base is $102$ , so the volume is $\frac{102*12}{3}=\boxed{408}$

Answer:  408
Problem: Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides.  A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle.  What is the volume of this pyramid?

Solution: Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$ $(8, 12, 0)$ , and $(25, 12, 0)$ . We can compute the area of this triangle as $102$ . Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$ . Clearly, the height of the pyramid is $z$ . The desired volume is thus $\frac{102z}{3} = 34z$
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$ $VP = PB$ , and $VQ = QC$ . We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$ . The desired volume is thus $34 \times 12 = \boxed{408}$

Answer:  408
Problem: Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides.  A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle.  What is the volume of this pyramid?

Solution: The formed tetrahedron has pairwise parallel planar and oppositely equal length ( $4\sqrt{13},15,17$ ) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces.  Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pythagoras)
$p^2+q^2=4^2\cdot{13}$
$q^2+r^2=15^2$
$r^2+p^2=17^2$
to find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\cdot{17},2^3\cdot{17},2^3\cdot{3^2}).$
Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\tfrac{1}{3}$ and then the volume is
$\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}$

Answer:  408
Problem: Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides.  A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle.  What is the volume of this pyramid?

Solution: Let $A = (0,0), B = (16, 24), C = (34,0).$ Then define $D,E,F$ as the midpoints of $BC, AC, AB$ . By Pythagorean theorem, $EF = \frac{1}{2} BC = 15, DE = \frac{1}{2}AB = 4 \sqrt{13}, DF = \frac{1}{2} AC = 17.$ Then let $P$ be the point in space which is the vertex of the tetrahedron with base $DEF$
Note that $\triangle DEP \cong \triangle EDF$ . Create point $F'$ on the plane of $DEF$ such that $\triangle DEP \cong \triangle DEF'$ (i.e by reflecting $F$ over the perpendicular bisector of $DE$ ). Project $F, P$ onto $DE$ as $X, Y$ . Note by the definition of $F'$ then $\angle PYF'$ is the dihedral angle between planes $DEP, DEF$
Now see that by Heron's, \[[DEP] = [DEF] = \sqrt{(16 + 2 \sqrt{13})(16 - 2 \sqrt{13})(1 + 2 \sqrt{13})(-1 + 2 \sqrt{13})} = 102.\] So $PY$ , the hypotenuse $DEP$ has length $\frac{102 \cdot 2}{4 \sqrt{13}} = \frac{51}{\sqrt{13}}$ . Similarly $F'Y = \frac{51}{\sqrt{13}}.$ Further from Pythagoras $DY = \sqrt{DP^2 - PY^2} = \frac{18}{\sqrt{13}}.$ Symmetrically $EX = \frac{18}{\sqrt{13}}.$ Therefore $XY = DE - DY - EX = \frac{16}{\sqrt{13}}.$
By Law of Cosines on $\triangle PYF'$ \begin{align*} PF'^2 &= PY^2 + F'Y^2 - 2 \cdot PY \cdot F'Y \cos{\angle PYF'} \\ PF^2 - XY^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ (4\sqrt{13})^2 - (\frac{16}{\sqrt{13}})^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'}  \\ \cos{\angle PYF'} &= \frac{9}{17} \\ \sin{\angle PYF'} &= \frac{4 \sqrt{13}}{17}. \end{align*}.
Therefore the altitude of the tetrahedron from vertex $P$ to base $DEF$ is $PY \sin{\angle PYF'} = \frac{51}{\sqrt{13}} \frac{4 \sqrt{13}}{17} = 12.$ So the area is $\frac{1}{3}bh = \frac{1}{3} 12 \cdot 102 = \boxed{408}.$

Answer:  408
Problem: For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ $8^8$ , and $k$

Solution: It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$
The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$ . Therefore $12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$ , and $b = 12$ .  Since $0 \le a \le 24$ , there are $\boxed{25}$ values of $k$

Answer:  25
Problem: The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions.  What is the area of the bounded region?

Solution: The equation given can be rewritten as:
We can split the equation into a piecewise equation by breaking up the absolute value
Factoring the first one: (alternatively, it is also possible to complete the square
Hence, either $y = -20$ , or $2x = 20 - y \Longrightarrow y = -2x + 20$
Similarily, for the second one, we get $y = 20$ or $y = -2x - 20$ . If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is $\boxed{800}$

Answer:  800
Problem: Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively.  Each of three players randomly selects and keeps three of the tiles, and sums those three values.  The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers .  Find $m+n.$

Solution: In order for a player to have an odd sum, he must have an odd number of odd tiles: that is, he can either have three odd tiles, or two even tiles and an odd tile.  Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ even tiles and $1$ odd tile.  We count the number of ways this can happen. (We will count assuming that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)
$\dbinom{5}{3} = 10$ choices for the tiles that he gets.  The other two odd tiles can be distributed to the other two players in $2$ ways, and the even tiles can be distributed between them in $\dbinom{4}{2} \cdot \dbinom{2}{2} = 6$ ways.  This gives us a total of $10 \cdot 2 \cdot 6 = 120$ possibilities in which all three people get odd sums.
In order to calculate the probability, we need to know the total number of possible distributions for the tiles.  The first player needs three tiles which we can give him in $\dbinom{9}{3} = 84$ ways, and the second player needs three of the remaining six, which we can give him in $\dbinom{6}{3} = 20$ ways. Finally, the third player will simply take the remaining tiles in $1$ way. So, there are $\dbinom{9}{3} \cdot \dbinom{6}{3} \cdot 1 = 84 \cdot 20 = 1680$ ways total to distribute the tiles.
We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is $\frac{360}{1680} = \frac{3}{14},$ so the answer is $3 + 14 = \boxed{017}$

Answer:  017
Problem: Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively.  Each of three players randomly selects and keeps three of the tiles, and sums those three values.  The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers .  Find $m+n.$

Solution: Let $O$ stand for an odd number and $E$ an even.
Therefore, one person must pick $OOO$ , the other person must pick $OEE$ and the last person must pick $OEE$ . Since any permutation of the order of who is picking or change in the order of the even numbers (e.g. $EOE$ instead of $OEE$ ) doesn't change the probability, we just need to multiply the probability of one case by $\binom{3}{2}^3 = 27$ as there are 27 such cases (by cases I mean ordered triples of ordered multisets $A, B, C$ such that one of them has 3 $O$ 's and the other two have two $E$ 's and an $O$ in them, respectively.) . Let's do the case $OOO$ $OEE$ $OEE$ $\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} = \frac{1}{126}$ . We now multiply by 27 to get $\frac{3}{14} \implies m + n = \boxed{017}$ - whatRthose

Answer:  017
Problem: Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively.  Each of three players randomly selects and keeps three of the tiles, and sums those three values.  The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers .  Find $m+n.$

Solution: For this problem, let's think about parity. There are $5$ odd numbers from $1-9$ and there are four even numbers from $1-9$ . Since this problem is asking for the probability that the each player gets an odd sum, we also have to calculate the total numeber of ways.
In this case, there are only two ways to get an odd sum. Either have the sequence $OOO$ or $OEE$ where the letters $O$ and $E$ stand for odd and even respectively.
Since we constrained to only $5$ odds, the only way to do the pairing is \[OOO-OEE-OEE\] There are of couse three ways to choose who gets the three odds.
Once we have chosen who has gotten the three odds, we can actually reorder the sequence like this: \[OEE-OEE-OOO\] Now since we are choosing in groups, we can ignor order of these terms.
For the first $OEE$ , there are $5$ ways to choose which odd to use, and $\dbinom{4}{2}$ or $6$ ways to choose the evens.
Great, let's move on to the second $OEE$ . There are $4$ odds left, so there are $4$ ways choose the odds. Since there are only two even numbers, they have to go here, so there is only $1$ way to choose the evens.
The three odds will now fall in place. We can now multiply all the numbers since they are independent, and we have $5*6*4*3$ or $360$
Since this is a probability question, we have to ask ourselves how many ways are there to distribute the 9 tiles equally among 3 players.
Fortunately for us, this is not hard as the first player has $\dbinom{9}{3}$ options and the second player has $\dbinom{6}{3}$ . When we multiply these, we get $1680$ . This is our denominator.
When we make the fraction, we have $\frac{360}{1680}$ . When we simplify it, we have: \[\frac{3}{14}\] We are asked to find the sum of the numerator and the denominator, so summing these, we have: \[\boxed{017}\] -Pi_is_3.14

Answer:  017
Problem: Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$

Solution: Though the problem may appear to be quite daunting, it is actually not that difficult. $\frac {k(k-1)}2$ always evaluates to an integer ( triangular number ), and the cosine of $n\pi$ where $n \in \mathbb{Z}$ is 1 if $n$ is even and -1 if $n$ is odd. $\frac {k(k-1)}2$ will be even if $4|k$ or $4|k-1$ , and odd otherwise.
So our sum looks something like:
$\left|\sum_{i=19}^{98} A_i\right| =- \frac{19(18)}{2} + \frac{20(19)}{2} + \frac{21(20)}{2} - \frac{22(21)}{2} - \frac{23(22)}{2} + \frac{24(23)}{2} \cdots - \frac{98(97)}{2}$
If we group the terms in pairs, we see that we need a formula for $-\frac{(n)(n-1)}{2} + \frac{(n+1)(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n$ . So the first two fractions add up to $19$ , the next two to $-21$ , and so forth.
If we pair the terms again now, each pair adds up to $-2$ . There are $\frac{98-19+1}{2 \cdot 2} = 20$ such pairs, so our answer is $|-2 \cdot 20| = \boxed{040}$

Answer:  040
Problem: Let $ABCD$ be a parallelogram .  Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$

Solution: We have $\triangle BRQ\sim \triangle DRC$ so $\frac{112}{RC} = \frac{BR}{DR}$ . We also have $\triangle BRC \sim \triangle DRP$ so $\frac{ RC}{847} = \frac {BR}{DR}$ . Equating the two results gives $\frac{112}{RC} =  \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$

Answer:  308
Problem: Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Solution: We want $x_1 +x_2+x_3+x_4 =98$ . This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$ , as for all integers $y_i$ $2y_i + 1$ will be odd. Substituting we get \[2y_1+2y_2+2y_3+2y_4 +4 = 98 \implies y_1+y_2+y_3+y_4 =47\] Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us $n= {50\choose3}$ . Computing this and dividing by 100 gives us an answer of $\boxed{196}$ .
~smartguy888

Answer:  196
Problem: Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Solution: Define $x_i = 2y_i - 1$ . Then $2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98$ , so $\sum_{i = 1}^4 y_i = 51$
So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is $n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600$ , and $\frac n{100} = \boxed{196}$

Answer:  196
Problem: Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Solution: Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into $4$ boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have $94$ stones left. Because we want an odd number in each box, we pair the stones, creating $47$ sets of $2$ . Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).
Our problem can now be restated: how many ways are there to partition a line of $47$ stones? We can easily solve this by using $3$ sticks to separate the stones into $4$ groups, and this is the same as arranging a line of $3$ sticks and $47$ stones. \[\frac{50!}{47! \cdot 3!} = 19600\] \[\frac{50 * 49 * 48}{3 * 2} = 19600\] Our answer is therefore $\frac{19600}{100} = \boxed{196}$

Answer:  196
Problem: Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Solution: Let $x = a + b$ and $y = c + d$ . Then $x + y = 98$ , where $x, y$ are positive even integers ranging from $2-98$
We quickly see that the total number of acceptable ordered pairs $(a, b, c, d)$ is:
\begin{align*} &\mathrel{\phantom{=}} 1 \cdot 48 + 2 \cdot 47 + 3 \cdot 46 + ... + 48 \cdot 1\\ &= (24.5 - 23.5)(24.5) + (24.5 - 22.5)(24.5 + 22.5) + ... + (24.5 + 23.5)(24.5 - 23.5)\\ &= 48(24.5)^2 - 2(0.5^2 + 1.5^2 + ... + 23.5^2)\\ &= 28812 - \frac{1^2 + 3^2 + ... + 47^2}{2}\\ &= 28812 - \frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2}\\ &= 28812 - \frac{\frac{47(47 + 1)(2(47) + 1)}{6} - \frac{4(23)(23 + 1)(2(23) + 1)}{6}}{2}\\ &= 19600 \end{align*}
Therefore, $\frac{n}{100} = \frac{19600}{100} = \boxed{196}$

Answer:  196
Problem: Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Solution: We write the generating functions for each of the terms, and obtain $(x+x^3+x^5\cdots)^4$ as the generating function for the sum of the $4$ numbers. We seek the $x^{98}$ coefficient, or the $x^{94}$ coefficient in $(1+x^2+x^4...)^4.$ Now we simplify this as $\left(\frac{1}{1-x^2}\right)^4=\binom{3}{3} +\binom{4}{3}x^2+\binom{5}{3}x^4 \cdots$ and in general we want that the coefficient of $x^{2k}$ is $\binom{k+3}{3}.$ We see the $x^{94}$ coefficient so we let $k=47$ and so the coefficient is $\binom{50}{3}=19600$ in which $\frac{n}{100}=\boxed{196}.$

Answer:  196
Problem: Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that.  The last term of the sequence is the first negative term encounted.  What positive integer $x$ produces a sequence of maximum length?

Solution: We can start to write out some of the inequalities now:
And in general,
It is apparent that the bounds are slowly closing in on $x$ , so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11):
$x < \frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}$
$x > \frac{F_{8}}{F_{9}} \cdot 1000 = \frac{21}{34} \cdot 1000 \approx 617.977$
Thus the sequence is maximized when $x = \boxed{618}.$

Answer:  618
Problem: Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that.  The last term of the sequence is the first negative term encounted.  What positive integer $x$ produces a sequence of maximum length?

Solution: It is well known that $\lim_{n\rightarrow\infty} \frac{F_{n-1}}{F_n} = \phi - 1 =\frac{1 + \sqrt{5}}{2} - 1 \approx .61803$ , so $1000 \cdot \frac{F_{n-1}}{F_n}$ approaches $x = \boxed{618}.$

Answer:  618
Problem: Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon .  A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres.  The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime .  Find $a + b + c$

Solution: The key is to realize the significance that the figures are spheres, not circles . The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.
1998 AIME-10a.png
Let us examine the relation between one of the outside 8 spheres and the center one (with radius $r$ ):
1998 AIME-10b.png
If we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid ; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$ . Then by the Pythagorean Theorem
\[x^2 + (r-100)^2 = (r+100)^2 \Longrightarrow x = 20\sqrt{r}\]
1998 AIME-10c.png
$x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40\sqrt{r}$ . We can draw another right triangle as shown above. One leg has a length of $200$ . The other can be found by partitioning the leg into three sections and using $45-45-90 \triangle$ s to see that the leg is $100\sqrt{2} + 200 + 100\sqrt{2} = 200(\sqrt{2} + 1)$ . Pythagorean Theorem:
\begin{eqnarray*} (40\sqrt{r})^2 &=& 200^2 + [200(\sqrt{2}+1)]^2\\ 1600r &=& 200^2[(1 + \sqrt{2})^2 + 1] \\ r &=& 100 + 50\sqrt{2} \end{eqnarray*}
Thus $a + b + c = 100 + 50 + 2 = \boxed{152}$

Answer:  152
Problem: Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon .  A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres.  The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime .  Find $a + b + c$

Solution: Isolate a triangle, with base length $200$ (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as $x$ . Since the interior angle is $45$ degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get: \begin{eqnarray*} 200^2 &=& 2x^2 - 2x^2*cos(45^\circ) \\ &=& 2x^2 - 2x^2*\frac{\sqrt{2}}{2} \\ &=& (2-\sqrt{2})x^2 \end{eqnarray*}
And thus \[x = \frac{200}{\sqrt{2-\sqrt{2}}}\]
From the above, $x = 20\sqrt{r}$ , so we get
\begin{eqnarray*} r &=& (\frac{200}{20(\sqrt{2-\sqrt{2}})})^2 \\ &=& (\frac{10}{\sqrt{2-\sqrt{2}})})^2 \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} \\ &=& \frac{200 + 100\sqrt{2}}{2} \\ &=& 100 + 50\sqrt{2} \end{eqnarray*}
And hence the answer is $100 + 50 + 2 \Rightarrow \boxed{152}$

Answer:  152
Problem: Let $ABC$ be equilateral , and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively.  There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + b\sqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime .  What is $a^{2} + b^{2} + c^{2}$

Solution: 1998 AIME-12.png
We let $x = EP = FQ$ $y = EQ$ $k = PQ$ . Since $AE = \frac {1}{2}AB$ and $AD = \frac {1}{2}AC$ $\triangle AED \sim \triangle ABC$ and $ED \parallel BC$
By alternate interior angles, we have $\angle PEQ = \angle BFQ$ and $\angle EPQ = \angle FBQ$ . By vertical angles, $\angle EQP = \angle FQB$
Thus $\triangle EQP \sim \triangle FQB$ , so $\frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y$
Since $\triangle EDF$ is equilateral, $EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1$ . Solving for $x$ and $y$ using $x^{2} = y$ and $x + y = 1$ gives $x = \frac {\sqrt {5} - 1}{2}$ and $y = \frac {3 - \sqrt {5}}{2}$
Using the Law of Cosines , we get
We want the ratio of the squares of the sides, so $\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}$ so $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}$

Answer:  083