Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$

Solution: Let $a_1=a,a_2=b.$ From the conditions, we have \[a_{n-1}+a_{n+1}=2a_n+1,\] for all $n>1.$ From this, we find that \begin{align*} a_3&=2b+1-a \\ a_4&=3b+3-2a\\ a_5&=4b+6-3a, \end{align*} or, in general, \[a_n=(n-1)b+\frac{(n-2)(n-1)}{2}-(n-2)a.\] Note: we can easily prove this by induction. Now, substituting $n=19,92,$ we find that \begin{align*} 0=&18b+\frac{17\cdot18}{2}-17a\\ 0=&91b+\frac{90\cdot91}{2}-90a\\ b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}. \end{align*} Now, cross multiplying, we find that \begin{align*} 91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\ 1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ 73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819}

Answer:  819
Problem: Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$

Solution: Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$
Then $XD=xy-70, XC=y(92-x)-50,$ thus \[\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},\] which we can rearrange, expand and cancel to get $120x=70\cdot 92,$ hence $AP=x=\frac{161}{3}$ . This gives us a final answer of $161+3=\boxed{164}$

Answer:  164
Problem: Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$

Solution: From $(1)$ above, $x = \frac{70r}{h}$ and $92-x = \frac{50r}{h}$ . Adding these equations yields $92 = \frac{120r}{h}$ . Thus, $x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}$ , and $m+n = \boxed{164}$

Answer:  164
Problem: Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$

Solution: The area of the trapezoid is $\frac{(19+92)h}{2}$ , where $h$ is the height of the trapezoid.
Draw lines $CP$ and $BP$ . We can now find the area of the trapezoid as the sum of the areas of the three triangles $BPC$ $CPD$ , and $PBA$
$[BPC] = \frac{1}{2} \cdot 50 \cdot r$ (where $r$ is the radius of the tangent circle.)
$[CPD] = \frac{1}{2} \cdot 19 \cdot h$
$[PBA] = \frac{1}{2} \cdot 70 \cdot r$
$[BPC] + [CPD] + [PBA] = 60r + \frac{19h}{2} = [ABCD] = \frac{(19+92)h}{2}$
$60r = 46h$
$r = \frac{23h}{30}$
From Solution 1 above, $\frac{h}{70} = \frac{r}{x}$
Substituting $r = \frac{23h}{30}$ , we find $x = \frac{161}{3}$ , hence the answer is $\boxed{164}$

Answer:  164
Problem: Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$ , inclusive. What is the integer that is nearest the area of $A$

Solution: Let $z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$ . Since $0\leq \frac{a}{40},\frac{b}{40}\leq 1$ we have the inequality \[0\leq a,b \leq 40\] which is a square of side length $40$
Also, $\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$ so we have $0\leq a,b \leq \frac{a^2+b^2}{40}$ , which leads to: \[(a-20)^2+b^2\geq 20^2\] \[a^2+(b-20)^2\geq 20^2\]
We graph them:
To find the area outside the two circles but inside the square, we want to find the unique area of the two circles. We can do this by adding the area of the two circles and then subtracting out their overlap. There are two methods of finding the area of overlap:
1. Consider that the area is just the quarter-circle with radius $20$ minus an isosceles right triangle with base length $20$ , and then doubled (to consider the entire overlapped area)
2. Consider that the circles can be converted into polar coordinates, and their equations are $r = 40sin\theta$ and $r = 40cos\theta$ . Using calculus with the appropriate bounds, we can compute the overlapped area.
Using either method, we compute the overlapped area to be $200\pi + 400$ , and so the area of the intersection of those three graphs is $40^2-(200\pi + 400) \Rightarrow 1200 - 200\pi \approx 571.68$
$\boxed{572}$

Answer:  572
Problem: Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$ , the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$ , and the resulting line is reflected in $l_2^{}$ . Let $R^{(1)}(l)=R(l)^{}_{}$ and $R^{(n)}(l)^{}_{}=R\left(R^{(n-1)}(l)\right)$ . Given that $l^{}_{}$ is the line $y=\frac{19}{92}x^{}_{}$ , find the smallest positive integer $m^{}_{}$ for which $R^{(m)}(l)=l^{}_{}$

Solution: Let $l$ be a line that makes an angle of $\theta$ with the positive $x$ -axis. Let $l'$ be the reflection of $l$ in $l_1$ , and let $l''$ be the reflection of $l'$ in $l_2$
The angle between $l$ and $l_1$ is $\theta - \frac{\pi}{70}$ , so the angle between $l_1$ and $l'$ must also be $\theta - \frac{\pi}{70}$ . Thus, $l'$ makes an angle of $\frac{\pi}{70}-\left(\theta-\frac{\pi}{70}\right) = \frac{\pi}{35}-\theta$ with the positive $x$ -axis.
Similarly, since the angle between $l'$ and $l_2$ is $\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}$ , the angle between $l''$ and the positive $x$ -axis is $\frac{\pi}{54}-\left(\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}\right) = \frac{\pi}{27}-\frac{\pi}{35}+\theta = \frac{8\pi}{945} + \theta$
Thus, $R(l)$ makes an $\frac{8\pi}{945} + \theta$ angle with the positive $x$ -axis. So $R^{(n)}(l)$ makes an $\frac{8n\pi}{945} + \theta$ angle with the positive $x$ -axis.
Therefore, $R^{(m)}(l)=l$ iff $\frac{8m\pi}{945}$ is an integral multiple of $\pi$ . Thus, $8m \equiv 0\pmod{945}$ . Since $\gcd(8,945)=1$ $m \equiv 0 \pmod{945}$ , so the smallest positive integer $m$ is $\boxed{945}$

Answer:  945
Problem: In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares . To take a bite, a player chooses one of the remaining squares , then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed form the original board in previous moves.)
AIME 1992 Problem 12.png
The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count.

Solution: By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts.
One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go down, and 7 that go across, with the shape on the right "carved" out by the path a possible subset.
Therefore, the total number of such paths is $\binom{12}{5}=\boxed{792}$

Answer:  792
Problem: In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares . To take a bite, a player chooses one of the remaining squares , then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed form the original board in previous moves.)
AIME 1992 Problem 12.png
The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count.

Solution: If any square is eaten, the squares to the right of it must also be eaten. Thus, if $a_i$ is the number of squares remaining on row $i$ , there is exactly one way the row can be configured (the leftmost $a_i$ squares are uneaten and the ones to the right are eaten.) Additionally, the squares above an eaten square must be eaten, so $\{a_i\}$ is nondecreasing. We can thus write $0 \leq a_1 \leq a_2 \leq \dots \leq a_5 \leq 7$ ; the number of sequences $\{a_i\}$ satisfying this inequality may be found by Stars and Bars to be $\binom{7+6-1}{6-1} = \boxed{792}$

Answer:  792
Problem: Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have?

Solution: First, consider the triangle in a coordinate system with vertices at $(0,0)$ $(9,0)$ , and $(a,b)$ . Applying the distance formula , we see that $\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}$
We want to maximize $b$ , the height, with $9$ being the base.
Simplifying gives $-a^2 -\frac{3200}{9}a +1600 = b^2$
To maximize $b$ , we want to maximize $b^2$ . So if we can write: $b^2=-(a+n)^2+m$ , then $m$ is the maximum value of $b^2$ (this follows directly from the trivial inequality , because if ${x^2 \ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \ge 0}$ ).
$b^2=-a^2 -\frac{3200}{9}a +1600=-\left(a +\frac{1600}{9}\right)^2 +1600+\left(\frac{1600}{9}\right)^2$
$\Rightarrow b\le\sqrt{1600+\left(\frac{1600}{9}\right)^2}=40\sqrt{1+\frac{1600}{81}}=\frac{40}{9}\sqrt{1681}=\frac{40\cdot 41}{9}$
Then the area is $9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}$

Answer:  820
Problem: Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have?

Solution: Let $A, B$ be the endpoints of the side with length $9$ . Let $\Gamma$ be the Apollonian Circle of $AB$ with ratio $40:41$ ; let this intersect $AB$ at $P$ and $Q$ , where $P$ is inside $AB$ and $Q$ is outside. Then because $(A, B; P, Q)$ describes a harmonic set, $AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{\frac{40}{9}}{BQ}\implies BQ=360$ . Finally, this means that the radius of $\Gamma$ is $\dfrac{360+\frac{40}{9}}{2}=180+\dfrac{20}{9}$
Since the area is maximized when the altitude to $AB$ is maximized, clearly we want the last vertex to be the highest point of $\Gamma$ , which just makes the altitude have length $180+\dfrac{20}{9}$ . Thus, the area of the triangle is $\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}$

Answer:  820
Problem: Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have?

Solution: We can apply Heron's on this triangle after letting the two sides equal $40x$ and $41x$ . Heron's gives
$\sqrt{\left(\frac{81x+9}{2} \right) \left(\frac{81x-9}{2} \right) \left(\frac{x+9}{2} \right) \left(\frac{-x+9}{2} \right)}$
This can be simplified to
$\frac{9}{4} \cdot \sqrt{(81x^2-1)(81-x^2)}$
We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.
We have that $-324x^3+13124x=0$ , so $x=\frac{\sqrt{3281}}{9}$
Plugging this into the expression, we have that the area is $\boxed{820}$

Answer:  820
Problem: Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have?

Solution: We can start how we did above in solution 4 to get $\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}$ .
Then, we can notice the inside is a quadratic in terms of $x^2$ , which is $-81(x^2)^2+6562x^2-81$ . This is maximized when $x^2 = \frac{3281}{81}$ .If we plug it into the equation, we get $\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} = \boxed{820}$

Answer:  820
Problem: In triangle $ABC^{}_{}$ $A'$ $B'$ , and $C'$ are on the sides $BC$ $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$

Solution: Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, \[\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.\] Therefore, we have \[\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}\] \[\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}\] \[\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.\] Thus, we are given \[\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.\] Combining and expanding gives \[\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}=92.\] We desire $\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.$ Expanding this gives \[\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}+2=\boxed{094}.\]

Answer:  094
Problem: In triangle $ABC^{}_{}$ $A'$ $B'$ , and $C'$ are on the sides $BC$ $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$

Solution: Using mass points , let the weights of $A$ $B$ , and $C$ be $a$ $b$ , and $c$ respectively.
Then, the weights of $A'$ $B'$ , and $C'$ are $b+c$ $c+a$ , and $a+b$ respectively.
Thus, $\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}$ $\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}$ , and $\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}$
Therefore: $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'} = \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}$ $= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =$
$2+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}+\frac{ab(a+b)}{abc} = 2 + \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}$ $= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}$

Answer:  094
Problem: In triangle $ABC^{}_{}$ $A'$ $B'$ , and $C'$ are on the sides $BC$ $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$

Solution: A consequence of Ceva's theorem sometimes attributed to Gergonne is that $\frac{AO}{OA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}$ , and similarly for cevians $BB'$ and $CC'$ .  Now we apply Gergonne several times and do algebra:
\begin{align*} \frac{AO}{OA'}\frac{BO}{OB'}\frac{CO}{OC'} &= \left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\right) \left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right)  \left(\frac{CB'}{B'A}+\frac{CA'}{A'B}\right)\\ &=\underbrace{\frac{AB'\cdot CA'\cdot BC'}{B'C\cdot A'B\cdot C'A}}_{\text{Ceva}} + \underbrace{\frac{AC'\cdot BA'\cdot CB'}{C'B\cdot A'C\cdot B'A}}_{\text{Ceva}} + \underbrace{\frac{AB'}{B'C} + \frac{AC'}{C'B}}_{\text{Gergonne}} + \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{\text{Gergonne}} + \underbrace{\frac{CA'}{A'B} + \frac{CB'}{B'A}}_{\text{Gergonne}}\\ &= 1 + 1 + \underbrace{\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'}}_{92} = \boxed{94}

Answer:  94
Problem: Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?

Solution: Let the number of zeros at the end of $m!$ be $f(m)$ . We have $f(m) = \left\lfloor \frac{m}{5} \right\rfloor + \left\lfloor \frac{m}{25} \right\rfloor + \left\lfloor \frac{m}{125} \right\rfloor + \left\lfloor \frac{m}{625} \right\rfloor + \left\lfloor \frac{m}{3125} \right\rfloor + \cdots$
Note that if $m$ is a multiple of $5$ $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$
Since $f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots  = \frac{m}{4}$ , a value of $m$ such that $f(m) = 1991$ is greater than $7964$ . Testing values greater than this yields $f(7975)=1991$
There are $\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$ , less than $1992$ . Thus, there are $1991-1595 = \boxed{396}$ positive integers less than $1992$ that are not factorial tails.

Answer:  396
Problem: Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?

Solution: After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$ , we find that $m \in \{7980, 7981, 7982, 7983, 7984\}$ . WLOG, we select $7980$ . Furthermore, note that every time $k$ reaches a multiple of $25$ $k!$ will gain two or more additional factors of $5$ and will thus skip one or more numbers.
With this logic, we realize that the desired quantity is simply $\left \lfloor \frac{7980}{25} \right \rfloor + \left \lfloor \frac{7980}{125} \right \rfloor \cdots$ , where the first term accounts for every time $1$ number is skipped, the second term accounts for each time $2$ numbers are skipped, and so on. Evaluating this gives us $319 + 63 + 12 + 2 = \boxed{396}$ . - Spacesam(edited by srisainandan6)

Answer:  396
Problem: Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*}

Solution: Define $a = x + y$ and $b = xy$ . Then $a + b = 71$ and $ab = 880$ . Solving these two equations yields a quadratic $a^2 - 71a + 880 = 0$ , which factors to $(a - 16)(a - 55) = 0$ . Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$ . For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $\le 16$ , no two factors of $16$ can sum greater than $32$ , and so there are no integral solutions for $(x,y)$ . The solution is $5^2 + 11^2 = \boxed{146}$

Answer:  146
Problem: Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*}

Solution: Let $a=x+y$ $b=xy$ then we get the equations \begin{align*} a+b&=71\\ ab&=880 \end{align*} After finding the prime factorization of $880=2^4\cdot5\cdot11$ , it's easy to obtain the solution $(a,b)=(16,55)$ . Thus \[x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}\] Note that if $(a,b)=(55,16)$ , the answer would exceed $999$ which is invalid for an AIME answer.
~ Nafer

Answer:  146
Problem: Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*}

Solution: First, notice that you can factor $x^2y + xy^2$ as $xy(x + y)$ . From this, we notice that $xy$ and $x + y$ is a common occurrence, so that lends itself to a simple solution by substitution. Let $xy = b$ and $x + y = a$ . From this substitution, we get the following system: \[a + b = 71\] \[ab = 880\] Solving that system gives us the following two pairs $(a, b)$ $(16, 55)$ and $(55, 16)$ . The second one is obviously too big as $55^2$ is clearly out of bounds for an AIME problem (More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$ . This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$ . Doing so, we get: \[(x + y)^2 = (x^2 + y^2) + 2xy\] We see that in this form, we can substitute everything in except for $(x^2 + y^2)$ , which is the desired answer. Substituting, we get: \[256 = (x^2 + y^2) + 110\] so $x^2 + y^2 = \boxed{146}$ . (If we were to go with the pair $(55, 16)$ , then the $(x + y)^2$ would be absurdly out of bounds)

Answer:  146
Problem: Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{} \le k \le 167$ , draw the segments $\overline {P_kQ_k}$ . Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$ , and then draw the diagonal $\overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn.

Solution: The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle ). For each $k$ $\overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 \cdot \frac{168-k}{168}, 4 \cdot \frac{168-k}{168}$ . Thus, its length is $5 \cdot \frac{168-k}{168}$ . Let $a_k=\frac{5(168-k)}{168}$ . We want to find $2\sum\limits_{k=1}^{168} a_k-5$ since we are over counting the diagonal. $2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 =2\frac{(0+5)\cdot169}{2}-5 =168\cdot5 =\boxed{840}$

Answer:  840
Problem: Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{} \le k \le 167$ , draw the segments $\overline {P_kQ_k}$ . Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$ , and then draw the diagonal $\overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn.

Solution: Using the above diagram, we have that $\Delta ABC \sim \Delta P_k B Q_k$ and each one of these is a dilated 3-4-5 right triangle (This is true since $\Delta ABC$ is a 3-4-5 right triangle). Now, for all $k$ , we have that $\overline{P_k Q_k}$ is the hypotenuse for the triangle $P_k B Q_k$ . Therefore we want to know the sum of the lengths of all $\overline{P_k Q_k}$ .This is given by the following: \[2 \cdot(\sum_{k=1}^{168} P_kQ_k) + 5\]
\[= 2 \cdot \frac{ 0+5+10+...+835}{168} +5\] Then by the summation formula for the sum of the terms of an arithmetic series, \[= \frac{835 \cdot 168}{168} +5 = 835+5 = \boxed{840}\]

Answer:  840
Problem: Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{} \le k \le 167$ , draw the segments $\overline {P_kQ_k}$ . Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$ , and then draw the diagonal $\overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn.

Solution: First, count the diagonal which has length $5$ . For the rest of the segments, think about pairing them up so that each pair makes $5$ . For example, the parallel lines closest to the diagonal would have length $\frac{167}{168}\cdot{5}$ while the parallel line closest to the corner of the rectangle would have length $\frac{1}{168}\cdot{5}$ by similar triangles. If you add the two lengths together, it is $\frac{167}{168}\cdot{5} + \frac{1}{168}\cdot{5} = 5.$ There are $\frac{335-1}{2}$ pairs of these segments, for a total of $5+(167)(5)=168(5)=\boxed{840}.$ ~justlearningmathog

Answer:  840