Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$ . What is $p+q$
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$

Solution: Let $w=\lfloor x \rfloor$ and $f=\{x\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\leq f<1$ and $x=w+f.$
We rewrite the given equation as \[w\cdot f=a\cdot(w+f)^2. \hspace{38.75mm}(1)\] Since $a\cdot(w+f)^2\geq0,$ it follows that $w\cdot f\geq0,$ from which $w\geq0.$
We expand and rearrange $(1)$ as \[af^2+(2a-1)wf+aw^2=0, \hspace{23mm}(2)\] which is a quadratic with either $f$ or $w.$
For simplicity purposes, we will treat $w$ as some fixed nonnegative integer so that $(2)$ is a quadratic with $f.$ By the Quadratic Formula, we have \[f=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3)\] If $w=0,$ then $f=0.$ We get $x=w+f=0,$ which does not affect the sum of the solutions. Therefore, we consider the case for $w\geq1:$
Recall that $0\leq f<1,$ so $\frac{1-2a\pm\sqrt{1-4a}}{2a}\geq0.$ From the discriminant, we require that $0\leq1-4a<1,$ or \[0<a\leq\frac14. \hspace{54mm}(4)\]
We consider each part of $0\leq f<1$ separately:
Now, we express $x$ in terms of $w$ and $k:$ \[x=w+f=w+wk=w(1+k).\] The sum of all solutions to the original equation is \[\sum_{w=1}^{W}w(1+k)=(1+k)\cdot\sum_{w=1}^{W}w=(1+k)\cdot\frac{W(W+1)}{2}=420. \hspace{10mm}(\bigstar)\] As $1+k<1+\frac1W,$ we conclude that $1+k$ is slightly above $1$ so that $\frac{W(W+1)}{2}$ is slightly below $420,$ or $W(W+1)$ is slightly below $840.$ By observations, we get $W=28.$ Substituting this into $(\bigstar)$ produces $k=\frac{1}{29},$ which satisfies $\frac{1}{W+1}\leq k<\frac1W,$ as required.
Finally, we solve for $a$ in $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}:$ \begin{align*} \frac{1}{29}&=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a} \\ \frac{2}{29}a&=1-2a-\sqrt{1-4a} \\ \frac{60}{29}a-1&=-\sqrt{1-4a} \\ \frac{60^2}{29^2}a^2-\frac{120}{29}a+1&=1-4a \\ \frac{60^2}{29^2}a^2-\frac{4}{29}a&=0 \\ a\left(\frac{60^2}{29^2}a-\frac{4}{29}\right)&=0. \end{align*} Since $a>0,$ we obtain $\frac{60^2}{29^2}a-\frac{4}{29}=0,$ from which \[a=\frac{4}{29}\cdot\frac{29^2}{60^2}=\frac{29}{900}.\] The answer is $29+900=\boxed{929}.$

Answer: C 929
Problem: The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$ . What is $p+q$
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$

Solution: Let $x_n$ be a root in the interval $(n,n+1)$ . In this interval, $\lfloor x_n \rfloor = n$ and $\{x_n\}=x_n-n$ , so we must have $ax_n^2 = nx_n-n^2$ , i.e., $ax_n^2-nx_n+n^2=0$ . We can homogenize this equation by setting $x_n=n\zeta$ ; then $x_1=\zeta$ , and $\zeta$ is a root of $a\zeta^2-\zeta+1=0$
Suppose $N$ is the largest integer for which there is such a root; we have, for $n=1,2,\ldots , N$ \[n < x_n = n\zeta < n+1\] Summing over $n\in \{1,2,\ldots , N\}$ we get \[\tfrac 12 N(N+1) < 420 = \tfrac 12 N(N+1)\zeta < \tfrac 12 N(N+3)\] From the right inequality we get $27< N$ and from the left one we get $N<29$ . Thus $N=28$ . Using this in the middle equality we get $\zeta = \tfrac{30}{29}$ . Since $\zeta$ satisfies $a\zeta^2-\zeta+1=0$ , we get \[a = \zeta^{-2}(\zeta-1)= \tfrac{29^2}{30^2}\cdot \tfrac 1{29}= \tfrac{29}{900}.\] The answer is $29+900=\boxed{929}.$

Answer: C 929
Problem: The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$ . What is $p+q$
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$

Solution: First note that $\lfloor x\rfloor \cdot \{x\}<0$ when $x<0$ while $ax^2\ge 0\forall x\in \mathbb{R}$ . Thus we only need to look at positive solutions ( $x=0$ doesn't affect the sum of the solutions). 
Next, we break $\lfloor x\rfloor\cdot \{x\}$ down for each interval $[n,n+1)$ , where $n$ is a positive integer. Assume $\lfloor x\rfloor=n$ , then $\{x\}=x-n$ . This means that when $x\in [n,n+1)$ $\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2$ . Setting this equal to $ax^2$ gives \[nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}\] We're looking at the solution with the positive $x$ , which is $x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)$ . Note that if $\lfloor x\rfloor=n$ is the greatest $n$ such that $\lfloor x\rfloor \cdot \{x\}=ax^2$ has a solution, the sum of all these solutions is slightly over $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ , which is $406$ when $n=28$ , just under $420$ . Checking this gives \begin{align*} \sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)&=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420 \\ \frac{1-\sqrt{1-4a}}{2a}&=\frac{420}{406}=\frac{30}{29} \\ 29-29\sqrt{1-4a}&=60a \\ 29\sqrt{1-4a}&=29-60a \\ 29^2-4\cdot 29^2a&=29^2+3600a^2-120\cdot 29a \\ 3600a^2&=116a \\ a&=\frac{116}{3600}=\frac{29}{900} \implies \boxed{929} ~ktong

Answer: C 929
Problem: What is the value in simplest form of the following expression? \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]
$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$

Solution: We have \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{10}.\] Note: This comes from the fact that the sum of the first $n$ odds is $n^2$

Answer: C 10
Problem: What is the value of the following expression?
\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80}$

Solution: Using difference of squares to factor the left term, we get \[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.\] Cancelling all the terms, we get $\boxed{1}$ as the answer.

Answer: A 1
Problem: Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution: Suppose team $A$ has played $g$ games in total so that it has won $\frac23g$ games.
It follows that team $B$ has played $g+14$ games in total so that it has won $\frac23g+7$ games.
We set up and solve an equation for team $B$ 's win ratio: \begin{align*} \frac{\frac23g+7}{g+14}&=\frac58 \\ \frac{16}{3}g+56&=5g+70 \\ \frac13g&=14 \\ g&=\boxed{42} ~MRENTHUSIASM

Answer: C 42
Problem: Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution: If we consider the number of games team $B$ has played as $x$ and the number of games that team $A$ has played as $y$ , then we can set up the following system of equations: \begin{align*} \frac{5}{8}x &= \frac{2}{3}y+7, \\ \frac{3}{8}x &= \frac{1}{3}y+7. \end{align*} The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by $2$ and solving the system, we get $y = 42$ or answer choice $\boxed{42}.$

Answer: C 42
Problem: Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution: First, let us assign some variables. Let \[A_w=2x, \ A_l=x, \ A_g=3x,\] \[B_w=5y, \ B_l=3y, \ B_g=8y,\] where $X_w$ denotes number of games won, $X_l$ denotes number of games lost, and $X_g$ denotes total games played for $X\in \{A, B\}$ . Using the given information, we can set up the following two equations: \begin{align*} B_w=A_w+7&\implies 5y=2x+7, \\ B_l=A_l+7&\implies 3y=x+7. \end{align*} We can solve through substitution, as the second equation can be written as $x=3y-7$ , and plugging this into the first equation gives $5y=6y-7\implies y=7$ , which means $x=3(7)-7=14$ . Finally, we want the total number of games team $A$ has played, which is $A_g=3(14)=\boxed{42}$

Answer: C 42
Problem: Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution: Using the information from the problem, we can note that team $A$ has lost $\frac{1}{3}$ of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for $A,$ represented in the form $(w, l)$ for convenience: \begin{align*} \textbf{(A)} &\implies (14, 7) \\ \textbf{(B)} &\implies (18, 9) \\ \textbf{(C)} &\implies (28, 14) \\ \textbf{(D)} &\implies (32, 16) \\ \textbf{(E)} &\implies (42, 21) \end{align*} Thus, we have $5$ matching $B$ scenarios, simply adding $7$ to $w$ and $l.$ We can then test each of the five $B$ scenarios for $\frac{w}{w+l} = \frac{5}{8}$ and find that $(35, 21)$ fits this description. Then working backwards and subtracting $7$ from $w$ and $l$ gives us the point $(28, 14),$ making the answer $\boxed{42}.$

Answer: C 42
Problem: Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution: Let's say that team $A$ plays $n$ games in total. Therefore, team $B$ must play $n + 14$ games in total (7 wins, 7 losses) Since the ratio of $A$ is \[\frac{2}{3} \implies n \equiv 0 \pmod{3}\] Similarly, since the ratio of $B$ is \[\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}\] Now, we can go through the answer choices and see which ones work: \begin{align*} \textbf{(A) } 21 &\implies 21 + 14 = 35 \not \equiv 0\pmod{8} \\ \textbf{(B) } 27 &\implies 27 + 14 = 41 \not \equiv 0\pmod{8} \\ \textbf{(C) } 42 &\implies 42 + 14 = 56  \equiv 0\pmod{8} \\ \textbf{(D) } 48 &\implies 48 + 14 = 62 \not \equiv 0\pmod{8} \\ \textbf{(E) } 63 &\implies 63 + 14 = 77 \not \equiv 0\pmod{8} \\ \end{align*} So we can see $\boxed{42}$ is the only valid answer.

Answer: C 42
Problem: Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution: Intersect at the origin and select a point on each line to define vectors $\mathbf{v}_{i}=(x_{i},y_{i})$ .
Note that $\theta=45^{\circ}$ gives equal magnitudes of the vector products \[\mathbf{v}_1\cdot\mathbf{v}_2 = v_{1}v_{2}\cos\theta \quad\mathrm{and}\quad |\mathbf{v}_1\times\mathbf{v}_2| = v_{1}v_{2}\sin\theta .\]
Substituting coordinate expressions for vector products, we find \[\mathbf{v}_1\cdot\mathbf{v}_2 = |\mathbf{v}_1\times\mathbf{v}_2| \ \implies\ x_{1}x_{2}+y_{1}y_{2} = x_{1}y_{2}-x_{2}y_{1} .\] Divide this equation by $x_{1}x_{2}$ to obtain \[1+m_{1}m_{2} = m_{2}-m_{1} ,\] where $m_{i}=y_{i}/x_{i}$ is the slope of line $i$ .
Taking $m_{2}=6m_{1}$ , we obtain \[6m_{1}^{2}-5m_{1}+1 = 0 \ \implies\ m_{1} \in \{\frac{1}{3},\frac{1}{2}\} .\] The latter solution gives the largest product of slopes $m_{1}m_{2} = 6m_{1}^2 = \frac{3}{2} . \quad \boxed{32}$

Answer: C 32
Problem: Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution: Place on coordinate plane.
Lines are $y=mx, y=6mx.$ The intersection point at the origin.
Goes through $(0,0),(1,m),(1,6m),(1,0).$ So by law of sines, $\frac{5m}{\sin{45^{\circ}}} = \frac{\sqrt{1+m^2}}{1/(\sqrt{1+36m^2})},$ lettin $a=m^2$ we want $6a.$ Simplifying gives $50a = (1+a)(1+36a),$ so $36a^2-13a+1=0 \implies 36(a-1/4)(a-1/9)=0,$ so max $a=1/4,$ and $6a=3/2 \quad \boxed{32}.$

Answer: C 32
Problem: Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution: Let one of the lines have equation $y=ax$ . Let $\theta$ be the angle that line makes with the x-axis, so $\tan(\theta)=a$ . The other line will have a slope of $\tan(45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}$ . Since the slope of one line is $6$ times the other, and $a$ is the smaller slope, $6a = \frac{1+a}{1-a} \implies 6a-6a^2=1+a \implies 6a^2-5a+1=0 \implies a=\frac{1}{2},\frac{1}{3}$ . If $a = \frac{1}{2}$ , the other line will have slope $\frac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3$ . If $a = \frac{1}{3}$ , the other line will have slope $\frac{1+\frac{1}{3}}{1-\frac{1}{3}} = 2$ . The first case gives the bigger product of $\frac{3}{2}$ , so our answer is $\boxed{32}$

Answer: C 32
Problem: Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution: Let the smaller slope be $m$ , then the larger slope is $6m$ . Since we want the greatest product we begin checking each answer choice, starting with (E).
$6m^2=6$
$m^2=1$
This gives $m=1$ and $6m=6$ . Checking with a protractor we see that this does not form a 45 degree angle.
Using this same method for the other answer choices, we eventually find that the answer is $\boxed{32}$ since our slopes are $\frac12$ and $3$ which forms a perfect 45 degree angle.

Answer: C 32
Problem: Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution: If you have this formula memorized then it will be very easy to do: If $\theta$ is the angle between two lines with slopes $m_1$ and $m_2$ , then $\tan(\theta)=\frac{m_1-m_2}{1+m_1m_2}$
Now let the smaller slope be $m$ , thus the other slope is $6m$ . Using our formula above: \[\tan(45^\circ)=\frac{6m-m}{1+6m^2} \implies 6m^2-5m+1=0 \implies (2m-1)(3m-1)=0.\] Therefore the two possible values for $m$ are $\tfrac{1}{2}$ and $\tfrac{1}{3}$ . We choose the larger one and thus our answer is \[6m^2=6 \cdot \frac{1}{4} = \frac{3}{2} \implies \boxed{32}.\]

Answer: C 32
Problem: Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$ . Let $CX=a$ and $EX=b$ . This implies that $DE = a - b$ . Since $CE = CX + EX = a + b$ , we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$ . Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50$ . Thus, our answer is $2\times50=\boxed{100}$

Answer: E 100
Problem: Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: Let $O$ be the center of the circle, and $X$ be the midpoint of $CD$ . Draw triangle $OCD$ , and median $OX$ . Because $OC = OD$ $OCD$ is isosceles, so $OX$ is also an altitude of $OCD$ $OE = 5\sqrt2 - 2\sqrt5$ , and because angle $OEC$ is $45$ degrees and triangle $OXE$ is right, $OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}$ . Because triangle $OXC$ is right, $CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}$ . Thus, $CD = 2\sqrt{15 + 10\sqrt{10}}$
We are looking for $CE^2$ $DE^2$ which is also $(CE + DE)^2 - 2 \cdot CE \cdot DE$
Because $CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}$ $(CE + DE)^2 = CD^2=4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}$
By Power of a Point, $CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20$ , so $2 \cdot CE \cdot DE = 40\sqrt{10} - 40$
Finally, $CE^2 + DE^2 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{100}$

Answer: E 100
Problem: Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: Let $O$ be the center of the circle. Notice how $OC = OD = r$ , where $r$ is the radius of the circle. By applying the law of cosines on triangle $OCE$ \[r^2=CE^2+OE^2-2(CE)(OE)\cos{45}=CE^2+OE^2-(CE)(OE)\sqrt{2}.\]
Similarly, by applying the law of cosines on triangle $ODE$ \[r^2=DE^2+OE^2-2(DE)(OE)\cos{135}=DE^2+OE^2+(DE)(OE)\sqrt{2}.\]
By subtracting these two equations, we get \[CE^2-DE^2-(CE)(OE)\sqrt{2}-(DE)(OE)\sqrt{2}=0.\] We can rearrange it to get \[CE^2-DE^2=(CE)(OE)\sqrt{2}+(DE)(OE)\sqrt{2}=(CE+DE)(OE\sqrt{2}).\]
Because both $CE$ and $DE$ are both positive, we can safely divide both sides by $(CE+DE)$ to obtain $CE-DE=OE\sqrt{2}$ . Because $OE = OB - BE = 5\sqrt{2} - 2\sqrt{5}$ \[(CE-DE)^2 = CE^2+DE^2 - 2(CE)(DE) = (OE\sqrt{2})^2 =2(5\sqrt{2} - 2\sqrt{5})^2 = 140 - 40\sqrt{10}.\]
Through power of a point, we can find out that $(CE)(DE)=20\sqrt{10} - 20$ , so \[CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{100}.\]

Answer: E 100
Problem: Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: [asy] /* Made by sofas103; edited by MRENTHUSIASM */ size(250);  pair O, A, B, C, D, E, D1; O = origin; A = (-5*sqrt(2),0); B = (5*sqrt(2),0); E = (5*sqrt(2)-2*sqrt(5),0); path p; p = Circle(O,5*sqrt(2)); C = intersectionpoint(p,E--E+10*dir(135)); D = intersectionpoint(p,E--E+10*dir(-45)); D1 = (D.x,-D.y); draw(p); dot("$O$",O,1.5*S,linewidth(4)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$D'$",D1,1.5*dir(D1),linewidth(4)); draw(A--B^^C--D^^C--D1--O--cycle^^D1--E); [/asy] Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$ , we have that $\angle BED'=45$ . Since $\angle AEC=45$ $\angle CED'=90$ . Since $DE=ED'$ , by the Pythagorean Theorem, our desired solution is just $CD'^2$ .
Looking next to circle arcs, we know that $\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45$ , so $\overarc{AC}+\overarc{BD}=90$ . Since $\overarc{BD'}=\overarc{BD}$ , and $\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180$ $\overarc{CD'}=90$ . Thus, $\angle COD'=90$ .
Since $OC=OD'=5\sqrt{2}$ , by the Pythagorean Theorem, the desired $CD'^2= \boxed{100}$

Answer: E 100
Problem: Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\  44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution: Basically, by PoP, you have that \[CE \times DE = (10\sqrt{2}-2\sqrt{5})(2\sqrt{5}) = 20\sqrt{10} - 20.\] Therefore, as $CE^2 + DE^2 = (CD)^2 - 2(CE \times DE),$ basically, once you find $CD^2,$ the problem is done. Now, this is an IMPORTANT concept: If you have a circle which you know the radius of and you want to find the length of a chord of that circle, drop an altitude from the center of the circle to the chord to find distance between the center of the circle and the chord.
In this case, let $M$ be the midpoint of chord $CD.$ Notice that now we can use our $45^\circ{}$ angle, since $OME$ is a $45^\circ{}-45^\circ{}-90^\circ{}$ triangle so that $ME = x$ and $OE = x\sqrt{2}.$ However, we have that $OE = 5\sqrt{2}-2\sqrt{5},$ so that $x = 5 - \sqrt{10}.$ Now, notice that $x^2 = 35 - 10\sqrt{10},$ so that \[CM^2 = 50 - x^2 = 50 - (35 - 10\sqrt{10}) = 15 + 10 \sqrt{10}\] and \[CD^2 = 60 + 4\sqrt{10}.\] Therefore, \[CE^2 + DE^2 = CD^2 - 2(CE \times DE) = (60+4\sqrt{10}) - 2(2\sqrt{10} - 20) = (60+4\sqrt{10}) + (40-4\sqrt{10}) = \boxed{100}.\]

Answer: E 100