Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Put $\triangle POQ$ and $\triangle POR$ with $O$ on the origin and the triangles on the $1^{st}$ quadrant.
The coordinates of $Q$ and $P$ is $(200 \cos^{2}a,200 \cos a\sin a )$ $(200\cos^{2}b,200\cos(b)\sin b)$ . So $PQ^{2}$ $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}$ , which we want to be less then $100^{2}$ .
So $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} <= 100^{2}$ \[(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4}\] \[\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(\cos^{2} b+\sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-\cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4}\] \[(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4}\] \[\sin^{2} (b-a) \le \frac{1}{4}\] So we want $-\frac{1}{2} \le \sin (b-a) \le \frac{1}{2}$ , which is equivalent to $-30 \le b-a \le 30$ or $150 \le b-a \le 210$ . The second inequality is impossible so we only consider what the first inequality does to our $75$ by $75$ box in the $ab$ plane. This cuts off two isosceles right triangles from opposite corners with side lengths $45$ from the $75$ by $75$ box. Hence the probability is $1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}$ and the answer is $16+25 = \boxed{41}$

Answer:  41
Problem: Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: [asy] pair O, P, Q, R; draw(circle(O, 10)); O = (10, 0); P = (-10, 0); Q = (10*cos(pi/3), 10*sin(pi/3)); R = (10*cos(5*pi/6), 10*sin(5*pi/6)); dot(Q); dot(O); dot(P); dot(R); draw(P--O--Q--P--R--O); draw(Q--R, red); label("$O$", O, 2*E); label("$P$", P, 2*W); label("$Q$", Q, NE); label("$R$", R, NW); label("$200$", (0,0), 2*S); label("$x$", (Q+R)/2, N); draw(rightanglemark(O, Q, P, 38)); draw(rightanglemark(O, R, P, 38)); [/asy] Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral . Thus, we are able to apply Ptolemy's Theorem to it: \[200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),\] \[x + 200 (\cos a \sin b) = 200 (\sin a \cos b),\] \[x = 200(\sin a \cos b - \sin b \cos a),\] \[x = 200 \sin(a-b).\] We want $|x| \le 100$ (the absolute value comes from the fact that $a$ is not necessarily greater than $b,$ so we cannot assume that $Q$ is to the right of $R$ as in the diagram), so we substitute: \[|200 \sin(a-b)| \le 100,\] \[|\sin(a-b)| \le \frac{1}{2},\] \[|a-b| \le 30 ^\circ,\] \[-30 \le a-b \le 30.\] By simple geometric probability (see Solution 2 for complete explanation), $\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},$ so $m+n = \boxed{041}.$

Answer:  041
Problem: Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Impose a coordinate system as follows:
Let the midpoint of $\overline{OP}$ be the origin, and let $\overline{OP}$ be the x-axis. We construct a circle with center at the origin with radius 100. Since $\angle OQP$ and $\angle ORP$ are both right angles, points $Q$ and $R$ are on our circle. Place $Q$ and $R$ in the first quadrant of the Cartesian Plane. Suppose we construct $Q'$ and $R'$ such that they are clockwise rotations of $Q$ and $R$ , respectively by an angle of $2b$ degrees. Thus, we see that $\overline{QR}=100\sqrt{2}\sqrt{\cos(2|a-b|)}$ . We want this quantity to be less than $100$ . This happens when $\cos(2|a-b|) \ge 1/2,$ or when $|a-b|\le 30^{\circ}$ . The probability that the last inequality is satisfied is $16/25$ . Therefore, the probability that $QR$ is less than $100$ is $16/25$ . Hence, $m+n=\boxed{41}$

Answer:  41
Problem: Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: WLOG, let $b\ge a$ . It does not actually matter, but it is necessary for this particular setup. It should be apparent that $\Delta RAQ\sim\Delta OAP$ . We write the equation
\[\dfrac{RA}{AO}=\dfrac{RQ}{OP}.\]
If we examine right triangle $\Delta ROA$ , we can see that $\sin(b-a)=\dfrac{RA}{AO}$ . Also, we are given $OP=200$ , so now we have
\[\sin(b-a)=\dfrac{QR}{200}.\]
We want $QR$ to be less than or equal to $100$ ; this is equivalent to $\dfrac{QR}{200}\le\dfrac12.$ We solve from there:
\begin{align*}
\dfrac{QR}{200}&\le\dfrac12 \\
\sin(b-a)&\le\dfrac12 \\
\arcsin(\sin(b-a))&\le\arcsin\left(\dfrac12\right) \\
b-a&\le30^\circ. \\
\end{align*}
(Notice that if $a>b$ , then this would become $a-b\le30^circ.$ As in Solution 1, we can write $|a-b|\le30$ .) One can now proceed as in Solution 1, but let us tackle the geometric probability for completeness.
We now have transformed this problem into another problem asking for the probability of two uniformly, randomly, and independently chosen real numbers between $0$ and $75$ being no more than $30$ from each other.
If the first number (let this be $x$ ) is between $30$ and $45$ , then the other number can be from $x-30$ to $x+30$ - a range of $60$ . Thus, the probability that this contributes is $\dfrac{45-30}{75}\cdot\dfrac{60}{75}=\dfrac4{25}$
If $x$ is between $0$ and $30$ or $45$ to $75$ (these two cases are equivalent), the chance is the same as that of the average value since the ranges are uniform. For $x=15$ (the average), the second number can be from $0$ to $15+30=45$ - a range of $45$ . The total range is $(30-0)+(75-45)=30+30=60$ . Thus, this case contributes $\dfrac{45}{75}\cdot\dfrac{60}{75}=\dfrac{12}{25}$
Adding the two, we get $\dfrac{16}{25}$ for an answer of $16+25=\boxed{041}$

Answer:  041
Problem: Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: Writing out the recursive statement for $a_n, a_{n-1}, \dots, a_{10}$ and summing them gives \[a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10\] Which simplifies to \[a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)\] Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$ , we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$ , we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{045}$

Answer:  045
Problem: Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: \[a_n \equiv a_{n-1} + n \pmod {99}\] By looking at the first few terms, we can see that \[a_n \equiv 10+11+12+ \dots + n \pmod {99}\] This implies \[a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}\] Since $a_n \equiv 0 \pmod {99}$ , we can rewrite the equivalence, and simplify \[0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}\] \[0 \equiv n(n+1) - 90 \pmod {99}\] \[0 \equiv 4n^2+4n+36 \pmod {99}\] \[0 \equiv (2n+1)^2+35 \pmod {99}\] \[64 \equiv (2n+1)^2 \pmod {99}\] The only squares that are congruent to $64 \pmod {99}$ are $(\pm 8)^2$ and $(\pm 19)^2$ , so \[2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}\] $2n+1 \equiv -8 \pmod {99}$ yields $n=45$ as the smallest integer solution.
$2n+1 \equiv 8 \pmod {99}$ yields $n=53$ as the smallest integer solution.
$2n+1 \equiv -19 \pmod {99}$ yields $n=89$ as the smallest integer solution.
$2n+1 \equiv 19 \pmod {99}$ yields $n=9$ as the smallest integer solution. However, $n$ must be greater than $10$
The smallest positive integer solution greater than $10$ is $n=\boxed{045}$

Answer:  045
Problem: Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: $a_n=a_{n-1} + n \pmod{99}$ .  Using the steps of the previous solution we get up to $n^2+n \equiv 90 \pmod{99}$ . This gives away the fact that $(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}$ so either $n$ or $n+1$ must be a multiple of 9.
Case 1 ( $9 \mid n$ ): Say $n=9x$ and after simplification $x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}$
Case 2: ( $9 \mid n+1$ ): Say $n=9a-1$ and after simplification $(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}$
As a result $a$ must be a divisor of $10$ and after doing some testing in both cases the smallest value that works is $x=5 \implies \boxed{045}$

Answer:  045
Problem: Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: We just notice that $100 \equiv 1 \pmod{99}$ , so we are just trying to find $10 + 11 + 12 + \cdots + n$ modulo $99$ , or $\dfrac{n(n+1)}{2} - 45$ modulo $99$ . Also, the sum to $44$ is divisible by $99$ , and is the first one that is. Thus, if we sum to $45$ the $45$ is cut off and thus is just a sum to $44$
Without checking whether there are other sums congruent to $45 \pmod{99}$ , we can just write the answer to be $\boxed{045}$

Answer:  045
Problem: Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: Let $b_n = 2a_{n+10}$ . We can find a formula for $b_n$
$b_n = (20+n)(n+1)$
Notice that both can't have a factor of 3. Thus we can limit our search range of n to $n \equiv 7,8 \pmod{9}$ . Testing values for n in our search range (like 7,8,16,17,25,26...), we get that 35 is the least n. But, don't write that down! Remember, $b_n = 2a_{n+10}$ , so, the 35th term in b corresponds to the 45th term in a. Thus our answer is $\boxed{045}$

Answer:  045
Problem: Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: The first thing you should realize is that each term after the tenth is another two-digit number chained to the last number. $10, 1011, 101112, \dots$ . Now the fact that the sequence starts at $10$ can be completely discarded for this solution. Just consider $a(10)$ then same as $a(1)$ , and we can add nine to the answer at the end.
The second step is to split $99$ as $9$ and $11$ and solve for divisibility rules individually. Let's start with $11$ because it gives us the most information to continue.
In any number generated, if the numbers don't go beyond $20$ , then the highest number we can get is $10111213141516171819$ , with every odd digit being $1$ . This is a little risky because we are assuming that it doesn't exceed $20$ . If someone wanted to be absolutely sure they could continue, but this is unnecessary later and a big hassle. Anyways, now we write an equation to check for divisibility by $11$ . The expression being $\frac{((n-1) + 0)\cdot n)}{2}-n$
The concept here is to add $0$ to the $n-1^{th}$ term altogether, then subtract the number of ones in it, which is $n$ . Simplify to $\frac{n(n-3)}{2}$ congruent to $0 \pmod{11}$ . Now notice the divide by two can be discarded because one of $n$ or $(n-3)$ will be even. So if $n$ or $n-3$ is to be divisible by $11$ , we can make a simple list.
\[n = 0, 3, 11, 14, 22, 25, 33, 36, 44, 47, \dots\]
Now we test each $n$ for divisibility by $9$ .
This is done by making a list that ultimately calculates the sum of every digit in the large number. $n(1)$ to $n(10)$ has the first digit $1$ $n(11)$ to $n(20)$ has the first digit $2$ , and so on.
The necessary thing to realize is that the sum of all digits $0-9$ is divisible by $9$ , so we only have to solve for the sum of the first digits, and then the short list of second digits.
For example, let's test $n=25$
So we know that $25$ include both $1-10$ and $11-20$ , so that's $10 + 20$ right away. $21-25$ contains $5$ numbers that have the first digit $3$ , so $+15$ . Then we add $0-4$ together, which is $10$ $10+20+15+10=55$ , which is not divisible by $9$ , so it is not the answer.
Do this for just a minute you get that $36$ sums to $99$ , a multiple of nine! So $n(36)$ is the answer, right? Don't forget we have to add $9$ because we translated $n(10)$ to $n(1)$ at the very beginning!
Finally, after a short bash, we get $\boxed{045}$

Answer:  045
Problem: Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: Taking the recurrence mod $99$ , we have \[a_n=a_{n-1}+n\] for $a_{10}=10$ . Then, we have \[a_n=10+11+12+\cdots+n \implies a_n=\frac{n(n+1)}2-45 \equiv 0 \pmod{99} \implies n(n+1)-90 \equiv 0 \pmod{99} \implies n(n+1)+9 \equiv 0 \pmod{99} \implies n \equiv 0 \pmod{9}.\] Then, we simply can test these values of $n$ to find that $n=\boxed{045}$ produces a value that is also divisible by $11$

Answer:  045
Problem: Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$

Solution: Now that we’ve proven this fact, we know that all four points lie on a circle (intuitively one can also observe this because $z=z_1$ and $z=z_2$ satisfy the property in the question, and $z=z_3$ techincally gives no imaginary part), so let’s draw that in;
[asy]  size(1200,300);   real xMin = 0; real xMax = 100; real yMin = 0; real yMax = 120;  draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$R$",(xMax,0),(2,0)); label("$Im$",(0,yMax),(0,2));  pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56,104.908997); dot(A); dot(B); dot(C); dot(D);  draw(A--C);  draw(A--B);  draw(D--C);  draw(C--B);  draw(D--B);  label("$Z_1$", A, W);  label("$Z_2$", B, W); label("$Z_3$", C, N);  label("$Z$", D, N);  draw(circle((56,61), 43.9089968));  [/asy]
While $Z_1, Z_2, Z_3$ are fixed, $Z$ can be anywhere on the circle because those are the only values of $Z$ that satisfy the problem requirements. However, we want to find the real part of the $Z$ with the maximum imaginary part. This Z would lie directly above the center of the circle, and thus the real part would be the same as the x-value of the center of the circle. So all we have to do is find this value and we’re done.
Consider the perpendicular bisectors of $Z_1Z_2$ and $Z_2Z_3$ . Since any chord can be perpendicularly bisected by a radius of a circle, these two lines both intersect at the center. Since $Z_1Z_2$ is vertical, the perpendicular bisector will be horizontal and pass through the midpoint of this line, which is (18, 61). Therefore the equation for this line is $y=61$ $Z_2Z_3$ is nice because it turns out the differences in the x and y values are both equal (60) which means that the slope of the line is 1. The slope of the perpendicular bisector is therefore -1 and it passes through the midpoint, (48,69), so the equation of this line is $y=-x+117$ . Finally, equating the two yields \[61=-x+117 \implies x = \boxed{056}\]

Answer:  056
Problem: Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$

Solution: We know that \[z_3-z_1 = (78+99i)-(16+83i) = 60 + 16i.\] \[z_2-z_1=(18+39i)-(18+83i) = -44i.\] Hence, \[\frac{z_3-z_1}{z_2-z_1} = \frac{60 + 16i}{-44i} = \frac{15i-4}{11} = \frac{c}{15i+4}\] where $c \in R$ .
Let $z = a+bi$ . Then, \[\frac{z-z_2}{z-z_3} = \frac{(a+bi)-(18+39i)}{(a+bi)-(78+99i)} =\] \[\frac{z-z_2}{z-z_3} = \frac{(a-18)+i(b-39)}{(a-78)+i(b-99)} =\] \[\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{((a-78)+i(b-99))((a-78)+i(99-b)} =\] \[\frac{z-z_2}{z-z_3} = \frac{((a-18)+i(b-39)((a-78)+i(99-b))}{k}\] The numerator is: \[(a-18)(a-78)-(b-39)(99-b)+i((b-39)(a-78) + (99-b)(a-18))=\] \[a^2+b^2-96a-138b+18 \cdot 78 + 39 \cdot 99 + i(60a - 60b + 39 \cdot 78 + 18 \cdot 99)\] The ratio of the imaginary part to the real part must be $\frac{15}{4}$ because $\frac{z_3-z_1}{z_2-z_1} = \frac{c}{15i+4}.$ Hence, \[\frac{60a - 60b + 39 \cdot 78 - 18 \cdot 99}{a^2+b^2-96a-138b+18 \cdot 78 + 39 \cdot 99} = \frac{15}{4} \implies\] \[\frac{4a-4b+84}{a^2+b^2-96a-138b+9(2 \cdot 78 + 39 \cdot 11)} = \frac{1}{4} \implies\] \[16a - 16b + 336 = a^2 + b^2 -96a - 138b +5265 \implies\] \[0 = a^2 + b^2 -112a-122b+4929.\] Evidently, $b$ is maximized when $112a-a^2$ is maximized or when $a = \boxed{056}.$

Answer:  056
Problem: Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$

Solution: Algebra Bash
First we calculate $\frac{z_3 - z_1}{z_2 - z_1}$ , which becomes $\frac{15i-4}{11}$
Next, we define $z$ to be $a+bi$ for some real numbers $a$ and $b$ . Then, $\frac {z-z_2}{z-z_3}$ can be written as $\frac{(a-18)+(b-39)i}{(a-78)+(b-99)i}.$ Multiplying both the numerator and denominator by the conjugate of the denominator, we get:
$\frac {[(a-18)(a-78)+(b-39)(b-99)]+[(a-78)(b-39)-(a-18)(b-99)]i}{(a-78)^2+(b-99)^2}$
In order for the product to be a real number, since both denominators are real numbers, we must have the numerator of $\frac {z-z_2}{z-z_3}$ be a multiple of the conjugate of $15i-4$ , namely $-15i-4$ So, we have $(a-18)(a-78)+(b-39)(b-99) = -4k$ and $(a-78)(b-39)-(a-18)(b-99) = -15k$ for some real number $k$
Then, we get: $(a-18)(a-78)+(b-39)(b-99) = \frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)]$
Expanding both sides and combining like terms, we get:
$a^2 - 112a +b^2 - 122b + 4929 = 0$
which can be rewritten as:
$(a-56)^2 + (b-61)^2 = 1928$
Now, common sense tells us that to maximize $b$ , we would need to maximize $(b-61)^2$ . Therefore, we must set $(a-56)^2$ to its lowest value, namely 0. Therefore, $a$ must be $\boxed{056}.$

Answer:  056
Problem: Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$

Solution: We see that $\frac{z_3-z_1}{z_2-z_1}=\frac{60+16i}{-44i}=\frac{15i-4}{11}$
Now, let $z-z_3=a+bi$ , in which case $z=(a+78)+(b+99)i$ and $z-z_2=(a+60)+(b+60)i$
We now have that $\left(\frac{(a+60)+(b+60)i}{a+bi}\right)\left(\frac{15i-4}{11}\right)$ is real, meaning that: \[((a+60)+(b+60)i)\left(\frac{15i-4}{a+bi}\right)\] is also real.
We see that: \[\frac{15i-4}{a+bi}=\frac{(15i-4)(a-bi)}{a^2+b^2}=\frac{(15b-4a)+(15a+4b)i}{a^2+b^2},\] so therefore, $x=((a+60)+(b+60)i)((15b-4a)+(15a+4b)i)$ is real.
This means that $Im(x)=0$ , so we now have that: \[(a+60)(15a+4b)+(b+60)(15b-4a)=15a^2+15b^2+660a+1140b=0 \Rightarrow a^2+b^2+44a+76b=0.\]
This can be rewritten as: \[(a+22)^2+(b+38)^2=22^2+38^2.\] In order to maximize $Im(z)$ we want to maximize $b$ , and in order to maximize $b$ we want $a+22=0$ and $a=-22$ , so $Re(z)=a+78=-22+78=\boxed{056}$ . 
(Note: $Im(\omega)$ is the imaginary part of $\omega$ , and $Re(\omega)$ is the real part of $\omega$ )
~Stormersyle

Answer:  056
Problem: Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$

Solution: We will just bash. Let $z=a+bi$ where $a,b\in\mathbb{R}$ . We see that $\frac{z_3-z_1}{z_2-z_1}=\frac{-4+15i}{11}$ after doing some calculations. We also see that $\frac{[(a-18)+(b-39)i][(a-78)-(b-99)i]}{\text{some real stuff}}.$ We note that $[(a-18)+(b-39)i][(a-78)-(b-99)i]$ is a multiple of $-4-15i$ because the numerator has to be real. Thus, expanding it out, we see that $(a-18)(a-78)+(b-39)(b-99)=-4k \\ (a-78)(b-39)-(a-18)(b-99)=-15k.$ Hence, $(a-18)(a-78)+(b-39)(b-99)=\frac{4}{15}[(a-78)(b-39)-(a-18)(b-99)] \implies a^2-96a+b^2-138b+5265=16a-16b+336 \\ (a-56)^2+(b-61)^2=1928.$ To maximize the imaginary part, $(a-56)^2$ must equal $0$ so hence, $a=\boxed{56}$

Answer:  56
Problem: Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$

Solution: Assume that $5 \in \{a_1, a_2, a_3\}$ $m \neq 5$ , and WLOG, $\max{(a_1, a_2, a_3)} = 5$ . Then we know that the other two medians in $\{a_1, a_2, a_3\}$ and the smallest number of rows 1, 2, and 3 are all less than 5. But there are only 4 numbers less than 5 in $1, 2, 3, \dots, 9$ , a Contradiction. Thus, if $5 \in \{a_1, a_2, a_3\}$ , then $m = 5$
WLOG, assume $5$ is in the upper left corner. One of the two other values in the top row needs to be below $5$ , and the other needs to be above $5$ . This can be done in $4\cdot4\cdot2=32$ ways.
The other $6$ can be arranged in $6!=720$ ways.
Finally, accounting for when $5$ is in every other space, our answer is $32\cdot720\cdot9$ , which is $207360$ . But we only need the last $3$ digits, so $\boxed{360}$ is our answer.

Answer:  360