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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Jane street swag package? USA(J)MO
arfekete   5
N 2 minutes ago by bebebe
Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.
5 replies
arfekete
Yesterday at 4:34 PM
bebebe
2 minutes ago
USAMO Medals
YauYauFilter   41
N 3 minutes ago by Inaaya
YauYauFilter
Apr 24, 2025
Inaaya
3 minutes ago
how prestigious is hsmc
ConfidentKoala4   1
N 8 minutes ago by Ilikeminecraft
been wonderin this for a while

how prestigious is it? ik its not as good as mathily (they rejected me :mad: ) but Idk how good it actually is
1 reply
ConfidentKoala4
21 minutes ago
Ilikeminecraft
8 minutes ago
9 Does Mental Health Actually Matter?
heheman   9
N 20 minutes ago by maxamc
Looking at the goals I once had, it was all just so silly and stupid

I didn't even reach my "Low" goal for AIME... so pathetic

Missed JMO by a huge margin, after missing by only 12.5 points last year

(BTW i didn't slack off one bit)

I guess the most important thing is just to keep my head up and keep going. I can't let failures stop me. Honestly I don't care about setting goals anymore. They only give me a lot of internal pressure to do well. I think the most important thing is to focus on what I do everyday, consistently, and pay attention to the beautiful things in life (like math).

I'm going to try getting more involved in real life. After coming back from COVID, I had trouble to make as many friends with non-math people. But I was reconnecting with some of my friends that I had prepandemic and I realized how precious those friendships really were.

Now the last thing to do is grind my last bit of nonexistent ego to dust and focus on the present, stop looking back

(Note: This doesn't mean I'm going to quit, I just mean I'm going to do math on my own and try to not feel any pressure to do well. Cause i feel like that pressure really beat me a lot.)

I love this community and am happy for everyone who qualified olympiad but at this point competition math just reminds me only of my failures. (Even if it's my own fault.) So I'm probably going to take a break for a while. Thanks everyone for being nice to me and stuff. Sorry if this sounds cringe (it will in a week)

9 replies
heheman
Mar 8, 2024
maxamc
20 minutes ago
No more topics!
complex number geo
zhoujef000   32
N Apr 18, 2025 by Jaxman8
Source: 2025 AIME I #8
Let $k$ be a real number such that the system \begin{align*} &|25+20i-z|=5\\ &|z-4-k|=|z-3i-k| \\ \end{align*}has exactly one complex solution $z.$ The sum of all possible values of $k$ can be written as $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Here $i=\sqrt{-1}.$
32 replies
zhoujef000
Feb 7, 2025
Jaxman8
Apr 18, 2025
complex number geo
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Source: 2025 AIME I #8
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zhoujef000
316 posts
#1 • 1 Y
Y by mathfan2020
Let $k$ be a real number such that the system \begin{align*} &|25+20i-z|=5\\ &|z-4-k|=|z-3i-k| \\ \end{align*}has exactly one complex solution $z.$ The sum of all possible values of $k$ can be written as $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Here $i=\sqrt{-1}.$
This post has been edited 1 time. Last edited by zhoujef000, Feb 8, 2025, 5:44 PM
Reason: think this was how it was aligned on the test
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MathXplorer10
163 posts
#2
Y by
I got 73/4 —> 77
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plang2008
337 posts
#3 • 2 Y
Y by bjump, AtharvNaphade
First equation is circle with center $(25, 20)$ and radius $5$, second equation is perpendicular bisector of line through $(4+k, 0)$ and $(k, 3)$. This is $y - 1.5 = \frac 43 (x - (2 + k))$. Now notice we want this line to be tangent to the circle. There are two such lines, symmetric across $(25,20)$. Plugging this point in gives $k = \frac{73}{8}$ so the answer is $\frac{73}{4} \implies \boxed{077}$.
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KenWuMath
58 posts
#4 • 9 Y
Y by bjump, AlexWin0806, OlympusHero, MathPerson12321, roribaki, EpicBird08, ihatemath123, Sedro, aidan0626
i forgor the geometric interpretation :blush:
and did perhaps the ugliest bash of my life
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ninjaforce
95 posts
#5
Y by
Clean solution that requires next to no insight
This post has been edited 1 time. Last edited by ninjaforce, Feb 11, 2025, 6:18 AM
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exp-ipi-1
1074 posts
#6
Y by
Who else though the circle is centered at (25,-20) and put 317
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OlympusHero
17020 posts
#7 • 1 Y
Y by razormouth
I put 73/8 after bashing for 20 min
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Elephant200
1472 posts
#8
Y by
I got the insight but then I totally screwed up after that :(
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EaZ_Shadow
1267 posts
#9 • 1 Y
Y by Ad112358
Elephant200 wrote:
I got the insight but then I totally screwed up after that :(

same :(
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xHypotenuse
778 posts
#10
Y by
This bash took me 30 MINUTES

stupid vietabash

still got it right -> 73/4 -> 77.
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Bluesoul
894 posts
#11
Y by
Let $z=x+yi, (x-(4+k))^2+y^2=(x-k)^2+(y-3)^2$, which simplifies to $8x-6y-8k-7=0$

Now we want this line's distance to $(25,20)$ is $5$ since the first expression is a circle centered at $(25,20)$ with radius $5$, so $\frac{|73-8k|}{10}=5, k=\frac{23}{8}, \frac{123}{8}$, sum up to $\frac{73}{4}\implies \boxed{077}$
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GrantStar
821 posts
#12 • 2 Y
Y by sami1618, remedy
Oops I forgot to subtract 4 :sob:
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williamxiao
2514 posts
#13
Y by
Didn’t simplify the fraction , put 154 :)

The key idea is to interpret the problem geometrically. The first equation says that z lies on a circle with radius 5 centered at $(25,20)$ and the second equation says that z lies on the perpendicular bisector of (k+4,0) and (k,3). The perpendicular bisector of this line has slope 4/3. We want only one solution for z, so we are simply looking for the lines tangent to the circle with slope 4/3. Once we have these two lines, it must pass through $(k+2, \frac{3}{2})$. Solve and sum the values of k to get $\frac{146}{8}$ (:sob:) $=\frac{73}{4} \rightarrow \boxed{77}$
This post has been edited 1 time. Last edited by williamxiao, Feb 7, 2025, 6:13 PM
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Math4Life7
1703 posts
#14
Y by
don't enjoy the amount of computation in this problem
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cappucher
94 posts
#15 • 2 Y
Y by anduran, ninjaforce
Bashing time

Let $z = a + bi$. Then our equations become

\[(25 - a)^2 + (20 - b)^2 = 25\]\[(a - 4 - k)^2 + b^2 = (a - k)^2 + (b - 3)^2\]
We expand out the second equation:

\[a^2 + 16 + k^2 - 8a - 2ak + 8k + b^2 = a^2 - 2ak + k^2 + b^2 - 6b + 9\]\[16 - 8a  + 8k = -6b + 9\]\[a = \frac{8k + 7 + 6b}{8}\]
We now plug this into the first equation:

\[\left(25 - \frac{8k + 7 + 6b}{8}\right)^2 + (20 - b)^2 = 25\]\[\left(\frac{193}{8} - k - \frac{3}{4}b\right)^2 + (20 - b)^2 = 25\]
Let $v = \frac{193}{8} - k$ to simplify calculations.

\[\left(v - \frac{3}{4}b\right)^2 + (20 - b)^2 = 25\]\[v^2 - \frac{3}{2}bv + \frac{9}{16}b^2 + 400 - 40b + b^2 = 25\]\[\frac{25}{16}b^2 - b\left(\frac{3}{2}v + 40\right) + (v^2 + 375)\]
In order for there to be exactly one solution for $z$, the discriminant of this quadratic must be $0$:

\[\left(\frac{3}{2}v + 40\right)^2 - 4\left(\frac{25}{16}\right)\left(v^2 + 375\right) = 0\]
Since we are asked for the sum of all $k$, we just use Vieta's relations on this quadratic.

\[\frac{9}{4}v^2 + 120v + 1600 - \frac{25}{4}v^2 - 4\left(\frac{25}{16}\right)\left(375\right) = 0\]\[-4v^2 + 120v + c = 0\]
We can ignore the constant term because it's irrelevant to the sum of the roots. Since we have $v_1 + v_2 = 30$, we conclude that

\[2\left(\frac{193}{8}\right) - (k_1 + k_2) = 30\]\[k_1 + k_2 = \frac{73}{4}\]
So the answer is $73 + 4 = \boxed{77}$.
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BS2012
1033 posts
#16
Y by
Algebraic non-bash solution

The first equation implies
$$z-k=(25-5\cos(a)-k)+(20-5\sin(a))i$$for some real $a.$ Then, let $z-k=x+yi.$ By the second equation,
$$(x-4)^2+y^2=x^2+(y-3)^2$$which simplifies to
$$8x=6y+7.$$Plugging in the real parts for $x$ and $y,$ we have
$$200-40\cos(a)-8k=127+7-30\sin(a).$$This means that
$$73-8k=40\cos(a)-30\sin(a).$$Dividing both sides by $50,$ we have
$$\dfrac{73-8k}{50}=\dfrac{4}{5}\cos(a)-\dfrac{3}{5}\sin(a).$$Let $m$ be an angle such that $\cos(m)=\frac{4}{5}$ and $\sin(m)=\frac{3}{5}.$ Recognizing the cosine angle addition formula, we have
$$\dfrac{73-8k}{50}=\cos(a+m).$$This has one solution for $a$ precisely when
$$\dfrac{73-8k}{50}=\pm 1,$$and simple algebra gets the answer
$$k=\dfrac{73\pm 50}{8}\implies \dfrac{73-50+73+50}{8}=\dfrac{73}{4}\implies\boxed{077}.$$
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Iwowowl253
133 posts
#17
Y by
Rip, i put $146+8=154$
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newhydepark
10 posts
#18
Y by
You can use calculus to solve this problem.
If we let $z=a+bi$, expanding/simplifying the system, we get
$$(25-a)^2+(20-b)^2=25$$$$-8a+8k+6b+7=0$$It's worthwhile to note, this boils down to observing where a tangent line occurs on the circle with slope $\frac{4}{3}$. If we let $25-a=x$ and $20-b=y$, using implicit differentiation on the circle, we get
$$ 2x + 2y (y') = 0 $$Solving for $y'$, we get
$$ \frac{dy}{dx} = \frac{-x}{y} = \frac{4}{3}$$If we substitute the derivative into the circle equation, we will get the solutions $(x,y) = (-4,3)$ and $(4,-3)$. Do not forget we sub $x$ and $y$ for $a$ and $b$, so the points $(a,b)$ are $(29,17)$ and $(21, 23)$. If we plug it into the tangent line equation, and solve for $k$ for both, we get
$$ \frac{8(50)-6(40)-14}{8} = \frac{146}{8} = \frac{73}{4} => 73 + 4 = \boxed{077} $$
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xTimmyG
265 posts
#19
Y by
this problem almost cooked my score :wallbash_red:
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megarnie
5606 posts
#20
Y by
Bash is not bad, especially if you eliminate fractions

Let $z = a + bi$. We have $(a-25)^2 + (b-20)^2 = 25$ and $(a - 4 - k)^2 + b^2 = (a - k)^2 + (b - 3)^2$.

The second equation can be easily simplified (through expansion or difference of squares) as \[ 6b = 8a - (8k + 7) \]
Now let $x = 8k + 7$.

The first equation becomes $a^2 - 50a + b^2 - 40b  + 1000 = 0$, or \[ 36a^2 - 1800a + (6b)^2 - 240(6b) + 36000 = 0 \]
Setting $6b = 8a - x$ gives that the LHS is equal to \[ 36a^2 - 1800a + (8a)^2 - 16xa + x^2 - 1920a + 240x + 36000\]
This can be rewritten as \[ 100a^2 - (16x + 3720) a + x^2 + 240x + 36000\]There is one solution iff the discriminant is $0$, so $x$ works iff \[ (16x + 3720)^2 = 400(x^2 + 240x + 36000),\]or \[ (4x + 930)^2 = 25(x^2 + 240x + 36000)\]If $x$ is an integer divisible by $4$, then taking modulo $8$ gives a contradiction. Thus, if $x$ is an integer, $4\nmid x$.

The sum of the roots (counting double roots twice) of this is just $\frac{8 \cdot 930 - 25 \cdot 240 }{9} = \frac{8 \cdot 310 - 25 \cdot 80}{3} = 160$. Since $4$ doesn't divide any root, there is not a double root, and therefore the sum of all possible values of $k$ is just \[ \frac{160 - 2 \cdot 7}{8} = \frac{73}{4} \implies \boxed{077} \]
(note: one can also just compute the roots to be $30$ and $130$)
This post has been edited 1 time. Last edited by megarnie, Feb 7, 2025, 10:40 PM
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Mr.Sharkman
500 posts
#21
Y by
Bruh $x= a+bi,$ then $(a-(4+k))^{2}+b^{2} = (a-k)^{2}+(b-3i)^{2},$ which becomes $8k+7 = 8a-6b.$ Then, $c = 20-a,$ $d = 20-d,$ and then $8k+7 =80-8c+6d,$ and $(-8c+6d)^{2} \le (8^{2}+6^{2})(25)$ by Cauchy, then $-50 \le -8c+6d \le 50,$ so $30 \le 8k+7 \le 130,$ giving a final answer of $\frac{73}{4} \implies \boxed{073}.$

EDIT: yeah I got $77.$ Typo lol
This post has been edited 1 time. Last edited by Mr.Sharkman, Feb 21, 2025, 1:58 PM
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RedFireTruck
4223 posts
#22 • 1 Y
Y by bjump
The perpendicular bisector of $(k+4,0)$ and $(k,3)$ is $8x-6y=8k+7$.

Note that $8(25)-6(20)=80$.

Therefore, we want $\frac{|8k+7-80|}{\sqrt{8^2+6^2}}=5$, or $|8k-73|=50$.

Therefore, the sum of all possible values of $k$ is $\frac{73}{8}\cdot 2=\frac{73}4$ and the answer is $73+4=\boxed{077}$.
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Happyllamaalways
478 posts
#23
Y by
Anyone else did this with calculus? :skull:
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lovematch13
662 posts
#24
Y by
I accidently multiplied by $2$ again and got $\dfrac{73}{2}$ :sob:
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ilikemath247365
253 posts
#25
Y by
cappucher wrote:
Bashing time

Let $z = a + bi$. Then our equations become

\[(25 - a)^2 + (20 - b)^2 = 25\]\[(a - 4 - k)^2 + b^2 = (a - k)^2 + (b - 3)^2\]
We expand out the second equation:

\[a^2 + 16 + k^2 - 8a - 2ak + 8k + b^2 = a^2 - 2ak + k^2 + b^2 - 6b + 9\]\[16 - 8a  + 8k = -6b + 9\]\[a = \frac{8k + 7 + 6b}{8}\]
We now plug this into the first equation:

\[\left(25 - \frac{8k + 7 + 6b}{8}\right)^2 + (20 - b)^2 = 25\]\[\left(\frac{193}{8} - k - \frac{3}{4}b\right)^2 + (20 - b)^2 = 25\]
Let $v = \frac{193}{8} - k$ to simplify calculations.

\[\left(v - \frac{3}{4}b\right)^2 + (20 - b)^2 = 25\]\[v^2 - \frac{3}{2}bv + \frac{9}{16}b^2 + 400 - 40b + b^2 = 25\]\[\frac{25}{16}b^2 - b\left(\frac{3}{2}v + 40\right) + (v^2 + 375)\]
In order for there to be exactly one solution for $z$, the discriminant of this quadratic must be $0$:

\[\left(\frac{3}{2}v + 40\right)^2 - 4\left(\frac{25}{16}\right)\left(v^2 + 375\right) = 0\]
Since we are asked for the sum of all $k$, we just use Vieta's relations on this quadratic.

\[\frac{9}{4}v^2 + 120v + 1600 - \frac{25}{4}v^2 - 4\left(\frac{25}{16}\right)\left(375\right) = 0\]\[-4v^2 + 120v + c = 0\]
We can ignore the constant term because it's irrelevant to the sum of the roots. Since we have $v_1 + v_2 = 30$, we conclude that

\[2\left(\frac{193}{8}\right) - (k_1 + k_2) = 30\]\[k_1 + k_2 = \frac{73}{4}\]
So the answer is $73 + 4 = \boxed{77}$.

Tried to do this, still failed, and put 73/8 :sob:
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Antonyliao
15 posts
#26
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Different solution:

we first let $z = a+bi$ and ignore k. Then $|(a-4)+bi|=|a+(b-3)i| \Rightarrow b=\frac{4}{3}a-\frac{7}{6}$. Now let's consider an a-b plane. $|25+20i-z|=5$ is another way of saying that a point on the line $-z$, transformed 25 units to the right and 20 units up, lies on the circle with radius 5 centered at the origin. Since $-z$ is linear, the fact that there is only one and only one solution to $z$ means that the line $-z$ is tangent to the circle. But $-z$ is the graph of $z$ rotated around the origin by 180 degrees, so $-z$ also has slope $\frac43$. Thus, the tangency points are $(-4,3)$ and$ (4,-3)$. transform these points back to their original positions, and we get $(-29,-17)$ and $(-21,23)$. But the a-coordinates of the intersection of $-z=\frac43a+\frac76$ with $y=-17$ and $y=-23$ is $-\frac{109}8, -\frac{145}8$, which is a contradiction. Now let's consider k. The added $-k$ in $|z-4-k|=|z-3i-k|$ means that $-z$ is moved $k$ units to the left. thus, $k =-\frac{145}8 + 23$ or $-\frac{109}8 + 17$, which gives $k_1=\frac{23}8$, $k_2=\frac{123}8 \Rightarrow k_1+k_2=\frac{73}4 \Rightarrow$ the answer is $73 + 4 = \boxed{077}$.


In this graph, $k_1=A'D$, $k_2=B'C$.
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This post has been edited 1 time. Last edited by Antonyliao, Feb 9, 2025, 12:09 PM
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sadas123
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#27
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Happyllamaalways wrote:
Anyone else did this with calculus? :skull:

Nope but I spent to much time but got the answer of $\boxed{077}$
This post has been edited 2 times. Last edited by sadas123, Feb 9, 2025, 10:08 PM
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jasperE3
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#28
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zhoujef000 wrote:
Let $k$ be a real number such that the system \begin{align*} |25+20i-z|&=5\\ |z-4-k|&=|z-3i-k| \\ \end{align*}has exactly one complex solution $z.$ The sum of all possible values of $k$ can be written as $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ Here $i=\sqrt{-1}.$

Write $z=x+yi$. We have:
$$5^2=|(25-x)+(20-y)i|^2=(25-x)^2+(20-y)^2=(x-25)^2+(y-20)^2$$and
$$|(x-4-k)+yi|^2=|(x-k)+(y-3)i|^2\Rightarrow(x-4-k)^2+y^2=(x-k)^2+(y-3)^2.$$Rearranging the second equation, we have $y=\frac43x-\frac{8k+7}6$.
-8k-7)/6

The equation $(x-25)^2+(y-20)^2=5^2$ can be interpreted geometrically to mean that $(x,y)$ lies on a circle in the Cartesian plane with center $(25,20)$ and radius $5$, while the equation $y=\frac43x-\frac43k-\frac76$ can be interpreted to mean that $(x,y)$ lies on a line in the Cartesian plane with slope $\frac43$ and a height that is based on $k$.

For this system to have exactly one solution, the line must intersect the circle at exactly one point, i.e. the line must be tangent to the circle.

Solution 1:
The radius that intersects the point of tangency must then be perpendicular to the line $y=\frac43x-\frac43k-\frac76$, so it must have slope $-\frac34$. Since the radius goes through the center of the circle, it lies on $y=-\frac34x+\frac{155}4$. The intersection of the line $y=-\frac34x+\frac{155}4$ with the circle yields the two possible points of tangency. To find these, we substitute into the equation of the circle:
\begin{align*}
(x-25)^2+\left(-\frac34x+\frac{155}4-20\right)^2&=25\\
(4x-100)^2+(-3x+75)^2&=400\\
16x^2-800x+10000+9x^2-450x+5625&=400\\
x^2-50x+609&=0\\
x&=25\pm4\\
y&=-\frac34x+\frac{155}4=20\mp3
\end{align*}Therefore, after rearranging $y=\frac43x-\frac43k-\frac76$ into $k=x-\frac34y-\frac78$, we get that $k$ is either:
$$k=29-\frac34\cdot17-\frac78=\frac{123}8$$or
$$k=21-\frac34\cdot23-\frac78=\frac{23}8$$so the sum of these is $\frac{146}8=\frac{73}4\Rightarrow\boxed{077}$.
alternate finish
[asy]
import graph;

size(400);
real r = 5;

draw((0,0)--(50,0), Arrow);
draw((0,0)--(0,40), Arrow);

label("$x$", (50, 0), NE);
label("$y$", (0, 40), NE);

draw(Circle((25,20), r));
dot((25,20));
label(scale(0.9)*"$(25,20)$", (25,20), NNE);

draw((3,36.5)--(47,3.5),Arrows);
label("$y=-\frac34x+\frac{155}4$",(46,4.25),NE);

dot((29,17));
label(scale(0.9)*"$(29,17)$",(25+4*r/5,-3*r/5+20),E);
dot((21,23));
label(scale(0.9)*"$(21,23)$",(25-4*r/5,3*r/5+20),W);

draw((6,3)--(31.5,37),Arrows);
label(scale(0.9)*"$k=\frac{23}8$",(31.5,37),NE);
draw((18.5,3)--(37,83/3),Arrows);
label(scale(0.9)*"$k=\frac{123}8$",(37,83/3),NE);

label(scale(0.9)*"$y=\frac43x-\frac43k-\frac76$",(12.27,3));
[/asy]

Solution 2:
Let the two possible points of tangency on the circle be $(x_1,y_1)$ and $(x_2,y_2)$. Note that the midpoint of these is the center of the circle, $(25,20)$, so $\left(\frac{x_1+y_1}2,\frac{x_2+y_2}2\right)=(25,20)$, that is, $x_1+x_2=50$ and $y_1+y_2=40$. Then the sum of the possible values for $k$ will be:
\begin{align*}
k_1+k_2&=x_1-\frac34y_1-\frac78+x_2-\frac34y_2-\frac78\\
&=(x_1+x_2)-\frac34(y_1+y_2)-\frac74\\
&=\frac{73}4\Rightarrow\boxed{077}
\end{align*}
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dragoon
1947 posts
#29
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Wait I got 68/9, I think I got really lucky lol
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Jaxman8
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#31
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I got 73/4, and also got 077 for q10 so I changed changed my answer.
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Gapple
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#32
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Let $z = x + yi$, and the system can be transformed into this:

$$
\begin{cases}
x^2 + y^2 - 50x - 40y + 1000 = 0 \\
y = \dfrac{4}{3} x - \dfrac{8k + 7}{6}
\end{cases}
$$
It has 1 solution, so we want the line to be tangent to the circle, so we get $(x, y) = (21, 23), (29, 17)$. Plugging these into the second equation, and we get $k_1 = 23/8, k_2 = 123/8$. So the answer is $k_1 + k_2 = 146/8 = 73/4 \implies \boxed{77}$.
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MathRook7817
683 posts
#33
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orz question
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Jaxman8
120 posts
#34
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I have no idea how I got this question right in the first place, but when I got #10 as 077 aswell I changed my answer to 047 on #8 bc I got that before.
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