Y by mathfan2020
Let 
be a real number such that the system 
has exactly one complex solution 
The sum of all possible values of can be written as 
where 
and 
are relatively prime positive integers. Find 
Here 
be a real number such that the system 
has exactly one complex solution 
The sum of all possible values of can be written as 
where 
and 
are relatively prime positive integers. Find 
Here 
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
Reason: think this was how it was aligned on the test
and radius 
,
and 
.
.
.
so the answer is 
.
lies on the circle centered at 
with radius 
in the complex plane.
lies on another circle of radius 5 centered at 
,
lies on a different circle of the same radius centered at 
.
where
which simplifies to
that gives only one possible value of 
is when 
is maximized or minimized, which by Cauchy-Schwarz is 
.
and
giving us a final answer of
,
,
.
(:sob:) 
.![\[(25 - a)^2 + (20 - b)^2 = 25\]](http://latex.artofproblemsolving.com/1/4/2/142960c40b7c7d6d44605edd9be1a328ff1971b8.png)
![\[(a - 4 - k)^2 + b^2 = (a - k)^2 + (b - 3)^2\]](http://latex.artofproblemsolving.com/e/2/8/e281a0c9bdc0100b4c3b63446a410278d30ce18a.png)
![\[a^2 + 16 + k^2 - 8a - 2ak + 8k + b^2 = a^2 - 2ak + k^2 + b^2 - 6b + 9\]](http://latex.artofproblemsolving.com/b/1/a/b1ab57289ee9d2242cd9ac66bfbb5ec59b8945d5.png)
![\[16 - 8a + 8k = -6b + 9\]](http://latex.artofproblemsolving.com/9/a/a/9aa289e32f56c44e336fac24c506a71b999a041f.png)
![\[a = \frac{8k + 7 + 6b}{8}\]](http://latex.artofproblemsolving.com/6/8/7/68720b95ac3b2239286047d7d65bbf570672d159.png)
![\[\left(25 - \frac{8k + 7 + 6b}{8}\right)^2 + (20 - b)^2 = 25\]](http://latex.artofproblemsolving.com/9/5/e/95e0d3b6ab7a15e72a228c04f7679fd838670eeb.png)
![\[\left(\frac{193}{8} - k - \frac{3}{4}b\right)^2 + (20 - b)^2 = 25\]](http://latex.artofproblemsolving.com/b/b/f/bbf1e29961ffa47425048fdc7637a9db95d70705.png)
to simplify calculations.![\[\left(v - \frac{3}{4}b\right)^2 + (20 - b)^2 = 25\]](http://latex.artofproblemsolving.com/5/5/9/559ea1ee9593dc0c842e207e62561d3e1f685c20.png)
![\[v^2 - \frac{3}{2}bv + \frac{9}{16}b^2 + 400 - 40b + b^2 = 25\]](http://latex.artofproblemsolving.com/0/4/e/04ebd76a8a0bf1b05d9c77e73641407c8fe6ace3.png)
![\[\frac{25}{16}b^2 - b\left(\frac{3}{2}v + 40\right) + (v^2 + 375)\]](http://latex.artofproblemsolving.com/6/4/a/64ad72ef7d2a7e5d5934cf0f8ffe7aad6d4268b1.png)
,
:![\[\left(\frac{3}{2}v + 40\right)^2 - 4\left(\frac{25}{16}\right)\left(v^2 + 375\right) = 0\]](http://latex.artofproblemsolving.com/a/5/a/a5ad41b8c2c1c6b536ac2a60e77a6fa19c0310f2.png)
![\[\frac{9}{4}v^2 + 120v + 1600 - \frac{25}{4}v^2 - 4\left(\frac{25}{16}\right)\left(375\right) = 0\]](http://latex.artofproblemsolving.com/8/f/6/8f610fc6a7bcc2ec65d41db71238e00064182fd7.png)
![\[-4v^2 + 120v + c = 0\]](http://latex.artofproblemsolving.com/a/9/2/a9292af701dc4c2470e9e5cc1b1375a60177a831.png)
,![\[2\left(\frac{193}{8}\right) - (k_1 + k_2) = 30\]](http://latex.artofproblemsolving.com/a/e/0/ae0996ab214582bb45d556c7410330a79935e8c9.png)
![\[k_1 + k_2 = \frac{73}{4}\]](http://latex.artofproblemsolving.com/0/a/6/0a62e0b5a3d4eaebb05aba581c08f896864f4b9b.png)
.
for some real 
Then, let 
By the second equation,
which simplifies to
Plugging in the real parts for 
and 
we have
This means that
Dividing both sides by 
we have
Let 
and 
Recognizing the cosine angle addition formula, we have
This has one solution for 
precisely when
and simple algebra gets the answer
,
It's worthwhile to note, this boils down to observing where a tangent line occurs on the circle with slope 
.
and 
,
Solving for 
,
If we substitute the derivative into the circle equation, we will get the solutions 
and 
.
for 
,
are 
and 
.
and 
.![\[ 6b = 8a - (8k + 7) \]](http://latex.artofproblemsolving.com/1/d/e/1de8052f2ae05872d0bb3e4f123b80999a937e3e.png)
.
,![\[ 36a^2 - 1800a + (6b)^2 - 240(6b) + 36000 = 0 \]](http://latex.artofproblemsolving.com/8/9/c/89c6cfbdbdda3d84851333466710f6f86b3a7286.png)
gives that the LHS is equal to ![\[ 36a^2 - 1800a + (8a)^2 - 16xa + x^2 - 1920a + 240x + 36000\]](http://latex.artofproblemsolving.com/3/a/b/3ab28f11476e3ffb6bc3e0df31c2a8abfc0b6a42.png)
![\[ 100a^2 - (16x + 3720) a + x^2 + 240x + 36000\]](http://latex.artofproblemsolving.com/3/b/7/3b75c85f0b8b921852c69c563aa136af02cc9673.png)
There is one solution iff the discriminant is ![\[ (16x + 3720)^2 = 400(x^2 + 240x + 36000),\]](http://latex.artofproblemsolving.com/1/d/9/1d96edc29b461337b9ca5c597b152c95758973e7.png)
or ![\[ (4x + 930)^2 = 25(x^2 + 240x + 36000)\]](http://latex.artofproblemsolving.com/f/d/2/fd2f843c0336a529c58f3c89adb0f3e50680c3f3.png)
If 
,
gives a contradiction. Thus, if 
.
.![\[ \frac{160 - 2 \cdot 7}{8} = \frac{73}{4} \implies \boxed{077} \]](http://latex.artofproblemsolving.com/a/b/3/ab3f9cb31eed9a53ea09a5dcd4d8cc312e611731.png)
and 
)
then 
which becomes 
Then, 
and then 
and 
by Cauchy, then 
so 
giving a final answer of 
Typo lol
and 
is 
.
.
,
.
and the answer is 
.
again and got 
:sob:
and ignore k. Then 
.
is another way of saying that a point on the line 
,
.
an
.
and 
.
with 
and 
is 
,
in 
means that 
or 
,
,
the answer is 
.
,
.
has exactly one complex solution 
.
and
Rearranging the second equation, we have 
.
can be interpreted geometrically to mean that 
lies on a circle in the Cartesian plane with center 
can be interpreted to mean that 
.
.
Therefore, after rearranging 
,
or
so the sum of these is 
.
we know that the sum of possibilities for 
.
(multiplying the constant term by 
.![[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]](http://latex.artofproblemsolving.com/0/6/d/06d138ade5e7d61c22ded9c2943b7aabfd80f774.png)
and 
.
,
and 
.
,
.
.
.
