## Chord Lines

```              ----C-----
P----/          \-----
---/                      \---
-/                              \-
A/-------------Q--M-----------------\B
```
Let AB be a chord of a circle with center O and radius R. M is the midpoint of AB and line OM intersects the circle at point C. Solve each of the following in terms of R.
1. Find AM and CM if AM equals 3*CM, 2*CM, CM, CM/2, or CM/3. Can you solve the general case for AM=X*CM?
2. Line segments are drawn from C to point Q on AM, from Q to point P on the circle, and from P to A. Find CQ if CQ=QP=PA. I believe there are multiple answers, so solve for CQ in terms of CM (or AB). What is CQ if CM=R/2?

Source: Original. Based upon a design on the side of a building in Rocklin, CA on I-80.

Solutions were received from Joseph DeVincentis, Philippe Fondanaiche and Denis Borris. Philippe's solutions follow (though I'll admit part 2 is more complicated than I'd expected):
```Q1
To simplify,let R=1.
Let D diametrically opposed to C on the circle.
From the identity MA*MB = MC*MD, we can infer: AM^2 = CM*(2-CM)
As AM=X*CM, therefore CM = 2/(1+X^2) and AM = 2*X/(1+X^2)
It is easy to check that for X>=1, M is between O and C whereas if X<1, M is
between D and O.
For:
X=3, then CM=1/5 and AM=3/5
X=2, then CM=2/5 and AM=4/5
X=1, then CM=AM=1. The point M is at the center O of the circle.
X=1/2, CM=8/5 and AM=4/5
X=1/3, CM=9/5 and AM=3/5

Q2
Let angle (COA) = a with 0=0 if pi<=u<=pi+a
QC^2 = QM^2+CM^2 = [1-cos(a)]^2 + [2*sin(u)+sin(a)]^2
If CQ=QP=PA (E), then:[1-cos(a)]^2 + [2*sin(u)+sin(a)]^2 = 4*cos^2(t) =
2*[2*cos(2*t)+1] = 2*cos(u-a)+2
Finally, by simplifying this relation,
(E) is satisfied if and only if: f(u) = 2*sin^2(u) - cos(u+a) -cos(a) = 0
with p<=u<=pi+a

Let A(u) = 2*sin^2(u) and B(u)=cos(u+a)+cos(a)
On the interval [pi,pi+a], A(u) is uniformly increasing from 0 to 2*sin^(2a)
while B(U) is uniformly increasing from 0 to -cos(2*a)+cos(a)
The graphs representing A(u) and B(u) have two common points:
-u=pi,that is to say t=(pi-a)/2 which corresponds to P in C and Q in B.
-u=u_0 included between pi and pi=a which is the requested solution
From the development of f(u),we can also demonstrated that sin(u_0) is also
the real solution of the following cubic equation:
4*sin^3(u) + 4*sin(a)*sin^2(u) +[1 - 4*cos(a)]*sin(u) - 2*sin(a)*cos(a)
Then,knowing u_0, we can calculate t_0=(u_0-a)/2 and CQ=QP=PA=2*cos(t_0)

If CM=R/2, then a=pi/3.
The equation f(u)=0 has the solution u_0=7*pi/6. Then t_0=5*pi/12 and
CQ=QP=PA=[sqr(6) - sqrt(2)]/2 = 0.51763809...

For different values of a such as:
a=pi/9=20x, then t_0=86,4...x, CM=0,123..
a=pi/6=30x, then t_0=84,3...x, CM=0,198..
a=2*pi/9=40x, then t_0=81.8...x, CM=0,282...
a=pi/4=45x, then t_0=80,3...x, CM=0,336...
a=5*pi/18=50x, then t_0=78,7..x; CM=0,389..
a=7*pi/18=70x, then t_0=70,5...], CM=0,664..
a=4*pi/9=.80x, then t_0=65.5..x, CM=0,827...
etc...
```

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