----C----- P----/ \----- ---/ \--- -/ \- A/-------------Q--M-----------------\B

- 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?
- 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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