Thanks for making that applet on the parabola, and noticing that M and K also lie on CD. This whole thing is just a projection of the same situation in a circle, which is how the idea came to me in the first place(while looking at Archimedes Book of Lemmas, Proposition 12). I have since realized, though, that one should be very wary of parabolas in composing problems, since generally it's so easy to deal with the equations y=x^2 and y'=2x, and that usually one can just mess around with equations to prove almost anything true. This problem is not too hard if you just write down equations and crank. Oh well.

As for the second one, probably it's no surprise that I noticed that while reading about Sangaku. It's roughly equivalent to an old Sangaku problem.

Greg

Thank you for posting here. They both are truly nice problems.
The second one combines quantities that I never before even saw used. But it is pretty simple.

Configuration of the first one admits a generalization that I placed at

https://www.cut-the-knot.org/Curriculum/Geometry/Markowsky2.shtml

I expect there to be some generalization anglewise, perhaps for a different direction. I am away right now. Will be looking into that when back.

Alex

Greg

It's trig, but rather simple and straightforward.

I posted a solution at

https://www.cut-the-knot.org/triangle/Markowsky.shtml

It appears the quantity is well known and is called versine (as a piece of since rotated 90°).

If AB is a segment in a (unit) circle with center O and midpoint M. If angle AOM = α and the radius through M is OS, then MS is the versine of α:

MS = vers(α) MS = 1 - cos(α) MS = 2·sin2(α/2).

I ran into this term in an article in Math Horizons (Sept., 2006).

Here's a hole in my education.

k = sqrt(q*r)

I have my doubts about existence of a synthetic proof, but, even with trigonometry, the above needs some perseverance.

No, it would have been nice, but is unlikely to be true.

I have found one following proof, please check for me if I make some mistakes.
Let A, B, C, A', B', C' are concyclic on one circle (O), radius R
Let A'', B'', C'' are orthogonal projection of A'B'C' on BC, CA, AB respectively.
Let A*, B*, C* are orthogonal projection of ABC on B'C', C'A', A'B' respectively.

We can easy show some following concyclic point sets:
A, A*, B', B'' are concyclic on circle (Ab) (center Ab = midpoint of AB'), radius = Rab
A, A*, C', C'' are concyclic on circle (Ac) (center Ac = midpoint of AC'), radius = Rac
B, B*, C', C'' are concyclic on circle (Bc) (center Bc = midpoint of BC'), radius = Rbc
B, B*, A', A'' are concyclic on circle (Ba) (center Ba = midpoint of BA'), radius = Rba
C, C*, A', A'' are concyclic on circle (Ca) (center Ca = midpoint of CA'), radius = Rca
C, C*, B', B'' are concyclic on circle (Cb) (center Cb = midpoint of CB'), radius = Rcb
Moreover:
A, Ac, Ab, O are concyclic on circle (Oa), (center Oa = midpoint of OA), radius R/2
B, Bc, Ba, O are concyclic on circle (Ob), (center Ob = midpoint of OB), radius R/2
C, Ca, Cb, O are concyclic on circle (Oc), (center Oc = midpoint of OC), radius R/2
By some similar triangles we can also easy show following fact:
The common chord d of two given circles with radius R1, R2 can be calculated as:
d = R1*R2/R3 where R3 = radius of the circle passing through two centers of given circle and one their intersection point.
We now consider three circles at vertex A: (Ac), (Ab), (Oa), we can show:
AA* = 2*Rab*Rac/R (1)
Similarly we can have
BB* = 2*Rbc*Rba/R (2)
CC* = 2*Rca*Rcb/R (3)
Do the same with triangle A'B'C' we can have:
A'A'' = 2*Rca*Rba/R (1')
B'B'' = 2*Rab*Rcb/R (2')
C'C'' = 2*Rbc*Rac/R (3')
From (1), (2), (3), (1'), (2'), (3') we have the result:
AA* x BB* x CC* = A'A'' x B'B'' x C'C'' or
k*l*m = p*q*r

Best regards,
Bui Quang Tuan

Dear Greg and Alex,

From my proof we can show one interesting fact for any inscribed hexagon (not necessary convex) AB'CA'BC'.
From (1), (2), (3), (1'), (2'), (3')
k*l*m = p*q*r = 8*Rab*Rac*Rbc*Rba*Rca*Rca =
8*(AB'/2)*(AC'/2)*(BC'/2)*(BA'/2)*(CA'/2)*(CB'/2)/R^3
If we denote six sides of inscribed hexagon AB'CA'BC' as a, b, c, d, e, f then

k*l*m = p*q*r = a*b*c*d*e*f/(2*R)^3 = a*b*c*d*e*f / W^3 or

a*b*c*d*e*f = k*l*m*W^3 = p*q*r*W^3 (4)
where
W = diameter of circumcircle of the hexagon
p, q, r = distances from A, B, C to sidelines of A'B'C'
k, l, m = distances from A', B', C' to sidelines of ABC

If can we somehow generalize (4) for other inscribed polygon?

Best regards,
Bui Quang Tuan

I just have generalized (and it is easy) the result as following:

n >=3 is one integer.
X1, Y1, X2, Y2, X3, Y3, ... Xn, Yn are any 2n points concyclic on one circle O with diameter W
y1, y2, y3, ... yn are distances from Y1, Y2, Y3, ... Yn to X1X2, X2X3, X3X4, ... XnX1 respectively.
x1, x2, x3, ... xn are distances from X1, X2, X3, ... Xn to YnY1, Y1Y2, Y2Y3, ... Yn-1Yn respectively.
a1, a2, a3, ... a2n are sides of polygon X1Y1X2Y2X3Y3 ... XnYn (by this order). It means:
a1 = X1Y1
a2 = Y1X2
a3 = X2Y2
a4 = Y2X3
...
a(2n-1) = XnYn
a2n = YnX1
We can use the same my posted proof to prove the following result:

a1*a2*a3* ... *a2n = x1*x2*x3* ... *xn*W^n = y1*y2*y3* ... *yn*W^n

Best regrads,
Bui Quang Tuan

This is just outright beautiful and as synthetic as could be. Outstanding.

Thank you for posting this here.

Alex

Using this simple lemma and the same proof method we can show also following interesting facts:

Given point P and n points P1, P2,… Pn (n>1) on the circle (O) with diameter d.
s1, s2, … sn are lengths of segments PP1, PP2, … PPn respectively
d1, d2, … dn are distances from P to segments P1P2, P2P3, … PnP1 respectively
The result:
(s1*s2*… *sn)^2 = d1*d2*… *dn*d^n
(When n=2 so we have our lemma.)
Please note that P1P2, P2P3, … PnP1 can be see as one close path passing all n point and each point one time (if I am not wrong it is Hamilton / Euler path in graph theory). For short we name it as "HE path". We can name each point P as a "vertex" and each segment PiPj as a "side" of this path.
Now our result can be formulated as:

Theorem 1:
If P, P1, P2, ... Pn (n>1) are concyclic on one circle with diameter d then
(s1*s2*… *sn)^2 = d1*d2*… *dn*d^n
where
si = segments from P to all vertices of one HE path from Pi (i=1 to n)
di = distances from P to all n sides of this HE path.
or by words
The square of product of all segments from one point to all vertices of one HE path is equal product of all distances from this point to all sides of this HE path multiplied by nth power of diameter of the cycle on which the point and the HE path are concyclic.

Of course, there are a lot of HE paths from n vertices and they have different distances from P to their sides. But they have the same one set of segments PP1, PP2, ... PPn or they have the same product of segments from P to n vertices: s1*s2*s3*...*sn. So we can have:

Theorem 2:
The product of all distances from one point to any HE path from given points are constant if all points are concyclic.
d1*d2*… *dn = (s1*s2*… *sn)^2 / (d^n) = constant

If we choose two HE paths with some common sides then they have also some common distances, therefore we can get some other equal products of distances.

One special example when n = 4
s1 = PP1, s2 = PP2, s3 = PP3, s4 = PP4
From four points P1, P2, P3, P4 we can make three HE paths
1. P1P2P3P4
- sides: P1P2, P2P3, P3P4, P4P1
- distances from P to these sides: d12, d23, d34, d41
2. P1P2P4P3
- sides: P1P2, P2P4, P4P3, P3P1
- distances from P to these sides: d12, d24, d43, d31
3. P2P3P1P4
- sides: P2P3, P3P1, P1P4, P4P2
- distances from P to these sides: d23, d31, d14, d42
(by these dij notations, of course dij = dji)
By our theorems:
d12*d23*d34*d41 = d12*d24*d43*d31 = d23*d31*d14*d42 = (s1*s2*s3*s4)^2/d^4
From these we can have following other equal products of distances:
d12*d34 = d14*d23 = d13*d24

This result can be formulated:
Theorem 3:
The products of two distances from one point on circumcircle of one inscribed quadrilateral to two opposite sides or two diagonals are equal.

Best regards,
Bui Quang Tuan

I am going to expand that page.

Alex

I continue study these distances from one points to segments of other points on one circle.
Using the same lemma and Ptolemy's theorem I can prove following fact:

Given one inscribed convex polygon X1X2X3...Xn-1Xn (n>2) and one point P on its circumcircle
We use following notations (i, j can be interchanged)
aij = length of side XiXj
dij = distance from P to sidelines XiXj
Rij = ratio aij/dij
then the set of ratios R12, R23, R34,...R(n-1)n, Rn1 hold a property: one is sum of others

First, we prove for n = 3, the triangle case
Given triangle ABC and one point P on circumcircle of ABC
a, b, c = sides of ABC
da, db, dc = distances from P to sides BC, CA, AB
The result:
Three ratios a/da, b/db, c/dc hold a property: one is sum of other two.
Proof:
We use following lemma: in the triangle ABC h = b*c/(2*R)
where h = altitude from A, b = AC, c = AB, R = circumradius of ABC
With triangle PBC
da = PB*PC/(2*R)
Similarly we have:
a/da = a/(PB*PC)*(2*R) = a*PA*(2*R)/(PA*PB*PC)
b/db = b/(PC*PA)*(2*R) = b*PB*(2*R)/(PA*PB*PC)
c/dc = c/(PA*PB)*(2*R) = c*PC*(2*R)/(PA*PB*PC)
Triangle ABC divides its circumcircle into three arc, say (BC), (CA), (AB)
If P on arc (BC) then by Ptolemy's theorem:
a*PA = b*PB + c*PC therefore a/da = b/db + c/dc. It is what should be proved.

For n>3, we use mathematical induction method.

Suppose now our theorem is true for n-1 and convex polygon X1X2...Xn divides circumcircle into n arcs and P is on arc X1X2. Consider triangle X1X2X3. This triangle devides circumcircle into three arcs and P is on arc X1X2 (It is true because polygon is convex).

X1X3X4...Xn is n-1 side convex polygon and P is on arc X1X3 (It is true because polygon is convex) so:
R13 = R34 + R45 + ... + Rn1 (our theorem for n-1)
In the triangle X1X2X3
R12 = R23 + R13 = R23 + R34 + R45 + ... + Rn1
So the theorem is true for any n>3
We can also know which ratio is sum of other: the ratio of side respective with the arc where P is.

Please correct for me if I have made any mistake!

Best regards,
Bui Quang Tuan

1. To non-equilateral triangles
2. To n-gons with n > 3.

Beautiful. I do not see any errors.

P.S.

For the next post, may I ask you to start a new thread. Thank you.