practice problems

Q1. (ABC) is a triangle. (D, E, F) are arbitrary points on (BC, AC,) and (AB) respectively (or on their extensions).

Draw the three circumcircles of triangles (AEF), (DBF), and (DEC). Prove that these three circles intersect at a single point (M).

S1. This is known as Miquel's theorem and M is known as Miquel's point.
Case 1.
When AEF,DBF intersect at 2 different points : F and M.


Proof 1:
Angle AFM = theta
=> MFB = 180 - theta (supplementary angles)
and AEM = 180 - theta (opposite angles in cyclic quadrilateral AEMF)
=> MEC = theta (supplementary)
and BDM = theta (opposite in cyclic quadrilateral BDMF)
=> CDM = 180 - theta (supplementary)

=> CEM + CDM = 180 => CEMD is cyclic quadrilateral H.P.

Proof 2:
AFME is cyclic => FME = 180 - A = B + C
Similarly, FMD = A + B
Now angles around M should add up to 360.
=> FME + EMD + DMF = 360
=> EMD = 360 - (B+C) - (A+B) = 

So we showed that M is the Miquel's point.
But could it have been F?
I mean, is it possible that CED passes through F?
If that were the case then the circle CEMD will pass through F as well.
But a circle passing through three distinct points is unique.
=> the circle AFME is same as CEMD.
=> points A,E,C will lie on this circle.
But 3 distinct points on a straight line can't line on a circle.
Why?
Proof by contradiction.
Let O(0,0) be the center, 'r' be the radius. Let AB be a chord. Let B = (x,h), A = (-x,h).
Let C be a point on AB => C = (x1,h)
If C lies on the circle then x1^2 + h^2 = x^2 + h^2 => x1 = +-x which means that C is either A or B.

Case 2: When AEF, DBF touch at a single point F. In this case F and M are same points.



Proof: 
We will use alternate segment theorem.

AFB and DFB touch at F.
Using alternate segment theorem angles AFT = AEF = theta => FEC = 180 - theta

Now BFT = 180 - theta
=> BDF = 180 -theta => FDC = theta
=> FDC + CEF = 180 => quadrilateral CDFE is cyclic.
H.P.

Q2.Let (ABCD) be a convex quadrilateral. Consider (C1, C2, C3, C4), each of which touches three sides of this quadrilateral:

  • (C1) touches (AB, BC, CD)

  • (C2) touches (BC, CD, DA)

  • (C3) touches (CD, DA, AB)

  • (C4) touches (DA, AB, BC)

Prove that the centers (O1, O2, O3, O4) form a cyclic quadrilateral.



S2.
Case 1: All circles touch the quadrilateral from outside:

C1 touches 2 lines from B.
=> O1 lies at external angle bisector of B. (Prove by congruence).
Similarly, O1 also lies at external angle bisector of C.

Similarly, each of the centers, lies on intersection of 2 angle bisectors.

=> In triangle O1BC,
Angle BO1C = 180 - (180-B)/2 - (180-C)/2 = (B+C)/2 = angle O4O1O2

Similarly,
Angle O4O3O2 = (A+D)/2

=> Angles O4O3O2 + O4O1O2 = 360/2 = 180 H.P.

Case 2: All circles touch the quadrilateral from inside:



Let's again look at O1.
O1 would lie on internal angle bisectors of B,C and inside the quadrilateral.

Angle O1BC = B/2, O1CB = C/2, BO1C = 180 - (B+C)/2 = O2O1O4

Similarly AO3D = 180 - (A+D)/2 = O2O3O4

=> O2O1O4 + O2O3O4 = 360 - 360/2 = 180 H.P.

Case 3: Mix and match, some circles touch outside and some inside

Both the proofs above rely on 2 of the centers colinear with the common vertex.
For e.g. C,O1,O2.

Now if one center is inside and one is outside => one is on the internal angle bisector and the other is on the external angle bisector then they won't be colinear, instead they will make a right angle.

For e.g. O1CO2 is a right angle here rather than 180.
So we won't be able to prove in this case.



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