Thursday, July 27, 2017

Geometry Problem 1339: Circle, Diameter, Tangent, Secant, Parallelogram

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Peter Tran.

Details: Click on the figure below.

1. Extend AF to meet the circle at P, join AD and consider the quadrilateral APBD
m(APB) = 90
Hence APBD is a rectangle and AF//GB ------ (1)
Similarly extend AG to meet the circle at Q, join AC and consider the quadrilateral ACQA
m(ACB) = 90
m(AQB) = 90
Hence APBD is a rectangle and FB//AG ------ (2)

If m(EAF) = m(EAP) = x => m(ABP) = x and m(ABD) = m(ABG) = 90-x
Similarly if m(EAC) = y => m(ABC) = m(ABF) = y
Since ACF is a right triangle, m(FAC) = x-y, m(ACF) = 90 and m(AFC) = 90-x+y => m(AFB) = 90+x-y
So the angles in triangle ABF are
m(A) = 90-x, m(F) = 90+x-y and m(B) = y ---------(3)

As AF//BG, m(AGB) = 90+x-y, m(ABG) = 90-x ---------(4)
From (3) and (4) the triangles AFB anhd AGB are similar (AAA) and since have the same side AB, are congruent
Hence AF = BG and FB = AG ----- (5)

From (1),(2) and (5) we can say AFBG is a parallelogram

1. To Anonymous
Referring to first 4 lines of your solution:
" Extend AF to meet the circle at P, join AD and consider the quadrilateral APBD
m(APB) = 90
Hence APBD is a rectangle and AF//GB "

Note that quadrilateral APBD with m(APB) = 90
m(ADB) = 90 is not enough to conclude that APBD is a rectangle .

Peter Tran

2. In your proof after u deduce (1) and (2) above, the proof need not go any further - for then AFBG is obviously a parellogram

Issue is, in your proof it is not clear how < ADB = < APB = 90 gives u that APBD is a rectangle. To prove this we must show that DOP are collinear. This is by no means clear.

Similarly for ACQA

2. m(APB) = 90
does not imply APDB is a rectangle :(

3. PROBLEM 1339
Let OK perpendicular in CD (K on CD) so CK=KD. Let CL//EG (L on BD) so CM=ML.
But FG//CL so FO=OG. Is AO=OB therefore the AFBG is parallelogram.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

1. To Apostolis
It is not clear how CM=ML in your solution.

See link below for the sketch per your solution

https://photos.app.goo.gl/37QP7aP3Mt6rI62d2

Peter Tran

4. PROBLEM 1339
Let OK perpendicular in CD (K on CD) so CK=KD. Let the cycle (C,A,K) she intercepts her AB in M then <ACK=<AMK. But <ACK=<ACD=<ABD then KM//DB. Si the CM intercepts
the DB in L ,then CM=ML.Is <EOA=<EKA=<CKA=<CMA so CM//FG.So FO=OG. Is AO=OB therefore the AFBG is parallelogram.
( <OKE=90=<OAE then EAKO=cyclic ).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

1. Why <EOA = <EKA , Can you reply me ?

5. Good solution Apostolis

We can simplify it

Draw CML // to FOG, M on AB and L on BD.
Draw KM // to DLGB, K on CD.

< ABD = < AMK = < ACD hence AKMC is concyclic and so < KAM = < KCM = < KEO and in turn AKOE is concyclic

Therefore < OKC = 90 and so K is the mid point of CD and in turn M is the midpoint of CL.

Now since CL // FG and CM = ML, FO = OG.

Finally in quadrilateral AFBG, the diagonals bisect each other - hence the same is s //ogram.

6. Good solution Apostolis

See below for the link to the Apostolis's solution

https://goo.gl/photos/ZRvSEA7SRHqddXoU6

Peter Tran

7. Thank you very much Peter and Sumith.

8. Only need to know that AB is the diameter then AB together with any two points F and G that are colinear with O form a parallelogram. Extend AF to A' on the perimeter of O. The AA' bisector F' passes through O's center as all cord bisectors do. Let the BD cord's bisector be G' then it follows that FF' = GG' since FG and F'G' are both colinear with O.

1. Clarification - perpendicular bisectors of all cords pass through O. And any two points F and G colinear with O not on AB.

2. My assertion that any two colinear points form a parallelogram is false.

9. I should have stated perpendicular cord bisector instead of any cord bisector. Also any points F and G colinear with O and not on AB.

10. The assertion that any two points F' an G' colinear with O with F' on CB and G' on BD form a parallelogram is true by my original argument.