# Thread: Another quiz - geometry

1. ## Another quiz - geometry

Here's another quiz. Please refer to the drawing below:

On a piece of paper, draw 3 non-overlapping circles of different sizes, and let's name the circles a, b, and c. Draw the tangents from circle a to circle b and extend them until they meet at point AB. Do the same for all three circles such that there are 3 intersection points AB, AC and BC.

From the example shown here it is apparent that points AB, AC and BC fall on a straight line. How to prove that the 3 points always fall on a straight line as long as the 3 circles are of different sizes and do not overlap each other?

2. Eh Roy, you trying to take over ClubSNAP and turn it into a science forum issit?

3. Roy,

Ok this might sound stupid...

How about if I drew the circles and reverse engineer? That would prove it, no?

4. Originally Posted by Jed
Eh Roy, you trying to take over ClubSNAP and turn it into a science forum issit?
this sort of thing should go to + Plus

5. Originally Posted by espn
Roy,

Ok this might sound stupid...

How about if I drew the circles and reverse engineer? That would prove it, no?
in mathematics you need to prove that ALL cases is true, OR prove that the reverse CANNOT exist.

6. Originally Posted by Jed
Eh Roy, you trying to take over ClubSNAP and turn it into a science forum issit?
Hehe...ClubMath?

7. Originally Posted by espn
Roy,

Ok this might sound stupid...

How about if I drew the circles and reverse engineer? That would prove it, no?
The problem is, there is an infinite number of combinations of circles that meet the criteria of a)different sizes b)non-overlaping. So if you construct 1 million different 3-circle diagrams and show that the 3 intersection points fall on a straight line, that does not prove that it will be true for the next 3-circle diagram.

8. i think your question is too difficult to attempt... got clue as to where to start?

9. OK, i'll bite. Forgive me, I last did geometry in my pre-university days (almost 15 years ago for me now!!!)

Here are a few observations:

(1) AC + AB + BC = 90 degrees.

(2) angle of AB = 2x sin (rA/hA)
where rA = radius of circle A, hA = hypotenuse)

(3) sin (rA/hA) = sin (rB/hB)

Furthermore:

(4) hA/hB = rA/rB (the ratio of the hypotenuse formed by circle A and B is the same as the ratio of the radii of circles A and B).

Putting equation (4) into (3) proves this observation is correct:

(4-1) hA = (rA*hB)/rB

(4-1 into 3) sin (rA*rB)/(rA*hB) = sin (rB/hB) ... therefore sin (rB/hB) = sin (rB/hB)!

Therefore,

(5) 2x (sin rA/hA) + 2x (sin rB/hB) + 2x sin (rC/hC) = 90 degrees.

My maths is probably flawed somewhere. What's your answer?

10. I didn't even understand the question

11. Originally Posted by ecc35
Problem with your proof:

eqn 4: "b = d + e" is only true if x+y+z = 180, which is what you are trying to proof in the first place.

So you will definitely arrive at the required answer because you assumed it in the first place.

Furthermore, this particular diagram is only 1 example out of the infinite possible scenarios.

Good try, nonetheless.

Thanks!

Roy

12. I have a hard time following your proof...

Originally Posted by Amfibius
OK, i'll bite. Forgive me, I last did geometry in my pre-university days (almost 15 years ago for me now!!!)

Here are a few observations:

(1) AC + AB + BC = 90 degrees.
AC, AB and BC are names of the intersection points of the tangents, not angles. Even if I follow your writing here by assuming that you are refering to the internal angles formed by the tangent intersections (based on your following caclculations), how do you derive at the statement that they add up to 90 degrees? Care to elaborate?

Originally Posted by Amfibius
(2) angle of AB = 2x sin (rA/hA)
where rA = radius of circle A, hA = hypotenuse)
My guess is you are referring to the inverse SIN function of (rA/hA), right?

Originally Posted by Amfibius
(3) sin (rA/hA) = sin (rB/hB)
Again, this I can follow if your are talking of ASIN instead of SIN.

Originally Posted by Amfibius
Furthermore:

(4) hA/hB = rA/rB (the ratio of the hypotenuse formed by circle A and B is the same as the ratio of the radii of circles A and B).

Putting equation (4) into (3) proves this observation is correct:

(4-1) hA = (rA*hB)/rB

(4-1 into 3) sin (rA*rB)/(rA*hB) = sin (rB/hB) ... therefore sin (rB/hB) = sin (rB/hB)!

Therefore,

(5) 2x (sin rA/hA) + 2x (sin rB/hB) + 2x sin (rC/hC) = 90 degrees.

My maths is probably flawed somewhere. What's your answer?
The rest here I could not follow anymore and I don't understand what have you proven...

13. upz....anyone?

14. Originally Posted by roygoh
Let the centres of the circles be A, B and C respectively.

Label AB as 'D', AC as 'E', BC as 'F'.

We end up with the figure below.

Let the radii of the three circles A, B and C be r1, r2 and r3 resp.

Using similar triangles, from the ratio of the radii, we get:

BD/AD = r2/r1 ----(1)

CE/AE = r3/r1 ----(2)

CF/BF = r3/r2 ----(3)

Also,

AB = AD-BD --- (4)

Subst (1) into (4):

AB = AD - AD(r2/r1)
= AD(1-r2/r1)
= AD[(r1-r2)/r1] ---- (5)

Let d12 = r1 - r2 and subst into (5):

AB = AD(d12/r1)

Similarly, we let d23 = r2-r3 and d13 = r1-r3, we get:

AB = AD(d12/r1) ----(6)
AC = AE(d13/r1) ----(7)
BC = BF(d23/r2) ----(8)

Also we note that

DE = AE - AD ----(9) (Note, these are vectors. It's vector maths from here onwards.)

We start with DF = DB + BF, and using the identities above, eventually arrive at:

DF = k.DE, hence proving that D, E and F are collinear. (k is scalar, const.)

Ok, here goes:

DF = DB + BF

= (r2/r1)DA + (r2/d23)BC

= - (r2/r1)AD + (r2/d23)(BA+AC)

= (r2/d23)[(r2.d13/d23.r1)AE - (d12/r1)AD] - (r2/r1)AD

= .... (do your own reaaranging, just algebra) ....

= (r2.d13/d23.r1) * (AE - AD) - [y]AD

where y = (d12r2 + r2d23 - d13r2)/(r1.d23)

Evaluating y using (d12=r1-r2, etc) we find that y reduces to zero. (Serious.)

We also note that (r2.d13/d23.r1) is a constant, and since

DF = (r2.d13/d23.r1) DE,

the points D, E and F are collinear.

QED.

(Thanx Roy, it was an interesting question.)

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