Some Questions from AIMO 2009 (1 Viewer)

[PC]

Here are two questions which I'm stuck on. You might like to have a go, and perhaps post the solutions.

8. In ∆ABC, angle ABC = 138° and angle ACB = 24°. Point D is on AC so that angle BDC = 60° and point E is on AB so that angle ADE = 60°. If angle DEC = x°, find the value of x.

9. Let ∆ABC be an equilateral triangle with AB = x. On the extension of BC, we define points A' (on the same side as B) and A" (on the same side as C) such that A'B = CA" = y. Similarly, on the extension of side CA, we define B' (on the same side as C) and B" (on the same side as A) such that B'C = AB" = y, while on the extension of side AB, we define C' (on the same side as A) and C" (on the same side as B) such that C'A = BC" = y.
(a) Prove that the points A', B", C', A", B' and C" all lie on a circle.
(b) If x and y are positive integers, determine the smallest integer value for R², where R is the radius of that circle.

I've done the remaining eight questions, and surprisingly the other question that I got most stuck on was Q1!

 

[mirakon]

A solution of Q8 courtesy of a friend of mine Tim Large (note that he got gold in AMO and high distinction in AIMO as well as perfect score in AMOC)

Anyway. Set EC = 1 unit (this doesnt really matter but whatever. Just set it to be 1, or you can just set it to be equal to say k, because it cancels out anyway).


Then DC = sin[x]/sin[60] (sin rule in EDC)
and also BC = sin[78 - x]/sin[138] (sin rule in EBC)


But also, DC/BC = sin[96]/sin[60] (sin rule in BDC)


So, sin[x]sin[138] / sin[78-x]sin[60] = sin[96]sin[60]
=> sin[x]/sin[78-x] = 2sin[48]cos[48] / sin[42] (sin[180 -t] = sin[t], sin[2t] = 2sin[t]cos[t])
=> sin[x]/sin[78-x] = 2sin[48]sin[42] / sin[42] (cos[t] = sin[90-t]) = 2sin[48]


The obvious solution to this is x = 48. Then we consider if 48 < x < 60, then sin[x] > sin[48], sin[78-x] < sin[30] so LHS > RHS contradiction. Likewise is 0<x<48 LHS < RHS. So as 0<x<60, the only solution for x is x=48.

Hope I helped (or rather Tim).

btw, what did you get in AIMO?

 

[PC]

I didn't do it. I'm just trying the questions for fun (possibly not the best word).

 

[mirakon]

I didn't do it. I'm just trying the questions for fun (possibly not the best word).

LOL okay, yeah the problems are somewhat of a nice challenge to do. Esepcially AMO and Senior Contest as well.

 

[Archman]

8. Straight from the question
angle DBC = 180-60-24 = 96
angle EBC = 138-96 = 42
Extend CB to produce another point F (closer to B than C).
angle EBF = 180-138 = 42
So EB bisects angle DBF, ED bisects angle BDA
Hence E is an excentre (look it up of triangle BDC and EC bisects angle BCD
So angle ECD = 24/2 = 12
Therefore angle DEC = 180 - 120 - 12 = 48.

9. By intesecting chord theorem (or power of a point, or secant chord theorem, or whatever you call it)
A'B . BA'' = C'B . BC'' = (x+y)y
Hence A'A''C'C'' is cyclic. The centre of that circle is the intesection of perpendicular bisectors of A'A'' and C'C'', which is also the centre of the equiliateral triangle ABC. call it O. Rotate the whole diagram 120 about O in any direction gives you B', B'' also lie on the same circle.

AO=BO=CO can be caluculated to be x/sqrt{3}. Again by power of a point,

(x+y)y = R^2 - (x/sqrt{3})^2 = R^2 - x^2/3
So R^2 = xy+y^2+x^2/3. Since x, y, R^2 are integers you'd want x to be a multiple of 3. And to keep everything as small as possible, let x=3, y=1 gives R^2 = 7.

Be sure to look things like "power of a point" up if you aren't sure what it is.

 

[addikaye03]

8. Straight from the question
angle DBC = 180-60-24 = 96
angle EBC = 138-96 = 42
Extend CB to produce another point F (closer to B than C).
angle EBF = 180-138 = 42
So EB bisects angle DBF, ED bisects angle BDA
Hence E is an excentre (look it up of triangle BDC and EC bisects angle BCD
So angle ECD = 24/2 = 12
Therefore angle DEC = 180 - 120 - 12 = 48.

9. By intesecting chord theorem (or power of a point, or secant chord theorem, or whatever you call it)
A'B . BA'' = C'B . BC'' = (x+y)y
Hence A'A''C'C'' is cyclic. The centre of that circle is the intesection of perpendicular bisectors of A'A'' and C'C'', which is also the centre of the equiliateral triangle ABC. call it O. Rotate the whole diagram 120 about O in any direction gives you B', B'' also lie on the same circle.

AO=BO=CO can be caluculated to be x/sqrt{3}. Again by power of a point,

(x+y)y = R^2 - (x/sqrt{3})^2 = R^2 - x^2/3
So R^2 = xy+y^2+x^2/3. Since x, y, R^2 are integers you'd want x to be a multiple of 3. And to keep everything as small as possible, let x=3, y=1 gives R^2 = 7.

Be sure to look things like "power of a point" up if you aren't sure what it is.

:O! Archman returns! You're Ivan Guo right?

 

[mirakon]

8. Straight from the question
angle DBC = 180-60-24 = 96
angle EBC = 138-96 = 42
Extend CB to produce another point F (closer to B than C).
angle EBF = 180-138 = 42
So EB bisects angle DBF, ED bisects angle BDA
Hence E is an excentre (look it up of triangle BDC and EC bisects angle BCD
So angle ECD = 24/2 = 12
Therefore angle DEC = 180 - 120 - 12 = 48.

9. By intesecting chord theorem (or power of a point, or secant chord theorem, or whatever you call it)
A'B . BA'' = C'B . BC'' = (x+y)y
Hence A'A''C'C'' is cyclic. The centre of that circle is the intesection of perpendicular bisectors of A'A'' and C'C'', which is also the centre of the equiliateral triangle ABC. call it O. Rotate the whole diagram 120 about O in any direction gives you B', B'' also lie on the same circle.

AO=BO=CO can be caluculated to be x/sqrt{3}. Again by power of a point,

(x+y)y = R^2 - (x/sqrt{3})^2 = R^2 - x^2/3
So R^2 = xy+y^2+x^2/3. Since x, y, R^2 are integers you'd want x to be a multiple of 3. And to keep everything as small as possible, let x=3, y=1 gives R^2 = 7.

Be sure to look things like "power of a point" up if you aren't sure what it is.

Score! So I was right in second party of Q9, but I used the centroid divides perpendicular bisector of equilateral triangle into ratio 2:1 then made calculations. It was confusing but kinda forgot how to do it.

Thanks anyway +1 as your proofs were much, much easier.

 

[Reanna]

can a cyclic quad have 3 vertices on the circumference and one on the centre

 

[gurmies]

Nope