A Simple Trigonometric Problem | International Mathematical Olympiad 1963 Problem 5

#Math #IMO #Trigonometry

In this video we are going to solve Problem 5 in IMO 1963.

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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.

[davidbrightly]

Geometrical proof: consider a unit-sided regular heptagon precariously balanced with one vertex at the origin and its opposite side horizontal above it. The vectors (cos nπ/7, sin nπ/7) n=1, 3, 5 represent the three sides on the right. The required sum of cosines is the x-coord of the right vertex of the top side and hence is 1/2.

[engjayah]

Adopt the approach 7th roots of -1; where w = cos(pi/7) + i sin(pi/7) and w^7+1 = 0
Now the real part Re(w^3-w^2+w) = 0.5 which could be derived by factorizing (W^7+1) = 0

[f-th]

There is another way using complex roots of unity. I wrote an explanation, but halfway through, it was too long and I got tired.

Use Euler's formula to get the 7th roots of unity, negate them, then plot them and look at the top and bottom parts, and think about their real parts. Top part and bottom part each have real parts summing to 1/2. The LHS in the question is basically the top part (sorta), so that's how you get 1/2.

[wrecker]

observe that the angles are in ap with common difference 2pi/7. Then u can use the formula for that summation. After that u need to do some manipulation to get the result(1 step). Hence, it becomes a 2 liner problem

[dovidglass]

Alternative solution:

Let


Then


which is a geometric series with common ratio -e^(i*pi/7). Using the formula for geometric series, the result follows after some manipulation.

[crustyoldfart]

When a problem contains angles based around 2*pi/7 this should be a clue that the geometry is based on a regular heptagon. David Brightly and Jayah Eng are therefore right in bringing this out. Eng's use of complex roots is also a nice touch. If I had been an examiner in the 1963 IMO I would have favoured these solutions over the clever but rather plodding technique given here.
As an aside this kind of problem involving regular polyhedra can sometimes be solved very quickly knowing certain relations, for example in the case of a pentagon we can deduce that sin(pi/10) = (sqrt(5)-1)/4 = phi/2 where phi is the golden ratio and 1/phi=phi+1.
In the case of a heptagon sin(2*pi/7) is tantalizingly close to sqrt(3)/2 which if it had been exact then the problem is simplified.

[dhriti]

On the LHS, we have

cos(π/7) - cos(2π/7) + cos(3π/7)
= cos(π/7) + cos(5π/7) + cos(3π/7)

Now consider the following sum

S = e^iπ/7 + e^i3π/7 + e^i5π/7 (Geometric progression with common ratio e^i2π/7)

On simplifying
S =

Re(S) = cos(π/7) + cos(3π/7) + cos(5π/7)
=
=
= (1/2)sin(6π/7)/sin(π/7)
= 1/2

Which is equal to the RHS. One of the first IMO problems I could solve fully on my own 😊

[amiyancandol]

I did it like this : π/7=ø so 7ø=π and hence 4ø=π-3ø hence cos3ø+cos4ø=0 from this 8c⁴+4c³-8c²-3c+1, dividing by (c+1) we have 8c³-4c²-4c+1=0
After manipulating 8c³-6c=2cos3ø and 2c²-1=cos2ø and hence we got cosπ/7-cos2π/7+cos3π/7=1/2

[Dharmarajan-ctld]

Two more solutions for solving it will be uploaded in my channel today. Kindly consider. Though it has 3 or 4 more steps . If one tool fails plan B is useful !!

[yoav]

Amazing proof!!. There is also the methot of using complex numbers like michael penn does in his videos

[ferraneb]

Consider w = cos(pi/7) + i*sin(pi/7). This is a fourteenth root of unity, in fact, a primitive one. The others are w^3, w^5, w^9, w^11 and w^13. You can write cos(n*pi/7) as 1/2(w^n + w^-n) = 1/2(w^n + w^(14-n)) Therefore, the whole expression becomes 1/2((w + w^13) + (w^9 + w^5) + (w^3 + w^11)), which is a half of the sum of all primitive 14th roots of unity, that is, the sum of the roots of the cyclotomic polinomial of order 14 (x^6-x^5+x^4-x^3+x^2-x+1).

You can obtain this polynomial by considering all the 14th roots of unity, that is roots of the polynomial x^14 - 1. Then you should divide by the polynomial x^7 - 1 to exclude the roots w^2, w^4... Which are precisely the 7th roots of unity (w^(2n))^7 = (w^14)^n = 1^n = 1. Finally divide by x+1 to get rid of w^7 = -1. You're only left with the roots mentioned above.

The sum of the roots of a polynomial of degree n is just the negative of the coefficient of degree n-1, namely -(-1)) = 1, and therefore the value of the sum is 1/2 * 1 = 1/2

[math]

solved, very well done, thanks for sharing

[theevilmathematician]

I used a different method:
Let cos(π/7)- cos(2π/7) + cos(3π/7) = x. Then, cos(π/7)-cos(2π/7) + cos(3π/7)= cos(π/7)+ cos(3π/7) + cos(5π/7)= x. By the Product Sum Formulae, 2x*sin(π/7)= sin(2π/7)+ sin(4π/7) - sin(2π/7) + sin(6π/7)- sin(4π/7)= sin(6π/7) because of the telescoping sum, which simplifies to sin(π/7). Hence, x=1/2 after some simplification. This completes the proof!

[nalapurraghavendrarao]

In first step you multplied the LHS by sin(pi/7)and forgot to divide by the same quantity, then it is possible to cancel it in the last step.

[sallytwotrees]

Why could you, at the beginning, turn 3п/7 and 5п/7 into п/7 ?

[praiseonionrings]

Once we get to the step there is another (less rigorous) solution from here
Note that cos(7pi/7)=-1 and cos(2pi-x)=cos(x)
So this is equivalent to saying (1/2+1/2-1)
How do we know this?
Well imagine 7 unit vectors pointing outward from the origin at the angles pi/7, 3pi/7, 5pi/7, etc...
If you imagine them as force vectors pulling on an object, they're equally spaced and equal magnitude so the object won't move
The sum of the cosines of the angles is the sum of the x-components of these vectors, i.e. the x-component of the net force which we know is 0
Then it's shown
Alternatively for a geometric solution consider you can connect the vectors tip to tail to make a heptagon (closed loop) showing that the sum of the vectors is zero vector
This also gives rise to other problems like:
Show that all angles in degrees

[touhami]

Cos(2pi/7)=- cos(5pi/7)
So
Cos(a)+cos(3a)+cos(5a) ** , a=pi/7
** = sin(6a) /[2sin(a)]
Sin(6a)=

[smrtghosh]

You cannot bring sin pi/7 out of nowhere

[Dharmarajan-ctld]

Is it an identity? Is it even a result ? An identity is... Kindly re consider

[jyoteshjaiswal]

Hey there can be another approach just change -cos2π/7 into cos5π/7 so the equation will be cos π/7+cos3π/7+cos5π/7 now apply formula where x=π/7, y=2π/7,n =2 and u will get the prove !