Don't Hug Me I'm Scared

Geometry Level 5

x 9 3 x 7 30 x 6 + 3 x 5 30 x 4 + 299 x 3 30 x 2 + 600 x 990 = 0 x^9 - 3x^7 - 30x^6 + 3x^5 - 30x^4 + 299x^3 - 30x^2 + 600x -990=0

Consider a triangle whose sides are a , b , c a,b,c , the real roots of the equation above (a>b>c). For nomenclature, consider that opposed to the side x x lies the angle X X .

Evaluate cos ( A C 2 ) cos ( A + C 2 ) \dfrac{\cos\left( \dfrac{A-C}{2} \right) }{\cos \left(\dfrac{A+C}{2} \right)}


This problem is not original.


The answer is 2.

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3 solutions

Lu Chee Ket
Oct 9, 2015

1.04895796084157 OR 2 OR 4.59867450788155 could be answer.

[(10)^(1/3)-1, (10)^(1/3), (10)^(1/3)+1] or [(10)^(1/3)+1, (10)^(1/3), (10)^(1/3)-1] gives 2 but this is only simplest.

As long as [12.70352421, 24.2296145, 143.0668613] is concerned, there should not be restriction to determine which angle is A, B or C respectively. May be this is a guess to mind of Guilherme Dela Corte.

The side a a is opposed to angle A A , the side b b is opposed to angle B B and the side c c is opposed to angle C C .

"For nomenclature, consider that opposed to the side x x lies the angle X X ."

Guilherme Dela Corte - 5 years, 7 months ago
Som Ghosh
Jul 23, 2015

From the plot we can see that we have clearly three real solutions for this problem and they are separated by 1, i.e. roots are α 1 , α , α + 1 \alpha-1,\alpha,\alpha+1 . Bit of algebraic manipulation will reveal that c o s A C 2 c o s A + C 2 = a + c b \frac { cos\frac { A-C }{ 2 } }{ cos\frac { A+C }{ 2 } } =\frac { a+c }{ b } I believe rest is you guys easily can manage.

From the information given in the question, it is feasible to assume that a = α + 1 , b = α 1 , c = α a=\alpha+1,b=\alpha-1,c=\alpha

This gives the desired ratio as 2 α + 1 α 1 \frac{2\alpha+1}{\alpha-1}

Janardhanan Sivaramakrishnan - 5 years, 10 months ago

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Are you seriously a college professor... Can't you see a + b c = 2 \frac{a+b}{c} = 2

Som Ghosh - 5 years, 10 months ago

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Dear Mr. Som Ghosh, whenever you post something in a public forum, you need to be objective and not subjective. Your question is incomplete in addition to being not original.

Nowhere in your question, which I see now at 11:32 AM, on 24-07-2015 IST, does it say that a < b < c a<b<c OR a > b > c a>b>c . With the lack of that particular piece of information, any of the three real values can be assigned to a , b , c a,b,c .

In addition to this, your graphical solution is also inexact. Though it is obvious that there are three solutions, it is not clear how you can conclude, without any further derivations, that they are separated by unity?

Further, you should also note that the question should yield an answer as a + c b \frac{a+c}{b} and NOT a + b c \frac{a+b}{c} (as you had put in your derogatory remark against me).

It is mandatory to be clear and precise in your questions. The interpretation should be unique and any argument that you put should be based on objective arguments, than ones like "it is obvious a,b,c means a < b < c a<b<c ," etc., This is what I had tried to point out in my solution, which, unexpectedly gives the correct answer. But, unfortunately, I was not clear enough.

I hope that this clarifies my original comment.

Janardhanan Sivaramakrishnan - 5 years, 10 months ago

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@Janardhanan Sivaramakrishnan Are you going to justify your claim that a = b = c a=b=c ? , which is not only wrong but also preposterous. I have claimed that the real roots are separated by unity, I have checked it by numerical solutions also. Anyway you don't understand as you are stubborn to go with your solution. However, if the solution is α 1 , α , α + 1 \alpha-1,\alpha ,\alpha+1 , anyone will make the corret choice of the answer. You are assuming the roots are equal whereas I am assuming a sequence only, you decide which one is going loss the generality. Before going into any further discussion just tell me can you prove that all the roots are equal ? If you can't, then accept your fault and don't point finger to other's solution.

Som Ghosh - 5 years, 10 months ago

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@Som Ghosh I do not assume that the roots are equal. Hence, my solution is NOT the correct one. But, it surprisingly works out.

What I am pleasantly surprised is that the answer is the same even when the roots are equal (or in arithmetic progression).

My ONLY point of contention is that it is NEVER specified in the question that " b b is neither the smallest or the largest root". Please do read my comment completely before responding. I too do not want to drag it on and had written the comment ONLY because of the words , " Are you seriously a college professor... ", in your comment...

Janardhanan Sivaramakrishnan - 5 years, 10 months ago

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@Janardhanan Sivaramakrishnan In your college ask your students that if a , b , c a, b,c are in arithmetic progression with common difference 1 and then what is the value of sum of any two divided by the third, and check how many of them come up with the answer 2. I can see that it is not mentioned about the sequence(which also a wrong statement in your problem) then I readily assumed a sequence a > b > c a>b>c , which is not going to loss the generality. Anyway what is about your mistake, how can a person like you can made such mistake, actually this is the reason students in our country are losing interest from mathematics because of teacher like you, you made a mistake but your so much full of your pride that you are not going to accept your flaw.

Som Ghosh - 5 years, 10 months ago

By Viete, the sum of roots is zero (as the given equation is a polynomial of degree 9 and coefficient of x 8 x^8 is zero). So, how can all the roots be positive?

Deeparaj Bhat - 5 years, 3 months ago

My answer is more of logic than algebra or geometry.

The question does not specify any restriction like a > b > c a>b>c etc. Hence, it may be implied that the order does not matter. OR a = b = c a=b=c . This gives A = B = C = π 3 A=B=C=\frac{\pi}{3}

From this

cos ( A C 2 ) cos ( A + C 2 ) = cos ( 0 ) cos ( π 6 ) = 2 \frac{\cos\left(\frac{A-C}{2}\right)}{\cos\left(\frac{A+C}{2}\right)} = \frac{\cos(0)}{\cos(\frac{\pi}{6})} = \boxed{2}

Sir, I think you made a mistake, I believe a b c a \neq b \neq c . Check my solution.

Som Ghosh - 5 years, 10 months ago

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Indeed a b c a a \ne b \ne c \ne a . But, where in the question have you mentioned which is the smallest and which is the largest side? Please read my original comment.

Janardhanan Sivaramakrishnan - 5 years, 10 months ago

Guess 2 to 3 times for one correct answer. But I am sure that most people shall make the correct guess.

Lu Chee Ket - 5 years, 8 months ago

Pi/ 6 --> Pi/ 3

Lu Chee Ket - 5 years, 8 months ago

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