Please help me find two more Equation (to prove that α=β=γ=30)

navneet9431

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I have been trying to solve this question(please see the image)

The question asks to prove that α=β=γ=30 .

I am trying to solve this question by discovering at least three different Equations. As the number of unknown variables is 3.So,i must find at least 3 different Equations in order to find the values of the three different variables (α,β and γ).

I have successfully found the first Equation

α+β+γ=90......(i)

Please help me find out the other two Equations!

I will be thankful for help!

Note: English is my second language.
84db753cdd6627ca373fb0691ec18a16.jpg


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State the meaning of the tiny, curved marks in each angle. Do they mean we know already that all these angles are congruent?
 
State the meaning of the tiny, curved marks in each angle. Do they mean we know already that all these angles are congruent?
No,those tiny curve marks are just made to indicate the angle measure.
They do not mean that we know that all these angles are congruent.

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Actually there are SIX unknowns. You have not labeled every angle and considered every triangle.

\(\displaystyle \angle AKB = \delta ,\ \angle BKC = \epsilon ,\ \angle AKC = \zeta.\)

And there are four obvious equations, one for each triangle.

\(\displaystyle ( \alpha + 30) + ( \beta + 30) + ( \gamma + 30) = 180 \implies \alpha + \beta + \gamma = 90.\)

\(\displaystyle \alpha + \delta + 30 = 180.\)

\(\displaystyle \beta + \epsilon + 30 = 180.\)

\(\displaystyle \gamma + \zeta + 30 = 180.\)

There is a less obvious equation of

\(\displaystyle \delta + \epsilon + \zeta = \text { WHAT and why.}\)

Getting the sixth equation seems to require some information about triangle ABC or about K or about AK, BK, or CK. Have you given us all the information from the problem?

EDIT: Depending on the geometric information given, you may not need all those equations. For example, if you are told that triangle ABC is equilateral, then you know that each angle equals 60 degrees.
 
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My expectation would be that you will need at least one equation involving side lengths, which means trigonometry. I say this because angles alone can't ensure that lines meet at a given place, which will be necessary if you try to construct the figure.

Does the source of the problem imply that you should use only synthetic geometry methods, or that trigonometry would be needed? I assume you stated the entire problem as given to you, and didn't omit anything.

Another thing to observe is that what you are told to prove is the "obvious" solution; so the problem becomes one of showing that it is the only solution. You might be able to start with that (an equilateral triangle) and show that any deviation from it will be impossible.
 
We could be severely overthinking this. How about \(\displaystyle \alpha = 30\;and\;\beta = 30\)? Seems like two equations to me.

I'm just not real clear on the nature of the assignment if we start with what is to be proven. Rather like Petitio Principii, I think.
 
Draw 3 lines from K, one perpendicular to AB, another to BC and the third to AC. Can we calculate the angles of all 6 triangles?
 
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Draw 3 lines from K, one perpendicular to AB, another to BC and the third to AC. Can we calculate the angles of all 6 triangles?
No, because that just gives us more unknowns, 12 I suspect without doing the work.

The whole problem is silly because incomplete. If, and only if, the three unknown angles are each 30 degrees, then the triangle is equilateral. But, so far, we have been given no information that allows us to make either conclusion.
 
No,those tiny curve marks are just made to indicate the angle measure.
They do not mean that we know that all these angles are congruent.

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Sorry. Marking all six angles with the same marking MEANS all six angles are congruent. If you don't want it to mean that, you will have to use some other markings.
 
No, because that just gives us more unknowns, 12 I suspect without doing the work.
Not really because it is easy to see the value of all newly created angles in terms of a, b and c (I can't do the Greek characters!).
I think there is enough information, drawing the figure seems to show that triangle ABC is equilateral. I think any proof will demonstrate that the three angles are equal using similar triangles showing that there is a symmetry to the figure. Whether this would be regarded as a rigorous proof, I don't know.
 
No, JeffM, the problem is perfectly clear:

The question asks to prove that α=β=γ=30 .

The goal is to prove that the ONLY triangle that satisfies the conditions in the picture (namely, that the three angles marked as 30° are so) is the equilateral triangle.

It appears to be a very interesting problem, not a standard geometry problem. In playing with the figure in GeoGebra, I find that it appears to be an extremum problem in disguise, because if two of the angles marked as 30° are so, the other is always less than 30° unless the triangle is equilateral.

When I follow Jonathan's suggestion, I obtain the following equation: \(\displaystyle sin\alpha \cdot sin\beta \cdot sin\gamma = \frac{1}{8}\). I have not dug into this, but I think 1/8 is the maximum value, and is attained only when the three angles are equal (to 30°).

So, Navneet, can you tell us the source of the problem, and whether a multivariable calculus solution seems appropriate? There may be a geometric inequality that can be applied, but I haven't seen it yet.
 
Sorry. Marking all six angles with the same marking MEANS all six angles are congruent. If you don't want it to mean that, you will have to use some other markings.

No, these little arcs are labeled with measurements or variables. In that case, they just mean "this is the angle I'm referring to".

You're thinking of marks that stand alone without labels, to indicate congruent pairs by using the same marking. Don't confuse the two usages. (And even if what he is using is not standard on your country, why couldn't it be standard in his? Accept what someone says on his own terms, rather than insist that he is saying nonsense when it is perfectly possible to interpret it to make sense.)

Incidentally, GeoGebra, which I used to check the problem, marks angles in exactly this way, with an arc together with a label.
 
No, JeffM, the problem is perfectly clear:



The goal is to prove that the ONLY triangle that satisfies the conditions in the picture (namely, that the three angles marked as 30° are so) is the equilateral triangle.

It appears to be a very interesting problem, not a standard geometry problem. In playing with the figure in GeoGebra, I find that it appears to be an extremum problem in disguise, because if two of the angles marked as 30° are so, the other is always less than 30° unless the triangle is equilateral.

When I follow Jonathan's suggestion, I obtain the following equation: \(\displaystyle sin\alpha \cdot sin\beta \cdot sin\gamma = \frac{1}{8}\). I have not dug into this, but I think 1/8 is the maximum value, and is attained only when the three angles are equal (to 30°).

So, Navneet, can you tell us the source of the problem, and whether a multivariable calculus solution seems appropriate? There may be a geometric inequality that can be applied, but I haven't seen it yet.
Ahh, I see.

The geometric fact that I was looking for is that AK, BK, and CK meet at a common point. Now of course the angle bisectors also meet at a common point, call it L. It may be possible to find a purely geometric proof by contradiction of the assumption that K and L are distinct points. I last studied plane geometry in 1962 and cannot remember enough to attempt a proof by contradiction.
 
No, JeffM, the problem is perfectly clear:



The goal is to prove that the ONLY triangle that satisfies the conditions in the picture (namely, that the three angles marked as 30° are so) is the equilateral triangle.

It appears to be a very interesting problem, not a standard geometry problem. In playing with the figure in GeoGebra, I find that it appears to be an extremum problem in disguise, because if two of the angles marked as 30° are so, the other is always less than 30° unless the triangle is equilateral.

When I follow Jonathan's suggestion, I obtain the following equation: \(\displaystyle sin\alpha \cdot sin\beta \cdot sin\gamma = \frac{1}{8}\). I have not dug into this, but I think 1/8 is the maximum value, and is attained only when the three angles are equal (to 30°).

So, Navneet, can you tell us the source of the problem, and whether a multivariable calculus solution seems appropriate? There may be a geometric inequality that can be applied, but I haven't seen it yet.
Thank you Dr.Peterson for making other people here understand this question!
Actually, i have found this question on this website. The website also provides many solutions of this question. But,i was trying to solve it my own way by discovering at least three different equations in order to get the value of α,β and γ.
I now realise that it is impossible to solve this question in the way i was trying to solve.

Thanks!
 
No, these little arcs are labeled with measurements or variables. In that case, they just mean "this is the angle I'm referring to".

You're thinking of marks that stand alone without labels, to indicate congruent pairs by using the same marking. Don't confuse the two usages. (And even if what he is using is not standard on your country, why couldn't it be standard in his? Accept what someone says on his own terms, rather than insist that he is saying nonsense when it is perfectly possible to interpret it to make sense.)

Incidentally, GeoGebra, which I used to check the problem, marks angles in exactly this way, with an arc together with a label.

Yes, it is incidental nonstandard notation. It should be explained explicitly if that is the intent. Does GeoGebra make them different colors or use some other method to avoid the standard interpretation as a congruence?
 
Yes, it is incidental nonstandard notation. It should be explained explicitly if that is the intent. Does GeoGebra make them different colors or use some other method to avoid the standard interpretation as a congruence?

I don't think there is anything nonstandard about it. For another example, see this picture from a Wikipedia page.

The arcs are all the same style, but the labels distinguish them. GeoGebra likewise labels them; you can change the colors, but they are not so by default, because it is not needed. I see no way to imagine that the three angles here are being called congruent.
 
I like the picture, but it is clear in the picture that they are three entirely different angles. It is not clear at all in the usage seen in this thread.
 
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