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What a math problem !!!

The math problem which I have managed to solve for a long time is: - Divide an angle into 3 same angles with compass and straight ruler !!!

Note by Tran Trung Nguyen
4 years, 4 months ago

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I'm sorry Nguyen, but that's not possible unless you have a marked straightedge. It's been proved that you can't trisect an arbitrary angle with a compass and an unmarked straightedge. But people still refuse to believe that. Why? It's human nature. You tell someone that they can't do a certain thing and it'll be the very thing they'll ever want to do.

There are a lot of people who believe that mathematicians have just gave up on angle trisection and declared it impossible to do. That is not the case. The cool thing about the impossibility of angle trisection is that it can be proved.

If you present your construction here, it can be shown that there are a lot of (unintentional) loopholes. A lot of mathematical fallacies are based on such loopholes.

If you want to know about the impossibility proof, you'll have to learn about group theory and group extensions. These topics are too big to discuss here.

But if you learn about them, you'll find out that a \(60\) degree angle can't be trisected and therefore there is no general way to trisect an arbitrary angle with an unmarked straightedge and a compass.

PS: It's also interesting why people refuse to believe that angle trisection is impossible with an unmarked straightedge and a compass. You can't construct an ellipse with those things too. You can't cut boards with a screwdriver, so why is it a shock that you can't trisect an arbitrary angle with the tools mentioned above? People who still try to trisect angles with these are expecting the ruler and compass to do something beyond their capabilities.

Mursalin Habib - 4 years, 4 months ago

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The simplest explanation, is that a number \(n\) is constructible if and only if the minimal polynomial of \(n\) over the rationals, has degree that is a power of 2. This will take some work, but you can understand why it is true by looking at the equations involved.

Then, using the trigonometric identity of \( \cos 3 \alpha \), we can show that the minimal polynomial of \(n\) over the rations is \( 8X^3 - 6X + 1 \), which has degree not a power of 2. Hence, \( \cos 20^\circ \) and thus \( 20^\circ \) cannot be constructed.

I referenced this in my post on construction.

Calvin Lin Staff - 4 years, 4 months ago

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