equilateral triangle on a parabola

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2 жыл бұрын

equilateral triangle on a parabola. Find two points on the parabola y = 1-x^2 such that the triangle formed by the x axis and going through those points is equilateral. It is a nice problem that combines calculus and geometry, more specifically tangent lines, intercepts, distances, and inscribed triangles. This was inspired by a instagram post by UC Irvine math.
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Пікірлер: 49

@afjalchowdhury4079 2 жыл бұрын

I did it by finding the derivative of the curve (-2x), finding when the line has 30 degrees slope (gradient of +-root3) and solving for x: 2x = root 3, x = root3 /2

@mehmetemin903 2 жыл бұрын

I like this type of questions very much. Thank you. 👏

@txikitofandango 2 жыл бұрын

I started from the fact that the slopes of the tangent lines must equal plus or minus √3, found the points of tangency by setting y' = -2x = plus or minus √3. But I didn't see that the equation was 1 - x^2, so I generalized it to b - x^2. Still the same x-coordinates, just different y-coordinates by adding a constant.

@andreynogueira7419 2 жыл бұрын

Awesome video, Dr Peyam! I must share that you made my day with that "just to waste your time" around 5:50. I didn't see it coming. Watching you having fun while teaching us clever and truly beautiful math facts is never a waste of time.

@tiarkrezar 2 жыл бұрын

Wouldn't it be simpler to just find points where the slope equals +-sqrt(3)?

@advaykumar9726 2 жыл бұрын

I did it the same way

@drpeyam 2 жыл бұрын

How so? I don’t follow

@llunaecy 2 жыл бұрын

@@drpeyam In an equilateral triangle all of the angles have to be pi/3, so the tangent lines have to form angles of pi/3 with the x-axis, and therefore their slopes should be +/- tan(pi/3) = +/- sqrt(3). Slope on y=1-x^2 at point x=a is - 2a, so at the points we're looking for you should get - 2a = +/- sqrt(3), or a = +/- sqrt(3) / 2.

@hotlatte1222 2 жыл бұрын

Same way.

@adamforte9530 2 жыл бұрын

@@llunaecy interesting solution

@AnonimityAssured 2 жыл бұрын

I seem to remember being told to move radicals up into the numerator, although I've no idea why. Presumably, it's just a convention. That would make the intercepts (-7√3)/12 and (7√3)/12.

@reeeeeplease1178 2 жыл бұрын

To calculate the fraction more easily (using long divison)

@AnonimityAssured 2 жыл бұрын

@@reeeeeplease1178 Ah, thanks. Not all conventions have a logical motive, but I'm glad that one does.

@mrlordsaif5708 2 жыл бұрын

I just set the derivitive equal to tan60 since thats the angle of an equilateral triangle and solved for the derivitive=tan60 and -tan60 and got sqrt(3)/2 and -sqrt(3)/2 and plugged it into the original function to find the y coordinates of the two points tangent to the parabola

@cherkicherki2286 Жыл бұрын

Le secret ou la clee du probleme c'est : F'(Xo)= F(X)-F(Xo)/ X- Xo . C'est ce que vous avez fait Monsieur . Bravoooo

@michaelempeigne3519 2 жыл бұрын

you can also use the fact that the area = (1 / 4 )*s^2 * sqrt 3

@EssentialsOfMath 2 жыл бұрын

Nice meeting you and getting a picture with you Dr. Peyam!! haha

@drpeyam 2 жыл бұрын

Hahaha same!!!! 🤩

@adamforte9530 2 жыл бұрын

Great problem!

@jesuisravi 11 ай бұрын

Looks good.

@nedmerrill5705 2 жыл бұрын

Good one!

@frozenmoon998 2 жыл бұрын

So much cooler than what they make geometry to be in regular classes in school. Usually, students are required to memorize formulas and rely on them, rather than to navigate through the problem and find a solution that way.

@harmsc12 2 жыл бұрын

The only thing I can predict about the location of A is that it's going to be where the derivative calculates out to -sqrt(3). I don't remember exactly how the power rule works, so I can't say beyond that.

@mmattoso1 2 жыл бұрын

It would have been much faster just to calculate the derivative y'=-2x and to find the points where it would be +tg(60°) = sqrt(3) and - tg(60°) = - sqrt(3)...therefore we immediately find x=+-sqrt(3)/2 and y=1/4, being these the 2 requested points... we don't need to find anything about the triangle itself.

@wallstreetoneil 2 жыл бұрын

That's what I did. Set Tan (60-degrees) = to Derivative -2X. This gives [X=Tan(60)/2, Y=1-(Tan(60)/2)^2)].

@slavinojunepri7648 Жыл бұрын

Very cool indeed

@mouskatoodle 2 жыл бұрын

very nice problem

@codahighland 2 жыл бұрын

I wonder how generalizable it is. If it hadn't required that the third line be the x axis, I think you'd have to think about it based on angles -- what are all the possible equilateral triangles that have two sides tangent to the parabola?

@jursamaj 2 жыл бұрын

It doesn't matter where the base is. It only matters that the slope of the sides is ±√3, and the slope of the parabola is -2x. Solve for those 2 being equal, and there's your x's, then put those in the parabola to solve for y. Peyam made this way too complicated.

@mathisnotforthefaintofheart 2 жыл бұрын

I looked at the title and I thought what the problem is about. I was wrong. I thought the problem is like there is an equilateral triangle with points (-1,0) and (1,0) forming one of its sides, and the third vertex on the positive y-axis Find the equation of the parabola that is tangent to the equilateral sides in quadrant I and II where (-1,0) and (1,0) are the points of tangencies. This problem would also fall in an introductory Calculus course, and as I found out, isn't that hard to do

@codahighland 2 жыл бұрын

Why didn't you expect it to be irrational? An equilateral triangle has angles of 60 degrees, and the sine of 60 degrees is irrational. You would have needed an irrational number in the equation to cancel that out.

@dschaegkarthur1093 2 жыл бұрын

Hi Iove the "Körper" T-Shirt. Is that custom made or somewhere buyable? And I also Love the Video lol

@drpeyam 2 жыл бұрын

Yes it is, check out my teespring store (link in bio)

@dschaegkarthur1093 2 жыл бұрын

@@drpeyam Amazing thx a lot^^

@gerdkah6064 2 жыл бұрын

5:53 just to waste more time .. i wrote that comment ^v^

@bscutajar 2 жыл бұрын

I dont understand where the equation at 1:05 came from

@drpeyam 2 жыл бұрын

It’s the equation of the tangent line of a function at a point a

@liehanjiang3640 2 жыл бұрын

Hi Dr. Peyam! There’s a question confusing me five years, why we learn algebra in university while no geometry in university?

@drpeyam 2 жыл бұрын

You actually learn geometry in university 😁 But we call it linear algebra and analysis! Notice how both deal with distances and things being parallel. If you take differential geometry, you’ll see the similarities

@liehanjiang3640 2 жыл бұрын

@@drpeyam OMG that’s right cuz vectors are in a plane?!

@drpeyam 2 жыл бұрын

Exactly!

@mathisnotforthefaintofheart 2 жыл бұрын

Maybe the question is more about why Euclidean Geometry isn't taught in university? (Or non Euclidean geometry for that matter) and that is because a lot of degree programs don't require it. Unless you want to become a math major. For most disciplines outside math, algebra has more relevance. It may not sound fair, but that is what it is. For example: Finance needs a decent amount of algebra but parallelograms or the SAS congruence statement aren't important for a finance manager working at a car dealership

@bobh6728 2 жыл бұрын

I took advanced geometry in 10th grade, about 15 years old. The book was set up to develop Euclidean geometry from the ground up. So it started with: UNDEFINED TERMS: point, line, plane. Then went on with axiom 1, theorem 1, etc. It was very theoretical and based on proving theorems. Then we took trigonometry and advanced algebra and calculus. We were expected to know formulas for areas, and names of solids, but that was not part of our geometry class. The non-advanced geometry students had all that so we had to learn those ourselves.