Micromass' big high school challenge thread

[Mastermind01]

Regarding #9

I don't know what alternative method @micromass has in mind but a good way of doing it without calculus would be:

Noticing that the two intersections form an asymmetric lens. We can find the points of intersection of two circles but placing one of them at the origin and another at $(0,a)$ . We can calculate the angle subtended by the common chord on each of the circles using similarity and then find out the area of their respective segments by subtracting the area of the triangle from that of the segment. A formula can be found out for the area of intersection.

Noticing that d (the distance between two centres) is r (the given radius) we get the following equation

$r^2\arccos(\frac{2r^2-a^2}{2r^2}) + a^2\arccos(\frac{a}{2r}) + \frac{1}{2}\sqrt{a^2(4r^2-a^2)} = \frac{\pi r^2}{2}$

Graphing the two sides of the equation independently for various values of $r$ we find that the ratio of the point of intersection (y-coordinate) to that of the original radius $r$ is always $\approx 1.1587$

 

[Charles Link]

Regarding #9

I don't know what alternative method @micromass has in mind but a good way of doing it without calculus would be:

Noticing that the two intersections form an asymmetric lens. We can find the points of intersection of two circles but placing one of them at the origin and another at $(0,a)$ . We can calculate the angle subtended by the common chord on each of the circles using similarity and then find out the area of their respective segments by subtracting the area of the triangle from that of the segment. A formula can be found out for the area of intersection.

Noticing that d (the distance between two centres) is r (the given radius) we get the following equation

$r^2\arccos(\frac{2r^2-a^2}{2r^2}) + a^2\arccos(\frac{a}{2r}) + \frac{1}{2}\sqrt{a^2(4r^2-a^2)} = \frac{\pi r^2}{2}$

Graphing the two sides of the equation independently for various values of $r$ we find that the ratio of the point of intersection (y-coordinate) to that of the original radius $r$ is always $\approx 1.1587$

A very minor correction, but I think the 3rd term in the equation needs a minus sign. If you convert arccosine to arcsine, it agrees precisely with @MAGNIBORO of post #51 (and #68) and also the result that I got.

 

[Mastermind01]

A very minor correction, but I think the 3rd term in the equation needs a minus sign. If you convert arccosine to arcsine, it agrees precisely with @MAGNIBORO of post #51 and also the result that I got.

You're right. Unfortunately I can't change it now.

 

[Charles Link]

You're right. Unfortunately I can't change it now.

It's neat that we're all getting the same answer. (and same equation.) It should be interesting to see what @micromass has for a solution.

 

[late347]

based on private message conversation, it is fairly reasonable to say that no further attempts will be made upon the problems 6, and problem 10.

These two have already been completed by high-schooler participants, so they themselves are not eligible to solve those problems. It seems that those two were actually common textbook exercises, therefore they were common math problems.

problem 4b is the only remaining problem, I think.

 

[ProfuselyQuarky]

@micromass Would you mind posting that "more methodical solution" for #8? I'd be happy with a link.

 

[micromass]

@micromass Would you mind posting that "more methodical solution" for #8? I'd be happy with a link.

I'm sorry, I forgot to respond. Anyway, you should start reading here: https://www.jstor.org/stable/27960002?seq=6#page_scan_tab_contents It's free but you'll need to register. If you read it and are interested in the math behind this, it's projective geometry and finite geometry. It's a popular field of research in mathematics.

 

[ProfuselyQuarky]

I'm sorry, I forgot to respond. Anyway, you should start reading here: https://www.jstor.org/stable/27960002?seq=6#page_scan_tab_contents It's free but you'll need to register. If you read it and are interested in the math behind this, it's projective geometry and finite geometry. It's a popular field of research in mathematics.

Ah! Thanks!

 

[member 587159]

But the root goes to infnity too. So you have $0^{1/\infty} = 0^0$ which is an undetermined form.

Well, I looked at this again, and, when you fill in p -> infinity, you get 1^0 = 1 which is not an indetermined form so I don't think that another calculation of the limit was necessary. If I would be missing something, please enlighten me :)

 

[micromass]

Well, I looked at this again, and, when you fill in p -> infinity, you get 1^0 = 1 which is not an indetermined form so I don't think that another calculation of the limit was necessary. If I would be missing something, please enlighten me :)

Indeed, that is correct.

 

[member 587159]

What's the solution for 3d as I think that nobody found it. I hope I don't ask this question too soon.

 

[Swapnil Das]

My solution to Question 6:

We can start forming a nested radical expression for $x$:

$$|x|=\sqrt{x^2}\\=\sqrt{1+(x-1)(x+1)}\\=\sqrt{1+(x-1)\sqrt{1+x(x+2)}}\\=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}}$$
In the same way if we progress indefinitely, we get the following nested radical expression for $x$:
$$|x|=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{\cdots}}}}}}$$
This expression matches with the form of the given problem , so put $x=3$ to get:
$$\boxed{3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}}$$

 

[ddddd28]

10. The same donkey is now tied with a 50m long rope. The rope is tied to the corner of a 20m by 10m barn. What is the total area that the donkey is capable of grazing?

This is a sketch of the donkey's trail:

The donkey starts from point o.
When the donkey goes to left, it covers the pink area + the blue one.
When the donkey goes to right, it covers the green area+ the blue one.
Therefore, the donkey covers three quarters of 50 m- circle and an additional odd part.
To my opinion, the easiest way to find this area is to calculate the sectors ADC, BEC, the area of the two triangles AOB and ABC.
S ABC = √11*100
By Heron's theorem
(AB= √50 AC=10 BC=20)
S ABO= 100
In order to find the sectors, we need to know the angles DAC and CBE, which can be found by arctan(slope BC) and arctan (AC).
We need to find point c:
The circles' equations:
$(x-2)^2+y^2=9$
$x^2+(y-1)^2=16$
$c( \frac{24+4√11}{10}), \frac{-1+4√11}{5} )$ *** I reduced by 10
$\arctan( \frac{\frac{-1+4√11}{5}-0}{\frac{24+4√11}{10}-2})=54.86°$
Similarly, we get 90°- 21.3°=68.7° for ∠DAC
S ADC= 1600π 68.7°/360°=305.3π
S CBE= 900π 54.86°/360°=137π

S total= S CBE + S ADC +S ABO+S ABC +0.75 2500π= 137π+305.3π+√11*100+100+0.75 2500π = 7711.7 m²

 

[ddddd28]

based on private message conversation, it is fairly reasonable to say that no further attempts will be made upon the problems 6, and problem 10.

These two have already been completed by high-schooler participants, so they themselves are not eligible to solve those problems. It seems that those two were actually common textbook exercises, therefore they were common math problems.

problem 4b is the only remaining problem, I think.

I solved the tenth problem just for fun, please let me know if my solution is correct, and do these problems have a common name? I think such problems of a wrapping rope are very interesting, in sense of trying to generalize them.