FIFA World Cup earthquake

Geometry Level 1

At the 1920 FIFA World Cup, there was an earth quake during one of the games and a cone shaped hole was created with a diameter of 4 feet and sloping sides, both 4 feet long from ground-level to the bottom of the pit. The soccer ball, with a radius of one foot, fell into the hole. Find the distance from the center of the Soccer ball to the bottom of the pit in feet.

Assumptions:
- The sides of the hole are perfectly straight and smooth.
- The ball falls as far as it can with out changing shape.
- FYI, there wasn't actually an earthquake at the 1920 FIFA World Cup.

The answer is 2.

6 solutions

A Former Brilliant Member
Jul 16, 2017

By similar triangles, we have

$\dfrac{x}{1}=\dfrac{4}{2}$ $\color{#3D99F6}\large \implies$ $x=$ $\color{#D61F06}\boxed{\large 2}$

Zandra Vinegar Staff
Oct 11, 2015

This problem can be solved using symmetry alone, by inscribing the ball (a circle in the picture) in two equilateral triangles:

STEPS:
(1) Since the angle at the bottom of the hole is 60 degrees, a line parallel with the ground and touching the ball will make an equilateral triangle.
(2) Consider a congruent triangle surrounding the ball, 180 degrees rotated from the one in step 1, by symmetry, it will also be touching the soccer ball on all 3 sides.
(3) Each of the little triangles in this image has height $1ft$ . This is given in the problem statement: the radius of the soccer ball is $1ft$ .
(4) The distance from the center of the ball to the point of the large downward pointing triangle is therefore $\fbox{2ft}$

How did you know the triangles all had a side length of one?

ibadassninja . - 4 years, 10 months ago

Sadie Robinson
Jun 22, 2014

For this solution, $O$ is the circumcenter of the ball, $V$ is a vertex of the cone, and $T$ is the point where the ball touches the edge of the cone. The reason $T$ is called $T$ , is that at $T$ the ball is tangent to the inside of the cone. The line we are looking for is $OV$ .

Now, looking at the triangle $OVT$ , the angle at $T$ is $90^{\circ}$ and the angle at $V$ is half of the vertex angle, i.e. $30^{\circ}$ . As OVT is a right-angled triangle, we can do this: $\frac{OT}{OV} = \frac{1}{OV} = \mbox{sin}(30^{\circ}) = \frac{1}{2}$ Thus, $OV$ is $\boxed{2}$ feet long.

Sathya Nc
Jan 14, 2016

Easy one

BUT STILL THERE WAS NO FIFA WORLD CUP IN 1920!!!!!!!

Agastya Chandrakant
Jan 14, 2015

Didn't even solve it!

First, saw that answer was only an integer. then using trigonometry, found the height of the pit! About 3.46 ft. Now, put 1 then 2 or 3! three tries and 3 answers! got correct in 2nd try! see, it was pure luck and thus i agree i don't own this 50 points! And i actually searched for that earthquake as i didn't read the question completely!

One Advice

if Trevor Arashiro you are posting a question like this make sure the earthquake has greater effect and answer isn't easy to guess! I ain't mocking you, couldn't even solve it properly; but don't let the guessing get easy ^_^

Trevor Arashiro
Jun 19, 2014

This problem is better looked at in 2D. The radius of any circle inscribed in a triangle is 2A/P, or 2 time the area over the perimeter. However, that is only necessary if you work it out the long way. It is irrelevant in this problem because the circumcenter of both an equilateral triangle and the circle it circumscribes will be located in the exact same spot. So for there to be an equilateral triangle that exists that circumscribes the circle, a line must be drawn tangent to the circle and parallel to the ground. In this case, the radius of the circle is actually the apothem of the equilateral triangle, and the apothem of any equilateral triangle = 1/3 of the height. All that is left now is to subtract the radius of the circle from the height and that will yield the distance from the circumcenter to the vertex of the triangle.

@Brian Charlesworth

Salve Brian. I and a several others are making a collaboration set on infinite and finite rectangular spirals and stair cases (I think that's what they're called). I would be very honored if you could take the time to make a problem or two regarding these topics to add to the set. I don't know when the set will be released just as yet.

An example of a finite rectangular spiral would be

Trevor begins at his house and runs 2015 miles north, 2014 miles east, 2013 miles south, 2012 miles west, 2011 miles north.... 2 miles east, and 1 mile south. What is the straight line distance from where he began to where he ended? If I'm right, the answer here is $1008\sqrt{2}$ .

To turn this into a stair case we could travel the same distances but in the pattern north, east, north, east, north...

A harder one which is infinite could be traveling distances of $cos(1)$ east, $\cos^2(1)$ north, $cos^3(1)$ west... (I honestly have no idea how to solve this, but it just crossed my mind lol). Or maybe using directions such as northeast and northwest

If you don't want want to take part in this that's completely fine. However, if you do, you can either reply here or send me an email at tma1701@iolani.org.

P.S. Sorry for using a problem so far back, just wanted to use one that no one would see our discussion on.

Trevor Arashiro - 6 years, 1 month ago

Haha. I like your under-the-radar communication method. :D

Sure thing; I'll try and dream up a couple of problems and report back here. Using trig functions might produce some novel questions, as well as logs, exponentials, etc.. Can I throw in some non-right angle "turns"? Perhaps a random direction function as well?

And yes, the answer to your first example is correct. For your second example, I believe the answer is $\dfrac{\cos(1)}{\sqrt{1 + \cos^{2}(1)}}$ at an angle of $\tan^{=1}(\cos(1))$ north of west.

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth Of course, anything you can dream up!

Btw, after I got some food in me, I realized how u were able to solve that problem. It's quite a neat result and for the stair case, the result is very similar $\dfrac{\cos(1)\sqrt{1+\cos^2(1)}}{\sin^2(1)}$ .

P.S. I also just realized that you need to use the @mention thingy every time you leave a comment or I won't get a notification because this problem is so old lol

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – O.k., I'm not sure if I'm supposed to post the questions myself and then notify you or just pass my ideas along to you through this black site, but for now I'll just run one idea by you.

Suppose a particle, starting at the origin, moves in six positive integral steps, (i.e., each step is of an integral length $\ge 1$ ), in the pattern right, up, right, up, right, up, such that the combined lengths of the steps is $10,$ and that each possible sequence of steps is equally likely to occur. What is the expected (magnitude of the) distance between the origin and the particle after it has completed the six steps?

The answer, (I think), is $\dfrac{5}{21}\sqrt{58} + \dfrac{20}{21}\sqrt{13} + \dfrac{10}{7}\sqrt{2} = 7.267$ to 3 decimal places.

I suppose this is more of a probability/combinatorics question with a side order of staircase, but I just wanted to run it by you and see what the protocol was. :)

Edit: A couple of other ideas:

(i) The series $\dfrac{x^{n}}{n!}$ starting at $n = 0$ : as a spiral the displacement is $1$ for any real $x,$ and as a staircase the displacement is $\sqrt{\cosh(x)}.$

(ii) The series $\dfrac{1}{n}$ starting at $n = 1$ : as a spiral the displacement is $\sqrt{\left(\dfrac{\pi}{4}\right)^{2} + (\ln(\sqrt{2}))^{2}},$ and as a staircase it does not cease.

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – Brian Charlesworth sure, here is probably the best place to discuss this. The view counter hasn't gone up by even 1 person since the start of our "top secret" conversation. You can either post the problems yourself or just post them here and I'll post it. Some of the other people who haven't been on this site as long as I have and don't have as many followers are sending their problems to me so that they gain more publicity.

For the first problem, I'm getting a very similar answer (it doesn't look similar but the decimal values are 0.13 off). I dunno if I'm including a few extraneous cases, but here's my work (I'm not certain if it's right either).

By stars and bars, we have the number of ways to go 3-7 units right and 7-3 units up is

$\displaystyle \sum_2^6 \dbinom{n}{2}\dbinom{8-n}{2}=126$

The total length traveled by each of these will be

$\sqrt{n^2+(10-n)^2}$

So we evaluate the sum and divide by the total number of cases to get to be

$\dfrac{\displaystyle \sum_2^6 \dbinom{n}{2}\dbinom{8-n}{2}\sqrt{n^2+(10-n)^2}}{\displaystyle \sum_2^6 \dbinom{n}{2}\dbinom{8-n}{2}}\approx 7.397$

Just curious, if we had non integral side lengths, would the answer be

$\cfrac{1}{10}\displaystyle \int_0^{10} \sqrt{x^2+(10-x)^2}$

I'm not sure if this accounts for every case because it is a 6 part step but the equation above assumes 1. Though I'm not sure if the result will change because they will be so close to each other numerically.

As for (i), I have no clue how to do that. It looks like the Taylor series expansion of $\cosh(x)+\sinh(x)$ but that's about as much as I can say. What I do know, however, is that it looks like an idea with large potential to become a great problem ;).

As for (ii) that is another good problem. It was one of the first ones that I thought of while doing this set. I got a slightly different answer of $\sqrt{\left(\dfrac{\pi}{4}\right)^{2} + \left(\dfrac{\ln(\sqrt{2})}{2}\right)^{2}}$ by using the sum

$\displaystyle \sum_0^{\infty} \dfrac{\cos(\frac{\pi}{2}n)+i\sin(\frac{\pi}{2}n)}{n}$

And some integration of the Laurent series expansion of

$\displaystyle \sum_1^{\infty} \frac{x^n}{n}$

To get

$\left|\dfrac{\ln(1-e^{i\pi/2})}{e^{i\pi/2}}\right|$

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – Trevor Arashiro Since we have roughly the same number of followers I think I'll just post the problems myself and then give you the links here. (We'll both have to post special "1000 follower" questions when we hit that mark, (looks like you'll get there first). :) )

I'll recheck my first problem before posting; I did it quickly so I could have easily made a mistake. I'll also see if I can make the non-integral variation into a question as well.

For (i) it's the Taylor series for the sine and cosine functions that come into play. For (ii), your answer is correct; I just forgot to divide the second term by 2. :P

Now are you posting the $\cos^{n}(1)$ and $\dfrac{1}{n}$ questions? I'm happy to write those up but it sounds like you've already got them in the pipeline.

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – Hopefully I don't fall asleep before I finish this comment :3 haha so I'll leave it rather curt.

That sounds good. I'll just add it to the set from your feed.

Thanks for clarifying (i) and (ii)

As for who posts the $cos$ spiral And the 1/n problem, you can post the two if you like, and I'll post a solution for the two since I already have full solutions typed. I'll post the stair case version of the cos problem and tag you in it for helping me on it.

Btw, the release DATE will probably be Saturday of next week (May 9). So no need rush. A lot of people have exams including myself.

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – Exams!! I still have nightmares about not having studied or not knowing where the exam room is. :O Anyway, I hope they go well. :)

I may have jumped the gun a bit, then, as I just posted my first question, "Up right and away". I went with my initial answer, but we'll see if it stands up to scrutiny.

Brian Charlesworth - 6 years, 1 month ago

@Trevor Arashiro – @Trevor Arashiro I've just posted another problem as discussed called "A controlled spiral". I found it hard to phrase this problem elegantly, so before I post any more of this type of problem I'd greatly appreciate some feedback, (when you have the time, of course). Thanks. :)

P.S.. Hope that the studying is going well.

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – Wow, yet another good problem! And don't worry, while the problem is confusing after 1 read over, it's definitely understandable when you thoroughly read it.

Which problems are you gon a post on next week Saturday? I'll still add this one and the previous to this set.

And exams are going well, thanks for asking. Except that I have a D-chem exam. (More commonly known as O-chem but I call it death chem).

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – @Trevor Arashiro Sorry I've been tardy in getting back to you. Yes, organic chemistry does have a nasty reputation. I notice that you have chemistry as one of your "Interests", yet from your remarks here and in the past it would seem that this subject is the bane of your existence. In any event, don't die. :P

As for problems I'll be posting, I was thinking of doing a staircase version of "A controlled spiral", and possibly a spiral version of "Right up and away", (although this last one could be wickedly difficult). You seem to have answers all set for the $\cos(1)$ and spiral-harmonic series questions we've discussed above, so I'm guessing that you've planned to post those ones yourself. I have in mind a spiral-harmonic variation with each "turn" being $45^{\circ}$ rather than a right angle, but I haven't worked out the details. I was also thinking of posting a staircase/probability problem where there are different probabilities for going right and going up, with the question asking to calculate the probability of passing through a certain point.

I'll pass along more details when I have these problems, (and\or possibly others), worked out. Is the plan to post all these questions at once? I was thinking of posting them Friday night and Saturday morning and then sending you the titles, but if you have another plan in mind let me know. :)

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – @Brian Charlesworth Sounds good. I'll post a full response later so this will be quite curt. Just got off the plane from the state golf Tournament. Sorry I haven't been able to reply sooner. I can post the cos staircase and u can post the spiral if u like. I'll post the harmonic spiral.

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – @Trevor Arashiro That's funny; I was just about to send you a note. Hope the tournament went well. :)

I'll post the cosine spiral, then, along with three others. These will be:

(i) the staircase version of "A controlled spiral",
(ii) a 3D version of "Right up and away", and
(iii) a staircase problem involving the series $\sum_{k=1}^{n} \frac{1}{\sqrt{kn}}.$ (Yes, this will be the weird one).

I'll send along the titles once I've posted them.

Brian Charlesworth - 6 years, 1 month ago

@Trevor Arashiro – @Trevor Arashiro So I've posted "Thrown for a loop" and "Step on it", and I'll wait until morning to post my other two problems. (and you already have "A controlled spiral" and "Right up and away" to add to the set).

If you think my two new problems require some editing just let me know. I used a tan rather than cos rectangular spiral in "Thrown for a loop", as it made the final result much more elegant; hope that doesn't mess things up with your staircase question.

Brian Charlesworth - 6 years, 1 month ago

@Trevor Arashiro – @Trevor Arashiro O.k., I've posted "This way and that" and "Stairway to heaven" now, for a total of six problems. I look forward to seeing the set that you and Jake put out, and to giving both your problems a try. I still need to post a full solution to "Stairway to heaven", and as usual I'll be a little nervous until someone actually gets the answer I've posted as confirmation. (Still no one has solved "Thrown for a loop"; does that question seem cumbersome to you or does it just look more difficult than it actually is?) Anyway, any comments are welcome, as always. :)

Update: All my problems have been solved by others now, which is good news. :)

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – @Brian Charlesworth I'm currently super busy posting everything, but I can say your problems look amazing. Not sure if I'll finish posting everything today. I'll start posting spiral problems soon. If you want, you can try the beast from hell I just posted named "now there's an n???". If this gets anything less than a 375 point rating id be surprised.

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – @Trevor Arashiro I have a quick question about your new question "Angle of displacement". I'm getting an answer of $\arctan(\cos(\frac{\pi}{10})) = 0.7603$ radians, (about $43.56^{\circ}$ ), but this is considered as incorrect. Have I misinterpreted the question somehow?

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – Eeeeyy, oops. That's correct, but I put the answer in degrees :|

My bad. If you still have two attempts, would you mind leaving the question unreported for convenience. However, if not, please do report it

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – No worries; I still had two attempts left, but I didn't want to waste them before checking in with you first. All is well now, :)

Brian Charlesworth - 6 years, 1 month ago

@Trevor Arashiro – @Trevor Arashiro This thread is getting so long. :P O.k., I really enjoyed reading your 1000 follower question, but I'm getting an earlier date. I have the three following equations:

(i) $a + c = 0$ , (initial conditions);
(ii) $ae^{25b} + c = 1000000$ , namely the December 11th, 2014 condition, and
(iii) $ae^{30b} + c = 1400000$ , namely the present-day condition.

After using WA to solve for the constants, I got $a = -c = 396195$ and $b = 0.05038396716326.$ Then for the 10000000 follower date I found $x = 64.85$ months, which corresponds to April 11th, 2018.

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – @Brian Charlesworth Arghhh. A careless mistake on my part again. We had the same values of x but I measured from today rather than Nov. 11th so I originally had the date 2.5 years off.

Haha, true this thread is pretty long :)

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – Great! Being a moderator came in handy there for changing the answer options. :) Definitely worth a reshare; you'll probably get a few new followers because of this question too.... grrrrr. Haha. :P

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – Haha, glad u like it :). I'm surprised at the number of people who actually read it. Didn't think this many people would take the time to go through that essay. Although I'm sure the brilliant staff will get a kick out of this :3

P.S. You should become a moderator, it would be great to have you on the mod team. We all look up to you, Michael, and Calvin sort of like Gods there :P

Trevor Arashiro - 6 years, 1 month ago

@Trevor Arashiro – Yeah, this one is going on the Home page for sure. :D

Calvin did approach me about being a moderator some time ago, but we decided that it was better if I just remained as a member-at-large and just kept doing my thing. I'm happier this way, but thanks for the invite and the compliment, nonetheless. :)

Brian Charlesworth - 6 years, 1 month ago

FIFA World Cup earthquake

The answer is 2.

6 solutions

Didn't even solve it!

One Advice

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