Shamekeeper12 In reply to Frostiln [2018-03-25 06:44:36 +0000 UTC]

Thank you! Means a lot. ^^

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Shamekeeper12 In reply to ArbitraryRenaissance [2018-03-25 06:43:52 +0000 UTC]

Thanks, Arbitrary!

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Shamekeeper12 In reply to ArbitraryRenaissance [2017-10-06 02:21:24 +0000 UTC]

Oh nice! I'll definitely take a look. Sorry for the late reply—haven't quite found enough free time these past few weeks.

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Shamekeeper12 In reply to ArbitraryRenaissance [2017-04-01 05:31:49 +0000 UTC]

Ha, if it was that much fun for you, I have to look into it!

That said, I have my own challenge for you.
I stumbled across this one on YouTube, so you might've heard of it before.



Jasper's Wobbly Desk

Jasper the rabbit has a square desk with four legs of equal length.
Problem is, it's wobbly since the ground in his burrow is uneven.

This means Jasper can't make his calculus
homework look all neat and tidy. Disaster!

Jasper, however, is a smart rabbit. He slowly rotates
his desk until all four legs are level. Amazing!

Challenge: Use the Intermediate Value Theorem to show
how Jasper could do this using less than a 90-degree turn.

(Intermediate Value Theorem: To get from A to B, you must first go through everything in between.)



As a side note, I'm almost through the first four chapters of Aura of Storms, and will be posting comments in a few days. ^^

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ArbitraryRenaissance In reply to Shamekeeper12 [2017-04-02 01:48:11 +0000 UTC]

We must first assume that the wobbling is because of the unevenness in the terrain or ground rather than a disparity in heights between the legs of the table.  If the ground is perfectly planar and the prongs are uneven, then there is no rotation of the table that will map all four points to the plane.

With that being said, we may now safely assume that the four legs of the table are planar.  We will also assume that they are the vertices of a square in this plane, otherwise I suspect the necessary rotation may need to be more than 90 degrees.  (It also might not; my method for solving this simply isn't strong enough to accommodate for a more general form of the question).

If the legs are the vertices of a square, then they exhibit rotational symmetry of 90 degrees.  That is, if we rotate the table by 90 degrees, then the footprints will map onto themselves.

Label the surface points of the four legs 1, 2, 3, and 4 in a counterclockwise fashion.  For some angle a, we will define four functions: f_1(a), f_2(a), f_3(a), and f_4(a).  The function f_i(a) will equal the vertical distance from leg i to the ground.  Due to the rotational symmetry of the legs, we know that f_i(a-90) = f_i-1(a) for all a and i, and where i-1 is taken modulo 4.  Additionally, these functions are all continuous.

We also know that planar stability will force at least three of these functions to equal 0 for any given a.  Now, by the pigeonhole principle, there must be two opposite legs that are equal to 0.  WLOG, let us assume that at a=0, these are legs 1 and 3.  Now let us define F(a) = |f_1(a) - f_2(a)| and G(a) = |f_3(a) - f_4(a)|.  Then F(0) = 0 and G(0) = b for some non-negative real number b.  Likewise, due to the rotational symmetry, F(90) = b and G(90) = 0.  Finally, these functions are both continuous, since they're the difference of two continuous functions.  Let us now define H(a) = F(a) - G(a).  Then H(a) is also continuous.  Additionally, H(0) = F(0) - G(0) = 0 - b = -b, and H(90) = F(90) - G(90) = b - 0 = b.  Since 0 ϵ [-b, b], it then follows by the IVT that there exists an angle R such that H(R) = 0.

Note now that 0 = H(R) = F(R) - G(R).  At least one of F(R) or G(R) is equal to zero, since they are defined as the differences of our original functions -- three of which must amount to zero.  If F(R) = 0, then 0 = H(R) = 0 - G(R) which implies G(R) = 0.  Similarly, if G(R) = 0, then F(R) = 0.  Therefore, both G(R) and F(R) equal zero.

But this implies that all of our original functions are equal to zero as well.  By the same argument, f_1(R) = 0 implies f_2(R) = 0 and vice versa, and f_3(R) = 0 implies f_4(R) = 0 and vice versa.  Hence, since at least three of our functions amount to zero at R, and since F and G amount to zero at R, that forces the fourth function to go to zero at R as well.  This implies that the table is stable at R, and so we're done.  Q.E.D.

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Shamekeeper12 In reply to ArbitraryRenaissance [2017-04-02 04:23:20 +0000 UTC]

Absolutely brilliant! Your level of understanding here is way beyond mine for sure. I was completely clueless when I first saw this problem—even after I learned the solution. I knew it had something to do with IVT, but seeing your proof explicitly use "0 ϵ [-b, b]" and relating these altitudes to functions made so much more intuitive sense. ^^

Again, excellent work! —and thanks a bundle!

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ArbitraryRenaissance In reply to Shamekeeper12 [2017-04-02 04:50:28 +0000 UTC]

You're welcome!  I actually heard of this problem before you showed it to me, but I never went through the steps of formally solving it myself.  Also, I've taken both real analysis and discrete math, so I have a bit of a background in proof writing and an intuition for continuous functions.  This is nothing brilliant on my end.  Nonetheless, I'm glad you got something out of my solution, and I thank you for your kind words.

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Shamekeeper12 In reply to ArbitraryRenaissance [2017-03-21 23:36:51 +0000 UTC]

Hey, thanks a lot!—and thanks for the presents. ^^
I'll take a closer look at both your puzzles in my off hours. And as for the olympiad contest, I'll consider it! 

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Shamekeeper12 In reply to Frostiln [2017-03-21 23:36:55 +0000 UTC]

Thanks a bundle! Means a lot. ^^

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Shamekeeper12 In reply to ArbitraryRenaissance [2017-04-02 04:51:14 +0000 UTC]

At first, I tried writing a program to brute-force this problem (I initially figured solving algebraically would be impossible with so many variables), but after a couple iterations, I began to realize what each setting actually did.

Let Δω be the net change in weed population.

Setting 3: Δω = +9
Setting 5: Δω = +12
Setting 14: Δω = -6
Setting 17: Δω = -15

I figured if I could get the weed population to be divisible by one of these Δw's, I'd have the solution! (although, I did realize later that it's more likely for the (ω - someSetting) to be divisible by Δω)

Furthermore, I saw that all Δω's were divisible by 3. However, not matter what I tried, I was always off by 1 or 2 weeds in getting everything to "line up" so-to-speak—which leads us to the following: 

Let n be any integer.

For any setting combination, the net change can be expressed as 3n
since all Δω's are divisible by 3.

To be solvable, 3, 5, 14, or 17 weeds must be reachable by the expression 100 + 3n

If any one of 5, 14, or 17 is reachable, the rest are also, since 14 and 17 can be expressed as 5 + 3n

No solution since 100 + 3n cannot equal 3 or 5 unless n is a non-integer.


To be honest, I'm not sure if these statements hold up since I'm sure there is some solution that I just can't see.

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ArbitraryRenaissance In reply to Shamekeeper12 [2017-04-02 05:15:50 +0000 UTC]

Yep, you have the right idea, and nothing you said was wrong.  The spirit of this problem is utilizing modular arithmetic, and the conclusion that you came to -- plus the justification behind it -- is absolutely correct.  So as far as I'm concerned, you're completely right.

I actually recently wrote up my solution in a journal the other day.  I go in a lot more detail just to make the answer clearer, but it utilizes all of the same ideas that you proposed here.  arbitraryrenaissance.deviantar…

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Shamekeeper12 In reply to ArbitraryRenaissance [2017-04-11 19:26:42 +0000 UTC]

Oh flip! It was right there, wasn't it? xD Either way, glad I was on the right track.

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Shamekeeper12 In reply to ArbitraryRenaissance [2016-12-23 03:01:15 +0000 UTC]

How's this?


Definition of Incircle:
1. Tangent to all triangle's sides
2. Centered @ intersection of all angle bisectors

sta.sh/0sxb2ejih9r

O is collinear with AI     Euler’s Line
EI ≅ DI                        definition of circle (radius)
AI bisects ∠DIE             definition of incircle (2)
ΔOIE ≅ ΔOID                SSA
∠OEI ≅ ∠ODI                corresponding angles

AC ⟂ EI & AB ⟂ DI        definition of incircle (1)
∴   m∠OEI + m∠CEO = 90° = m∠ODI + m∠BDO
∴   m∠CEO = 90° - m∠OEI = 90° - m∠ODI = m∠BDO
∴   ∠CEO ≅ ∠BDO   that is,   ∠OEC ≅ ∠ODB

Thus, it has been demonstrated.


Now, I'm not sure about the Euler's Line statement since Euler's Line applies to a single triangle. This figure involves triangles ABC and BIC.

I'm not quite sure how else to prove this statement though, so I'll roll with it. (^^ゞ

It's been a while since I've done geometric proofs (about three—four years). Admittedly, I do geometric constructions more to appreciate mathematical phenomena than prove them. xD

That said, I'm pretty sure everything else holds up—unless I read the question wrong again. xD

For a cleaner proof, I made a GoogleDoc: docs.google.com/document/d/1Ab…

Credit where it's due, figure made using SolidWorks 2015.

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ArbitraryRenaissance In reply to Shamekeeper12 [2016-12-23 06:53:38 +0000 UTC]

I'm afraid the Euler line of triangle ABC is not the line that passes through A and I.  In fact, the Euler Line in general does not pass through the incenter of a triangle.  However, it is true that O, A, and I are collinear.  If you're interested, here's my proof of it on good paper.  You'll definitely want a pen and paper with you as you read through it:

Proof

And here's the diagram I was looking at:

Diagram

Let me know if you have any questions.

(Credit where it's due, the diagram was made with GeoGebra.)

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Shamekeeper12 In reply to ArbitraryRenaissance [2016-12-24 01:43:42 +0000 UTC]

Then ∠BIC = 90 + α ∠BIC = 90 + β Why? I see neither a right angle, nor understand how α or β can be used in this regard.
I'm sure this has to do with arc angles, but it's unapparent how this is true.

∠OBC = α'∠IBC = β
∠OBC + ∠ICB = α' + βThere seems to be contradiction.
Assuming these to be true, I can substitute:

∠OBC + ∠ICB = α' + β
α' + ∠ICB = α' + β
∠ICB = β
∠ICB = ∠IBC

However, we already said that:

∠IBC = β

and

∠ICB = γ

with the understanding that, for most cases,

β != γ (not equal)

Am I missing something here?

That said, I'm pretty comfortable with everything else. I think it's quite interesting. ^^

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ArbitraryRenaissance In reply to Shamekeeper12 [2016-12-24 02:16:32 +0000 UTC]

First of all, I made a pretty bad typo in the second line you mentioned.  It should say∠BIO = ∠OIB == ∠OBI = ∠OBC + ∠CBI = α' + β.I should have caught that in proofreading.  Sorry.

As for why ∠BIC = 90 + α and ∠AIC = 90 + β, I'll leave that as a significantly easier exercise for you.  Consider, in general, triangle ABC with angles 2α, 2β, and 2γ and incenter I, and see if you can write ∠BIC in terms of α, β, and γ.  (HINT: consider all of the angles in triangle BIC.)

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Shamekeeper12 In reply to ArbitraryRenaissance [2016-12-24 06:32:19 +0000 UTC]

Ah, I got it!
∠BIC + β + γ = 180

Let β + γ = u, & ∠BIC = x:
∴   x + u = 180

2α + 2β + 2γ = 180
2(α + β + γ) = 2(90)
α + u = 90
u = 90 - α

Recall: x + u = 180
x = 180 - u
x = 180 - (90 - α)
x = 180 - 90 + α
x = 90 + α = ∠BIC

Corresponding process shows ∠AIC = 90 + β

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ArbitraryRenaissance In reply to Shamekeeper12 [2016-12-24 07:30:47 +0000 UTC]

Excellent!  Ain't that a nifty little property?

I'm personally more a fan of geometric proofs than you are.  In fact, the proofs kind of stem off of my interest in the constructions themselves.  I'll create a geometric structure with circles and triangles and what have you, and I'll notice, "Hey, these are collinear" or "Hey, these are always perpendicular" or "Oh, hey, that's a cyclic quadrilateral" and then I start to wonder why that's the case.  And to work out why it's the case, you then get into proof thinking mode.  Chase these angles, add up these side lengths, apply these theorems, and bam! you've got the result you wanted.

Just for fun, here's another cool property.  I won't ask you to prove it, but if it interests you, you may end up working it out yourself anyway:

Consider triangle ABC.  The perpendicular feet of A, B, and C to their opposite sides are points D, E, and F respectively.  (In other words, D is on the line BC such that AD is perpendicular to BC, and likewise for the other points.)  As you may know, the three altitudes of triangle ABC intersect at a point P known as the orthocenter of the triangle.  As it turns out, this point P is also the incenter to triangle DEF!

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Shamekeeper12 In reply to ArbitraryRenaissance [2016-12-27 00:17:36 +0000 UTC]

Say, that is pretty nifty!

I've often thought the same things myself; 'why is this true?', 'is this always true?' and the like. But as I've said, it's been a long time since I've made geometric proofs, so I've forgotten a lot of the tools that I could otherwise use.

While I could use something like GeoGebra or SolidWorks all the time, I relish the traditional practice of compass and straightedge. I was practicing constructing tangents to a circle when I realized some interesting patterns. It appears the center of any hypotenuse is always the triangle's circumcenter, and apparently this is an actual theorem! That's what I love about constructions. You teach yourself.

You're right in that I use these more for practical applications—such as working out a challenging feature in a working drawing—but I'm still a huge fan of the proofs themselves—given
they're not needlessly verbose. xD That said, I should probs look into a proof for this one some other time.

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ArbitraryRenaissance In reply to Shamekeeper12 [2016-12-27 00:33:03 +0000 UTC]

By the way, the geometric proof that you produced is less formal of a proof and more of an outline.  The statement/reason columns that you're taught in grade school actually transform into paragraphs simply stating what's going on with less formality to it.  Here's an example of a problem and a solution: www.artofproblemsolving.com/wi…

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Shamekeeper12 In reply to ArbitraryRenaissance [2016-12-27 00:51:10 +0000 UTC]

So it is! Interesting, I wish they had told us the true nature of using columns, but I guess that's arbitrary for grade school learners. I think, what matters is less the form and more the meaning of what's there—since truth is truth, formal or not. Though, it would be a satisfying exercise to write one formally. I'm curious, have you ever thought to be a teacher? I think you'd be a good fit. ^^

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ArbitraryRenaissance In reply to Shamekeeper12 [2016-12-27 01:35:08 +0000 UTC]

I'm terrible at verbal communication, so I'd do better to do something like write a textbook.  (Plus, from what I've seen, teaching is stressful as hell, so I'm not particularly eager to do it for a living.)

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Shamekeeper12 In reply to ArbitraryRenaissance [2016-12-27 17:15:20 +0000 UTC]

Indeed, it really is stressful—especially with some crowds. 
But yeah, thanks for the challenge! I might have to hand you a puzzle of my own one of these days.

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Shamekeeper12 In reply to ArbitraryRenaissance [2016-12-22 19:44:08 +0000 UTC]

Hmm...this one's got me. I'll have to answer at a later time.
Thanks for the watch, by the way. ^^

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Shamekeeper12 In reply to Meemie7 [2015-10-14 05:35:26 +0000 UTC]

AH AMBUSH *melts*

Unfortunately, my attempt to reciprocate has been denied because 11pm.
Hugs will become contagious at a later date.

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Meemie7 In reply to Shamekeeper12 [2015-10-16 00:39:05 +0000 UTC]

Reciprocate when you are well rested and when you feel like it!

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Shamekeeper12 In reply to Frostiln [2015-08-25 00:17:19 +0000 UTC]

Well, I don't see why not. ^^ It might just take a while on my end, but that shouldn't be too much of a problem. How do you suggest we go about it? (・・?)

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Frostiln In reply to Shamekeeper12 [2015-08-25 00:22:02 +0000 UTC]

well, lets brainstorm a little together, that seems like a good start

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Shamekeeper12 In reply to Frostiln [2015-08-25 01:19:04 +0000 UTC]

Right! I hope you're familiar with dA's Note system? Maybe I'll try to send you one tomorrow, and let's talk it out?

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Frostiln In reply to Shamekeeper12 [2015-08-25 01:24:54 +0000 UTC]

kk sounds good!

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Shamekeeper12 In reply to RaoKurai [2015-08-23 03:10:10 +0000 UTC]

No problem! Figured I may as well, since you'll be posting frequently. ^^

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Shamekeeper12 In reply to DuskLugia [2015-08-22 14:27:34 +0000 UTC]

No problem!

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Shamekeeper12 In reply to Frostiln [2015-08-08 05:29:56 +0000 UTC]

Wow, that's... that's really kind of you to ask. Sure! Why not? X3

I haven't been pursuing cameos really, but I guess it wouldn't hurt to mention that in the App description.

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Frostiln In reply to Shamekeeper12 [2015-08-08 14:57:00 +0000 UTC]

Thanks! Your team just looked so fun to draw!

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Shamekeeper12 In reply to Frostiln [2015-07-22 23:33:58 +0000 UTC]

No problem. ^^

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