Math game (fun quizzes) ^^

[Kyriakos]

nhytgbvfeco2 never wants to reply, unless he is given a cookie But I am interested (question in post 20).

Anyway, here is a new problem:

You are fighting in the Trojan war, as one of the Greeks. High above you, Aphrodite -who famously sided with the Trojans, due to the apple - appears. Diomedes is planning to strike her, but he is on the opposite side to you (=Aphrodite is between you, high up). Your angle of elevation (the angle in which you raise your head to see Aphrodite) is 20 degrees. Diomedes' angle is 25 degrees. You do know that the distance between you and Diomedes is 500 meters, so the question is: how high above you is Aphrodite?

[nhytgbvfeco2]

nhytgbvfeco2 never wants to reply, unless he is given a cookie But I am interested (question in post 20).

Oh, real classy calling a Jew greedy
I wouldn't consider my knowledge of mathematics wholly insufficient to answer the question in your previous post. I think if the first cut is too close to one of the edges then the next cut has to be in the middle of what's left for it to be a viable triangle, but I'd have no idea on how to calculate the odds on any of that.

Anyway, here is a new problem:

You are fighting in the Trojan war, as one of the Greeks.

That's quite the transition

High above you, Aphrodite -who famously sided with the Trojans, due to the apple - appears. Diomedes is planning to strike her, but he is on the opposite side to you (=Aphrodite is between you, high up). Your angle of elevation (the angle in which you raise your head to see Aphrodite) is 20 degrees. Diomedes' angle is 25 degrees. You do know that the distance between you and Diomedes is 500 meters, so the question is: how high above you is Aphrodite?

Just to clarify, is she in the middle between us, or just somewhere between us?

[Kyriakos]

Depends on what you mean: she is in the same line, just higher. She is NOT at the same distance from you and Diomedes (otherwise you'd both be looking at the same angle). I guess I should have made the fight take place in a deep schism in the mountain (furthermore, due to the relative distance, you can take the two people pretty much as being at the same height as the ground - they aren't tens of meters tall ^^)

[nhytgbvfeco2]

Here is my professional solution:

[Kyriakos]

I don't have more cookies,currently :/
Correct. Other methods could be used, but tangent is the briefest - eg law of sines is another method.

Do you plan to at least look for a puzzle this time?
(btw, your pixel drawing looks nasty :o )

[nhytgbvfeco2]

I don't have more cookies,currently :/
Correct. Other methods could be used, but tangent is the briefest - eg law of sines is another method.

Do you plan to at least look for a puzzle this time?

Hey, I provided one last time. It might not have been very good, but I did!

(btw, your pixel drawing looks nasty :o )

Now that's rude and uncalled for

Alright then:
2_1_6_6 = 48
Fill in with: + - : *
Make the equasion work.

[Septentrionalis]

Now that's rude and uncalled for

Given what an absolute beauty it is, Kyriakos is way out of line.

[nhytgbvfeco2]

Given what an absolute beauty it is, Kyriakos is way out of line.

Thank you. It feels good to be appreciated.

[Kyriakos]

Thank you. It feels good to be appreciated.

I am starting to think you work in a matrix factory or something
I got the solution, but not with any generalized method, so pls share that (I suppose one with... matrices? ^^)

[nhytgbvfeco2]

You.. jusrt want me to share the answer without answering it yourself? Nah man, post an answer xD
No matrices or anything complicated involved. Though be aware, it doesn't follow PEMBAS, just goes by order of left to right.

[Kyriakos]

Eh, I mean intuitively one can find that it is [2(1)+6]6. But I am sure there are generalized methods (this wouldn't work easily if you had... 100 numbers ).
I did spend 5 min trying to work out a general method on my own, using the fact this has 4(4)4 combinations, but although I couldn't, I am sure such do exist, and matrices appear linked very directly (a second matrix would be the directions).
But unlike you, I haven't studied them at all ^^

I also suspect it might be possible to work it out using nested functions and a floating C value, but again couldn't do it in a few min and gave up.

[nhytgbvfeco2]

That is correct
I don't know if there's any fancy shmansy way to solve it, i think it's just clear you need to multiply by at least one of the sixes, then you just figure out how to get an 8,

[Kyriakos]

"Fancy shmansy" is the very reason math exists ^^ Otherwise you'd be stuck with the Babylonians and non-theorem-based math.
Or at least that used to be the prevalent view - that proof-based math originated with ancient Greece in the west. The opposing view is (afaik) still not well-documented, so might as well be false

[Kyriakos]

Anyway...
Here is a different riddle (though based again on geometry and simple procedures there; just requires a bit of imagination if you have never seen it...)
Construct a line which would be equal to the square root of another line You only have ruler and compass at your disposal.

(obviously you can draw any line you want, as that to find the square root of, so that first step is merely to draw any line; BUT your way of finding the square root should be workable for any line length anyone can pick)

[nhytgbvfeco2]

[Kyriakos]

Come on, man, that's a crap single case solution
Obviously I mean to present a method which works for any line length ^^

It's actually a very neat bit of geometry. And (in the end), that if (something)=sqr1, then x=sqroriginal line, is used

[Kyriakos]

Anyway, let me give you a hint, using the case nhyt did



So, the general solution should have to do with the above, but altered in a way. Can you find how? ^^

[nhytgbvfeco2]

Well, I did figure it probably has something to do with radius's and circles and whatnot, as I know the circle's.. circumferance? has to do with R^2. Well, also because what on earth else would you need the compass for.
But anyway I've already given a good solution

[Kyriakos]

No you haven't and you should feel ashamed of yourself for that post
Anyway, I will give it a day or two and then post the answer if no one gets it by then.

[Kyriakos]

Two days later, as promised, here is the solution:



The first part(above) is in Greek - but so was the Ptolemaic Empire's translation of the Old Testament, nhy -, but it is just the Thales1 (above angle is 90 degrees) followed by applying the pyth theorem to show that in those conditions (similar triangles) you end up with the height being sqr BD(sqrDC).
Of course this doesn't just mean you can replace (eg BD) with 1, and then measure the height, because DB=1 has to satisfy the equation (this is the second part of the construction) BC=f(DC)=DC+1 => your BC has to be measured by 1 or vice-versa.

You can also take this to be a proof for the (obvious) fact that there is only one square root of each number, for there is only one x for every f(x)=x+C.

Open floor