# Math forum game ^^

#### Kyriakos

##### Creator
Worth trying, I think ^^
If the thread has to be moved to science, it's ok, though I am of the view it will have more of a purpose in this subforum which is strictly about games/quizzes Goes without saying (or does it) that this isn't a math homework thread; meant to be used for fun little quizzes, of the type hopefully the first post can be an example of.

Also I have to note this isn't a quiz I came up with myself (you can do that, if you want to). I found it on youtube. I am providing, in the spoiler, a solution to it (but there are more than just one or two solutions).

Have fun! If you get stuck, or just wish to see an answer, click on the spoiler:
Spoiler : Summoning a few possible players: @Takhisis @Grendeldef @red_elk @Samson @Synobun @The_J @El_Machinae @Perfection

• Henri Christophe

#### Henri Christophe

##### L'empereur
If you like Mathematic as a hobby, I would recommend you this book:
Is written by a Brazilian, but make you fell in Arabia with a lot of mathematics envolved.

• Kyriakos

#### Aiken_Drumn

##### Deity
Supporter
Likely I misunderstand the question. But a 3 -sided object is a triangle.. so 100%?

• Kyriakos

#### Kyriakos

##### Creator
Likely I misunderstand the question. But a 3 -sided object is a triangle.. so 100%?
Yes, but not all lengths of three parts will form a triangle That is because in a triangle the sum of length of (any) two sides has to be smaller than the length of the third (otherwise there won't be enough to close the thing up).

#### Kyriakos

##### Creator
Ok, a year has passed ( ), so let's try a new quiz. This time easier. To win, calculate x. "O" is the center of the diameter. If stuck or just interested, look in the spoiler for (one of possible methods) the answer ^^

Spoiler : (note: once again, this isn't a quiz I came up with, but at least my solution is -imo- more informative/inclusive of theorems than what its creator used )

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#### Kyriakos

##### Creator
I wish there was some participation, but maybe this time I can not post the answer and see if anyone is lured in ^^ Besides, the above result is historically significant, because it is the geometric way of calculating a square root=> Α,Γ are the edge points of the diameter of a circle, with E on its periphery (Δ is the center of the circle) @Samson ,@red_elk

#### Kyriakos

##### Creator
^^ Damn it. Enticement failed.
Anyway, it is a neat result, since it allows you to study what square roots are, geometrically. It can also be generalized to n roots.
Trivia: the result is a special case of the height of the triangle (EB) equaling sqrxsqry, when (of course) you have an angle looking at a diameter.

#### Kyriakos

##### Creator
Ok, 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? (note: assume Aphrodite is at the same line as you and Diomedes, just a lot higher).

Since this is a Civ site, I suppose that if even this doesn't lead to any interest, the thread simply had no chance from the start ^^

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#### Harv

##### Emperor
Hi! I cannot draw a picture from your description in Post 8. Since I do not feel like doing work this early in the morning, I will take a look at what's on Post 6. Your post should draw people eventually, because the forum draws from very smart, mathematical thinkers. I wonder if the forums are just less active.

Added later. I can't draw very good pictures. Text only. Here is what I have.
Spoiler :

I ended up drawing 4 Triangles. Here is a text version of them. I will put facts derived from the triangles in bold.
My knowledge of math language and convention is not good. I will make it a point to make fun of it so it is very obvious.

Triangle 1:
Exactly as per the original problem. I changed your Gamma to G, because of my keyboard. I do not know how to make a Gamma. My definitions are as follows:
EAG is a right-angle triangle. The point B forms two more right angle triangles. (terminology is off here)
The length EB you are solving for I defined as h.
Alpha is associated with A. It is the angle of EAG.
Gamma is associated with G. It is the angle of EGA.
Gamma = 90 degrees - Alpha.

Triangle 2:
This is Triangle 1 redrawn. I redrew all of the triangles so that the right angle is in the bottom left corner like the origin of Cartesian Coordinates.
So E is at (0,0). A is at (0,g). G is at (a,0).
I defined g as the length of AE and a as the length of EG. I used small letters as lengths opposite of the big letters.
I redrew it, because rotating the triangle in my head and thinking is hard. It is easier to look at all of these on one piece of paper.
A
E----G

Triangle 3:
This is EBG
E
B----G
I defined h as the length of BE.
I defined y as the length of BG.

Triangle 4:
This is ABE
A
B----E
The length h shows up again in line BE.
The length AB is defined as 1.

All of these facts are very easily drawn on the original diagram. However, it is so pretty when drawn on a piece of paper. All triangles are k-k-k-congruent. The angles are all the same. 90 degrees, Alpha, Gamma. Triangle 3 EBG is h times bigger than Triangle 4 ABE. Therefore y=h² and h is the square root of y.

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#### Kyriakos

##### Creator
Thanks, @Harv , sorry for not being aware there was a reply until now!
Hm, illustrating it would be of use (so pls do) Because you redrew the triangles, starting from 0,0, and now I can't quickly follow which letter corresponds to which, and/or if other things changed. Particularly this causes an issue with your final sentence: "Triangle EBG is h times bigger than triangle ABE", but in my original image triangle EBG is 1/2hBG, and triangle ABE is 1/2h, so BG times larger.
The triangles are congruent, and for arbitrary values of x you'd get height=sqrxsqry (which can be established by the system of equations for the height in those two triangles along with the original one which they form together).
The important thing with x=1 is that this is the geometric way to construct a square root of a length (here y). Of course it also has to satisfy AG=F(BG)=BG+1, ie BG,1 have to be measured together. It's also a proof for the (obvious) fact that there can only be one square root for each number, since F(x)=x+C is linear.

Here's one solution, along those lines (just with even more Greek in it; also mentions the simple case for AD=2,DG=4) Last edited:

#### Harv

##### Emperor
The only issue I have with the Greek letters is I do not know how to make them in ASCII and I would do well to look it up. I have a similar issue with Spanish and the extra modified letters and accents. It appears I am getting a lesson in Linguistics as well as Mathematics. Here is what I get:

Ta trigona = The triangle BAG and ADG
einal omoia... not certain and too lazy to try to Google Translate, probably means are congruent
dioti exoun duo omiez goniez

I'm going to write it all down along with the diagram, because this is the new challenge!
(Added - This is fun trying to pick out words from Greek letters! I will need a couple of days. Or maybe I copy and print the diagram. Done. I will also do well to look up the Greek function words, prepositions and function words.
I picked out ortho = orthagonal? diametro, perifereia

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• Kyriakos

#### Kyriakos

##### Creator
^^ Translation:
Triangles BAG and ADG are similar, because they have two equal angles (an orthogonal and theta). The angle a (inferred: from point A) is a right angle because it is opposite a diameter of a circle in which point A is on the periphery. (note: this is Thales' theorem of right angle).
The AG/BG=DG/AG follows from the corresponding side analogies in similar triangles.
The only other Greek term is in the last sentence, where "riza" just means "(square) root".

#### Kyriakos

##### Creator
Ok, here is a new one... If the two blue areas are parts of semicircles with their diameter being the two sides of the orthogonal triangle, show that the red area equals the blue areas Last edited:
• Harv

#### Harv

##### Emperor
^^I have not started yet and looked at it for a single minute. Is this a graphical proof of Pythagoras Theorem?

• Kyriakos

#### Kyriakos

##### Creator
^^I have not started yet and looked at it for a single minute. Is this a graphical proof of Pythagoras Theorem?
No, it's the other way around: it follows from the pythagorean theorem #### Harv

##### Emperor
No, it's the other way around: it follows from the pythagorean theorem Okay, vamos! This will test if I can formulate and articulate a thought clearly and correctly.
Spoiler :

The orthogonal triangle has sides a,b,c. I will define c as the longest and a as the shortest.
The area of a semicircle with radius r, I will use Ar = 0.125pi r^2.
The big semicircle with the triangle is Ac = 0.125pi c^2.
The two other semicircles are Aa = 0.125pi a^2 and Ab = 0.125pi b^2.
From Pythagorean Theorem, it follows that the area of the two smaller semicircles is equal to the large semicircle. That is, Aa + Ab = Ac.

The white area of the large semicircle not covered by the triangle (Area 0.125pi c^2 - 0.5ab) must be equal to the white shaded area of each blue semicircle not covered by the large semicircle.

If I define x as the white area enclosed by both the large semicircle C and the small semicircle A, and if I define y as the white area enclosed by both the large semicircle C and the small semicircle B....

The red area is equal to Ac - x - y.
The blue area is equal to Aa + Ab - x - y.
We already established that Aa + Ab = Ac.
Therefore the red and blue areas are equal.

• Kyriakos

#### Kyriakos

##### Creator
Not sure where the 0.125 comes from (arithmetic error and you took half of the 0.25side^2 times pi for the semicircle? But it doesn't matter what the percentage is, as long as it is kept to the same for all sides). Otherwise, yes, since the areas of circles are proportional to the square of the radius, they are also proportional to the square of double the radius, and likewise for the pyth theorem. Then, as you did, you just compare what is left.

#### Harv

##### Emperor
Not sure where the 0.125 comes from (arithmetic error and you took half of the 0.25side^2 times pi for the semicircle? But it doesn't matter what the percentage is, as long as it is kept to the same for all sides). Otherwise, yes, since the areas of circles are proportional to the square of the radius, they are also proportional to the square of double the radius, and likewise for the pyth theorem. Then, as you did, you just compare what is left.

For circle:
pi r^2 - r is radius
0.25 pi D^2 - D is diameter

For semicircle I took half of that and I was looking at diameters of semicircles. Does that make more sense?

• Kyriakos

#### Kyriakos

##### Creator
@Harv , if you want to, you can post the next riddle Replies
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