### Small Geometry Puzzle

Feb. 13th, 2015 11:11 pm**ch3cooh**

Challenge: write up your initial instinct, then try to prove it and let me know what happens. :D

I really like this puzzle. And I'm sure that there's a simpler proof than the one I'm using right now.

Puzzle:

Solution:

The orange is

Proof:

Step 1: The Orange circle is exactly 1/4 of the big circle. Proof by picture:(Draw another equlateral triangle inscribed in the circle, with a 180 degree rotation relative to the original triangle. These two triangles overlap in a regular hexagon that can be split into 6 triangles like a pie. Each of these triangles is the same size as any of the 6 points of the star now inscribed in the circle. So it is clear that the radius of the orange circle is half the radius of the large circle. Since A = pi*r^2, this means that the area of the orange circle is 1/4 the area of the whole circle.

Step 2: Blue area is exactly 1/3 of the remaining area in the large circle that is not part of the orange circle. Proof by picture:The remaining area has three semetrical components.

Step 3: Since the orange circle is 1/4 of the large circle, the remaining interior of the large circle is 3/4 of the area. The blue area is 1/3 of that, and 1/3 of 3/4 is 1/4. So both the orange region and the combined two blue regions are 1/4 the area of the whole circle.

Boom! Done!

Using this to draw a very pretty infinite summation picture. :)

1/4 + 1/16 + 1/64 + ... 1/4^n as n approaches infinity = 1/3

I really like this puzzle. And I'm sure that there's a simpler proof than the one I'm using right now.

Puzzle:

Solution:

The orange is

**exactly the same**area as the sum of the two blue areas.Proof:

Step 1: The Orange circle is exactly 1/4 of the big circle. Proof by picture:(Draw another equlateral triangle inscribed in the circle, with a 180 degree rotation relative to the original triangle. These two triangles overlap in a regular hexagon that can be split into 6 triangles like a pie. Each of these triangles is the same size as any of the 6 points of the star now inscribed in the circle. So it is clear that the radius of the orange circle is half the radius of the large circle. Since A = pi*r^2, this means that the area of the orange circle is 1/4 the area of the whole circle.

Step 2: Blue area is exactly 1/3 of the remaining area in the large circle that is not part of the orange circle. Proof by picture:The remaining area has three semetrical components.

Step 3: Since the orange circle is 1/4 of the large circle, the remaining interior of the large circle is 3/4 of the area. The blue area is 1/3 of that, and 1/3 of 3/4 is 1/4. So both the orange region and the combined two blue regions are 1/4 the area of the whole circle.

Boom! Done!

Using this to draw a very pretty infinite summation picture. :)

1/4 + 1/16 + 1/64 + ... 1/4^n as n approaches infinity = 1/3

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Date: 2015-02-26 03:32 am (UTC)greensword.livejournal.comIf the triangle is equilateral, then each of the blue regions is equal to the equivalently-shaped white regions. The blue area is therefore 1/3rd of the area of the larger circle, excluding the area of the orange circle.

The question then becomes, how big is the orange circle? If it is exactly 1/4 the size of the larger circle, then it will be exactly the same size as the blue area. If it is larger than 1/4 of the large circle, it will be bigger than the blue area, and if smaller, then smaller. :P

Okay, here's where I'm running into difficulty. How can I tell the size of the smaller circle? Let's call the radius of the larger triangle 2. That makes the area of the larger circle 4pi. That means that we can figure out the area of the triangle by splitting it up into six triangles which use the radius as their hypotenuse. Those six triangles will be 30-60-90 triangles, since they'll have 90 degree angles along the actual sides of the large triangle, and 60 degree angles around the center of the circle (360/6 being of course sixty). I had to look the formula up, but if the hypotenuse is 2 then the other sides are 1 and √3. We multiply them to get the area of the mini triangle of 1√3/2. That makes the total area of the triangle 6√3/2 or 3√3. The triangle is therefore about 2/3rds the total area of the larger circle.

* I actually started with radius 1 but changed it later for ease of calculating.

There is probably more math I can do to figure out the size of the inner circle but I don't know it offhand. :/ If the inner circle is a similar proportion to the triangle as the triangle was to the outer circle, then it is 4/9thsish-sized, which is way more than 1/4 the size of the full circle. Even if it's half of the triangle, it's more than 1/4 the size of the full circle. So I'm going to guess that the orange circle is bigger than the blue parts. But I haven't proved anything and I could believe the circle is more like 1/3 of the triangle and my brain hurts, so I'm going to stop typing and see what the answer is. :)

And my brain hurts. So I'm going to give up and see what the answer is.

## no subject

Date: 2015-02-26 03:37 am (UTC)greensword.livejournal.com## no subject

Date: 2015-03-03 07:37 am (UTC)ch3cooh.livejournal.comIn the site I linked to below, I haven't yet figured out how to construct a square in 8 moves, and I've tried at it for more than an hour ;-P Also, for the proof above, as I mentioned, there's probably an easier way to see the 1/4 relationship than the star I thought of, but I don't know it. Who knows, maybe squares are involved somehow. :)

Construction game:

http://sciencevsmagic.net/geo/

Or, this is a simpler but less 'visually clean' version:

http://euclidthegame.com/

Squares are so hard! My TA told me today that there's a compass & straight edge method for, given a rectangle of any area (even one with an irrational area) constructing a square that has the same area. Still trying to figure it out...

Edited Date: 2015-03-03 09:02 am (UTC)## no subject

Date: 2015-03-05 03:30 pm (UTC)cesium12.livejournal.comGb pbafgehpg gur fvqr bs lbhe fdhner, lbh arrq gur trbzrgevp zrna bs gur fvqrf bs lbhe erpgnatyr.

Va nal pvepyr, vs pubeqf no naq pq vagrefrpg ng r, gura nr gvzrf ro rdhnyf pr gvzrf rq. V'z fher gurer'f n anzr sbe guvf cebcregl ohg V qba'g erzrzore vg.

Gurersber, sbe lbhe erpgnatyr jvgu fvqrf k naq l, pbafgehpg gur pvepyr jvgu qvnzrgre k cyhf l, naq qenj n crecraqvphyne pubeq guebhtu vg yvxr n puevfgvna pebff. Gur arj pubeq vf qvivqrq vagb gjb rdhny frtzragf bs yratgu fdeg bs k gvzrf l. Qbar!

## no subject

Date: 2015-03-03 07:35 am (UTC)ch3cooh.livejournal.comComponents of problem solving - like picking which length to set as the 'unit' of measurement are relevant not only in math, right? Hmm... in fact, it strikes me as something that could even be relevant to analysis approaches in the humanities. :)

And in a lot of geometry problems, your first strategy would have worked out ---

"A is equal to B, which is equal to C, which is greater than D, so therefore A is greater than D as well..." or something like that.

I like turning this proof around into a construction too. Aka, now you know one way to draw a smaller circle inside a larger circle so that they have the same center and the small circle is exactly 1/4 the area of the large circle. Although... maybe there's an easier way than inscribing a triangle inside the circle and then a circle inside the triangle. ;-P

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Date: 2015-03-03 02:24 pm (UTC)greensword.livejournal.comAlthough... maybe there's an easier way than inscribing a triangle inside the circle and then a circle inside the triangle. ;-PWell, along the way we proved that the smaller circle's radius is half the size of the larger one's. So to construct it without the triangle, you'd just make a circle, draw a radius, and then draw the smaller circle so its perimeter intersected that first radius exactly half way.

Components of problem solving - like picking which length to set as the 'unit' of measurement are relevant not only in math, right?Oh yeah. "What do I measure and how?" is perhaps the biggest practical recurring problem in my life.