/sci/ - Science & Math

[Anonymous No.10925209]

I can't read this shit

[Anonymous No.10925221]

>>10925209
and I can't build muscle ; to each his lot

[Anonymous No.10925227]

>>10925209
Part of the problem is reading the problem.

[Anonymous No.10925251]

The octagon P1P2P3P4P5P6P7P8 is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon P1P3P5P7 is a square of area 5, and the polygon P2P4P6P8 is a rectangle of area 4, find the maximum possible area of the octagon.

[Anonymous No.10925289]

>>10925227
[math] \color{#781b86}{\tt{The~}}\color{#6a2c98}{\tt{octa}}\color{#5b3cab}{\tt{gon~P_1}}
\color{#4d4dbd}{\tt{P_2 P_3}}\color{#3e5dcf}{\tt{P_4 P_5}}\color{#4d74ba}{\tt{P_6 P_7}}\color{#5c8aa5}{\tt{P_8}}\color{#6ba18f}{\tt{~is~}}\color{#7ab77a}{\tt{ins}}\color{#88b96f}{\tt{crib}}
\color{#96ba64}{\tt{ed}}\color{#a3bc58}{\tt{~in}}\color{#b1bd4d}{\tt{~a~cir}}\color{#bcb749}{\tt{cle,}}
\color{#c8b244}{\tt{~with}}\color{#d3ac40}{\tt{~the}}\color{#dea63b}{\tt{~vert}}\color{#dd8535}{\tt{ices}} [/math]
[math] \color{#6234a1}{\tt{arou}}\color{#5444b4}{\tt{nd~th}}
\color{#4555c6}{\tt{e~cir}}\color{#4668c4}{\tt{cumf}}\color{#557faf}{\tt{eren}}
\color{#64959a}{\tt{ce}}\color{#73ac85}{\tt{~in}}\color{#81b874}{\tt{~the~gi}}
\color{#8fb969}{\tt{ven~o}}\color{#9cbb5e}{\tt{rder.~Gi}}\color{#aabc53}{\tt{ven}}\color{#b7ba4b}{\tt{~tha}}\color{#c2b446}{\tt{t~the}}
\color{#cdaf42}{\tt{~poly}}\color{#d8a93d}{\tt{gon~P_1}}\color{#de9538}{\tt{P_3}}\color{#dd7431}{\tt{P_5}}\color{#dc532b}{\tt{P_7}} [/math]
[math] \color{#4d4dbd}{\tt{is~a}}\color{#3e5dcf}{\tt{~squa}}\color{#4d74ba}{\tt{re~of~a}}\color{#5c8aa5}{\tt{rea~5,~}}\color{#6ba18f}{\tt{and~t}}\color{#7ab77a}{\tt{he~pol}}\color{#88b96f}{\tt{ygon~}}\color{#96ba64}{\tt{P_2}}\color{#a3bc58}{\tt{P_4}}\color{#b1bd4d}{\tt{P_6}}\color{#bcb749}{\tt{P_8}}\color{#c8b244}{\tt{~is~a~}}\color{#d3ac40}{\tt{recta}}\color{#dea63b}{\tt{ngle~}}\color{#dd8535}{\tt{of~}}\color{#dc642e}{\tt{area~4,~}}\color{#db4228}{\tt{find}} [/math]
[math] \color{#4668c4}{\tt{th}}\color{#557faf}{\tt{e~m}}\color{#64959a}{\tt{axi}}\color{#73ac85}{\tt{mum}}\color{#81b874}{\tt{~pos}}\color{#8fb969}{\tt{sib}}\color{#9cbb5e}{\tt{le~ar}}\color{#aabc53}{\tt{ea~of~}}\color{#b7ba4b}{\tt{the~}}\color{#c2b446}{\tt{octa}}\color{#cdaf42}{\tt{gon.}} [/math]
no kidding

[Anonymous No.10925304]

>>10925227
>>10925289
these putnam exams just get better every year huh

[Anonymous No.10925304,1]

Bump because I'm making your homework. And I don't want it to go to the trash.

[Anonymous No.10925304,2]

>>10925304
>"putnam exams"
topkek
I don't come here often, and I don't even know how to post an image, but be glad I'm trying to help. Seriously though...
How can you guys block at this? You can prove this with what you've learned concerning Pythagoras, Thales and triangles. I'd murder your exam so hard I'd wake greek geometers out of their tombs.
I'm european, if you wonder.
Here it goes... Don't present this to your exam, i tend to forget shit like proving an angle bisector is perpendicular to a square's side. You'll have to do the shit yourself. AND I'M DOING THIS ON MY OWN. NO SOURCES ASIDE FROM MY FUCKING BRAIN AND UNDISPUTABLE AXIOM-BASED PROPERTIES.

Let's first calculate using ONLY squares, not your shitty rectangles; P2P4P6P8 IS A SQUARE FOR NOW.
Let's rename P1P3P5P7 to ACEG and P2P4P6P8 to BDFH, because I defecate on that type of long notation.
Let's spit some facts straight first:
-Octagon ABCDEFGH is regular because it is composed of purely right angles , because the (which is a property of squares, that can be used to imply with octagons, not going to explain but it should be obvious.).
-Since Octagon ABCDEFGH is regular, the center of the squares ACEG and BDFH and octagon ABCDEFGH is the same point (point O.)
-All seven other triangles on the octagon's edge such as AHG are two rectangle triangles that form one isometric triangle , because their sides are the ones of a *regular* octagon, therefor the sides are of isometric lengths.Their side that is perpendicular to the base is the same (eg. on AHG, triangle AHJ (whereas J is the intersection point of angle bisector HD and segment AG)
Now let's do the bothering shit, aka hard word that i have no idea why i'm doing for you:
Calculate area of AHG or any other wide ass isometric triangle by multiplying the double of the rectangle halves (such as AHJ)'s area, which are of some annoying height.. BUT THATS NOT EVEN NECESSARY! Because AHG IS a isometric triangle, we don't give a shit about AHJ ! So, the area of AHG is of ((JH*PQ)/2) . Same for every single fucking triangle in a regular octagon on this sheet of paper! Now you have octagon PQRSTUV, which you can calculate as the sum of the square's area (here, both squares are of area Ä) MINUS JH*PQ/2:
((2Ä)-((JG*PQ)/2)
And look! YOU'VE GOT THE AREA OF THAT OCTAGON! Now, use some Thales Interception Theorem by taking, by example, for triangle JOG (and for this REGULAR, mind the REGULAR, octagon's other triangles), the coefficient will be of JO/HO . Now multiply every bissector angle segment (ex. JL,KM...) of said octagon,and LOOK AT IT! IT'S THE SAME OCTAGON! Now, knowing Thales's Interception Theorem that you TOTALLY didn't forget because you kept drawing on the margin of your notebook some weird stickman story in which they end up doing a mass bunny genocide for whatever lightning strike involving reason while your teacher was asking you about exercise 24 a), IT TURNS OUT THAT , WONDERFULLY, the coefficient for the AREA (which is of (2Ä)-((JG*PQ)/2), is coefficient² (squared)! Wonderful ain't it?
If your area is of 8, and your k is of 2, then your k² is of 4 and your area is of 8*4, WHICH IS 32!!!
And look even harder! AREA FOUND. IT'S OF THOSE LITTLE SHITTY TRIANGLES' COEFFICIENT that helped calculating it.

[Anonymouse, again No.10925304,3]

>"buh anon, ho you maek for rectungul? u useless incel"
First, I'd tell you I'm not an incel, at least i PRACTICALLY COULD'VE NOT BEEN AN INCEL, am I right Juliett? (The name was changed for safety reasons.) And second, YOU DON'T GIVE A SHIT! Because , in the :

"If your area is of 8, and your k is of 2, then your k² is of 4 and your area is of 8*4, WHICH IS 32!!!"

I DIDNT SPECIFY IF IT WAS A REGULAR OCTAGON! In fact, I didn't even precise it was a OCTAGON! It could've been a rectangle!
So all you have to do is calculate the non-so-regular octagon (which will still tend to be partly regular, because it's a rectangle and all angles will STILL be the same), and you'll end up with a smaller, known-size copy, AND with a coefficient that will be the same EVERYWHERE!
Just look at the pics I am sending if you don't think so.
Have fun clearing this of all the junk I added in , making it unbrided and presenting it to your exam; I think taking in the concepts I said would be more useful.
Also the side of a square is sqrt(Area), if you can't concentrate on basic geometric equations (because the area is side²). Therefor you can find the perimeter by doing side4, etc... and end up in the way I explained.
I just realized I screwed up somewhere; the thing is inscribed in a circle. Which means that the biggest possible size is likely using two squares. YOU CAN DO THAT BECAUSE A SQUARE IS A RECTANGLE AND A RECTANGLE *CAN* BE A SQUARE. Except you actually can't because of areas being different, which makes no sense if both are inscribed squares, therefor you must find the rectangle with an area of 5 that is the CLOSEST to a square. I'm NOT bothering with this, but you should if you care about your exam. Also that's a demonstration-ish reply, you can apply this to calculate the area.

I could post an image but I'm not going to bother seeking how.
If you really want one just tell me how to post one. I'll be glad to show you , anon.