Puzzles

[Thebobo]

I applaud you for failing the reading comprehension test. Or... it's possible you never heard of FB, but using occam's razor, it's simpler to assume my initial assumption.

Ok I have one for you.

What year long event ended in 1635?

 

[[DHT]Osiris]

Ok I have one for you.

What year long event ended in 1635?

1634?
Or 1636 in Bizarro Earth.

Reverse both for B.C. (C.B. for Bizarro Earth)

 

[dullard]

Ok I have one for you.

What year long event ended in 1635?

Are your interval brackets inclusive or exclusive?
https://en.wikipedia.org/wiki/Bracket_(mathematics)#Intervals

Also, will you accept any of the infinite answers or just the one you were thinking about?

 

[Thebobo]

1634?
Or 1636 in Bizarro Earth.

Reverse both for B.C. (C.B. for Bizarro Earth)

ding ding!

 

[Thebobo]

Are your interval brackets inclusive or exclusive?
https://en.wikipedia.org/wiki/Bracket_(mathematics)#Intervals

Also, will you accept any of the infinite answers or just the one you were thinking about?

Not my puzzle and I don't know the answer because I can't even figure out how the first horizontal is possible. But I was never great at math.

 

[Cozarkian]

You can't logically conclude that one paw represents half of the number represented by two paws. Thus, the answer is

One paw = (? - 5)/10

 

[dullard]

You can't logically conclude that one paw represents half of the number represented by two paws. Thus, the answer is

One paw = (? - 5)/10

The zoomed in photo is worth half that of a zoomed out photo, so therefore, one paw must be half of two paws. Or something like that.

 

[tynopik]

ding ding!

false, years don't overlap, so any previous year would have ended BEFORE said year

 

[[DHT]Osiris]

false, years don't overlap, so any previous year would have ended BEFORE said year

Unless it's a fiscal year!

 

[Thebobo]

 

[[DHT]Osiris]

Spoiler: Spoilz

30, unless this is super-sekret-hidden-answer of 360 (the 'X' in the middle of the triangle)

 

[tynopik]

the bottom 2 angles are 80 degrees (20+60 & 70+10), which means the top angle is 0 degrees (160-80-80) which means it's not a valid triangle

 

[[DHT]Osiris]

the bottom 2 angles are 80 degrees (20+60 & 70+10), which means the top angle is 0 degrees (160-80-80) which means it's not a valid triangle

Triangle is 180 degrees, half of a circle (360)...

 

[tynopik]

Triangle is 180 degrees, half of a circle (360)...

durr, not sure why i got 160 on the brain

 

[[DHT]Osiris]

durr, not sure why i got 160 on the brain

Because geometry was bullshit and doesn't matter if you aren't a professional billiards player, or astronomer?

 

[IronWing]

I can see how to solve it but that would involve remembering trig and I'd rather not.

1) Fill in all the easy angles.
2) assign an arbitrary length to the bottom side (1 would work nicely)
3) Solve for the length of the two other sides of the outside triangle using trig that I don't want to remember
4) Solve for the length of the right side of the 60-80-40 triangle which also yields the length of the right side of the top triangle
5) Solve the rest of the top triagle
6) Solve for x

 

[[DHT]Osiris]

I can see how to solve it but that would involve remembering trig and I'd ratther not.

I thought it'd take longer than it did, it didn't. I didn't use any formulas or anything, just logic'd out 180 degrees for each triangle. I imagine a trig teacher would thwack me in the back of the head for not showing my work though.

 

[IronWing]

I thought it'd take longer than it did, it didn't. I didn't use any formulas or anything, just logic'd out 180 degrees for each triangle. I imagine a trig teacher would thwack me in the back of the head for not showing my work though.

I don't think you can. You end up with two angles that are unconstrained without using side lengths. I'll look again to be sure.

One can solve the big triangle, no problem
One can solve the lower three triangles
One can solve for the angle at the bottom of the triangle containing angle x
One can constrain angle x + the angle above x to 150 degrees
One can constrain the angle across from x + the angle above it to 140 degrees

I think that's it for simple visual inspection.

 

[[DHT]Osiris]

I don't think you can. You end up with two angles that are unrestrained without using side lengths. I'll look again to be sure.

You get to a point where you've got two angles to figure out inside the primary triangle where x is, x and one more I labeled y, you also have the angles for complete triangles outside of those angles, so you can work out two formula where x and y need to fit to make 180 degree triangles between them (as well as the triangle housing x itself), only one pair of numbers works for all 3 equations.

Again, I'm sure there's a better way to do that where a smarter person than me has already created a formula for, I just slogged through it.

 

[IronWing]

Okay, using your method I have four equations and four variables so we should be good.

Spoiler

x + x' = 150
y + y' = 140
x' + y' = 160
x + y = 130

x = 30
x' = 120
y = 100
y' = 40

works

 

[[DHT]Osiris]

Okay, using your method I have four equations and four variables so we should be good.

x + x' = 150
y + y' = 140
x' + y' = 160
y + y = 160

I mean, I don't know where those plug into what, but if you got the same answer as me it's 'right enough'

 

[jiy78]

I am sure answer is 15.