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Serious Daily math problem thread

Over for mathematicacels
I just don't know Mathematica well enough yet. I should have remembered there is a function to define a triangle from its side lengths, and then there is a function to get the angles.

1610477091461


There's a function for almost everything in Mathematica.

I'm slightly disappointed that Mathematica can't pick up on remarkable ratios of ArcCosines, though. It should have simplified ArcCos(1/8)/ArcCos(3/4) into 2.

My earlier demonstration still works, though :

2*(3/4)² - 1 = 2*9/16 - 1 = 9/8 - 1 = 1/8
 
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I mean, yeah. I'm pretty sure, unless I missed something about the problem statement.

Code:
    cosineRule[a_, b_, c_] := (a^2 - b^2 - c^2)/(2 b c)
    ArcCos /@ Simplify@Table[cosineRule @@ RotateLeft[{n, n + 1, n + 2}, i], {i, 0, 2}]
    % /. n -> 2
    {Max[%],Min[%]}

does return {ArcCos[-7/8], ArcCos[1/4]}

Mathematica then fails to see that the ratio between those two values is exactly 2, but the numerical value suggests it strongly, and there is the demonstration I gave about it which I think is fairly simple.
There must be a mistake!
If the largest angle is twice the smallest angle, then the interior angles of the triangle are x , 180-3x , and 2x.
But x < 180-3x < 2x
This gives that 36<x<45

arccos(1/4) is approximately 76 degrees
 
Yes there was, as I wrote. You were right about a sign mistake. In the cosine rule I divided by 2 b c instead of -2 b c
Sorry, i left this tab open for like half an hour, so I couldn't catch up on new posts to catch up
 
Problem #0014

A general point P(c,d) is selected on the circle x^2+y^2=a^2+b^2 ; from P, tangents are drawn to the ellipse

(b^2)(x^2)+(a^2)(y^2)-(a^2)(b^2)=0​

Show that, no matter where P is selected on the circle, the two tangents from P to the ellipse are always perpendicular.

Here is my solution to Problem #0013
Three side lengths of a triangle are consecutive integers. The largest angle of the triangle is two times the smallest angle. What are the side lengths of the triangle?

It may be helpful to draw a useful diagram.
Screenshot 7

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A general point P(c,d) is selected on the circle x^2+y^2=a^2+b^2 ; from P, tangents are drawn to the ellipse

(b^2)(x^2)+(a^2)(y^2)-(a^2)(b^2)=0​
Show that, no matter where P is selected on the circle, the two tangents from P to the ellipse are always perpendicular.
This problem sounds poorly formulated.
 
Pre cal is getting hard to follow
Calc is hard to follow
 
Literally fundamental complex algebra. How hard cohld it be, just multiply the damn things
Its the i that im having trouble with, the other stuff is easy enough
 
PROBLEM #0015

Normal

In the diagram, the largest semi-circle has radius 4.
Two identical semi-circles are tangent to each other and the larger semi-circle as shown.
Also, two more identical circles are tangent to each other, the two smaller semi-circles and the large semi-circle.
If the centers of the small semi-circles and circles are joined, determine the exact area of the trapezoid?

I'll share my solution tomorrow. Enjoy.


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PROBLEM #0015

View attachment 402099
In the diagram, the largest semi-circle has radius 4.
Two identical semi-circles are tangent to each other and the larger semi-circle as shown.
Also, two more identical circles are tangent to each other, the two smaller semi-circles and the large semi-circle.
If the centers of the small semi-circles and circles are joined, determine the exact area of the trapezoid?

I'll share my solution tomorrow. Enjoy.


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I’m too stupid to solve most of these problems you post:feelsbadman: I wish I was smarter
 
I’m too stupid to solve most of these problems you post:feelsbadman: I wish I was smarter
Ok. Do you want me to stop tagging u? or are u ok with it?
 
I wish I still had a compass to draw circles with. My mom probably lost my old cheap one. She can't do math at all. Maybe they can be bought cheap online.
 
I would only learn math if I had access to creating weapons but of course I don't
 
You should post some number theory problems, those are more fun
 
You should post some number theory problems, those are more fun
Ok, I am big on geometry, but not too knowledgeable on number theory. I can post a diophantine equation for the next problem. Is that ok?
 
Ok, I am big on geometry, but not too knowledgeable on number theory. I can post a diophantine equation for the next problem. Is that ok?
Yeah, it’s just a personal dislike for geometry anyway
 
I am trying to find the radius of the smaller circle in order to calculate the length of the base of the trapezoid. It is apparent from the picture that the radius in question is half the radius of the next larger semicircle and a quarter of the largest semicircle, but I don't know how to prove this. Basically, the center of the right small circle appears to be located at (1,3) considering the origin as the center of the large circle.

1611952586408
 
I am trying to find the radius of the smaller circle in order to calculate the length of the base of the trapezoid. It is apparent from the picture that the radius in question is half the radius of the next larger semicircle and a quarter of the largest semicircle, but I don't know how to prove this. Basically, the center of the right small circle appears to be located at (1,3) considering the origin as the center of the large circle.
You may find it helpful to consult the Pythagorean Theorem here. You can try and find the ratio of the radii from there.
Here is a useful fact: The radius at the point of tangency of mutually tangent circles passes through both their centers.
 
1611947728410

So the equations are:
h^2 + r^2 = (R-r)^2
h^2 + (R/2 - r)^2 = (R/2 + r)^2

We get :
r^2 - (R/2 - r)^2 = (R-r)^2 - (R/2 + r)^2
- (R^2)/4 + rR = 3/4 * (R^2) - 3rR
(R^2) = 4Rr
r = R/4
 
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Problem #016

We have a cat, and a mouse.
The cat moves at speed 2, while the mouse moves at speed 1.

The mouse starts running from (1,0) in the direction (0,1) at t = 0.
The cat runs towards the mouse at all times starting at (0,0) and t = 0.

When and where does the cat catch the mouse.

gif : cat - red, mouse - green

Cat and mouse

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seems like 2/3, but idk the proof
 
i wouldn’t know how to do this problem.
i assume it has to do with differential eqns
 
the general formula seems to be x/(x^2 - 1), idk the proof
 
PROBLEM #0015

View attachment 402099
In the diagram, the largest semi-circle has radius 4.
Two identical semi-circles are tangent to each other and the larger semi-circle as shown.
Also, two more identical circles are tangent to each other, the two smaller semi-circles and the large semi-circle.
If the centers of the small semi-circles and circles are joined, determine the exact area of the trapezoid?

I'll share my solution tomorrow. Enjoy.


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1613329880706


This is to get the two bases I think (sizes 2 and 1).


Then the diagonals must have length 2 + 1 = 3, making the height sqrt(3^2 - 1) = sqrt(8).

This makes the area 4sqrt(8) - sqrt(8) = 3sqrt(8) = 8.485
 
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yeah this shit is too hard wheres the kiddie math area?
 
Got triggered remembering I have a Calc 2 Exam tomorrow.
 
View attachment 409774

This is to get the two bases I think (sizes 2 and 1).


Then the diagonals must have length 2 + 1 = 3, making the height sqrt(3^2 - 1) = sqrt(8).

This makes the area 4sqrt(8) - sqrt(8) = 3sqrt(8) = 8.485
Good job, thats the right answer
 
Problem #0017
Desmos graph

In the diagram, a circle of radius 1 is internally tangent to y=x^2.
Determine the black area between the circle and parabola.

I will share a written solution tomorrow, same time. Happy problem-solving


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Problem #0017
View attachment 413879

In the diagram, a circle of radius 1 is internally tangent to y=x^2.
Determine the black area between the circle and parabola.

I will share a written solution tomorrow, same time. Happy problem-solving


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Brute forcing this

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2
contact point y coord has only one value.

The equation
y + (y - c)^2 = 1^2
has only one solution. (b^2 = 4ac)

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
(1 - 2c)^2 = 4 * (c^2 -1)
4c^2 -4c + 1 = 4c^2 - 4
4c = 5
c = 5/4
 
Brute forcing this

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2
contact point y coord has only one value.

The equation
y + (y - c)^2 = 1^2
has only one solution. (b^2 = 4ac)

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
(1 - 2c)^2 = 4 * (c^2 -1)
4c^2 -4c + 1 = 4c^2 - 4
4c = 5
c = 5/4
i got “h” (i used “h” instead of “c”) the same way.

i am just evaluating an integral rn and im done.

this problem is sort of just algebraic bashing. next time, i’ll post a problem that requires more of an insightful construction or idea
 
Brute forcing this

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2
contact point y coord has only one value.

The equation
y + (y - c)^2 = 1^2
has only one solution. (b^2 = 4ac)

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
(1 - 2c)^2 = 4 * (c^2 -1)
4c^2 -4c + 1 = 4c^2 - 4
4c = 5
c = 5/4
...cont.
Now we make a rectangle around the whole thing.

putting c into the equation

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
y^2 + (1 - 5/2) *y + (25/16 - 1) = 0
y^2 - 3/2 * y + 9/16 = 0
(y - 3/4)^2 = 0
y = 3/4

1614019670374
1614019740159

1614019717274

Area(purple) = tri (green) + rect(red) - curve(blue) - circle(yellow)

Area =
+ 1/2 * 1/2 * sqrt(3/4)
+ sqrt(3/4) * 3/4
- (sqrt(3/4)^3) / 3
- pi/3 * 1/2 * 1^2
i am just evaluating an integral rn and im done.
what integral?

(doubled to give full black area)
(3/4) * sqrt(3) - pi/3
0.25184055448
 
Last edited:
...cont.
Now we make a rectangle around the whole thing.

putting c into the equation

y^2 + (1 - 2c) * y + (c^2 - 1) = 0
y^2 + (1 - 5/2) *y + (25/16 - 1) = 0
y^2 - 3/2 * y + 9/16 = 0
(y - 3/4)^2 = 0
y = 3/4

View attachment 414017
View attachment 414020
View attachment 414019
Area(purple) = tri (green) + rect(red) - curve(blue) - circle(yellow)

Area =
+ 1/2 * 1/2 * sqrt(3/4)
+ sqrt(3/4) * 3/4
- (sqrt(3/4)^3) / 3
- pi/3 * 1/2 * 1^2

what integral?

(doubled to give full black area)
(3/4) * sqrt(3) - pi/3
0.25184055448
im impressed, you did the question the “elegant” way. i just brainlessly did the area under the circle minus the area under the parabola.
0752B766 DC6E 4F14 A475 C21565132737

this integral.
 
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