Serious Daily math problem thread

[grondilu]

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.

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

 

[Deleted member 31630]

I mean, yeah. I'm pretty sure, unless I missed something about the problem statement.

    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

 

[grondilu]

There must be a mistake!

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

 

[Deleted member 31630]

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

 

[mNFwTJ3wz9]

arccos(1/4) is approximately 76 degrees

did you do that in your head?

 

[Deleted member 31630]

did you do that in your head?

No, I used desmos and punched it in.

 

[mNFwTJ3wz9]

No, I used desmos and punched it in.

The "approximate" made me think that lol

 

[Deleted member 31630]

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.

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[grondilu]

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.

 

[Deleted member 31630]

This problem sounds poorly formulated.

Why?

 

[LDARbeforeROPE]

Pre cal is getting hard to follow

 

[Caesercel]

Pre cal is getting hard to follow
View attachment 398489

Lol wtf is this childhood level shit.

 

[LDARbeforeROPE]

Lol wtf is this childhood level shit.

Precalculus

 

[Caesercel]

PrecalculusView attachment 400718View attachment 400717

Literally fundamental complex algebra. How hard cohld it be, just multiply the damn things

 

[LDARbeforeROPE]

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

 

[Caesercel]

Its the i that im having trouble with, the other stuff is easy enough

i*i = - 1.

Here, I just saved you thousands of dollars in tution

 

[mNFwTJ3wz9]

i*i = - 1.

Here, I just saved you thousands of dollars in tution

base

 

[PPEcel]

Its the i that im having trouble with, the other stuff is easy enough

u a highschoolcel?

 

[Deleted member 28148]

meanwhile chads measuring foids ass size with their pp

brutal

 

[Deleted member 29003]

i*i = - 1.

Here, I just saved you thousands of dollars in tution

beyond based

 

[LDARbeforeROPE]

u a highschoolcel?

First semester college

 

[Deleted member 31630]

PROBLEM #0015

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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[Deleted member 31093]

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 I wish I was smarter

 

[Deleted member 31630]

I’m too stupid to solve most of these problems you post I wish I was smarter

Ok. Do you want me to stop tagging u? or are u ok with it?

 

[Deleted member 31093]

Ok. Do you want me to stop tagging u? or are u ok with it?

No keep tagging me Id like to try but I fail or give up most of the time is all

 

[Deleted member 6657]

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.

 

[waste matter]

I would only learn math if I had access to creating weapons but of course I don't

 

[Deleted member 29529]

You should post some number theory problems, those are more fun

 

[Deleted member 31630]

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?

 

[Deleted member 29529]

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

 

[Deleted member 6657]

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.

 

[Deleted member 31630]

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.

 

[mNFwTJ3wz9]

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

 

[mNFwTJ3wz9]

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

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[BANG]

seems like 2/3, but idk the proof

 

[Deleted member 31630]

i wouldn’t know how to do this problem.
i assume it has to do with differential eqns

 

[BANG]

the general formula seems to be x/(x^2 - 1), idk the proof

 

[nasser22]

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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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

 

[Cyrrow]

yeah this shit is too hard wheres the kiddie math area?

 

[Deleted member 32725]

Got triggered remembering I have a Calc 2 Exam tomorrow.

 

[LDARbeforeROPE]

Got triggered remembering I have a Calc 2 Exam tomorrow.

Good luck bro

 

[Deleted member 31630]

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

 

[based_meme]

Good job, thats the right answer

That was a good one. Glad this thread is stickied.

 

[zangano1]

me right now

 

[Deleted member 31630]

That was a good one. Glad this thread is stickied.

thanks, i’ll post some more problems soon.

 

[Deleted member 31630]

Problem #0017

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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[mNFwTJ3wz9]

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

Spoiler: 1

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2

Spoiler: 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

 

[Deleted member 31630]

Brute forcing this

Spoiler: 1

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2

Spoiler: 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

 

[mNFwTJ3wz9]

Brute forcing this

Spoiler: 1

equations :
parabola : y = x^2
cicle : x^2 + (y - c)^2 = 1^2

Spoiler: 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

Spoiler: 3

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

Spoiler: 4

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?

Spoiler: Final ans

(doubled to give full black area)
(3/4) * sqrt(3) - pi/3
0.25184055448

 

[Deleted member 31630]

...cont.
Now we make a rectangle around the whole thing.

putting c into the equation

Spoiler: 3

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

Spoiler: 4

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?

Spoiler: Final ans

(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.

this integral.