Serious Daily math problem thread

Divergent_Integral

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For your edification and intellectual stimulation I will post a math problem on here daily (or at least a few times a week). I will try to make the problems as accessible as possible, in that I will assume no really advanced mathematical knowledge. Just a bit of cleverness will do.

You may post your solutions in the thread. If none are received, I will give the solution myself the next day. Good luck, and happy problem solving!

Problem #0001

Let a, b, c be positive numbers such that a + b + c = 1. Prove that (1 - a)(1 - b)(1 - c) >= 8abc. Under what conditions on a, b, c does the equal sign hold?
 
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Shut up GrAYboi
 
WorldYamataizer

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Yes. Thank you for this.
Can I share some problems as well?
 
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Yes. Thank you for this.
Can I share some problems as well?
Sure, that'd be fun. It would be best I think if we used the same problem number sequence, for ease of reference. (So your next problem would be #0002, etc.)
 
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the eternal maths major strikes again
 
Divergent_Integral

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What does this even mean? @ me when you post the solution.
I asked to prove that (1 - a)(1 - b)(1 - c) is greater than or equal to 8abc, given a + b + c = 1. Typically, the first option, >, is true. My second question, then, is: when is the non-typical situation true, with (1 - a)(1 - b)(1 - c) exactly equal to 8abc; which extra condition(s) on a, b, c do we need for that to happen?

Sure, I'll @ you.
 
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WorldYamataizer

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@Diocel
The maximum values is 8/27.
Achieved when a=b=c=1/3.

 
Divergent_Integral

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WorldYamataizer

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Problem #0002

In the diagram, ABC is right-angled at A, with AB=3 and AC=4.
A semicircle with diameter AD, with A,D,B being collinear, is tangent to BC at E.
What is the radius of the semicircle?

I will post my solution tomorrow at 21:00-ish. Feel free to share your solutions!
 
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Its over for me, I couldnt solve it

Probably could in an hour or so but I cant be bothered to math outside school
 
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Rlly?
graycels
 
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Problem #0002
View attachment 386354
In the diagram, ABC is right-angled at A, with AB=3 and AC=4.
A semicircle with diameter AD, with A,D,B being collinear, is tangent to BC at E.
What is the radius of the semicircle?

I will post my solution tomorrow at 21:00-ish. Feel free to share your solutions!
EC = AC = 4, so EB = BC - EC = 5 - 4 = 1

Let's name O the center of the circle, and r its radius.

OEB is right-angled at E, so OB² = OE² + EB² = r² + 1. Besides OB = AB - r, so OB² = AB² - 2rAB + r².

Thus AB² - 2rAB = 1, 9 - 6r = 1, therefore r = 8/6 = 4/3
 
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WorldYamataizer

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EC = AC = 4, so EB = BC - EC = 5 - 4 = 1

Let's name O the center of the circle, and r its radius.

OEB is right-angled at E, so OB² = OE² + EB² = r² + 1. Besides OB = AB - r, so OB² = AB² - 2rAB + r².

Thus AB² - 2rAB = 1, 9 - 6r = 1, therefore r = 8/6 = 4/3
Good job! Yes the radius is 4/3. It would have been nice to explain why EC=AC (congruent triangles) though.
 
Divergent_Integral

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Proceeds to ask a question no one who hasn't done a maths degree would know about :feelsUgh:. I did physics and never learned about AM GM shit
You learned physics at which university? Bumblefuck State Tech? JFL
 
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im a CScel doe
 
Divergent_Integral

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Good job! Yes the radius is 4/3. It would have been nice to explain why EC=AC (congruent triangles) though.
I will give a (slightly more complicated and messier) analytical solution to Problem #0002.

Let the midpoint of AD be the origin of a cartesian coordinate system, with x-axis parallel to AB and y-axis parallel to AC, giving the axes their usual orientations (positive is right and up, respectively). Then the equation of the circle is y = sqrt(r^2 - x^2), with r to be determined, and the equation of the tangent line BC is y = -4/3 x + 4 - 4/3 r. This line is tangent to the circle if and only if it has exactly one point in common with it. That is to say, the equation - 4/3 x + 4 - 4/3 r = sqrt(r^2 - x^2) should have exactly one solution. This is a quadratic equation in x, whose discriminant equation works out to -64 + 128 r/3 + 4r^2 = 0. The only valid solution to this is r = 4/3.
 
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I will give a (slightly more complicated and messier) analytical solution to Problem #0002.

Let the midpoint of AD be the origin of a cartesian coordinate system, with x-axis parallel to AB and y-axis parallel to AC, giving the axes their usual orientations (positive is right and up, respectively). Then the equation of the circle is y = sqrt(r^2 - x^2), with r to be determined, and the equation of the tangent line BC is y = -4/3 x + 4 - 4/3 r. This line is tangent to the circle if and only if it has exactly one point in common with it. That is to say, the equation - 4/3 x + 4 - 4/3 r = sqrt(r^2 - x^2) should have exactly one solution. This is a quadratic equation in x, whose discriminant equation works out to -64 + 128 r/3 + 4r^2 = 0. The only valid solution to this is r = 4/3.
Nice analytic approach! It's truly a beautiful thing when multiple people can approach and solve a problem in different ways.
 
Divergent_Integral

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Nice analytic approach! It's truly a beautiful thing when multiple people can approach and solve a problem in different ways.
Thanks brocel. I will set the next problem tomorrow!
 
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EC = AC = 4, so EB = BC - EC = 5 - 4 = 1

Let's name O the center of the circle, and r its radius.

OEB is right-angled at E, so OB² = OE² + EB² = r² + 1. Besides OB = AB - r, so OB² = AB² - 2rAB + r².

Thus AB² - 2rAB = 1, 9 - 6r = 1, therefore r = 8/6 = 4/3
Can you explain why
AB² - 2rAB = 1 and 9 - 6r = 1?

I could follow only up to that point.
 
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comes from equating the two different expressions of OB² and eliminating r² from both sides of the equation.
I see now tbh
The analytical solution posted by @Divergent_Integral was only something I could follow until he posted the equation for the tangent line ngl
Just applying the known numerical value for AB.
tbh I'm always succumbing to that slip up of forgetting preliminary information stated in the problem.
 
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Notice that there is a very simple solution to #prob0002, if one admits the formula for the inradius of a triangle. We just need to consider the big triangle CC'B, where C' is the point symmetric to C with respect to (AB).

s = (8+5+5)/2 = 9
r = sqrt((9-8)(9-5)(9-5)/9) = sqrt(16/9) = 4/3
 
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Divergent_Integral

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I havent heard of it either, but surely you must have heard of arithmetic mean and geometric mean. I have a feeling that we Americans are worse at math because public schools teach students to pass standardized tests while catering to the worst students. I am excited to see the next problem.
This. AM-GM merely means that the arithmetic mean of a set of numbers is >= its geometric mean. Or in the case of two variables: (a + b)/2 >= sqrt(ab). I learnt about this in the European equivalent of high school, but apparently not everyone does anymore.
 
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I havent heard of it either, but surely you must have heard of arithmetic mean and geometric mean. I have a feeling that we Americans are worse at math because public schools teach students to pass standardized tests while catering to the worst students. I am excited to see the next problem.
Arithmetic mean obviously, never heard of the geometric mean though, guess it never came up.
 
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I will give a (slightly more complicated and messier) analytical solution to Problem #0002.

Let the midpoint of AD be the origin of a cartesian coordinate system, with x-axis parallel to AB and y-axis parallel to AC, giving the axes their usual orientations (positive is right and up, respectively). Then the equation of the circle is y = sqrt(r^2 - x^2), with r to be determined, and the equation of the tangent line BC is y = -4/3 x + 4 - 4/3 r. This line is tangent to the circle if and only if it has exactly one point in common with it. That is to say, the equation - 4/3 x + 4 - 4/3 r = sqrt(r^2 - x^2) should have exactly one solution. This is a quadratic equation in x, whose discriminant equation works out to -64 + 128 r/3 + 4r^2 = 0. The only valid solution to this is r = 4/3.
How did you get the equation of the tangent line BC?
 
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How did you get the equation of the tangent line BC?
It goes down by 4 units when you move 3 units in the positive direction (to the right). So its slope is -4/3. It hits the x-axis at a distance 3 to the right of -r, i.e., for x = 3 - r. Now we substitute x = 3 - r into -4/3 x + b = 0, which gives -4 + 4/3 r + b = 0. So b = 4 - 4/3 r and you get the equation as stated.
 
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It goes down by 4 units when you move 3 units in the positive direction (to the right). So its slope is -4/3.
That's understandable.
It hits the x-axis at a distance 3 to the right of -r, i.e., for x = 3 - r.
Because the length of AB is 3 but you said that the value of r is to be determined I don't really understand what you mean by "at a distance 3 to the right of -r".
Now we substitute x = 3 - r into -4/3 x + b = 0, which gives -4 + 4/3 r + b = 0. So b = 4 - 4/3 r and you get the equation as stated.
I recognize the y = mx + b format.
Plugging in x = 3-r into the equation -4/3x +b to solve for the y-intercept? is straightforward enough.
 
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That's understandable.

Because the length of AB is 3 but you said that the value of r is to be determined I don't really understand what you mean by "at a distance 3 to the right of -r".

I recognize the y = mx + b format.
Plugging in x = 3-r into the equation -4/3x +b to solve for the y-intercept? is straightforward enough.
I've defined the x-axis in such a way that the origin is the midpoint of AD, i.e., the center of the circle. Thus A corresponds to x = -r and D to x = r. But we also know that AB = 3, so B's position on the x-axis is 3 units to the right of -r. Which means that B's x-coordinate is x = 3 - r.

 
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I've defined the x-axis in such a way that the origin is the midpoint of AD, i.e., the center of the circle. Thus A corresponds to x = -r and D to x = r. But we also know that AB = 3, so B's position on the x-axis is 3 units to the right of -r. Which means that B's x-coordinate is x = 3 - r.

View attachment 386640
So in essence the x position of B is -r + 3 or x = 3 - r and that's how you arrived to that expression?
 
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So in essence the x position of B is -r + 3 or x = 3 - r and that's how you arrived to that expression?
Yes. The beauty of that approach is that you don't need to know beforehand the numerical value of r.
 
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Yes. The beauty of that approach is that you don't need to know beforehand the numerical value of r.
Variables that can take on many possible values are good for versatility ngl
You see the same thing in programming. Instead of coding in a specific value to verify a conditional statement and having multiple such conditional statements, it's better to have a single general variable expression or equation in the conditional statement to account for a wider range of possible values. This is however one of the more difficult aspects of programming imo. Strangely enough being more general makes the code more concise. And in today's hyper NT focused world concise and to the point code and descriptions are highly sought after tbh
 
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Variables that can take on many possible values are good for versatility ngl
You see the same thing in programming. Instead of coding in a specific value to verify a conditional statement and having multiple such conditional statements, it's better to have a single general variable expression or equation in the conditional statement to account for a wider range of possible values. This is however one of the more difficult aspects of programming imo. Strangely enough being more general makes the code more concise. And in today's hyper NT focused world concise and to the point code and descriptions are highly sought after tbh
Exactly. Conciseness and generality are aesthetic qualities that are highly sought after in both math and programming. Today's mathematicians have taken this quest for generality to a whole new level btw, with stuff like abstract algebra and category theory. That way you can discover that mathematical theories which look quite distinct from one another on the surface level have in fact the same (or at least similar) underlying structure. One of the main concepts in modern math is that of an equivalence relation, which you may look up if you're interested.
 
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Today's mathematicians have taken this quest for generality to a whole new level btw, with stuff like abstract algebra and category theory. That way you can discover that mathematical theories which look quite distinct from one another on the surface level have in fact the same (or at least similar) underlying structure. One of the main concepts in modern math is that of an equivalence relation, which you may look up if you're interested.
I looked up equivalence relations and saw that logical matrices are often involved in depicting them. This is another common theme today imo in qualitatively and at the same time quantitatively representing data. Also relevant to programming because matrix multiplication and algebra have many uses in programming like plotting data in arrays and matrices and computing variables that use data from previously defined arrays tbh
 
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Written solution to problem #0002
 
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How is r/1 = 4/3? Does that come from how angle EBP can be also thought of as angle CBA?
It is a well-known property of tangents to circles that such a tangent is perpendicular to the radius that ends in the point of touching. So triangle EBP is a right-angled triangle with angle BEP being the right angle. So by the definition of tangent we have tan(angle EBP) = PE / EB = r/1 = r. But angle EBP is indeed equal to angle CBA, whose tangent is 4/3.
 
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It is a well-known property of tangents to circles that such a tangent is perpendicular to the radius that ends in the point of touching. So triangle EBP is a right-angled triangle with BEP the right angle. So by the definition of tangent we have tan(angle EBP) = PE / EB = r/1 = r.
I see. But can it be said that angle EBP is equal to angle CBA and if not, why not?
 
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How is r/1 = 4/3? Does that come from how angle EBP can be also thought of as angle CBA?
recall the the tangent function of an angle in right angled triangle is opposite/adjacent. Yes, this comes from the fact that angle EBP is angle CBA. Another way to think about this is with similar triangles.
 
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recall the the tangent function if an angle in right angled triangle in opposite/adjacent. Yes, this comes from the fact that angle EBP is angle CBA. Another way to think about this is with similar triangles.
Yeah @Divergent_Integral explained this in the previous post connecting point of tangency to a circle to orthoganality to the radius of a circle? But as with many math problems there are multiple such ways to approach and solve a problem but only a few possible correct answers at most, usually only one correct answer at best. That's another interesting aspect of math problems. Many ways to approach solving the problem but only a few/one possible correct answer(s).
 
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@Divergent_Integral
It looks like this thread got pinned.
Should I post my problem here tomorrow?
 
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Notice that there is a very simple solution to #prob0002, if one admits the formula for the inradius of a triangle. We just need to consider the big triangle CC'B, where C' is the point symmetric to C with respect to (AB).

s = (8+5+5)/2 = 9
r = sqrt((9-8)(9-5)(9-5)/9) = sqrt(16/9) = 4/3
Can you provide a drawing of big triangle CC'B? It's hard to visualize and I don't know where you are getting that side a = 8 and sides b and c = 5. Thanks.
 
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Can you provide a drawing of big triangle CC'B? It's hard to visualize and I don't know where you are getting that side a = 8 and sides b and c = 5. Thanks.

reflecting C over AB gives this isosceles triangle. his steps should be clear now.