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Prove that in every quadrilateral the segments connecting the midpoints its opposite sides, as well as the segment connecting the midpoints of the diagonals pass through one point and divide it in half.
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k a May Highlights and 2025 AoPS Online Class Information
jlacosta 0
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.
Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.
Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!
Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.
Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!
Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
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Prealgebra 1 Self-Paced
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0 replies
help me ..
exoticc 0
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
0 replies
Pls help I'm stuck
UtkarshSri 3
N
by lbh_qys
So I was solving this problem
Cbrt(x+1) - cbrt(x-1) = pow((x²-1),1/6)
Here, the square root is defined in the domain x≤-1 OR x≥1
I substituted a= cbrt(x+1) and b= cbrt(x-1)
And then the equation turned out to be
a-b= sqrt(ab)
Here, since the RHS is +ve, the LHS must be too, so a>b (Done so as to prevent extraneous roots)
Now, I can write the domain in terms of a and b as ab≥0
Squaring, we get a²+b²-3ab=0
On solving we get a/b = (3±sqrt(5))/2
Now, we know that a>b from the above condition, so the only root is a/b= (3+sqrt(5))/2 and on substituting for x, and applying componendo and dividendo after cubing, I get x= sqrt(5)/2
Now, my question is, since the equation is even i.e f(x)=f(-x), it's symmetric about the y-axis, so x= -sqrt(5)/2 must also satisfy the equation, and indeed it does, so where did I made some irreversible step that caused this loss of the root?
when I plugged the equation to wolframalpha, it showed that the "principle root" has the former solⁿ only where's for a "real valued root", it has both.I also found that on solving the equation
b-a= sqrt(ab), I get the latter root.But I mean where did that extra -ve sign come from?
I'm familiar that f(x)=sqrt(g(x)) is equivalent to f²(x)=g(x) AND f(x)≥0 AND g(x)≥0 but the g(x)≥0 condition is not necessary as f²(x) is already ≥0
and also that abs(f(x))=abs(g(x)) is equivalent to f²(x)=g²(x)
PS: Pardon me I don't know latex well, I'm tried to make this as comprehensive as possible and made sure there's no loss of equivalence in any step.
I hope someone clarifies this for me, it's really bothering me for a while.
Cbrt(x+1) - cbrt(x-1) = pow((x²-1),1/6)
Here, the square root is defined in the domain x≤-1 OR x≥1
I substituted a= cbrt(x+1) and b= cbrt(x-1)
And then the equation turned out to be
a-b= sqrt(ab)
Here, since the RHS is +ve, the LHS must be too, so a>b (Done so as to prevent extraneous roots)
Now, I can write the domain in terms of a and b as ab≥0
Squaring, we get a²+b²-3ab=0
On solving we get a/b = (3±sqrt(5))/2
Now, we know that a>b from the above condition, so the only root is a/b= (3+sqrt(5))/2 and on substituting for x, and applying componendo and dividendo after cubing, I get x= sqrt(5)/2
Now, my question is, since the equation is even i.e f(x)=f(-x), it's symmetric about the y-axis, so x= -sqrt(5)/2 must also satisfy the equation, and indeed it does, so where did I made some irreversible step that caused this loss of the root?
when I plugged the equation to wolframalpha, it showed that the "principle root" has the former solⁿ only where's for a "real valued root", it has both.I also found that on solving the equation
b-a= sqrt(ab), I get the latter root.But I mean where did that extra -ve sign come from?
I'm familiar that f(x)=sqrt(g(x)) is equivalent to f²(x)=g(x) AND f(x)≥0 AND g(x)≥0 but the g(x)≥0 condition is not necessary as f²(x) is already ≥0
and also that abs(f(x))=abs(g(x)) is equivalent to f²(x)=g²(x)
PS: Pardon me I don't know latex well, I'm tried to make this as comprehensive as possible and made sure there's no loss of equivalence in any step.
I hope someone clarifies this for me, it's really bothering me for a while.
3 replies
No more topics!
concurrency in a quadrilateral, midpoints (2010 Euler Teacher's MO I p11)
parmenides51 0
Prove that in every quadrilateral the segments connecting the midpoints its opposite sides, as well as the segment connecting the midpoints of the diagonals pass through one point and divide it in half.
0 replies
concurrency in a quadrilateral, midpoints (2010 Euler Teacher's MO I p11)
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