99% of people can't solve this math problem... Can you??🤔🤔🤔

I can’t explain my answer cuz I suck at explaining stuff but it’s 1/3.

Here's a good one: Prove that sqrt(2) is an irrational number.

1/infinity would be 0, also it is the probability that the chord is LESS THAN the RADIUS.

that's incorrect, because 50% of 180 is not 60, and thus the chord would be longer than the other sides of the isosceles triangle (and thus longer than the radius).

Crystal what was your method for the original question?

Wow

so let the center be C, select A and B on the circle. angle CBA = x and is congruent to angle BCA. angle BAC = y (originally I used theta and alpha but x and y are easier on chess.com). 2x+y=180

2x<180 thus x<90. y>0 otherwise x = 90 which is not in the domain. x>=y because y must be smaller than or equal to x for BC<r. y=180-2x (rearranged equation 1). thus x>=180-2x thus x>=60. the domain is {60<=x<90}, thus the range is {0<y<=60}. when you randomly chose point A, you must chose point B to make an arc that measures less than or equal to 60 degrees, thus there is a 120 degree arc of possible choices 120/360 = 1/3.

the probability is 1/3.

yes crystal0192 is correct

I checked his steps.

good, i though some of them might be hard to understand cuz im bad at explaining

i think its time to BUMP this tho

crystal or anyone try answer my question in #22

i used to know it but not anymore

Here is my proof:

let sqrt(2)=a/b where a and b are natural numbers, because sqrt(2) is positive.

a^2/b^2=2, thus a^2/2=b^2

a^2 < b^2 because for any positive numbers x=y+z, x>y and x>z, and since b^2=2a^2, b^2>a^2

sqrt(a^2)<sqrt(b^2), a<b. for any natural numbers x and y such that y<x, y/x<1. since we know that sqrt(2) is greater than 1. the sqrt(2) cannot be expressed as a/b, thus it is irrational.

i made it up so it probably is weak but it is my best try.

Here is a new one:

2^x+2x=5

its very difficult ill reveal the solution tomorrow

If the maximum chord length is 2r and the minimum chord length pretty much asymptotes at 0… you have an infinite amount of values in between 2 and zero. So one over infinity. (Not done yet). That’s not the answer in practice, though. Given there is only a certain amount of area on all sides of the circle inside of which the chord’s length (AB) is less than that of the radius, idk. This is getting too complicated… in summary, the probability is ~50%. (Slightly less than due to the small probability of AB= r) The radius essentially divides all possible lengths of the chords into two groups: AB>r and AB<2. Since there are only two groups for AB to fall into, the probability of it falling into any specified one of those halves would be 1/2 or 50%.

that is incorrect, the correct probability is 1/3. you can see the two proofs in previous posts.

good try though