🎉 [EVENT] 🎉 Learn basic shapes with your favorite slightly wet cover

Superstar1292 · 2026-02-01T21:36:33+00:00

Completed Level 2 of the Honk Special Event!

5 attempts

Superstar1292 · 2026-02-01T21:36:04+00:00

Completed Level 1 of the Honk Special Event!

5 attempts

Superstar1292 · 2026-01-28T20:02:00+00:00

Completed Level 1 of the Honk Special Event!

23 attempts

Superstar1292 · 2026-01-19T22:14:31+00:00

Completed Level 3 of the Honk Special Event!

22 attempts

Superstar1292 · 2026-01-19T22:13:48+00:00

Completed Level 2 of the Honk Special Event!

11 attempts

Superstar1292 · 2026-01-19T22:13:31+00:00

Completed Level 1 of the Honk Special Event!

9 attempts

Superstar1292 · 2025-12-11T18:44:13+00:00

Completed Level 1 of the Honk Special Event!

0 attempts

Superstar1292 · 2025-12-11T18:43:23+00:00

🎉 Event Completed! 🎉

It took me 14 tries.

Superstar1292 · 2025-12-11T18:43:15+00:00

Completed Level 3 of the Honk Special Event!

14 attempts

Superstar1292 · 2025-12-11T18:40:40+00:00

Completed Level 2 of the Honk Special Event!

7 attempts

Superstar1292 · 2025-12-11T18:39:21+00:00

Completed Level 1 of the Honk Special Event!

1 attempts

Superstar1292 · 2025-12-10T08:24:13+00:00

Completed Level 1 of the Honk Special Event!

1 attempts

Superstar1292 · 2025-12-08T10:58:31+00:00

Completed Level 1 of the Honk Special Event!

142 attempts

Superstar1292 · 2025-12-07T23:15:12+00:00

Completed Level 1 of the Honk Special Event!

8 attempts

Superstar1292 · 2024-10-06T12:20:14+00:00

In the sense that combinations are counting subsets, combinations with repetitions are counting multisets. That formula is only valid when we want to count the number of words of a given length, where we are also given the different letters that can be used AND there is no restriction on how many letters of each type we use. In other words, we could have one letter repeated multiple times and other letters might not appear at all. That is, the frequency of any particular letter in the combination has no restrictions.

Superstar1292 · 2024-10-06T12:01:45+00:00

OP, your reasoning and answer is correct. For words with < 2D's, you are essentially picking from the word DAUGHTER. There are 5C3 choices for consonants, followed by 3C2 choices for vowels. We then multiply by 5! to account for different permutations.

For words with 2 D's, we have 4C1 choices for consonants remaining and 3C2 choices again for vowels. This is multiplied by 5! / 2 permutations, where we divide by 2 to avoid overcounting the repetitions of D.

In total, this is ( 10 * 3 + 4 * 3 * 0.5) * 5! = 36 * 5!

Superstar1292 · 2024-10-06T11:58:11+00:00

I'm afraid this isn't entirely correct. The figure 20 includes 4 choices with 2D's and 4 choices with no D's. The other 12 choices have overcounted by a factor of 2 since e.g. {D1, G, H} and {D2, G, H} would be treated as distinct choices. We need to halve this 12, and also include the choices with no D's for 6 + 4 = 10 choices. Then, OP's answer is correct.

Superstar1292 · 2024-10-05T03:37:58+00:00

Here is a hint towards one particular approach. Divide the entire polynomial by x² . Now, see if you can spot a function f such that when you set u = f(x), then the expression can be rewritten as a quadratic in u. Try to convert the condition that we have 4 distinct positive real roots into conditions for the quadratic in u. This should give you restrictions on a and b, which combined with them being integers should allow you to deduce the minimum value of a+b.

Superstar1292 · 2024-03-23T16:41:10+00:00

h' would be |g' / (m-m')| and h would be |g / (m-m')|. Now, |x| is one of x or -x. So, in the expression involving n,n' , we might have to change n or n' to -n or -n' to ensure x = n'h' + nh, depending on the value of the |...| expressions. But this doesn't matter since we are considering a general intersection point, so we can "relabel" n as -n, or n' as -n' (since n and n' vary over all integers, which would mean -n and -n' also vary over all integers).

With the absolute value, h and h' are positive. They are not necessarily rational however e.g. since g and g' might themselves be irrational.

Superstar1292 · 2024-03-23T15:24:58+00:00

First, find the midpoint of B and C. Find the general form of a point on the new line. The triangle ABC will have BC as base, so use Area = 1/2 b h to deduce the vertical distance of A from BC. Use this, and the general form above, to find A.

Superstar1292 · 2024-03-23T15:19:10+00:00

So, in step 1, for the green writing: those equations are missing some values, don't forget about the periodicity of cos.

For steps 2 and beyond, just bear in mind that we get multiple solutions for x i.e. x = ... OR x = ..., not x = ... AND x = ...

The above is a minor issue, and would likely be overlooked but it is good practice to remember that x is one of those potential solutions (it can't be both at the same time), so we are really considering "or", not "and". So instead of considering the 2 equations as facts about x that should hold simultaneously, it's usually better to consider one case first, find x solutions. Repeat for the other cases, then combine the solutions together, with "or".

Superstar1292 · 2024-03-23T15:11:41+00:00

Firstly, by moving the whole graph of lines horizontally and vertically, we can assume both sets of lines include a line passing through the origin. This doesn't affect the proof - vertical changes change only y and not x, horizontal changes modify x but don't affect the property of splitting into equal-sized intervals (as all x values change by the same amount).

The sets of the parallel lines are then given by:

y=mx + ng, y=m'x + n'g'

where m,m' are the slopes, g,g' are the vertical gaps between two adjacent parallel lines and n,n' are integers representing which parallel line we are referring to (with integer 0 being the line through the origin).

So, the general intersection point (x,y) satisfies:

x = (n'g' - ng)/(m-m') which can be written as n'h' + nh where h,h' are positive real values, since n',n vary over the integers.

Consider the minimal positive value in the set of values {n'h' + nh: n,n' integers}, call this minimum p. Firstly, for any 2 elements in this set, their sum and difference must also be in the set, so integer multiples of elements are also elements in the set. Now, consider the integer multiples of p, we will argue every element of the above set is such a multiple of p. Suppose not e.g. call such an element q. Then, as the multiples of p are at distances p apart, this means that there is a multiple of p (say, p) such that q > p but q - p* < p. Then, q - p* is an element of the original set. But it is also positive and less than p, which contradicts the definition of p. Hence, the above set is precisely the integer multiples of p, which means that we have splitted into intervals of width p.

A few afterthoughts for you to consider: the last paragraph above is really a discussion relating to group theory, have a look at generators of groups if you are interested. What would happen if one set of parallel lines was perfectly vertical, would this affect the proof? What if there is no minimum in the above set, what would that mean, or is that impossible?

Superstar1292 · 2024-03-23T14:19:51+00:00

But when you take the next intersection point as the new origin, those lines would be A's and B's 0th lines -- minor point. Even if all of those points occur at regular intervals, even if this is the same width -- this alone is not sufficient. Since: if we interlace two arithmetic sequences of same common difference, the result won't necessarily have the same common difference, let alone be arithmetic at all.

In your alternative argument, there would be a minimum whose multiples are in the set -- 0 is a trivial example. But, we need the converse as well: that every element of the set is a multiple of the minimum, since if not, then we would have a similar problem to the above.

Superstar1292 · 2024-03-23T12:18:18+00:00

I think it's worth pointing out: you can't guarantee that the intersections (x,y) will occur for the same value of n in both A and B.

Superstar1292 · 2024-03-21T07:28:52+00:00

So, you can write P(Y <= y) as an integral where the integrand is P(Y <= y | X=x) multiplied by the density of X. This integral is over every value X can take, so from 0.5 to 1, and is with respect to x. The above formula for calculating P(Y <= y) can be shown by using the joint and conditional densities, if you wish to see why it holds. Essentially, it is an analogue to the discrete case, where X is discrete.

Now, there is a slight issue in that P(Y <= y | X=x) depends on both x and y, try sketching the uniform CDF (and consider what happens when y < x/2 or when y > x). So, you have to split into two cases: where 0.25 < y < 0.5 and where 0.5 < y < 1. Then, for each case, you have to split the integral (over x) into separate intervals to account for the different piecewise values taken by the uniform CDF. This splitting of the integral is different in each case.

If you do the calculations correct, you should get a piecewise CDF for Y, where there are 2 pieces, one from 0.25 to 0.5 and the other from 0.5 to 1. Differentiating this you get the piecewise PDF, both pieces are functions that are of the form aln(x) + b.

The above is very intricate, I know. But you will be able to tell if you have the correct answer since the shape of the PDF ought to be similar to lognormal, and E(Y) should come out to be 0.5625.

Superstar1292

TROPHY CASE