ka April Highlights and 2025 AoPS Online Class Information
jlacosta0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.
WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.
Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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)!
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NC State Math Contest Wake Tech Regional Problems and Solutions
mathnerd_1014
Nan hour ago
by Aaronjudgeisgoat
Problem 1: Determine the area enclosed by the graphs of Hint
The key to this problem is to just plug in some points in the LHS (I assume you guys already know how to graph the RHS). This lets you see what the graph is like, since sum of absolute values can be wonky sometimes.
Solution to P1
Funnily enough, I managed to silly P1 in-contest! Goes to show how well I did under a time crunch.
IMAGE
The solution to this is by graphing out both functions. Doing this gives us a pentagon, which we can find by complementary area. Note that we have a larger trapezoid and two smaller triangles. By finding the coordinates of the intersections, we get that our answer is
Problem 2: Calculate the sum of the real solutions to the equation Hint
Do the substitution and then factor. Easy as that.
Solution to P2
The key to this is to substitute Doing this, we get that Now, we can solve this by factoring. By using RRT, we get that is a factor. Upon division, we get is the factored form. Now, we know that our only positive solutions are so our values of are Summing these gives our answer of
Remark: In contest, I started doing some crazy algebraic manipulation that was far too complicated. I overthought and wasted a lot of time. How lovely, given time was my biggest restraint.
Problem 3: List the two transformations needed to convert the graph to Hint
I have no idea what to say here. Just... learn transformations...
Solution to P3
Wow so I got this VERY wrong. Looks like I need to brush up on 8th-grade math (as does virtually everyone else I know haha). Essentially you just plug in the answer choices and see where it gets you. The correct answer ends up being "Shift right by 3, Vertical dilation by a factor of 3" because you just substitute in or something like that idk my friend explained it pretty wonkily idk if he even solved it right but anyhow we press on. Transformations suck.
Problem 4: Let be positive integers such that and Determine the value of Hint
Remember your factorizations, kids. Also, it's helpful to always manipulate the desired expression first.
Solution for P4
By rewriting the denominator, we get that it is equal to Furthermore, note that Thus, the answer is just
Problem 5: Eve and Oscar are playing a game where they roll a fair, six-sided die. If an even number occurs on two consecutive rolls, then Eve wins. If an odd number is immediately followed by an even number, Oscar wins. The die is rolled until one person wins. What is the probability that Oscar wins? Hint
Don't overthink it like I did.
Solution to P5
If the first roll is odd, Oscar wins (Think about it. There is NO other way that there can be two evens before Oscar wins.) If the first roll is even, there's a 50% chance that the next roll is odd, and thus Oscar wins yet again. Thus, the probability is just
Problem 6: In triangle is on point such that and and is a point on such that and Given that calculate Hint
Similar triangles.
Solution to P6
By similar triangles, we get that so Cross multiplying and solving this quadratic, we get that our solution is Thus,
Problem 7: Determine the sum of the zeroes of the quadratic of polynomial given that Hint
Substitute and boom system of equations
Solution to Problem 7
Denote We know from that so substituting this gives us Now, plugging this in to and gives us Solving these equations gives us Thus, by Vieta's, the sum of the roots is
Problem 8: Hint
Remember median properties, unlike my clown ahh
Solution to P8
I WANT TO BEGIN BY SAYING THAT I SPENT 5 MINS ON THIS PROBLEM, DIDN'T SOLVE IT AND JUST SOLVED IT IN <1 MINUTE.
Ok, anyhow. IMAGE
Note that thus by median properties (the 2:1 ratio split thingy), we know that (For reference, I missed this part in-contest) Now, we can calculate by the Pythagorean Theorem to be and thus Now, by Pythagorean theorem, we know that the altitude from is so our area is just
Problem 9:
Find the sum of all real solutions to Hint
Spam log properties
Solution to P9
Note that we can simplify this to Thus, simplifying and substituting we get that Simple guess and check gives us as our solutions. My dumb self forgot the former solution. Thus, giving us as our sum.
Problem 10:
Define the function
Calculate .
Hint
Try values of that are close to 100
Solution to P10
Trying we get that Now if we try finding we will realize that this pattern continues until we once again reach thus Now return to the problem statement; but each inner function will just become hence the answer is
Problem 11:
Let be real numbers such that Find Hint
Abuse log properties
Solution to P11
By multiplying the first two equations, we get that Thus, simplifying, we get that or that Plugging this in to the first equation, we get that so this means that Now, thus
Note: HOW DID I MISS THIS IN CONTEST WHAT WAS I THINKING... WAS I EVEN THINKING?!
Problem 12: Points are on circle such that and Determine the path length from to formed by segment and arc
Hint
Draw a good diagram
Solution to P12
Yeah so basically we note that is the diameter, thus we can draw which is a right triangle with and so by Special Right Triangles we do NOT in fact get but rather
Problem 13: Determine the number of integers such that the expression is also an integer. Hint
Substitute .
Solution to Problem 13
Substituting gives us that we need to be an integer. Plugging in factors of we see that there are integers that work.
Problem 14: Determine the smallest positive integer such that is a multiple of
Hint
Legendre's
Solution to Problem 14
By Legendre's, we know that Plugging in numbers gives us as our smallest value.
Problem 15: Suppose and are real numbers such that and Calculate
Funnily enough, I guessed this question right in contest.
Hint
Algebraic Manipulation on Sum of Cubes
Solution to Problem 15
Note that we can rewrite Furthermore, so meaning that . Thus, and thus giving us
Problem 16: A sequence of points will follow the rules such that How many sequences are possible such that is the only point with equal coordinates?
Hint
It's essentially a complementary counting problem
Solution to P16
By the scenario in Bertrand's ballot theorem, we know that our solution is just Furthermore, we know that so our answer is just
Problem 17 (With a solution shamelessly stolen from akliu):
Calculate
Hint
Binomial Theorem
Solution to P17
The first thing here is to notice the in the summation. Essentially, we have . This is pretty recognizable if you get enough practice with choose functions. From here, notice that our sum is:
In other words, we want to find . Consider the polynomial . You might remember that the coefficient for in such a polynomial is . Since we only want the even terms, we can eliminate the odd terms by computing , and then dividing by . For a more simple demonstration, consider finding the even terms of . We have , and . Adding, we removed the term, and have double the sum of the coefficients of and , which both have even powers. This is also just a special case of Roots of Unity Filter, but that's completely unnecessary here. Now, just compute and divide by to get a total sum of . Since we don't want the terms and in our sum, subtract by and factor: . Now, doesn't that term look kind of weird? Well, we actually notice , or . Indeed, we can factor our summation value: , and dividing by , we have is our answer!
(Again, thank you akliu for both the problem statement and solution)
Problem 18: (Also stolen from akliu's blog post)
Calculate
Hint
Consider the cycles between even and odd values of
Solution to Problem 18
It's important to consider what plugging in gives us. Notice that the term on the inside, , just cycles in value every terms. Drawing it out on the unit circle (which I heavily advise), it cycles between and . Now, consider what each term individually is: For certain where , we have , or . For , we have for when is odd. Writing out these terms and caseworking on , we can manually compute the sum. It's not that bad, it's just terms! Anyways, the answer is .
Problem 19: Determine the constant term in the expansion of
Hint
Binomial Theorem
Solution to P19
We know that it'll just be the term where it's so our coefficient is just
Problem 20:
In a magical pond there are two species of talking fish: trout, whose statements are always true, and \emph{flounder}, whose statements are always false. Six fish -- Alpha, Beta, Gamma, Delta, Epsilon, and Zeta -- live together in the pond. They make the following statements:
Alpha says, "Delta is the same kind of fish as I am.''
Beta says, "Epsilon and Zeta are different from each other.''
Gamma says, "Alpha is a flounder or Beta is a trout.''
Delta says, "The negation of Gamma's statement is true.''
Epsilon says, "I am a trout.''
Zeta says, "Beta is a flounder.''
How many of these fish are trout?
Hint
Idk man this is just a standard logic problem
Solution to P20
By evaluating the logic, we get that Alpha, Delta, Epsilon, and Zeta are telling the truth. Thus, there are 4 trouts.
SHORT ANSWER QUESTIONS:
1. Five people randomly choose a positive integer less than or equal to The probability that at least two people choose the same number can be written as Find
Hint
Complementary counting
Solution to S1
By complementary counting, we get that the probability is just Thus, our answer is just
2. Define a function on the positive integers using the rule that for For all prime , and for all other Find the smallest possible value of such that
Hint
I honestly have no idea how to not bash this. Sorry.
Solution to S2
By bashing out every single value we get that
3. How many integers can be written as the sum of two distinct, non-negative integer powers of
Huge shoutout to OTIS for teaching me how to solve problems like this.
Hint
Think of it in base . We can have etc.
Solution to S3
By noting that we can rewrite the set of values that satisfy the condition in base we get that it is Thus, we note that for an integer the number of integers that are length or below is Since we know that all values of satisfy our conditions. Thus, our answer is just
4. Let be the set of positive integers of such that for some other positive integer Find the only three-digit value of in Hint
Pell equations
Solution to S4
We wish to find a fundamental solution. By plugging and chugging, we get that is our solution. For these Pell Equations, all solutions are of the form for Thus, we plug in giving us as our answer.
5. Let be a positive integer and let be the integer that is formed by removing the first three digits from Find the value of with least value such that Hint
I honestly have no idea what hint to give here... sorry.
Solution to S5
Let the number of digits of be and let be our three-digit number that we remove. Thus, we can write By our condition, we know that or that Furthermore, we know that does not work, so we try We know that must have the last two digits by the fact that it must be divisible by a Noting that doesn't work, we try or that This, miraculously, works, and thus our value of is just