Whether you landed here out of curiosity about the toughest problems in math, or you’re trying to get better at hard math problems for a class or competition, the same approach works every time: break the problem into steps, find the key idea, and follow it all the way through.
What Makes These Problems Hard
Ask ten mathematicians what the hardest math problem is and you’ll get ten different answers, because problems that are genuinely hard rarely look intimidating on the surface. Most of them are short, clean statements that hide a single trick. That’s exactly what makes them feel deceptively difficult the first time you read them, and it’s also what makes them satisfying once you find the way in.
Circle Geometry and the Extended Law of Sines
Many hard geometry problems come down to one relationship: the Extended Law of Sines. For any triangle with a side of length s opposite an angle θ, the circumradius R satisfies s / sinθ = 2R. Whenever a problem ties circumradii to changing angles at a moving point, this formula is almost always the way in.
Math tip
Write out the radius formula for every triangle in the problem before you try to compare them. Seeing all the expressions side by side usually reveals which angles need to match or add up to a fixed total.
Common mistake
Students often test one obvious configuration, confirm it works, and stop there. Locus problems like this frequently have more than one valid answer, including isolated special points such as the orthocenter, not just an entire curve.
Area Between Curves and Integration by Parts
Finding the area between two curves always starts the same way: find where the curves meet, decide which one sits on top across the interval, then integrate the difference. When one of the functions is a variable multiplied by an exponential, that final integral almost always needs integration by parts.
Math tip
Before integrating, plug in a test value from inside the interval to confirm which curve is actually larger. Guessing this incorrectly is one of the fastest ways to end up with a negative area or the wrong answer choice.
Common mistake
Sign errors are extremely common when integrating by parts with a negative exponential such as e to the negative x. Write each term out fully rather than trying to combine steps in your head.
Infinite Products and Geometric Series
An infinite product of roots looks intimidating, but converting every root into a fractional exponent turns the whole expression into a single power with a sum in the exponent. If that sum happens to be a geometric series, you can find its exact limiting value instead of guessing whether the product grows without bound.
Math tip
As soon as you see repeated roots like a square root, a fourth root, an eighth root, and so on, rewrite everything using fractional exponents immediately. The pattern underneath becomes far easier to spot.
Common mistake
Do not assume a product of infinitely many terms greater than 1 must diverge. If the exponents shrink fast enough, as they do in a geometric series, the product settles on a finite limit.
Tips and Tricks for Tackling Hard Math Problems
Beyond the specific techniques above, these habits separate students who successfully solve hard problems from those who get stuck.
| Tip | Description |
|---|---|
|
Rewrite before you compute |
Roots, radians, and repeated fractions often hide their structure. Convert them into a consistent form (like fractional exponents or a single variable) first. Half the difficulty disappears once the problem is written cleanly. |
|
Test a simple case first |
Plug in an easy number, a degenerate triangle, or an endpoint before solving the general case. This helps you understand what the answer should look like and catches sign errors early. |
|
Work backward from answer choices |
On multiple-choice questions, test promising answers directly in the original condition. This is often faster than solving from scratch, especially under time pressure. |
|
Separate setup from computation |
First decide the method (Extended Law of Sines, integration by parts, geometric series, etc.), then do the algebra. Mixing both steps is where most careless errors happen. |
|
Sanity-check the final answer |
A probability above 1, negative area, or a limit outside the possible range usually means an error earlier. Always do a quick reasonableness check. |
|
Don’t stop at the first valid case |
Especially in locus or “find all” problems — once you find one solution, ask if there are other possibilities (like the orthocenter in addition to the circumcircle). |
Practice Questions with Video Solutions
Work through each of these problems first, then watch the step-by-step video solution to check your reasoning. Together they cover geometry, calculus, and limits.
Question 1
Let ABC be a triangle in a plane Σ. Find the set of all points P (distinct from A, B, C) in the plane Σ such that the circumcircles of Δs ABP, BCP, and CAP have the same radii.
Solution
Let triangle ABC have circumradius R. For any point P, applying the Extended Law of Sines to each sub-triangle gives:
AB / sin(angle APB) = 2 x R1, BC / sin(angle BPC) = 2 x R2, CA / sin(angle CPA) = 2 x R3
The condition R1 = R2 = R3 means:
AB / sin(angle APB) = BC / sin(angle BPC) = CA / sin(angle CPA)
1. Points on the circumcircle
First, test P on the circumcircle of ABC. By the inscribed angle theorem, a point on the circumcircle sees each side under an angle equal to (or supplementary to) the angle at the opposite vertex of the original triangle:
- angle APB = angle ACB (or 180° – angle ACB)
- angle BPC = angle BAC (or 180° – angle BAC)
- angle CPA = angle ABC (or 180° – angle ABC)
Since sin(180° – x) = sin(x), substituting these in gives:
AB / sin C = BC / sin A = CA / sin B
This is just the Law of Sines for triangle ABC itself, so it is automatically true. Every point on the circumcircle other than A, B, and C satisfies the condition, because all three sub-triangles share the very same circumcircle as ABC.
2. The orthocenter H
Next, test the orthocenter H. A classical property of the orthocenter is that the circumradius of each of the triangles BHC, CHA, and AHB equals the circumradius R of the original triangle ABC.
This follows from the fact that reflecting H over any side of the triangle lands exactly on the circumcircle, which forces each sub-triangle’s circumscribed circle to have the same radius R as the original. So H also satisfies the condition.
3. Uniqueness
To confirm no other points work, consider that for a generic point P off both the circumcircle and the orthocenter, the three angles at P no longer relate to A, B, and C in a way that forces the three ratios to match. The equal-radius condition acts as three independent constraints, which pins down only the circumcircle and the single point H as solutions.
Conclusion
The set of all such points P is the circumcircle of ABC (excluding A, B, and C), together with the orthocenter H.
Key rule
Whenever a problem ties circumradii to a moving point, check the circumcircle of the original triangle first, then check the orthocenter separately.
Think10x.ai video explaining a circumcircle and orthocenter geometry problem
Question 2
The area of the region enclosed by the curves y = xeˣ and y = xe⁻ˣ and the line x = 1 is
A. 1/e B. 1 − 1/e C. 2/e D. 1 − 2/e
Solution
Start by finding where the curves intersect. Setting xeˣ = xe⁻ˣ gives x(eˣ − e⁻ˣ) = 0, so x = 0 or eˣ = e⁻ˣ. The second equation only holds at x = 0 as well, since eˣ and e⁻ˣ only coincide there. So the two curves meet just once, at the origin, and the enclosed region runs from x = 0 to x = 1.
Next, check which curve sits on top across that interval. Plugging in x = 1: e¹ ≈ 2.718 while e⁻¹ ≈ 0.368, so xeˣ is the larger function throughout (0, 1). The area is:
Area = ∫₀¹ (xeˣ − xe⁻ˣ) dx
Evaluate each integral by parts separately.
For ∫xeˣ dx, let u = x and dv = eˣ dx, so du = dx and v = eˣ:
∫xeˣ dx = xeˣ − ∫eˣ dx = xeˣ − eˣ
Evaluated from 0 to 1: (1·e − e) − (0 − 1) = 0 − (−1) = 1
For ∫xe⁻ˣ dx, let u = x and dv = e⁻ˣ dx, so du = dx and v = −e⁻ˣ:
∫xe⁻ˣ dx = −xe⁻ˣ − ∫(−e⁻ˣ) dx = −xe⁻ˣ − e⁻ˣ
Evaluated from 0 to 1: (−1·e⁻¹ − e⁻¹) − (0 − 1) = −2/e − (−1) = 1 − 2/e
Subtracting the two results:
Area = 1 − (1 − 2/e) = 2/e
Answer: C. 2/e
Key rule
Confirm which curve sits on top before subtracting, then write out each integration by parts step fully.
Think10x.ai video explaining the area between exponential curves (Step-by-Step)
Question 3
The limit of the product ²√5, ⁴√5, …, ²ⁿ√5 as n approaches infinity is
A. 1/5 B. log₁₀ 5 C. 1 D. 5
Solution
Rewrite each root as a fractional power of 5. The square root is 5^(1/2), the fourth root is 5^(1/4), the eighth root is 5^(1/8), and in general the term with a 2ⁿ root is 5^(1/2ⁿ). Multiplying terms with the same base means adding exponents, so the whole product becomes:
5^(1/2 + 1/4 + 1/8 + … + 1/2ⁿ)
The exponent is a geometric series with first term 1/2 and common ratio 1/2. As n approaches infinity, the sum of an infinite geometric series with |ratio| < 1 approaches a/(1−r), which here gives (1/2)/(1 − 1/2) = 1. So the exponent approaches 1, and the whole expression approaches 5¹ = 5.
Answer: D. 5
Key rule
Whenever you see a chain of nested roots, convert everything to fractional exponents first, then look for a geometric series hiding in the exponent.
Think10x.ai video explaining an infinite geometric series with exponents (Step-by-Step)
Key Takeaways
- The Extended Law of Sines connects a triangle’s side, its opposite angle, and its circumradius through s / sinθ = 2R.
- Locus problems can have more than one type of answer, including both a full curve and a separate isolated point.
- Always confirm which curve is larger before subtracting when finding the area between two curves.
- Integration by parts is the standard tool whenever a polynomial is multiplied by an exponential inside an integral.
- Converting roots to fractional exponents is the fastest way to simplify an infinite product.
- A geometric series in the exponent means the product converges to a finite value rather than growing without bound.
- Most hard math problems open up once you find the first solid foothold, not because you saw the entire path in advance.
Frequently Asked Questions
There’s no single agreed-upon answer, since it depends on the field you’re looking at. In research mathematics, unsolved Millennium Prize problems often get that title, while in a classroom setting, it might just mean the toughest question on your next test.
No. Plenty of hard math problems show up in ordinary high school and college courses. They’re hard because of a hidden trick or an easy-to-miss case, not because they require years of specialized training.
Difficulty in math rarely comes from length. Short statements often leave room for several possible approaches, and picking the right one is usually the hard part.
Focus on a small set of core techniques rather than memorizing individual problems. The Extended Law of Sines, integration by parts, and geometric series sums appear across many different-looking hard problems, so mastering these tools generalizes far better than memorizing solutions.
Completely normal. Even experienced problem solvers rarely see the entire path from the first step. Progress usually comes from testing a promising idea and adjusting from there.
Think10x.ai can turn a photo of almost any math problem into a narrated, animated video explanation in minutes, which is exactly how the video solutions in this guide were created.
Need Help with Your Own Question?
Upload a clear photo to Think10x.ai and get an instant video explanation. Built for tutors and students. Private by default.
Try Think10x.ai for free