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Problems Collection
This is a page where you can share the problems you made (try not to use past exams).
- 1 AMC styled
- 2 AIME styled
- 4.1 Problem 1
- 4.2 Solution 1
- 4.3 Problem 2
- 4.4 Solution 1 (Slow, probably official MAA)
- 4.5 Solution 2 (Fast)
- 4.6 Solution 3 (Faster)
- 4.7 Problem 3
- 4.8 Solution 1(Probably official MAA, lots of proofs)
- 4.9 Solution 2 (Fast, risky, no proofs)
- 4.10 Problem 4
- 4.11 Solution 1
- 4.12 Problem 5
- 4.13 Solution 1 (Euler's Totient Theorem)
- 4.14 Problem 6
- 4.15 Solution 1 (Recursion)
- 4.16 Problem 7
- 4.17 Solution 1 (Tedious Casework)
- 4.18 Solution 2 (Official)
- 4.19 Solution 3 (Official and Fastest)
- 4.20 Problem 8
- 4.21 Solution 1
- 4.22 Problem 9
- 4.23 Solution 1
- 4.24 Problem 10
- 4.25 Solution 1(Wordless endless bash)
- 4.26 Problem 11
- 4.27 Solution 1 (Analytic geo)
- 4.28.1 Solution 2a (Hard)
- 4.28.2 Solution 2b (Harder)
AIME styled
1. There is one and only one perfect square in the form
![art of problem solving promo code \[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]](https://latex.artofproblemsolving.com/b/1/1/b11bf3dce41593092737df0eea5f902de3989389.png)
3.The fraction,
![art of problem solving promo code \[\frac{ab+bc+ac}{(a+b+c)^2}\]](https://latex.artofproblemsolving.com/6/0/7/60701a0dfd9308e7ddc45f43e5380a0a359f23e1.png)
Someone mind making a diagram for this?
![art of problem solving promo code \[\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]=\frac{p}{q}\]](https://latex.artofproblemsolving.com/d/e/a/dea95e94bb54f5b043a09ff92662c3c99bbd3632.png)
There is one and only one perfect square in the form
![art of problem solving promo code $(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2 \cdot q^2 +p^2+q^2+1-p^2 \cdot q^2 +pq-1=p^2+q^2+pq$](https://latex.artofproblemsolving.com/b/0/7/b07b71756dbffa657a8077f8325a20d3de06f4b1.png)
Solution 1 (Slow, probably official MAA)
![art of problem solving promo code \[m^2=2^8+2^{11}+2^n\]](https://latex.artofproblemsolving.com/8/b/a/8baacd6b7665fb5a4e91932827462a952b1ae613.png)
Solution 2 (Fast)
![art of problem solving promo code $(a + b)^2$](https://latex.artofproblemsolving.com/f/6/1/f613327478b9be645ed8ce93b117b0464d4e2c82.png)
Solution 3 (Faster)
![art of problem solving promo code $256 + 2048 + 2^n = 2304 + 2^n = m^2$](https://latex.artofproblemsolving.com/b/d/b/bdbda63099b4e475014eb3ad981b16fa65812aa0.png)
~ (also) cxsmi
The fraction,
Solution 1(Probably official MAA, lots of proofs)
![art of problem solving promo code $\text{max} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{3}$](https://latex.artofproblemsolving.com/9/3/2/932ef12316c463ca3a734c1dcedc3d0247a2d148.png)
Proof: By the Triangle Inequality , we have
![art of problem solving promo code \[a+b>c\]](https://latex.artofproblemsolving.com/e/b/c/ebce22c7cc803ecbdc0c13c86aa51257a92b85f5.png)
Add them together gives
![art of problem solving promo code \[a^2+b^2+c^2<c(a+b)+a(b+c)+b(a+c)=2(ab+bc+ac)\]](https://latex.artofproblemsolving.com/6/3/1/6315aa9383e20276b54e0998d5640121638ed6ea.png)
Solution 2 (Fast, risky, no proofs)
![art of problem solving promo code $a,b,c$](https://latex.artofproblemsolving.com/a/5/b/a5b29b358e825defa3e11b7d903d43ec31e5909a.png)
To make things even simpler, let
![art of problem solving promo code \[a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}\]](https://latex.artofproblemsolving.com/9/f/8/9f833befc986bc6c40ad6e99219cef82c1025df7.png)
Note: If you don't know Newton's Sums , you can also use Vieta's Formulas to bash.~ Ddk001
Solution 1 (Euler's Totient Theorem)
![art of problem solving promo code $2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:$](https://latex.artofproblemsolving.com/8/5/f/85f7220e4801d3d01a1f3a5a5a02c1e9f91e641d.png)
where the last step of all 3 congruences hold by the Euler's Totient Theorem . Hence,
![art of problem solving promo code \[x \equiv 1 \pmod{5}\]](https://latex.artofproblemsolving.com/0/5/c/05c2d5aa9f1e6c39f502e7c33bc7f6aef1a8ef40.png)
(ii) No bigger rings are on top of smaller rings.
Solution 1 (Recursion)
![art of problem solving promo code $M_n$](https://latex.artofproblemsolving.com/e/b/1/eb12e06e542e73a4fa3b7237d8be00b2bef219c5.png)
Solution 1 (Tedious Casework)
![art of problem solving promo code $a>b$](https://latex.artofproblemsolving.com/0/8/2/082852571ecc48ed136ba13dcffa8f6e40064bbd.png)
In this case, we have
![art of problem solving promo code \[\overline{ab}^2=a! +b!=(1+a \cdot (a-1) \cdot \dots \cdot (b+1)) \cdot b! \implies b!|\overline{ab}^2=(10a+b)^2\]](https://latex.artofproblemsolving.com/d/6/6/d66bc68298f08f74f02e1d5c7a55f96f1385d389.png)
In this case, we have that
![art of problem solving promo code \[a! \equiv \overline{ab}^2-b! \equiv (10a+1)^2-1 \equiv 0 \pmod{10} \implies 10|a! \implies a \ge 5\]](https://latex.artofproblemsolving.com/d/a/9/da936f2fd27de182ddc41454512bb74d8c5df17e.png)
There is no apparent contradiction here, so we leave this as it is.
![art of problem solving promo code $a>b=2$](https://latex.artofproblemsolving.com/1/e/a/1ea77dbd5f4addcab72b4c32598b227431b14f87.png)
To simplify future calculations, note that
![art of problem solving promo code \[a!=\overline{ab}^2-b!=(10a+1)^2-1=100a^2+20a=10a(10a+2)\]](https://latex.artofproblemsolving.com/2/6/b/26b728a7da1ae0c1e0d4d8d8242f2392e3d2923d.png)
For this case, we must have
![art of problem solving promo code \[(11a)^2=\overline{ab}^2=a!+b!=2a! \implies 11|a!\]](https://latex.artofproblemsolving.com/9/b/3/9b389fcce14b2e05053d66793af4c945d55a81c0.png)
Solution 2 (Official)
![art of problem solving promo code $100(10^{2})$](https://latex.artofproblemsolving.com/c/2/1/c21aa708e299817871979cedb78693c2c5949393.png)
Hence unit digit of RHS is 0,1,2, 6 or 4. 0,2,4 and 6 are rejected as follows:-
1 . 2 can't be the unit digit of a perfect square.
3 . If 0 is the unit digit of LHS then 50 60 70 are the only cases (as one of the digits is greater than or equal to 5) that don't satisfy
![art of problem solving promo code $\boxed{008}$](https://latex.artofproblemsolving.com/e/3/5/e35637d1dc49de532ea171d5a8bf2ea9421f6a3d.png)
Solution 3 (Official and Fastest)
![art of problem solving promo code $(mod 4)$](https://latex.artofproblemsolving.com/0/b/a/0bab117f6c5e1b4d0741579828038453df5b7802.png)
which is very easy to calculate and get 71 as the only possible solution to the problem and
![art of problem solving promo code $f(n)-n=0$](https://latex.artofproblemsolving.com/6/f/c/6fc334f56a835d7c2b8d03b20d52161f52eab9af.png)
Now, notice that
![art of problem solving promo code \[m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})\]](https://latex.artofproblemsolving.com/d/e/c/dec580ee947ee9339df60788e9238cda0107a5f0.png)
Similarly, we have
![art of problem solving promo code \[(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}\]](https://latex.artofproblemsolving.com/b/c/7/bc7aa1aeb2bb9ed1154eb55e463c8be49cca21d4.png)
Now we state a few claims :
![art of problem solving promo code $\Delta O’IO$](https://latex.artofproblemsolving.com/b/9/a/b9adcf011ea516d87af91039a3d7eb52a4ab32bf.png)
where the last equality holds by the Power of a Point Theorem .
![art of problem solving promo code $IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2$](https://latex.artofproblemsolving.com/4/c/f/4cfe48f4a729e85b8628988ee45537abd50cd9f1.png)
With this in mind, we see that
![art of problem solving promo code \[2OJ=OO’=OI=OK+KI=OJ+GI=OJ+AC \implies OA=OJ=AC\]](https://latex.artofproblemsolving.com/6/8/8/6880c271ef3184c5dd9cde79b3d9a50047256207.png)
Here, we state another claim :
![art of problem solving promo code $BH$](https://latex.artofproblemsolving.com/4/b/f/4bfc6159e268a0f5d736eca73477c99a13209da9.png)
Now, apply Ptolemy’s Theorem gives
![art of problem solving promo code \[BH \cdot AC+BC \cdot AH=CH \cdot AB \implies BH \cdot AC+AC^2=3AC^2 \implies BH=2AC=2OA\]](https://latex.artofproblemsolving.com/9/2/7/9275ad2689eb3e6d9653fd9abe8db742f9785948.png)
Solution 1(Wordless endless bash)
![art of problem solving promo code \[\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}\]](https://latex.artofproblemsolving.com/6/d/9/6d93ad9118f7e604ce688ecee8b89f62354f719a.png)
Solution 1 (Analytic geo)
![art of problem solving promo code \[A=(0,0)\]](https://latex.artofproblemsolving.com/c/2/9/c29c3304210784e0db1a8233bf2a676e8e7013fd.png)
Now, we see that
![art of problem solving promo code \[\text{Slope} _ {AF}=\frac{b}{2a+2 \sqrt{a^2+b^2}}\]](https://latex.artofproblemsolving.com/3/a/0/3a025d1d9a54667a4a52e73f48b507f18fe3417d.png)
Solution 2 (Hard vector bash)
Solution 2a (hard).
![art of problem solving promo code \[\overrightarrow{AF} \cdot \overrightarrow{BE}\]](https://latex.artofproblemsolving.com/8/a/8/8a89684d1ca608a40dd5adce69102f5cc183176d.png)
Solution 2b (Harder)
![art of problem solving promo code \[\angle ACD=\angle ECD\]](https://latex.artofproblemsolving.com/7/f/9/7f92db2ec536f9e45f8c6d1b80fe7483eaffadfa.png)
Now come the coordinates. Let
![art of problem solving promo code \[B=(-a,-b)\]](https://latex.artofproblemsolving.com/c/0/5/c05deb9f221ecd31e16d9cf1b70ce79128e5b8d7.png)
Here's the source for the problems:
1,2,3,4,5,6,8,9,10,11: Ddk001 , credits given to Ddk001
7: SANSKAR'S OG PROBLEMS , credits given to SANSGANKRSNGUPTA
- Note: Problem 6 is based on the Tower of Hanoi Problem
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![art of problem solving promo code art of problem solving promo code](https://www.artofproblemsolving.org/images/logos/corner.png)
About the Art of Problem Solving Initiative
Founded in 2004, the Art of Problem Solving Initiative, Inc. was created by people who love math and love teaching to help students access the study of advanced mathematics.
The Initiative began by running the USA Mathematical Talent Search , a nationwide math contest sponsored by the National Security Agency which continues to run to this day. In each round, this contest gives students a full month to work on five challenging proof-based problems. Students then get individual feedback on their work, including comments on their mathematical reasoning and on their proof-writing skills. By taking away the usual time pressures seen in many math contests, students have the opportunity to go more deeply and to explore.
For many years, the Initiative also ran the Local Programs Initiative which provided fiscal sponsorship to math circles and clubs across the country.
Now, the main project of the Initiative is Bridge to Enter Advanced Mathematics (BEAM) , a project to help underserved students find a realistic pathway towards becoming scientists, mathematicians, engineers, and programmers. During the summer after 7th grade, students are invited to participate in a free three-week residential program on a college campus where they gain the academic and social/emotional preparation to succeed in future programs for advanced study. They then receive academic advising throughout 8th grade and high school to help them and their families find the best opportunities for their educations.
The Art of Problem Solving Initiative receives support from Art of Problem Solving (AoPS) , which develops resources for high-performing middle and high school students including the largest online community of avid math students in the English-speaking world. The AoPS online school has trained many winners of major national mathematics competitions, including several gold medalists at the International Math Olympiad, Davidson Fellows, and winners of the Intel and Siemens Talent Search competitions.
The Art of Problem Solving Initiative is a 501(c)(3) charitable organization. It continues to be led by expert mathematicians and educators to this day. Please consider joining our contributors by donating to support advanced math education.
Financial Information
The Art of Problem Solving Initiative, Inc. makes its financial statements available for public inspection in the interest of greater transparency. Please see below to learn more about the organization.
- Tax return (Form 990)
- Audited financial statements
- Tax return (Form 990) (Note that due to a change in fiscal year these cover a shortened period.)
![art of problem solving promo code Art of Problem Solving Initiative](https://www.artofproblemsolving.org/images/footers/bottom_ini.png)
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Art of Problem Solving offers two other multifaceted programs. Beast Academy is our comic-based online math curriculum for students ages 6-13. And AoPS Academy brings our methodology to students grades 2-12 through small, in-person classes at local campuses. Through our three programs, AoPS offers the most comprehensive honors math pathway ...
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Art of Problem Solving. 37,564 likes · 78 talking about this. AoPS produces books, classes, and other materials for outstanding math students.
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Our virtual campus combines Beast Academy curriculum with a live, video-based classroom. Students are guided by expert instructors and learn alongside 10-16 curious peers. Beast Academy is the elementary school math program created by Art of Problem Solving (AoPS), a global leader in advanced K-12 math education.
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Starting at $50/week. Enroll Today. As seen in. Since 1993, Art of Problem Solving has helped train the next generation of intellectual leaders. Hundreds of thousands of our students have gone on to attend prestigious universities, win global math competitions, and achieve success in highly competitive careers.
The training I received in math from AoPS was absolutely crucial to helping me develop problem-solving skills in physics. At the Olympiad level, most problems can't be solved with only one concept-cracking them usually requires putting multiple concepts together, and the WOOT program really helped me develop this essential skill.
Art of Problem Solving has been a leader in math education for high-performing students since 1993. We launched AoPS Academy in 2016 to bring our rigorous curriculum and expert instructors into classrooms around the United States. With campuses in 8 states (and growing!), our approach nurtures a love for complex problem solving, which is fully ...
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The Art of Problem Solving, Volume 2, is the classic problem solving textbook used by many successful high school math teams and enrichment programs and have been an important building block for students who, like the authors, performed well enough on the American Mathematics Contest series to qualify for the Math Olympiad Summer Program which trains students for the United States ...
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Since 1993, Art of Problem Solving has prepared hundreds of thousands of motivated students in grades 2-12 for college and career success. Through our innovative approach, students build a problem-solving foundation, an unparalleled skill set that helps them overcome obstacles in school and in life. ...
1. There is one and only one perfect square in the form. where and are prime. Find that perfect square. 2. and are positive integers. If , find . 3.The fraction, where and are side lengths of a triangle, lies in the interval , where and are rational numbers.
Founded in 2004, the Art of Problem Solving Initiative, Inc. was created by people who love math and love teaching to help students access the study of advanced mathematics. The Initiative began by running the USA Mathematical Talent Search, a nationwide math contest sponsored by the National Security Agency which continues to run to this day.
Small live classes for advanced math and language arts learners in grades 2-12.