Difference between revisions of "User:Ddk001"
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− | ==Problems== | + | ==Introduction== |
− | + | I am a 5th grader who likes making and doing problems, doing math, and redirecting pages (see [[Principle of Insufficient Reasons]]). I like geometry and don't like counting and probability. My number theory skill are also not bad. | |
+ | |||
+ | ==User Counts== | ||
+ | |||
+ | If this is your first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1) | ||
+ | |||
+ | <math>\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{18}}}}}}}}}}}}}}}}</math> | ||
+ | |||
+ | For those of you who want more boxes, me too. However, this is the max number of boxes. Also, I check the pages history so I know if someone edited something. | ||
+ | |||
+ | (Please don't mess with the user count) | ||
+ | |||
+ | Doesn't that look like a number on a pyramid | ||
+ | |||
+ | ==Cool asyptote graphs== | ||
+ | |||
+ | Asymptote is fun! | ||
+ | <asy>draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));</asy> | ||
+ | |||
+ | <asy>draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));</asy> | ||
+ | |||
+ | |||
+ | |||
+ | ==Problems Sharing Contest== | ||
+ | Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start: | ||
1. There is one and only one perfect square in the form | 1. There is one and only one perfect square in the form | ||
− | < | + | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> |
where <math>p</math> and <math>q</math> are prime. Find that perfect square. | where <math>p</math> and <math>q</math> are prime. Find that perfect square. | ||
− | 2. Suppose | + | <math>\textbf{Solution by cxsmi}</math> |
+ | |||
+ | 1. We can expand the product in the expression. <math>(p^2+1)(q^2+1)-((pq)^2-pq+1) = p^2q^2+p^2+q^2+1-((pq)^2-pq+1) = p^2 + q^2 + pq</math>. Suppose this equals <math>m^2</math> for some positive integer <math>m</math>. We rewrite using the square of a binomial pattern to find that <math>m^2 = (p + q)^2 - pq</math>. Through trial and error on small values of <math>p</math> and <math>q</math>, we find that <math>p</math> and <math>q</math> must equal <math>3</math> and <math>5</math> in some order. The perfect square formed using these numbers is <math>\boxed{49}</math>. | ||
+ | |||
+ | Note: I will be the first to admit that this solution is somewhat lucky. | ||
+ | |||
+ | |||
+ | 2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle <math>\triangle ABC</math> intersects <math>\triangle ABC</math> itself. <math>\triangle ABC</math> has leg length <math>2024</math>. The perimeter of this diamond is expressible as <math>a\sqrt{b}-c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers, and <math>c</math> is not divisible by the square of any prime. What is the remainder when <math>a + b + c</math> is divided by <math>1000</math>? | ||
+ | |||
+ | <asy> | ||
+ | unitsize(1inch); | ||
+ | draw((0,0)--(0,2)); | ||
+ | draw((0,2)--(2,0)); | ||
+ | draw((2,0)--(0,0)); | ||
+ | draw(circle((0.586,0.586),0.586)); | ||
+ | draw((0,0)--(0,1.172),red); | ||
+ | draw((0,1.172)--(1.172,1.172)); | ||
+ | draw((1.172,1.172)--(1.172,0)); | ||
+ | draw((1.172,0)--(0,0),red); | ||
+ | draw((0,1.172)--(0.828,1.172),red); | ||
+ | draw((0.828,1.172)--(1.172,0.828),red); | ||
+ | draw((1.172,0.828)--(1.172,0),red); | ||
+ | draw((0,0.1)--(0.1,0.1)); | ||
+ | draw((0.1,0.1)--(0.1,0)); | ||
+ | label("$A$",(0,2.1)); | ||
+ | label("$B$",(0,-0.1)); | ||
+ | label("$C$",(2,-0.1)); | ||
+ | label("$2024$",(-0.2,1)); | ||
+ | label("$2024$",(1,-0.2)); | ||
+ | </asy> | ||
+ | |||
+ | ==Contributions== | ||
+ | [[2005 AMC 8 Problems/Problem 21]] Solution 2 | ||
+ | |||
+ | [[2022 AMC 12B Problems/Problem 25]] Solution 5 (Now it's solution 6) | ||
+ | |||
+ | [[2023 AMC 12B Problems/Problem 20]] Solution 3 | ||
+ | |||
+ | [[2016 AIME I Problems/Problem 10]] Solution 3 | ||
− | <math>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</ | + | [[2017 AIME I Problems/Problem 14]] Solution 2 |
+ | |||
+ | [[2019 AIME I Problems/Problem 15]] Solution 6 | ||
+ | |||
+ | [[2022 AIME II Problems/Problem 3]] Solution 3 | ||
+ | |||
+ | Restored diagram for [[1994 AIME Problems/Problem 7]] | ||
+ | |||
+ | [[Divergence Theorem]] | ||
+ | |||
+ | [[Stokes' Theorem]] | ||
+ | |||
+ | [[Principle of Insufficient Reasons]] | ||
+ | |||
+ | ==Problems I made== | ||
+ | ===Aime styled=== | ||
+ | ====Introductory==== | ||
+ | 1. There is one and only one perfect square in the form | ||
+ | |||
+ | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> | ||
+ | |||
+ | where <math>p</math> and <math>q</math> are prime. Find that perfect square. | ||
+ | |||
+ | |||
+ | 2. <math>m</math> and <math>n</math> are positive integers. If <math>m^2=2^8+2^{11}+2^n</math>, find <math>m+n</math>. | ||
+ | |||
+ | ====Intermediate==== | ||
+ | 3.The fraction, | ||
+ | |||
+ | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath> | ||
+ | |||
+ | where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>. | ||
+ | |||
+ | |||
+ | 4. Suppose there is complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy | ||
+ | |||
+ | <cmath>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</cmath> | ||
Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>. | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>. | ||
− | |||
− | < | + | 5. Suppose |
+ | |||
+ | <cmath>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</cmath> | ||
+ | |||
+ | Find the remainder when <math>\min{x}</math> is divided by <math>1000</math>. | ||
+ | |||
+ | |||
+ | 6. Suppose that there is <math>192</math> rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are <math>2</math> other pegs positioned sufficiently apart. A <math>move</math> is made if | ||
+ | |||
+ | (i) <math>1</math> ring changed position (i.e., that ring is transferred from one peg to another) | ||
+ | |||
+ | (ii) No rings are on top of smaller rings. | ||
+ | |||
+ | Then, let <math>x</math> be the minimum possible number <math>moves</math> that can transfer all <math>192</math> rings onto the second peg. Find the remainder when <math>x</math> is divided by <math>1000</math>. | ||
− | |||
− | + | 7. Suppose <math>f(x)</math> is a <math>10000000010</math>-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that | |
− | < | + | <cmath>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</cmath> |
− | for an integer <math>m</math>. If <math>p</math> is the minimum possible value of | + | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of |
− | < | + | <cmath>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</cmath>. |
Find the number of factors of the prime <math>999999937</math> in <math>p</math>. | Find the number of factors of the prime <math>999999937</math> in <math>p</math>. | ||
− | + | ||
+ | ====Olympiad==== | ||
+ | |||
+ | 8. (Much harder) <math>\Delta ABC</math> is an isosceles triangle where <math>CB=CA</math>. Let the circumcircle of <math>\Delta ABC</math> be <math>\Omega</math>. Then, there is a point <math>E</math> and a point <math>D</math> on circle <math>\Omega</math> such that <math>AD</math> and <math>AB</math> trisects <math>\angle CAE</math> and <math>BE<AE</math>, and point <math>D</math> lies on minor arc <math>BC</math>. Point <math>F</math> is chosen on segment <math>AD</math> such that <math>CF</math> is one of the altitudes of <math>\Delta ACD</math>. Ray <math>CF</math> intersects <math>\Omega</math> at point <math>G</math> (not <math>C</math>) and is extended past <math>G</math> to point <math>I</math>, and <math>IG=AC</math>. Point <math>H</math> is also on <math>\Omega</math> and <math>AH=GI<HB</math>. Let the perpendicular bisector of <math>BC</math> and <math>AC</math> intersect at <math>O</math>. Let <math>J</math> be a point such that <math>OJ</math> is both equal to <math>OA</math> (in length) and is perpendicular to <math>IJ</math> and <math>J</math> is on the same side of <math>CI</math> as <math>A</math>. Let <math>O’</math> be the reflection of point <math>O</math> over line <math>IJ</math>. There exist a circle <math>\Omega_1</math> centered at <math>I</math> and tangent to <math>\Omega</math> at point <math>K</math>. <math>IO’</math> intersect <math>\Omega_1</math> at <math>L</math>. Now suppose <math>O’G</math> intersects <math>\Omega</math> at one distinct point, and <math>O’, G</math>, and <math>K</math> are collinear. If <math>IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2</math>, then <math>\frac{EH}{BH}</math> can be expressed in the form <math>\frac{\sqrt{b}}{a} (\sqrt{c} + d)</math>, where <math>b</math> and <math>c</math> are not divisible by the squares of any prime. Find <math>a^2+b^2+c^2+d^2+abcd</math>. | ||
Someone mind making a diagram for this? | Someone mind making a diagram for this? | ||
− | |||
− | + | 9. Suppose <cmath>\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}</cmath> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | |
+ | |||
+ | |||
+ | ===Proofs=== | ||
+ | 10. In <math>\Delta ABC</math> with <math>AB=AC</math>, <math>D</math> is the foot of the perpendicular from <math>A</math> to <math>BC</math>. <math>E</math> is the foot of the perpendicular from <math>D</math> to <math>AC</math>. <math>F</math> is the midpoint of <math>DE</math>. Prove that <math>AF</math> is perpendicular to <math>BE</math>. | ||
+ | |||
+ | I will leave a big gap below this sentence so you won't see the answers accidentally. | ||
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Line 103: | Line 249: | ||
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Line 114: | Line 259: | ||
− | |||
− | + | ==Answer key== | |
1. 049 | 1. 049 | ||
− | 2. | + | 2. 019 |
− | 3. | + | 3. 092 |
− | 4. | + | 4. 170 |
− | 5. | + | 5. 736 |
− | + | 6. 895 | |
− | ==Problem 1== | + | 7. 011 |
+ | |||
+ | 8. 054 | ||
+ | |||
+ | 9. 077 | ||
+ | |||
+ | ==Solutions== | ||
+ | |||
+ | *Note: Most the solutions so far have been made by me :) | ||
+ | |||
+ | <math>\textbf{I wrote a couple of solutions here. Hope it's okay :) - cxsmi (please feel free to delete this note and/or the solutions)}</math> | ||
+ | |||
+ | I like your solutions.~[[Ddk001]] | ||
+ | ===Problem 1=== | ||
There is one and only one perfect square in the form | There is one and only one perfect square in the form | ||
− | < | + | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> |
− | Find that perfect square. | + | where <math>p</math> and <math>q</math> is prime. Find that perfect square. |
− | ==Solution== | + | |
+ | ===Solution 1=== | ||
<math>(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</math>. | <math>(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</math>. | ||
Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>. | Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>. | ||
− | Then, < | + | Then, |
+ | |||
+ | <cmath>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)</cmath> | ||
+ | |||
+ | , so since <math>n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}</math>, <math>n>p,n>q</math> so <math>p+q-n</math> is less than both <math>p</math> and <math>q</math> and thus we have <math>p+q-n=1</math> and <math>p+q+n=pq</math>. Adding them gives <math>2p+2q=pq+1</math> so by [[Simon's Favorite Factoring Trick]], <math>(p-2)(q-2)=3 \implies (p,q)=(3,5)</math> in some order. Hence, <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}</math>.<math>\square</math> | ||
+ | |||
+ | ===Problem 2=== | ||
+ | <math>m</math> and <math>n</math> are positive integers. If <math>m^2=2^8+2^{11}+2^n</math>, find <math>m+n</math>. | ||
+ | ===Solution 1 (Slow, probably official MAA)=== | ||
+ | <cmath>m^2=2^8+2^{11}+2^n</cmath> | ||
+ | |||
+ | <cmath>\implies 2^n=m^2-2^8-2^{11}</cmath> | ||
+ | |||
+ | <cmath>\implies 2^n=(m+48)(m-48)</cmath> | ||
+ | |||
+ | Let <math>m+48=2^t</math> and <math>m-48=2^s</math>. Then, | ||
+ | |||
+ | <cmath>2^t-2^s=96 \implies 2^s(2^{t-s}-1)=2^5 \cdot 3 \implies 2^{t-s}-1=3,2^s=2^5 \implies (t,s)=(7,5) \implies m+n=80+12=\boxed{092}</cmath> <math>\square</math> | ||
+ | |||
+ | ===Solution 2 (Fast)=== | ||
+ | Recall that a perfect square <math>(a + b)^2</math> can be written as <math>a^2 + 2ab + b^2</math>. Since <math>m^2</math> is a perfect square, the RHS must be in this form. We substitute <math>2^4</math> for <math>a</math> to get that <math>2^8 + 2^5 \cdot 2^b + 2^{2b} = m^2</math>. To make the middle term have an exponent of <math>11</math>, we must have <math>b = 6</math>. Then <math>n = 12</math> and <math>m = (2^4 + 2^6)^2 = (16 + 64)^2 = 80^2</math>, so <math>m + n = \boxed{092}</math>. | ||
+ | |||
+ | ~ cxsmi | ||
+ | |||
+ | ===Solution 3 (Faster)=== | ||
+ | Calculating the terms on the RHS, we find that <math>256 + 2048 + 2^n = 2304 + 2^n = m^2</math>. We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that <math>4096 = 2^{12}</math> works, and it produces <math>6400 = 80^2</math>. Then <math>m + n = \boxed{092}</math>. | ||
+ | |||
+ | ~ (also) cxsmi | ||
+ | |||
+ | ===Problem 3=== | ||
+ | The fraction, | ||
+ | |||
+ | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath> | ||
+ | |||
+ | where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>. | ||
+ | ===Solution 1(Probably official MAA, lots of proofs)=== | ||
+ | '''Lemma 1: <math>\text{max} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{3}</math>''' | ||
+ | |||
+ | ''Proof:'' Since the sides of triangles have positive length, <math>a,b,c>0</math>. Hence, | ||
+ | |||
+ | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}>0 \implies \text{max} (\frac{ab+bc+ac}{(a+b+c)^2})= \frac{1}{\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})}</cmath> | ||
+ | |||
+ | , so now we just need to find <math>\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})</math>. | ||
+ | |||
+ | Since <math>(a-c)^2+(b-c)^2+(a-b)^2 \ge 0</math> by the [[Trivial Inequality]], we have | ||
+ | |||
+ | <cmath>a^2-2ac+c^2+b^2-2bc+c^2+a^2-2ab+b^2 \ge 0</cmath> | ||
+ | |||
+ | <cmath>\implies a^2+b^2+c^2 \ge ac+bc+ab</cmath> | ||
+ | |||
+ | <cmath>\implies a^2+b^2+c^2+2(ac+bc+ab) \ge 3(ac+bc+ab)</cmath> | ||
+ | |||
+ | <cmath>\implies (a+b+c)^2 \ge 3(ac+bc+ab)</cmath> | ||
+ | |||
+ | <cmath>\implies \frac{(a+b+c)^2}{ab+bc+ac} \ge 3</cmath> | ||
+ | |||
+ | <cmath>\implies \frac{ab+bc+ac}{(a+b+c)^2} \le \frac{1}{3}</cmath> | ||
+ | |||
+ | as desired. <math>\square</math> | ||
+ | |||
+ | To show that the minimum value is achievable, we see that if <math>a=b=c</math>, <math>\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{3}</math>, so the minimum is thus achievable. | ||
+ | |||
+ | Thus, <math>q=\frac{1}{3}</math>. | ||
+ | |||
+ | '''Lemma 2: <math>\frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}</math>''' | ||
+ | |||
+ | ''Proof:'' By the [[Triangle Inequality]], we have | ||
+ | |||
+ | <cmath>a+b>c</cmath> | ||
+ | |||
+ | <cmath>b+c>a</cmath> | ||
+ | |||
+ | <cmath>a+c>b</cmath>. | ||
+ | |||
+ | Since <math>a,b,c>0</math>, we have | ||
+ | |||
+ | <cmath>c(a+b)>c^2</cmath> | ||
+ | |||
+ | <cmath>a(b+c)>a^2</cmath> | ||
+ | |||
+ | <cmath>b(a+c)>b^2</cmath>. | ||
+ | |||
+ | Add them together gives | ||
+ | |||
+ | <cmath>a^2+b^2+c^2<c(a+b)+a(b+c)+b(a+c)=2(ab+bc+ac)</cmath> | ||
+ | |||
+ | <cmath>\implies a^2+b^2+c^2+2(ab+bc+ac)<4(ab+bc+ac)</cmath> | ||
+ | |||
+ | <cmath>\implies (a+b+c)^2<4(ab+bc+ac)</cmath> | ||
+ | |||
+ | <cmath>\implies \frac{(a+b+c)^2}{ab+bc+ac}<4</cmath> | ||
+ | |||
+ | <cmath>\implies \frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}</cmath> <math>\square</math> | ||
+ | |||
+ | Even though unallowed, if <math>a=0,b=c</math>, then <math>\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{4}</math>, so | ||
+ | |||
+ | <cmath>\lim_{b=c,a \to 0} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{4}</cmath>. | ||
+ | |||
+ | Hence, <math>p=\frac{1}{4}</math>, since by taking <math>b=c</math> and <math>a</math> close <math>0</math>, we can get <math>\frac{ab+bc+ac}{(a+b+c)^2}</math> to be as close to <math>\frac{1}{4}</math> as we wish. | ||
+ | |||
+ | <math>p+q=\frac{1}{3}+\frac{1}{4}=\frac{7}{12} \implies r+s=7+12=\boxed{019}</math> <math>\blacksquare</math> | ||
+ | |||
+ | ===Solution 2 (Fast, risky, no proofs)=== | ||
+ | By the [[Principle of Insufficient Reason]], taking <math>a=b=c</math> we get either the max or the min. Testing other values yields that we got the max, so <math>q=\frac{1}{3}</math>. Another extrema must occur when one of <math>a,b,c</math> (WLOG, <math>a</math>) is <math>0</math>. Again, using the logic of solution 1 we see <math>p=\frac{1}{4}</math> so <math>p+q=\frac{7}{12}</math> so our answer is <math>\boxed{019}</math>. <math>\square</math> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Expand the denominator. We now have <math>\frac{ab + bc + ac}{a^2 + b^2 + c^2 + 2ab + 2ac + 2bc}</math>. Consider its reciprocal; if this expression takes values on the interval <math>(p,q]</math>, then its reciprocal will take values on the interval <math>[\frac{1}{q},\frac{1}{p})</math>. This is important because we can now write the reciprocal of the expression as <math>\frac{a^2 + b^2 + c^2}{ab + ac + bc} + 2</math>. We attempt to maximize and minimize <math>\frac{a^2 + b^2 + c^2}{ab + ac + bc}</math>. To maximize the expression, we consider the triangle inequality. From it, we find the following. <cmath>a + b > c</cmath> <cmath>a + c > b</cmath> <cmath>b + c > a</cmath> We rewrite. <cmath>ac + bc > c^2</cmath> <cmath>ab + bc > b^2</cmath> <cmath>ab + ac > a^2</cmath> Add all of the inequalities. We find the following. <cmath>2ab + 2ac + 2bc > a^2 + b^2 + c^2</cmath> Considering the equality case and plugging into the expression, we find that the maximum value of the expression is <math>4</math>. However, since this "equality case" cannot actually happen, this part of the interval must be open. Now, we minimize the inequality by using the Power Mean Inequality (specifically, the QM-AM part of the inequality). Considering the terms <math>a</math>, <math>b</math>, and <math>c</math>, we find the following. <cmath>\sqrt{\frac{a^2 + b^2 + c^2}{3}} \geq \frac{a + b + c}{3}</cmath> Square both sides. <cmath>\frac{a^2 + b^2 + c^2}{3} \geq \frac{a^2 + b^2 + c^2 + 2ab + 2ac + 2bc}{9}</cmath> Rewrite as follows. <cmath>3(a^2 + b^2 + c^2) \geq a^2 + b^2 + c^2 + 2ab + 2ac + 2bc</cmath> <cmath>2a^2 + 2b^2 + 2c^2 \geq 2ab + 2ac + 2bc</cmath> <cmath>a^2 + b^2 + c^2 \geq ab + ac + bc</cmath> Considering the equality case and plugging into the expression, we find that the minimum value of the expression is <math>3</math>. Since the expression (which we said was the reciprocal of the original expression) takes values on the interval <math>[3, 4)</math>, the original expression must take values on the interval <math>(\frac{1}{4}, \frac{1}{3}]</math>. Then <math>p + q = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}</math>, so our final answer is <math>7 + 12 = \boxed{019}</math>. | ||
+ | |||
+ | ~ cxsmi | ||
+ | |||
+ | ===Problem 4=== | ||
+ | Suppose there are complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy | ||
+ | |||
+ | <cmath>(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}</cmath> | ||
+ | |||
+ | Find <math>x_{1}^3+x_{2}^3+x_{2}^3</math>. | ||
+ | ===Solution 1=== | ||
+ | To make things easier, instead of saying <math>x_i</math>, we say <math>x</math>. | ||
+ | |||
+ | Now, we have | ||
+ | <cmath>(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}</cmath>. | ||
+ | Expanding gives | ||
+ | |||
+ | <cmath>x^3-(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}) \cdot x^2+(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})x-(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})=0</cmath>. | ||
+ | |||
+ | To make things even simpler, let | ||
+ | |||
+ | <cmath>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}</cmath> | ||
+ | |||
+ | , so that <math>x^3-ax^2+bx-c=0</math>. | ||
+ | |||
+ | Then, if <math>P_n=x_{1}^n+x_{2}^n+x_{3}^n</math>, [[Newton's Sums]] gives | ||
+ | |||
+ | <cmath>P_1+(-a)=0</cmath> <math>(1)</math> | ||
+ | |||
+ | <cmath>P_2+(-a) \cdot P_1+2 \cdot b=0</cmath> <math>(2)</math> | ||
+ | |||
+ | <cmath>P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0</cmath> <math>(3)</math> | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <cmath>P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))</cmath> | ||
+ | |||
+ | <cmath>=a \cdot P_2-b \cdot P_1+3 \cdot c</cmath> | ||
+ | |||
+ | <cmath>=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c</cmath> | ||
+ | |||
+ | <cmath>=a(a^2-2b)-ab+3c</cmath> | ||
+ | |||
+ | <cmath>=a^3-3ab+3c</cmath> | ||
+ | |||
+ | Now, we plug in <math>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}:</math> | ||
+ | |||
+ | <cmath>P_3=(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})^3-3(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})+3(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})</cmath>. | ||
+ | |||
+ | We substitute <math>x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}</math> to get | ||
+ | |||
+ | <cmath>P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})</cmath> | ||
+ | |||
+ | <cmath>=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3(x^2y+y^2x+x^2z+z^2x+z^2y+y^2z+3xyz)+3xyz+1</cmath> | ||
+ | |||
+ | <cmath>=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3x^2y-3y^2x-3x^2z-3z^2x-3z^2y-3y^2z-9xyz+3xyz+1</cmath> | ||
+ | |||
+ | <cmath>=x^3+y^3+z^3+1</cmath> | ||
+ | |||
+ | <cmath>=13+53+103+1</cmath> | ||
+ | |||
+ | <cmath>=\boxed{170}</cmath>. <math>\square</math> | ||
+ | |||
+ | Note: If you don't know [[Newton's Sums]], you can also use [[Vieta's Formulas]] to bash. | ||
+ | |||
+ | ===Problem 5=== | ||
+ | Suppose | ||
+ | |||
+ | <cmath>x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}</cmath> | ||
+ | |||
+ | Find the remainder when <math>\min{x}</math> is divided by 1000. | ||
+ | |||
+ | ===Solution 1 (Euler's Totient Theorem)=== | ||
+ | We first simplify <math>2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:</math> | ||
+ | |||
+ | <cmath>2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6=42^4+6 \cdot 30^6=(\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)}</cmath> | ||
+ | |||
+ | so | ||
+ | |||
+ | <cmath>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)} \equiv 1 \pmod{5}</cmath> | ||
+ | |||
+ | <cmath>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \pmod{6}</cmath> | ||
+ | |||
+ | <cmath>x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 6 \cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)} \equiv 6 \pmod{7}</cmath>. | ||
+ | |||
+ | where the last step of all 3 congruences hold by the [[Euler's Totient Theorem]]. | ||
+ | Hence, | ||
+ | |||
+ | <cmath>x \equiv 1 \pmod{5}</cmath> | ||
+ | |||
+ | <cmath>x \equiv 0 \pmod{6}</cmath> | ||
+ | |||
+ | <cmath>x \equiv 6 \pmod{7}</cmath> | ||
+ | |||
+ | Now, you can bash through solving linear congruences, but there is a smarter way. Notice that <math>5|x-6,6|x-6</math>, and <math>7|x-6</math>. Hence, <math>210|x-6</math>, so <math>x \equiv 6 \pmod{210}</math>. With this in mind, we proceed with finding <math>x \pmod{7!}</math>. | ||
+ | |||
+ | Notice that <math>7!=5040= \text{lcm}(144,210)</math> and that <math>x \equiv 0 \pmod{144}</math>. Therefore, we obtain the system of congruences : | ||
+ | |||
+ | <cmath>x \equiv 6 \pmod{210}</cmath> | ||
+ | |||
+ | <cmath>x \equiv 0 \pmod{144}</cmath>. | ||
+ | |||
+ | Solving yields <math>x \equiv 2\boxed{736} \pmod{7!}</math>, and we're done. <math>\square</math> | ||
+ | |||
+ | ===Problem 6=== | ||
+ | Suppose that there is <math>192</math> rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are <math>2</math> other pegs positioned sufficiently apart. A <math>move</math> is made if | ||
+ | |||
+ | (i) <math>1</math> ring changed position (i.e., that ring is transferred from one peg to another) | ||
+ | |||
+ | (ii) No bigger rings are on top of smaller rings. | ||
+ | |||
+ | Then, let <math>x</math> be the minimum possible number <math>moves</math> that can transfer all <math>192</math> rings onto the second peg. Find the remainder when <math>x</math> is divided by <math>1000</math>. | ||
+ | ===Solution 1 (Recursion)=== | ||
+ | Let <math>M_n</math> be the minimum possible number <math>moves</math> that can transfer <math>n</math> rings onto the second peg. To build the recursion, we consider what is the minimum possible number <math>moves</math> that can transfer <math>n+1</math> rings onto the second peg. If we use only legal <math>moves</math>, then <math>n+1</math> will be smaller on the top, bigger on the bottom. Hence, the largest ring have to be at the bottom of the second peg, or the largest peg will have nowhere to go. In order for the largest ring to be at the bottom, we must first move the top <math>n</math> rings to the third peg using <math>M_n</math> <math>moves</math>, then place the largest ring onto the bottom of the second peg using <math>1</math> <math>move</math>, and then get all the rings from the third peg on top of the largest ring using another <math>M_n</math> <math>moves</math>. This gives a total of <math>2M_n+1</math>, hence we have <math>M_{n+1}=2M_{n}+1</math>. Obviously, <math>M_1=1</math>. We claim that <math>M_n=2^n-1</math>. This is definitely the case for <math>n=1</math>. If this is true for <math>n</math>, then | ||
+ | |||
+ | <cmath>M_{n+1}=2M_{n}+1=2(2^n-1)+1=2^{n+1}-1</cmath> | ||
+ | |||
+ | so this is true for <math>n+1</math>. Therefore, by [[induction]], <math>M_n=2^n-1</math> is true for all <math>n</math>. Now, <math>x=M_{192}=2^{192}-1</math>. Therefore, we see that | ||
+ | |||
+ | <cmath>x+1 \equiv 0 \pmod{8}</cmath>. | ||
+ | |||
+ | But the <math>\text{mod 125}</math> part is trickier. Notice that by the [[Euler's Totient Theorem]], | ||
+ | |||
+ | <cmath>2^{\phi (125)}=2^{100} \equiv 1 \pmod{125} \implies 2^{200} \equiv 1 \pmod{125}</cmath> | ||
+ | |||
+ | so <math>x+1=\frac{2^{200}}{256}</math> is equivalent to the inverse of <math>256</math> in <math>\text{mod 125}</math>, which is equivalent to the inverse of <math>6</math> in <math>\text{mod 125}</math>, which, by inspection, is simply <math>21</math>. Hence, | ||
+ | |||
+ | <cmath>x+1 \equiv 0 \pmod{8}</cmath> | ||
+ | |||
+ | <cmath>x+1 \equiv 21 \pmod{125}</cmath> | ||
+ | |||
+ | , so by the [[Chinese Remainder Theorem]], <math>x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}</math>. <math>\square</math> | ||
+ | |||
+ | ===Problem 7=== | ||
+ | Suppose <math>f(x)</math> is a <math>10000000010</math>-degrees polynomial. The [[Fundamental Theorem of Algebra]] tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that | ||
+ | |||
+ | <cmath>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</cmath> | ||
+ | |||
+ | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of | ||
+ | |||
+ | <cmath>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</cmath>. | ||
+ | |||
+ | Find the number of factors of the prime <math>999999937</math> in <math>p</math>. | ||
+ | |||
+ | ===Solution 1=== | ||
+ | Since all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>, we have that all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)-n=0</math>, so by the [[Factor Theorem]], | ||
+ | |||
+ | <cmath>n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n</cmath> | ||
+ | |||
+ | <cmath>\implies (n+1)n \dots (n-10000000008)|f(n)-n</cmath>. | ||
+ | |||
+ | <cmath>\implies f(n)=a(n+1)n \dots (n-10000000008)+n</cmath> | ||
+ | |||
+ | since <math>f(n)</math> is a <math>10000000010</math>-degrees polynomial, and we let <math>a</math> to be the leading coefficient of <math>f(n)</math>. | ||
+ | |||
+ | Also note that since <math>r_1, r_2, \dots, r_{10000000010}</math> is the roots of <math>f(n)</math>, <math>f(n)=a(n-r_1)(n-r_2) \dots (n-r_{10000000010})</math> | ||
+ | |||
+ | Now, notice that | ||
+ | |||
+ | <cmath>m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})</cmath> | ||
+ | |||
+ | <cmath>=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})</cmath> | ||
+ | |||
+ | <cmath>=\frac{f(-2)}{a}</cmath> | ||
+ | |||
+ | <cmath>=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}</cmath> | ||
+ | |||
+ | <cmath>=\frac{10000000010! \cdot a-2}{a}</cmath> | ||
+ | |||
+ | <cmath>=10000000010!-\frac{2}{a}</cmath> | ||
+ | |||
+ | Similarly, we have | ||
+ | |||
+ | <cmath>(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}</cmath> | ||
+ | |||
+ | To minimize this, we minimize <math>m</math>. The minimum <math>m</math> can get is when <math>m=10000000011</math>, in which case | ||
+ | |||
+ | <cmath>-\frac{2}{a}=10000000011!-10000000010!</cmath> | ||
+ | |||
+ | <cmath>=10000000011 \cdot 10000000010!-10000000010!</cmath> | ||
+ | |||
+ | <cmath>=10000000010 \cdot 10000000010!</cmath> | ||
+ | |||
+ | <cmath>\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})</cmath> | ||
+ | |||
+ | <cmath>=-\frac{1}{a}</cmath> | ||
+ | |||
+ | <cmath>=\frac{10000000010 \cdot 10000000010!}{2}</cmath> | ||
+ | |||
+ | <cmath>=5000000005 \cdot 10000000010!</cmath> | ||
+ | |||
+ | , so there is <math>\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}</math> factors of <math>999999937</math>. <math>\square</math> | ||
+ | |||
+ | ===Problem 8=== | ||
+ | <math>\Delta ABC</math> is an isosceles triangle where <math>CB=CA</math>. Let the circumcircle of <math>\Delta ABC</math> be <math>\Omega</math>. Then, there is a point <math>E</math> and a point <math>D</math> on circle <math>\Omega</math> such that <math>AD</math> and <math>AB</math> trisects <math>\angle CAE</math> and <math>BE<AE</math>, and point <math>D</math> lies on minor arc <math>BC</math>. Point <math>F</math> is chosen on segment <math>AD</math> such that <math>CF</math> is one of the altitudes of <math>\Delta ACD</math>. Ray <math>CF</math> intersects <math>\Omega</math> at point <math>G</math> (not <math>C</math>) and is extended past <math>G</math> to point <math>I</math>, and <math>IG=AC</math>. Point <math>H</math> is also on <math>\Omega</math> and <math>AH=GI<HB</math>. Let the perpendicular bisector of <math>BC</math> and <math>AC</math> intersect at <math>O</math>. Let <math>J</math> be a point such that <math>OJ</math> is both equal to <math>OA</math> (in length) and is perpendicular to <math>IJ</math> and <math>J</math> is on the same side of <math>CI</math> as <math>A</math>. Let <math>O’</math> be the reflection of point <math>O</math> over line <math>IJ</math>. There exist a circle <math>\Omega_1</math> centered at <math>I</math> and tangent to <math>\Omega</math> at point <math>K</math>. <math>IO’</math> intersect <math>\Omega_1</math> at <math>L</math>. Now suppose <math>O’G</math> intersects <math>\Omega</math> at one distinct point, and <math>O’, G</math>, and <math>K</math> are collinear. If <math>IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2</math>, then <math>\frac{EH}{BH}</math> can be expressed in the form <math>\frac{\sqrt{b}}{a} (\sqrt{c} + d)</math>, where <math>b</math> and <math>c</math> are not divisible by the squares of any prime. Find <math>a^2+b^2+c^2+d^2+abcd</math>. | ||
+ | |||
+ | Someone mind making a diagram for this? | ||
+ | ===Solution 1=== | ||
+ | Line <math>IJ</math> is tangent to <math>\Omega</math> with point of tangency point <math>J</math> because <math>OJ=OA \implies \text{J is on } \Omega</math> and <math>IJ</math> is perpendicular to <math>OJ</math> so this is true by the definition of tangent lines. Both <math>G</math> and <math>K</math> are on <math>\Omega</math> and line <math>O’G</math>, so <math>O’G</math> intersects <math>\Omega</math> at both <math>G</math> and <math>K</math>, and since we’re given <math>O’G</math> intersects <math>\Omega</math> at one distinct point, <math>G</math> and <math>K</math> are not distinct, hence they are the same point. | ||
+ | |||
+ | Now, if the center of <math>2</math> tangent circles are connected, the line segment will pass through the point of tangency. In this case, if we connect the center of <math>2</math> tangent circles, <math>\Omega</math> and <math>\Omega_1</math> (<math>O</math> and <math>I</math> respectively), it is going to pass through the point of tangency, namely, <math>K</math>, which is the same point as <math>G</math>, so <math>O</math>, <math>I</math>, and <math>G</math> are collinear. Hence, <math>G</math> and <math>I</math> are on both lines <math>OI</math> and <math>CI</math>, so <math>CI</math> passes through point <math>O</math>, making <math>CG</math> a diameter of <math>\Omega</math>. | ||
+ | |||
+ | Now we state a few claims : | ||
+ | |||
+ | '''Claim 1: <math>\Delta O’IO</math> is equilateral. ''' | ||
+ | |||
+ | ''Proof:'' | ||
+ | |||
+ | <cmath>\frac{3}{4} (IK+O’L)^2</cmath> | ||
+ | |||
+ | <cmath>=\frac{3}{4} IK^2+\frac{3}{2} IK \cdot O’L+\frac{3}{4} O’L^2</cmath> | ||
+ | |||
+ | <cmath>=IG^2+IG \cdot GC</cmath> | ||
+ | |||
+ | <cmath>=IG \cdot (IG+GC)</cmath> | ||
+ | |||
+ | <cmath>=IG \cdot IC</cmath> | ||
+ | |||
+ | <cmath>=IJ^2</cmath> | ||
+ | |||
+ | where the last equality holds by the [[Power of a Point Theorem]]. | ||
+ | |||
+ | Taking the square root of each side yields <math>IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2</math>. | ||
+ | |||
+ | Since, by the definition of point <math>L</math>, <math>L</math> is on <math>\Omega_1</math>. Hence, <math>IK=IL</math>, so | ||
+ | |||
+ | <math>IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2=\frac{\sqrt{3}}{2} (IL+O’L)^2=\frac{\sqrt{3}}{2} IO’^2</math>, and since <math>O’</math> is the reflection of point <math>O</math> over line <math>IJ</math>, <math>OJ=O’J=\frac{OO’}{2}</math>, and since <math>IJ=\frac{\sqrt{3}}{2} IO’^2</math>, by the [[Pythagorean Theorem]] we have | ||
+ | |||
+ | <math>JO’=\frac{IO’}{2} \implies \frac{OO’}{2}=\frac{IO’}{2} \implies OO’=IO’</math> | ||
+ | |||
+ | Since <math>IJ</math> is the perpendicular bisector of <math>OO’</math>, <math>IO’=IO</math> and we have <math>IO=IO’=OO’</math> hence <math>\Delta O’IO</math> is equilateral. <math>\square</math> | ||
+ | |||
+ | With this in mind, we see that | ||
+ | |||
+ | <cmath>2OJ=OO’=OI=OK+KI=OJ+GI=OJ+AC \implies OA=OJ=AC</cmath> | ||
+ | |||
+ | Here, we state another claim : | ||
+ | |||
+ | '''Claim 2 : <math>BH</math> is a diameter of <math>\Omega</math>''' | ||
+ | |||
+ | ''Proof:'' Since <math>OA=OC=AC</math>, we have | ||
+ | |||
+ | <cmath>\angle AOC =60^\circ \implies \angle ABC=\frac{1}{2} \angle AOC=30^\circ \implies AB=\sqrt{3} AC</cmath> | ||
+ | |||
+ | and the same reasoning with <math>\Delta CAH</math> gives <math>CH=\sqrt{3} AC</math> since <math>AH=IG=AC</math>. | ||
+ | |||
+ | Now, apply [[Ptolemy’s Theorem]] gives | ||
+ | |||
+ | <cmath>BH \cdot AC+BC \cdot AH=CH \cdot AB \implies BH \cdot AC+AC^2=3AC^2 \implies BH=2AC=2OA</cmath> | ||
+ | |||
+ | so <math>BH</math> is a diameter. <math>\square</math> | ||
+ | |||
+ | From that, we see that <math>\angle BEH=90^\circ</math>, so <math>\frac{EH}{BH}=\cos{BHE}</math>. Now, | ||
+ | |||
+ | <cmath>\angle BHE=\angle BAE=\frac{1}{2} \angle CAB=15^\circ</cmath> | ||
+ | |||
+ | , so | ||
+ | |||
+ | <cmath>\frac{EH}{BH}=\cos{15}=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{2}}{4} (\sqrt{3}+1)</cmath> | ||
+ | |||
+ | , so | ||
+ | |||
+ | <cmath>a=4, b=2, c=3, d=1 \implies a^2+b^2+c^2+d^2+abcd=1+4+9+16+24=\boxed{054}</cmath> | ||
+ | |||
+ | , and we’re done. <math>\blacksquare</math> | ||
+ | |||
+ | ===Problem 9=== | ||
+ | Suppose <cmath>\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}</cmath> where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | ||
+ | |||
+ | ===Solution 1(Wordless endless bash)=== | ||
+ | <cmath>\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}</cmath> | ||
+ | |||
+ | <cmath>=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn(m+n+2)}</cmath> | ||
+ | |||
+ | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{m(m+n+2)}</cmath> | ||
+ | |||
+ | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{n+2} (\frac{1}{m}-\frac{1}{m+n+2})</cmath> | ||
+ | |||
+ | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \sum_{m=1}^{\infty} (\frac{1}{m}-\frac{1}{m+n+2})</cmath> | ||
+ | |||
+ | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot [(1-\frac{1}{n+3})+(\frac{1}{2}-\frac{1}{n+4})+ \dots]</cmath> | ||
+ | |||
+ | <cmath>=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})</cmath> | ||
+ | |||
+ | <cmath>=\sum_{n=1}^{\infty} (\frac{\frac{1}{2}}{n}-\frac{\frac{1}{2}}{n+2}) \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2} [(1-\frac{1}{3})(1+\frac{1}{2}+\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+ \dots</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2} [[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\dots](1+\frac{1}{2}+\frac{1}{3})+[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{6})+\dots](1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+\dots](\frac{1}{4}+\frac{1}{5})+\dots]</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2} [(1+\frac{1}{2}+\frac{1}{3})+\frac{1}{2} (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+\frac{1}{3} (\frac{1}{4}+\frac{1}{5})+\dots]</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2} [\frac{11}{6}+\frac{1}{2} \cdot \frac{25}{12}+(\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\dots)+(\frac{1}{3 \cdot 5}+\frac{1}{4 \cdot 6}+\dots)]</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2} [\frac{69}{24}+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\dots ]+\frac{1}{2} [(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+\dots ]</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2} [\frac{69}{24}+\frac{1}{3}+\frac{1}{6}+\frac{1}{8}]</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2} \cdot \frac{84}{24}</cmath> | ||
+ | |||
+ | <cmath>=\frac{7}{4}</cmath> | ||
+ | |||
+ | <cmath>(1+\frac{1}{x})^x=e^{x \cdot \ln (1+\frac{1}{x})}</cmath> | ||
+ | |||
+ | <cmath>=e^{x \cdot [(\frac{1}{x})-\frac{(\frac{1}{x})^2}{2}+\frac{(\frac{1}{x})^3}{3}+\dots]}</cmath> | ||
+ | |||
+ | <cmath>=e^{1-\frac{1}{2} (\frac{1}{x})+\frac{1}{3} (\frac{1}{x})^2+\dots}</cmath> | ||
+ | |||
+ | <cmath>=e \cdot e^{-\frac{1}{2} (\frac{1}{x})} \cdot e^{\frac{1}{3} (\frac{1}{x})^2} \dots</cmath> | ||
+ | |||
+ | <cmath>=e \cdot [1-\frac{1}{2x}+\frac{1}{2!} (\frac{1}{2x})^2- \dots] \cdot [1+\frac{1}{3x^2}+\frac{1}{2!} (\frac{1}{3x^2})^2+ \dots]</cmath> | ||
+ | |||
+ | <cmath>=e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]</cmath> | ||
+ | |||
+ | <cmath>\implies \lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]</cmath> | ||
+ | |||
+ | <cmath>=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]}{e}-1]]</cmath> | ||
+ | |||
+ | <cmath>=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots-1)]</cmath> | ||
+ | |||
+ | <cmath>=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots)]</cmath> | ||
+ | |||
+ | <cmath>=\lim_{x\rightarrow \infty} (\frac{x}{2}-\frac{x}{2}+\frac{11}{24}+\dots)</cmath> | ||
+ | |||
+ | <cmath>=\frac{11}{24}</cmath> | ||
+ | |||
+ | <cmath>\implies \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]]</cmath> | ||
+ | |||
+ | <cmath>=\frac{7}{4}+\frac{11}{24}</cmath> | ||
+ | |||
+ | <cmath>=\frac{53}{24}</cmath> | ||
+ | |||
+ | <cmath>\implies p=53,q=24</cmath> | ||
+ | |||
+ | <cmath>\implies p+q=\boxed{077}</cmath> <math>\square</math> | ||
+ | |||
+ | ===Problem 10=== | ||
+ | In <math>\Delta ABC</math> with <math>AB=AC</math>, <math>D</math> is the foot of the perpendicular from <math>A</math> to <math>BC</math>. <math>E</math> is the foot of the perpendicular from <math>D</math> to <math>AC</math>. <math>F</math> is the midpoint of <math>DE</math>. Prove that <math>AF \perp BE</math>. | ||
+ | |||
+ | ===Solution 1 (Analytic geo)=== | ||
+ | Let | ||
+ | |||
+ | <cmath>A=(0,0)</cmath> | ||
+ | |||
+ | <cmath>B=(4a,4b)</cmath> | ||
+ | |||
+ | <cmath>C=(4 \sqrt{a^2+b^2},0)</cmath> | ||
+ | |||
+ | We set it this way to simplify calculation when we calculate the coordinates of <math>E</math> and <math>F</math> (Notice to find <math>E</math>, you just need to take the x coordinate of <math>D</math> and let the y coordinate be <math>0</math>). | ||
+ | |||
+ | Obviously, | ||
+ | |||
+ | <cmath>D=(\frac{4a+4 \sqrt{a^2+b^2}}{2},\frac{4b+0}{2})=(2a+2 \sqrt{a^2+b^2},2b)</cmath> | ||
+ | |||
+ | <cmath>\implies E=(2a+2 \sqrt{a^2+b^2},0)</cmath> | ||
+ | |||
+ | <cmath>\implies F=(\frac{2a+2 \sqrt{a^2+b^2}+2a+2 \sqrt{a^2+b^2}}{2},\frac{2b+0}{2})=(2a+2 \sqrt{a^2+b^2},b)</cmath> | ||
+ | |||
+ | Now, we see that | ||
+ | |||
+ | <cmath>\text{Slope} _ {AF}=\frac{b}{2a+2 \sqrt{a^2+b^2}}</cmath> | ||
+ | |||
+ | <cmath>\text{Slope} _ {BE}=\frac{0-4b}{2a+2 \sqrt{a^2+b^2}-4a}=\frac{-2b}{\sqrt{a^2+b^2}-a}</cmath> | ||
+ | |||
+ | <cmath>\implies \text{Slope} _ {AF} \cdot \text{Slope} _ {BE}=\frac{b}{2a+2 \sqrt{a^2+b^2}} \cdot \frac{-2b}{a+ \sqrt{a^2+b^2}-2a}=\frac{-2b^2}{2(a+\sqrt{a^2+b^2})(\sqrt{a^2+b^2}-a)}=\frac{-2b^2}{2b^2}=-1</cmath> | ||
+ | |||
+ | , so <math>AF \perp BE</math>, as desired. <math>\square</math> | ||
+ | ===Solution 2 (Hard vector bash)=== | ||
+ | ====Solution 2a (Hard)==== | ||
+ | <cmath>\overrightarrow{AF} \cdot \overrightarrow{BE}</cmath> | ||
+ | |||
+ | <cmath>=(\overrightarrow{AE}+\overrightarrow{EF}) \cdot (\overrightarrow{BD}+\overrightarrow{DE})</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{AE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{DE}+\overrightarrow{AE} \cdot \overrightarrow{DE}</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{AE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{DE}</cmath> | ||
+ | |||
+ | <cmath>=(\overrightarrow{AD}+\overrightarrow{DE}) \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD} + \overrightarrow{EF} \cdot \overrightarrow{DE}</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{DE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD} + \overrightarrow{EF} \cdot \overrightarrow{DE}</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{DE} \cdot \overrightarrow{DC}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{BD}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DE}</cmath> | ||
+ | |||
+ | <cmath>=\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DC}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DE}</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{DE} \cdot (\frac{\overrightarrow{DC}-\overrightarrow{DE}}{2})</cmath> | ||
+ | |||
+ | <cmath>=\frac{\overrightarrow{DE} \cdot \overrightarrow{EC}}{2}</cmath> | ||
+ | |||
+ | <cmath>=0</cmath> | ||
+ | |||
+ | Hence, <math>AF \perp BE</math>. <math>\square</math> | ||
+ | ====Solution 2b (Harder)==== | ||
+ | <cmath>\angle ACD=\angle ECD</cmath> | ||
+ | |||
+ | <cmath>\angle ADC=\angle DEC</cmath> | ||
+ | |||
+ | <cmath>\implies \Delta ADC \sim \Delta DEC</cmath> | ||
+ | |||
+ | <cmath>\implies \frac{EC}{DC}=\frac{DC}{AC}</cmath> | ||
+ | |||
+ | <cmath>\implies EC=\frac{DC^2}{AC}</cmath> | ||
+ | |||
+ | <cmath>\implies \overrightarrow{E}=\overrightarrow{C}+\overrightarrow{CE}</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{C}+\frac{CE}{AC} \cdot \overrightarrow{CA}</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{CA}</cmath> | ||
+ | |||
+ | <cmath>=\overrightarrow{C}+\frac{DC^2}{AC^2} (\overrightarrow{A}-\overrightarrow{C})</cmath> | ||
+ | |||
+ | <cmath>=\frac{AC^2-DC^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}</cmath> | ||
+ | |||
+ | <cmath>=\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}</cmath> | ||
+ | |||
+ | <cmath>\overrightarrow{D}=\frac{\overrightarrow{B}+\overrightarrow{C}}{2}</cmath> | ||
+ | |||
+ | Since <math>F</math> is the midpoint of <math>DE</math>, | ||
+ | |||
+ | <cmath>\overrightarrow{F}=\frac{\overrightarrow{D}+\overrightarrow{E}}{2}</cmath> | ||
+ | |||
+ | <cmath>=\frac{\frac{\overrightarrow{B}+\overrightarrow{C}}{2}+\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}}{2}</cmath> | ||
+ | |||
+ | <cmath>=\frac{\overrightarrow{B}}{4}+\frac{AC^2+2AD^2}{4AC^2} \overrightarrow{C}+\frac{DC^2}{2AC^2} \overrightarrow{A}</cmath> | ||
+ | |||
+ | <cmath>\overrightarrow{AF}=\overrightarrow{F}-\overrightarrow{A}=\frac{\overrightarrow{B}}{4}+\frac{AC^2+2AD^2}{4AC^2} \overrightarrow{C}+\frac{DC^2-2AC^2}{2AC^2} \overrightarrow{A}</cmath> | ||
+ | |||
+ | <cmath>\overrightarrow{BE}=\overrightarrow{E}-\overrightarrow{B}=\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}-\overrightarrow{B}</cmath> | ||
+ | |||
+ | Now come the coordinates. Let | ||
+ | |||
+ | <cmath>A=(0,0)</cmath> | ||
+ | |||
+ | <cmath>B=(-a,-b)</cmath> | ||
+ | |||
+ | <cmath>C=(a,-b)</cmath> | ||
+ | |||
+ | so that | ||
+ | |||
+ | <cmath>\overrightarrow{A}=\langle 0,0 \rangle</cmath> | ||
+ | |||
+ | <cmath>\overrightarrow{B}=\langle -a,-b \rangle</cmath> | ||
+ | |||
+ | <cmath>\overrightarrow{C}=\langle a,-b \rangle</cmath>. | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <cmath>\overrightarrow{AF}=\langle \frac{-a}{4},\frac{-b}{4} \rangle + \frac{(a^2+b^2)+2b^2}{4(a^2+b^2)} \langle a,-b \rangle=\langle \frac{ab^2}{2(a^2+b^2)},\frac{-a^2b-2b^3}{2(a^2+b^2)} \rangle</cmath> | ||
+ | |||
+ | <cmath>\overrightarrow{BE}=\frac{b^2}{a^2+b^2} \langle a,-b \rangle-\langle -a,-b \rangle=\langle \frac{2ab^2+a^3}{a^2+b^2},\frac{a^2b}{a^2+b^2} \rangle</cmath> | ||
+ | |||
+ | <cmath>\implies \overrightarrow{AF} \cdot \overrightarrow{BE}=\langle \frac{ab^2}{2(a^2+b^2)},\frac{-a^2b-2b^3}{2(a^2+b^2)} \rangle \cdot \langle \frac{2ab^2+a^3}{a^2+b^2},\frac{a^2b}{a^2+b^2} \rangle</cmath> | ||
+ | |||
+ | <cmath>=\frac{1}{2(a^2+b^2)^2}[ab^2(a^3+2ab^2)-a^2b(a^2b+2b^3)]</cmath> | ||
+ | |||
+ | <cmath>=\frac{ab}{2(a^2+b^2)^2} (a^3b+2ab^3-a^3b-2ab^3)</cmath> | ||
+ | |||
+ | <cmath>=0</cmath> | ||
+ | |||
+ | Hence, we have that <math>AF</math> is perpendicular to <math>BE</math>. <math>\square</math> | ||
+ | |||
+ | ==Vandalism area== | ||
+ | Here, you can add anything, delete anything, and do anything! (Don't delete this line since it's instruction and don't be inappropriate) Do not delete the see also. However, do NOT vandalize before this word (Feel free to delete this and the period that follows). | ||
+ | |||
+ | (ok :) :) this page is so cool!) | ||
+ | |||
+ | honestly i think your user page is very cool. :) | ||
+ | |||
+ | ==See also== | ||
+ | * My [[User talk:Ddk001|talk page]] | ||
+ | * [[Problems Collection|My problems collection]] | ||
+ | |||
+ | The problems on this page are NOT copyrighted by the [http://www.maa.org Mathematical Association of America]'s [http://amc.maa.org American Mathematics Competitions]. [[File:AMC_logo.png|middle]] | ||
+ | <div style="clear:both;"> |
Revision as of 23:23, 16 May 2024
Contents
- 1 Introduction
- 2 User Counts
- 3 Cool asyptote graphs
- 4 Problems Sharing Contest
- 5 Contributions
- 6 Problems I made
- 7 Answer key
- 8 Solutions
- 8.1 Problem 1
- 8.2 Solution 1
- 8.3 Problem 2
- 8.4 Solution 1 (Slow, probably official MAA)
- 8.5 Solution 2 (Fast)
- 8.6 Solution 3 (Faster)
- 8.7 Problem 3
- 8.8 Solution 1(Probably official MAA, lots of proofs)
- 8.9 Solution 2 (Fast, risky, no proofs)
- 8.10 Solution 3
- 8.11 Problem 4
- 8.12 Solution 1
- 8.13 Problem 5
- 8.14 Solution 1 (Euler's Totient Theorem)
- 8.15 Problem 6
- 8.16 Solution 1 (Recursion)
- 8.17 Problem 7
- 8.18 Solution 1
- 8.19 Problem 8
- 8.20 Solution 1
- 8.21 Problem 9
- 8.22 Solution 1(Wordless endless bash)
- 8.23 Problem 10
- 8.24 Solution 1 (Analytic geo)
- 8.25 Solution 2 (Hard vector bash)
- 9 Vandalism area
- 10 See also
Introduction
I am a 5th grader who likes making and doing problems, doing math, and redirecting pages (see Principle of Insufficient Reasons). I like geometry and don't like counting and probability. My number theory skill are also not bad.
User Counts
If this is your first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)
For those of you who want more boxes, me too. However, this is the max number of boxes. Also, I check the pages history so I know if someone edited something.
(Please don't mess with the user count)
Doesn't that look like a number on a pyramid
Cool asyptote graphs
Asymptote is fun!
Problems Sharing Contest
Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start:
1. There is one and only one perfect square in the form
where and are prime. Find that perfect square.
1. We can expand the product in the expression. . Suppose this equals for some positive integer . We rewrite using the square of a binomial pattern to find that . Through trial and error on small values of and , we find that and must equal and in some order. The perfect square formed using these numbers is .
Note: I will be the first to admit that this solution is somewhat lucky.
2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle intersects itself. has leg length . The perimeter of this diamond is expressible as , where , , and are integers, and is not divisible by the square of any prime. What is the remainder when is divided by ?
Contributions
2005 AMC 8 Problems/Problem 21 Solution 2
2022 AMC 12B Problems/Problem 25 Solution 5 (Now it's solution 6)
2023 AMC 12B Problems/Problem 20 Solution 3
2016 AIME I Problems/Problem 10 Solution 3
2017 AIME I Problems/Problem 14 Solution 2
2019 AIME I Problems/Problem 15 Solution 6
2022 AIME II Problems/Problem 3 Solution 3
Restored diagram for 1994 AIME Problems/Problem 7
Principle of Insufficient Reasons
Problems I made
Aime styled
Introductory
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 .
Intermediate
3.The fraction,
where and are side lengths of a triangle, lies in the interval , where and are rational numbers. Then, can be expressed as , where and are relatively prime positive integers. Find .
4. Suppose there is complex values and that satisfy
Find .
5. Suppose
Find the remainder when is divided by .
6. Suppose that there is rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are other pegs positioned sufficiently apart. A is made if
(i) ring changed position (i.e., that ring is transferred from one peg to another)
(ii) No rings are on top of smaller rings.
Then, let be the minimum possible number that can transfer all rings onto the second peg. Find the remainder when is divided by .
7. Suppose is a -degrees polynomial. The Fundamental Theorem of Algebra tells us that there are roots, say . Suppose all integers ranging from to satisfies . Also, suppose that
for an integer . If is the minimum possible positive integral value of
.
Find the number of factors of the prime in .
Olympiad
8. (Much harder) is an isosceles triangle where . Let the circumcircle of be . Then, there is a point and a point on circle such that and trisects and , and point lies on minor arc . Point is chosen on segment such that is one of the altitudes of . Ray intersects at point (not ) and is extended past to point , and . Point is also on and . Let the perpendicular bisector of and intersect at . Let be a point such that is both equal to (in length) and is perpendicular to and is on the same side of as . Let be the reflection of point over line . There exist a circle centered at and tangent to at point . intersect at . Now suppose intersects at one distinct point, and , and are collinear. If , then can be expressed in the form , where and are not divisible by the squares of any prime. Find .
Someone mind making a diagram for this?
9. Suppose where and are relatively prime positive integers. Find .
Proofs
10. In with , is the foot of the perpendicular from to . is the foot of the perpendicular from to . is the midpoint of . Prove that is perpendicular to .
I will leave a big gap below this sentence so you won't see the answers accidentally.
Answer key
1. 049
2. 019
3. 092
4. 170
5. 736
6. 895
7. 011
8. 054
9. 077
Solutions
- Note: Most the solutions so far have been made by me :)
I like your solutions.~Ddk001
Problem 1
There is one and only one perfect square in the form
where and is prime. Find that perfect square.
Solution 1
. Suppose . Then,
, so since , so is less than both and and thus we have and . Adding them gives so by Simon's Favorite Factoring Trick, in some order. Hence, .
Problem 2
and are positive integers. If , find .
Solution 1 (Slow, probably official MAA)
Let and . Then,
Solution 2 (Fast)
Recall that a perfect square can be written as . Since is a perfect square, the RHS must be in this form. We substitute for to get that . To make the middle term have an exponent of , we must have . Then and , so .
~ cxsmi
Solution 3 (Faster)
Calculating the terms on the RHS, we find that . We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that works, and it produces . Then .
~ (also) cxsmi
Problem 3
The fraction,
where and are side lengths of a triangle, lies in the interval , where and are rational numbers. Then, can be expressed as , where and are relatively prime positive integers. Find .
Solution 1(Probably official MAA, lots of proofs)
Lemma 1:
Proof: Since the sides of triangles have positive length, . Hence,
, so now we just need to find .
Since by the Trivial Inequality, we have
as desired.
To show that the minimum value is achievable, we see that if , , so the minimum is thus achievable.
Thus, .
Lemma 2:
Proof: By the Triangle Inequality, we have
.
Since , we have
.
Add them together gives
Even though unallowed, if , then , so
.
Hence, , since by taking and close , we can get to be as close to as we wish.
Solution 2 (Fast, risky, no proofs)
By the Principle of Insufficient Reason, taking we get either the max or the min. Testing other values yields that we got the max, so . Another extrema must occur when one of (WLOG, ) is . Again, using the logic of solution 1 we see so so our answer is .
Solution 3
Expand the denominator. We now have . Consider its reciprocal; if this expression takes values on the interval , then its reciprocal will take values on the interval . This is important because we can now write the reciprocal of the expression as . We attempt to maximize and minimize . To maximize the expression, we consider the triangle inequality. From it, we find the following. We rewrite. Add all of the inequalities. We find the following. Considering the equality case and plugging into the expression, we find that the maximum value of the expression is . However, since this "equality case" cannot actually happen, this part of the interval must be open. Now, we minimize the inequality by using the Power Mean Inequality (specifically, the QM-AM part of the inequality). Considering the terms , , and , we find the following. Square both sides. Rewrite as follows. Considering the equality case and plugging into the expression, we find that the minimum value of the expression is . Since the expression (which we said was the reciprocal of the original expression) takes values on the interval , the original expression must take values on the interval . Then , so our final answer is .
~ cxsmi
Problem 4
Suppose there are complex values and that satisfy
Find .
Solution 1
To make things easier, instead of saying , we say .
Now, we have . Expanding gives
.
To make things even simpler, let
, so that .
Then, if , Newton's Sums gives
Therefore,
Now, we plug in
.
We substitute to get
.
Note: If you don't know Newton's Sums, you can also use Vieta's Formulas to bash.
Problem 5
Suppose
Find the remainder when is divided by 1000.
Solution 1 (Euler's Totient Theorem)
We first simplify
so
.
where the last step of all 3 congruences hold by the Euler's Totient Theorem. Hence,
Now, you can bash through solving linear congruences, but there is a smarter way. Notice that , and . Hence, , so . With this in mind, we proceed with finding .
Notice that and that . Therefore, we obtain the system of congruences :
.
Solving yields , and we're done.
Problem 6
Suppose that there is rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are other pegs positioned sufficiently apart. A is made if
(i) ring changed position (i.e., that ring is transferred from one peg to another)
(ii) No bigger rings are on top of smaller rings.
Then, let be the minimum possible number that can transfer all rings onto the second peg. Find the remainder when is divided by .
Solution 1 (Recursion)
Let be the minimum possible number that can transfer rings onto the second peg. To build the recursion, we consider what is the minimum possible number that can transfer rings onto the second peg. If we use only legal , then will be smaller on the top, bigger on the bottom. Hence, the largest ring have to be at the bottom of the second peg, or the largest peg will have nowhere to go. In order for the largest ring to be at the bottom, we must first move the top rings to the third peg using , then place the largest ring onto the bottom of the second peg using , and then get all the rings from the third peg on top of the largest ring using another . This gives a total of , hence we have . Obviously, . We claim that . This is definitely the case for . If this is true for , then
so this is true for . Therefore, by induction, is true for all . Now, . Therefore, we see that
.
But the part is trickier. Notice that by the Euler's Totient Theorem,
so is equivalent to the inverse of in , which is equivalent to the inverse of in , which, by inspection, is simply . Hence,
, so by the Chinese Remainder Theorem, .
Problem 7
Suppose is a -degrees polynomial. The Fundamental Theorem of Algebra tells us that there are roots, say . Suppose all integers ranging from to satisfies . Also, suppose that
for an integer . If is the minimum possible positive integral value of
.
Find the number of factors of the prime in .
Solution 1
Since all integers ranging from to satisfies , we have that all integers ranging from to satisfies , so by the Factor Theorem,
.
since is a -degrees polynomial, and we let to be the leading coefficient of .
Also note that since is the roots of ,
Now, notice that
Similarly, we have
To minimize this, we minimize . The minimum can get is when , in which case
, so there is factors of .
Problem 8
is an isosceles triangle where . Let the circumcircle of be . Then, there is a point and a point on circle such that and trisects and , and point lies on minor arc . Point is chosen on segment such that is one of the altitudes of . Ray intersects at point (not ) and is extended past to point , and . Point is also on and . Let the perpendicular bisector of and intersect at . Let be a point such that is both equal to (in length) and is perpendicular to and is on the same side of as . Let be the reflection of point over line . There exist a circle centered at and tangent to at point . intersect at . Now suppose intersects at one distinct point, and , and are collinear. If , then can be expressed in the form , where and are not divisible by the squares of any prime. Find .
Someone mind making a diagram for this?
Solution 1
Line is tangent to with point of tangency point because and is perpendicular to so this is true by the definition of tangent lines. Both and are on and line , so intersects at both and , and since we’re given intersects at one distinct point, and are not distinct, hence they are the same point.
Now, if the center of tangent circles are connected, the line segment will pass through the point of tangency. In this case, if we connect the center of tangent circles, and ( and respectively), it is going to pass through the point of tangency, namely, , which is the same point as , so , , and are collinear. Hence, and are on both lines and , so passes through point , making a diameter of .
Now we state a few claims :
Claim 1: is equilateral.
Proof:
where the last equality holds by the Power of a Point Theorem.
Taking the square root of each side yields .
Since, by the definition of point , is on . Hence, , so
, and since is the reflection of point over line , , and since , by the Pythagorean Theorem we have
Since is the perpendicular bisector of , and we have hence is equilateral.
With this in mind, we see that
Here, we state another claim :
Claim 2 : is a diameter of
Proof: Since , we have
and the same reasoning with gives since .
Now, apply Ptolemy’s Theorem gives
so is a diameter.
From that, we see that , so . Now,
, so
, so
, and we’re done.
Problem 9
Suppose where and are relatively prime positive integers. Find .
Solution 1(Wordless endless bash)
Problem 10
In with , is the foot of the perpendicular from to . is the foot of the perpendicular from to . is the midpoint of . Prove that .
Solution 1 (Analytic geo)
Let
We set it this way to simplify calculation when we calculate the coordinates of and (Notice to find , you just need to take the x coordinate of and let the y coordinate be ).
Obviously,
Now, we see that
, so , as desired.
Solution 2 (Hard vector bash)
Solution 2a (Hard)
Hence, .
Solution 2b (Harder)
Since is the midpoint of ,
Now come the coordinates. Let
so that
.
Therefore,
Hence, we have that is perpendicular to .
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