Difference between revisions of "Problems Collection"
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− | + | ==Problems I made== | |
− | + | ===Aime styled=== | |
− | == | + | ====Introductory==== |
− | == | ||
1. There is one and only one perfect square in the form | 1. There is one and only one perfect square in the form | ||
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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>. | 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, | 3.The fraction, | ||
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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 | 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>. | 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 | + | 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 |
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<cmath>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</cmath> | <cmath>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</cmath> | ||
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− | + | ====Olympiad==== | |
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+ | 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>. | |
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+ | ===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>. | ||
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+ | I will leave a big gap below this sentence so you won't see the answers accidentally. | ||
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==Solutions== | ==Solutions== | ||
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+ | <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> | ||
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+ | I like your solutions.~[[Ddk001]] | ||
===Problem 1=== | ===Problem 1=== | ||
There is one and only one perfect square in the form | There is one and only one perfect square in the form | ||
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<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> | <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> | + | , 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> |
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===Problem 2=== | ===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>. | <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>. | ||
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Let <math>m+48=2^t</math> and <math>m-48=2^s</math>. Then, | 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> | + | <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)=== | ===Solution 2 (Fast)=== | ||
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~ (also) cxsmi | ~ (also) cxsmi | ||
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===Problem 3=== | ===Problem 3=== | ||
The fraction, | The fraction, | ||
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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. | 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> | + | <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)=== | ===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> | + | 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> |
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+ | ===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>. | ||
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+ | ~ cxsmi | ||
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===Problem 4=== | ===Problem 4=== | ||
Suppose there are complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy | Suppose there are complex values <math>x_1, x_2,</math> and <math>x_3</math> that satisfy | ||
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<cmath>=\boxed{170}</cmath>. <math>\square</math> | <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. | + | Note: If you don't know [[Newton's Sums]], you can also use [[Vieta's Formulas]] to bash. |
===Problem 5=== | ===Problem 5=== | ||
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<cmath>x \equiv 0 \pmod{144}</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> | + | Solving yields <math>x \equiv 2\boxed{736} \pmod{7!}</math>, and we're done. <math>\square</math> |
===Problem 6=== | ===Problem 6=== | ||
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, so by the [[Chinese Remainder Theorem]], <math>x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}</math>. <math>\square</math> | , so by the [[Chinese Remainder Theorem]], <math>x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}</math>. <math>\square</math> | ||
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===Problem 7=== | ===Problem 7=== | ||
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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 | 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 | ||
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<cmath>=5000000005 \cdot 10000000010!</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> | + | , so there is <math>\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}</math> factors of <math>999999937</math>. <math>\square</math> |
− | ===Problem | + | ===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>. | <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>. | ||
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<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> | <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> | + | , and we’re done. <math>\blacksquare</math> |
− | ===Problem | + | ===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>. | 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>. | ||
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<cmath>\implies p=53,q=24</cmath> | <cmath>\implies p=53,q=24</cmath> | ||
− | <cmath>\implies p+q=\boxed{077}</cmath> <math>\square</math> | + | <cmath>\implies p+q=\boxed{077}</cmath> <math>\square</math> |
− | ===Problem | + | ===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>. | 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>. | ||
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<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> | <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> | + | , so <math>AF \perp BE</math>, as desired. <math>\square</math> |
===Solution 2 (Hard vector bash)=== | ===Solution 2 (Hard vector bash)=== | ||
====Solution 2a (Hard)==== | ====Solution 2a (Hard)==== | ||
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<cmath>=0</cmath> | <cmath>=0</cmath> | ||
− | Hence, <math>AF \perp BE</math>. <math>\square</math> | + | Hence, <math>AF \perp BE</math>. <math>\square</math> |
====Solution 2b (Harder)==== | ====Solution 2b (Harder)==== | ||
<cmath>\angle ACD=\angle ECD</cmath> | <cmath>\angle ACD=\angle ECD</cmath> | ||
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<cmath>=0</cmath> | <cmath>=0</cmath> | ||
− | Hence, we have that <math>AF</math> is perpendicular to <math>BE</math>. <math>\square</math> | + | Hence, we have that <math>AF</math> is perpendicular to <math>BE</math>. <math>\square</math> |
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Revision as of 17:59, 24 June 2024
Contents
[hide]- 1 Problems I made
- 2 Solutions
- 2.1 Problem 1
- 2.2 Solution 1
- 2.3 Problem 2
- 2.4 Solution 1 (Slow, probably official MAA)
- 2.5 Solution 2 (Fast)
- 2.6 Solution 3 (Faster)
- 2.7 Problem 3
- 2.8 Solution 1(Probably official MAA, lots of proofs)
- 2.9 Solution 2 (Fast, risky, no proofs)
- 2.10 Solution 3
- 2.11 Problem 4
- 2.12 Solution 1
- 2.13 Problem 5
- 2.14 Solution 1 (Euler's Totient Theorem)
- 2.15 Problem 6
- 2.16 Solution 1 (Recursion)
- 2.17 Problem 7
- 2.18 Solution 1
- 2.19 Problem 8
- 2.20 Solution 1
- 2.21 Problem 9
- 2.22 Solution 1(Wordless endless bash)
- 2.23 Problem 10
- 2.24 Solution 1 (Analytic geo)
- 2.25 Solution 2 (Hard vector bash)
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.
Solutions
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 .