Problems Collection

This is a page where you can share the problems you made (try not to use past exams).

If you have problems or solutions to problems already on this page to contribute, please put it below. In the former case, please give at least one solution to that problem. Problems/solutions in the following section shall be inspected by Ddk001 and will be put in the section after that section. If you would like, put your user name to the list of contributors of this page (at the near-end).

Contributed Problems and Solutions

Please contribute whatever problems you have.

Problems

AMC styled

Nothing yet to show

AIME styled

1.There is one and only one perfect square in the form

$(p^2+1)(q^2+1)-((pq)^2-pq+1)$

where $p$ and $q$ is prime. Find that perfect square.

2. $m$ and $n$ are positive integers. If $m^2=2^8+2^{11}+2^n$ , find $m+n$ .

3.The fraction,

$\frac{ab+bc+ac}{(a+b+c)^2}$

where $a,b$ and $c$ are side lengths of a triangle, lies in the interval $(p,q]$ , where $p$ and $q$ are rational numbers. Then, $p+q$ can be expressed as $\frac{r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r+s$ .

4.Suppose there are complex values $x_1, x_2,$ and $x_3$ that satisfy

$(x_i-\sqrt[3]{13})(x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}$ Find $x_{1}^3+x_{2}^3+x_{2}^3$ .

5.Suppose

$x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}$ Find the remainder when $\min{x}$ is divided by 1000.

6.Suppose that there is $192$ rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are $2$ other pegs positioned sufficiently apart. A $move$ is made if

(i) $1$ 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 $x$ be the minimum possible number $moves$ that can transfer all $192$ rings onto the second peg. Find the remainder when $x$ is divided by $1000$ .

7.Let $\overline{ab}$ be a 2-digit positive integer satisfying $\overline{ab}^2=a! +b!$ . Find the sum of all possible values of $\overline{ab}$ .

8.Suppose $f(x)$ is a $10000000010$ -degrees polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \dots, r_{10000000010}$ . Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$ . Also, suppose that

$(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!$

for an integer $m$ . If $p$ is the minimum possible positive integral value of

$(1+r_1)(1+r_2) \dots (1+r_{10000000010})$ .

Find the number of factors of the prime $999999937$ in $p$ .

9. $\Delta ABC$ is an isosceles triangle where $CB=CA$ . Let the circumcircle of $\Delta ABC$ be $\Omega$ . Then, there is a point $E$ and a point $D$ on circle $\Omega$ such that $AD$ and $AB$ trisects $\angle CAE$ and $BE<AE$ , and point $D$ lies on minor arc $BC$ . Point $F$ is chosen on segment $AD$ such that $CF$ is one of the altitudes of $\Delta ACD$ . Ray $CF$ intersects $\Omega$ at point $G$ (not $C$ ) and is extended past $G$ to point $I$ , and $IG=AC$ . Point $H$ is also on $\Omega$ and $AH=GI<HB$ . Let the perpendicular bisector of $BC$ and $AC$ intersect at $O$ . Let $J$ be a point such that $OJ$ is both equal to $OA$ (in length) and is perpendicular to $IJ$ and $J$ is on the same side of $CI$ as $A$ . Let $O’$ be the reflection of point $O$ over line $IJ$ . There exist a circle $\Omega_1$ centered at $I$ and tangent to $\Omega$ at point $K$ . $IO’$ intersect $\Omega_1$ at $L$ . Now suppose $O’G$ intersects $\Omega$ at one distinct point, and $O’, G$ , and $K$ are collinear. If $IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2$ , then $\frac{EH}{BH}$ can be expressed in the form $\frac{\sqrt{b}}{a} (\sqrt{c} + d)$ , where $b$ and $c$ are not divisible by the squares of any prime. Find $a^2+b^2+c^2+d^2+abcd$ .

Olympiad styled

Putnam styled (Calculus version of AMC, AIME, and Olympiad)

1.Suppose $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]=\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

2. Let $x,y$ and $z$ be real numbers. Show that

$(x^2+z^2)^2+y^4 \ge 4xzy^2$

3. Common blood types are determined genetically by 3 alleles: $A$ , $B$ , and $O$ . A person whose blood genotype is $AB$ , $AO$ , $BO$ , $BA$ , $OA$ , or $OB$ is heterozygous while a person whose blood genotype is $AA$ , $OO$ , or $BB$ is homozygous. The Hardy-Weinberg Law states that the proportion P of heterozygous individuals in a certain population (in genetic equilibrium) is given by

$P(p,q,r)=2pq+2pr+2qr$

where $p$ is the percent of allele $A$ , $q$ the percent of allele $B$ , and $r$ the percent allele $O$ within the population. Find the maximum possible proportion of heterozygous individuals in a population (in genetic equilibrium).

Others (proofs & ect.)

1.In $\Delta ABC$ with $AB=AC$ , $D$ is the foot of the perpendicular from $A$ to $BC$ . $E$ is the foot of the perpendicular from $D$ to $AC$ . $F$ is the midpoint of $DE$ . Prove that $AF \perp BE$ .

2.Show that the series

$\sum_{n=2}^\infty \frac{1}{n^m (\log{n})^p}$

where p and m are real numbers converge if $m>1$ or $m=1$ but $p>1$ and diverge otherwise.

3. Let $F_n$ denote the $n$ th Fibonacci number. Prove that if $n$ is odd, then all odd prime divisors of $F_n$ are $1 \mod{4}$ .

Problem by Yiyj1 (talk) 23:44, 3 December 2024 (EST)

Solutions

Problem 1

There is one and only one perfect square in the form

$(p^2+1)(q^2+1)-((pq)^2-pq+1)$

where $p$ and $q$ is prime. Find that perfect square.

Solution 1

$(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2 \cdot q^2 +p^2+q^2+1-p^2 \cdot q^2 +pq-1=p^2+q^2+pq$ . Suppose $n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)$ . Then,

$n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)$

, so since $n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}$ , $n>p,n>q$ so $p+q-n$ is less than both $p$ and $q$ and thus we have $p+q-n=1$ and $p+q+n=pq$ . Adding them gives $2p+2q=pq+1$ so by Simon's Favorite Factoring Trick, $(p-2)(q-2)=3 \implies (p,q)=(3,5)$ in some order. Hence, $(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}$ . $\square$ ~Ddk001

Problem 2

$m$ and $n$ are positive integers. If $m^2=2^8+2^{11}+2^n$ , find $m+n$ .

Solution 1 (Slow, probably official MAA)

$m^2=2^8+2^{11}+2^n$

$\implies 2^n=m^2-2^8-2^{11}$

$\implies 2^n=(m+48)(m-48)$

Let $m+48=2^t$ and $m-48=2^s$ . Then,

$2^t-2^s=96 \implies 2^s(2^{t-s}-1)=2^5 \cdot 3 \implies 2^{t-s}-1=3,2^s=2^5 \implies (t,s)=(7,5) \implies m+n=80+12=\boxed{092}$ $\square$ ~Ddk001

Solution 2 (Fast)

Recall that a perfect square $(a + b)^2$ can be written as $a^2 + 2ab + b^2$ . Since $m^2$ is a perfect square, the RHS must be in this form. We substitute $2^4$ for $a$ to get that $2^8 + 2^5 \cdot 2^b + 2^{2b} = m^2$ . To make the middle term have an exponent of $11$ , we must have $b = 6$ . Then $n = 12$ and $m = (2^4 + 2^6)^2 = (16 + 64)^2 = 80^2$ , so $m + n = \boxed{092}$ .

~ cxsmi

Solution 3 (Faster)

Calculating the terms on the RHS, we find that $256 + 2048 + 2^n = 2304 + 2^n = m^2$ . We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that $4096 = 2^{12}$ works, and it produces $6400 = 80^2$ . Then $m + n = \boxed{092}$ .

~ (also) cxsmi

Problem 3

The fraction,

$\frac{ab+bc+ac}{(a+b+c)^2}$

where $a,b$ and $c$ are side lengths of a triangle, lies in the interval $(p,q]$ , where $p$ and $q$ are rational numbers. Then, $p+q$ can be expressed as $\frac{r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r+s$ .

Solution 1(Probably official MAA, lots of proofs)

Lemma 1: $\text{max} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{3}$

Proof: Since the sides of triangles have positive length, $a,b,c>0$ . Hence,

$\frac{ab+bc+ac}{(a+b+c)^2}>0 \implies \text{max} (\frac{ab+bc+ac}{(a+b+c)^2})= \frac{1}{\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})}$

, so now we just need to find $\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})$ .

Since $(a-c)^2+(b-c)^2+(a-b)^2 \ge 0$ by the Trivial Inequality, we have

$a^2-2ac+c^2+b^2-2bc+c^2+a^2-2ab+b^2 \ge 0$

$\implies a^2+b^2+c^2 \ge ac+bc+ab$

$\implies a^2+b^2+c^2+2(ac+bc+ab) \ge 3(ac+bc+ab)$

$\implies (a+b+c)^2 \ge 3(ac+bc+ab)$

$\implies \frac{(a+b+c)^2}{ab+bc+ac} \ge 3$

$\implies \frac{ab+bc+ac}{(a+b+c)^2} \le \frac{1}{3}$

as desired. $\square$

To show that the minimum value is achievable, we see that if $a=b=c$ , $\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{3}$ , so the minimum is thus achievable.

Thus, $q=\frac{1}{3}$ .

Lemma 2: $\frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}$

Proof: By the Triangle Inequality, we have

$a+b>c$

$b+c>a$

$a+c>b$ .

Since $a,b,c>0$ , we have

$c(a+b)>c^2$

$a(b+c)>a^2$

$b(a+c)>b^2$ .

Add them together gives

$a^2+b^2+c^2<c(a+b)+a(b+c)+b(a+c)=2(ab+bc+ac)$

$\implies a^2+b^2+c^2+2(ab+bc+ac)<4(ab+bc+ac)$

$\implies (a+b+c)^2<4(ab+bc+ac)$

$\implies \frac{(a+b+c)^2}{ab+bc+ac}<4$

$\implies \frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}$ $\square$

Even though unallowed, if $a=0,b=c$ , then $\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{4}$ , so

$\lim_{b=c,a \to 0} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{4}$ .

Hence, $p=\frac{1}{4}$ , since by taking $b=c$ and $a$ close $0$ , we can get $\frac{ab+bc+ac}{(a+b+c)^2}$ to be as close to $\frac{1}{4}$ as we wish.

$p+q=\frac{1}{3}+\frac{1}{4}=\frac{7}{12} \implies r+s=7+12=\boxed{019}$ $\blacksquare$ ~Ddk001

Solution 2 (Fast, risky, no proofs)

By the Principle of Insufficient Reason, taking $a=b=c$ we get either the max or the min. Testing other values yields that we got the max, so $q=\frac{1}{3}$ . Another extrema must occur when one of $a,b,c$ (WLOG, $a$ ) is $0$ . Again, using the logic of solution 1 we see $p=\frac{1}{4}$ so $p+q=\frac{7}{12}$ so our answer is $\boxed{019}$ . $\square$ ~Ddk001

Problem 4

Suppose there are complex values $x_1, x_2,$ and $x_3$ that satisfy

$(x_i-\sqrt[3]{13})(x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}$

Find $x_{1}^3+x_{2}^3+x_{2}^3$ .

Solution 1

To make things easier, instead of saying $x_i$ , we say $x$ .

Now, we have $(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}$ . Expanding gives

$x^3-(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}) \cdot x^2+(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})x-(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})=0$ .

To make things even simpler, let

$a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}$

, so that $x^3-ax^2+bx-c=0$ .

Then, if $P_n=x_{1}^n+x_{2}^n+x_{3}^n$ , Newton's Sums gives

$P_1+(-a)=0$ $(1)$

$P_2+(-a) \cdot P_1+2 \cdot b=0$ $(2)$

$P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0$ $(3)$

Therefore,

$P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))$

$=a \cdot P_2-b \cdot P_1+3 \cdot c$

$=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c$

$=a(a^2-2b)-ab+3c$

$=a^3-3ab+3c$

Now, we plug in $a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}:$

$P_3=(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})^3-3(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})+3(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})$ .

We substitute $x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}$ to get

$P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})$

$=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3(x^2y+y^2x+x^2z+z^2x+z^2y+y^2z+3xyz)+3xyz+1$

$=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3x^2y-3y^2x-3x^2z-3z^2x-3z^2y-3y^2z-9xyz+3xyz+1$

$=x^3+y^3+z^3+1$

$=13+53+103+1$

$=\boxed{170}$ . $\square$

Note: If you don't know Newton's Sums, you can also use Vieta's Formulas to bash.~Ddk001

Problem 5

Suppose

$x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}$

Find the remainder when $\min{x}$ is divided by 1000.

Solution 1 (Euler's Totient Theorem)

We first simplify $2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:$

$2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6=42^4+6 \cdot 30^6=(\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)}$

so

$x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)} \equiv 1 \pmod{5}$

$x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \pmod{6}$

$x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 6 \cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)} \equiv 6 \pmod{7}$ .

where the last step of all 3 congruences hold by the Euler's Totient Theorem. Hence,

$x \equiv 1 \pmod{5}$

$x \equiv 0 \pmod{6}$

$x \equiv 6 \pmod{7}$

Now, you can bash through solving linear congruences, but there is a smarter way. Notice that $5|x-6,6|x-6$ , and $7|x-6$ . Hence, $210|x-6$ , so $x \equiv 6 \pmod{210}$ . With this in mind, we proceed with finding $x \pmod{7!}$ .

Notice that $7!=5040= \text{lcm}(144,210)$ and that $x \equiv 0 \pmod{144}$ . Therefore, we obtain the system of congruences :

$x \equiv 6 \pmod{210}$

$x \equiv 0 \pmod{144}$ .

Solving yields $x \equiv 2\boxed{736} \pmod{7!}$ , and we're done. $\square$ ~Ddk001

Problem 6

Suppose that there is $192$ rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are $2$ other pegs positioned sufficiently apart. A $move$ is made if

(i) $1$ 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 $x$ be the minimum possible number $moves$ that can transfer all $192$ rings onto the second peg. Find the remainder when $x$ is divided by $1000$ .

Solution 1 (Recursion)

Let $M_n$ be the minimum possible number $moves$ that can transfer $n$ rings onto the second peg. To build the recursion, we consider what is the minimum possible number $moves$ that can transfer $n+1$ rings onto the second peg. If we use only legal $moves$ , then $n+1$ 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 $n$ rings to the third peg using $M_n$ $moves$ , then place the largest ring onto the bottom of the second peg using $1$ $move$ , and then get all the rings from the third peg on top of the largest ring using another $M_n$ $moves$ . This gives a total of $2M_n+1$ , hence we have $M_{n+1}=2M_{n}+1$ . Obviously, $M_1=1$ . We claim that $M_n=2^n-1$ . This is definitely the case for $n=1$ . If this is true for $n$ , then

$M_{n+1}=2M_{n}+1=2(2^n-1)+1=2^{n+1}-1$

so this is true for $n+1$ . Therefore, by induction, $M_n=2^n-1$ is true for all $n$ . Now, $x=M_{192}=2^{192}-1$ . Therefore, we see that

$x+1 \equiv 0 \pmod{8}$ .

But the $\text{mod 125}$ part is trickier. Notice that by the Euler's Totient Theorem,

$2^{\phi (125)}=2^{100} \equiv 1 \pmod{125} \implies 2^{200} \equiv 1 \pmod{125}$

so $x+1=\frac{2^{200}}{256}$ is equivalent to the inverse of $256$ in $\text{mod 125}$ , which is equivalent to the inverse of $6$ in $\text{mod 125}$ , which, by inspection, is simply $21$ . Hence,

$x+1 \equiv 0 \pmod{8}$

$x+1 \equiv 21 \pmod{125}$

, so by the Chinese Remainder Theorem, $x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}$ . $\square$

~Ddk001

Problem 7

Let $\overline{ab}$ be a 2-digit positive integer satisfying $\overline{ab}^2=a! +b!$ . Find the sum of all possible values of $\overline{ab}$ .

Solution 1 (Tedious Casework)

Case 1: $a>b$

In this case, we have

$\overline{ab}^2=a! +b!=(1+a \cdot (a-1) \cdot \dots \cdot (b+1)) \cdot b! \implies b!|\overline{ab}^2=(10a+b)^2$ .

If $b \ge 5$ , we must have

$10|b!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0$

, but this contradicts the original assumption of $b \ge 5$ , so hence we must have $b \le 4$ .

With this in mind, we consider the unit digit of $\overline{ab}^2$ .

Subcase 1.1: $a>b=1$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+1)^2-1 \equiv 0 \pmod{10} \implies 10|a! \implies a \ge 5$ .

There is no apparent contradiction here, so we leave this as it is.

Subcase 1.2: $a>b=2$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+2)^2-2 \equiv 2 \pmod{10} \implies a! \equiv 2 \pmod{10} \implies a=2$ .

This contradicts with the fact that $a>b$ , so this is impossible.

Subcase 1.3: $a>b=3$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+3)^2-6 \equiv 3 \pmod{10}$ .

However, this is impossible for all $a$ .

Subcase 1.4: $a>b=4$

In this case, we have that

$a! \equiv \overline{ab}^2-b! \equiv (10a+4)^2-24 \equiv 2 \pmod{10}$ .

Again, this yields $a=2$ , which, again, contradicts $a>b$ . $\square$

Hence, we must have $b=1$ .

Now, with $b$ determined by modular arithmetic, we actually plug in the values.

To simplify future calculations, note that

$a!=\overline{ab}^2-b!=(10a+1)^2-1=100a^2+20a=10a(10a+2)$ .

For $a=5$ , this does not hold.

For $a=6$ , this does not hold.

For $a=7$ , this does hold. Hence, $(a,b)=(7,1)$ is a solution.

For $a=8$ , this does not hold.

For $a=9$ , this does not hold.

Hence, there is no positive integers $a$ and $b$ between $1$ and $9$ inclusive such that $a!+b!=\overline{ab}^2$ .

Case 2: $a=b$

For this case, we must have

$(11a)^2=\overline{ab}^2=a!+b!=2a! \implies 11|a!$

which is impossible if a is a integer and $1 \le a \le 9$ .

Case 3: $a<b$

In this case, we have

$\overline{ab}^2=a! +b!=(1+b \cdot (b-1) \cdot \dots \cdot (a+1)) \cdot a! \implies a!|\overline{ab}^2=(10a+b)^2$ .

If $a \ge 5$ , we must have

$10|a!|\overline{ab}^2=(10a+b)^2 \implies 10|(10a+b)^2=100a^2+20ab+b^2 \implies 10|b \implies b=0 \implies 10|a!+b!$

which is impossible since $10|a!$ and $b!=1$ .

Hence, $a \le 4$ .

Subcase 3.1: $b>a=1$

$(10+b)^2=1+b!$

Testing cases, we can see that there is no such $b$ .

Subcase 3.2: $b>a=2$

$(20+b)^2=2+b!$

Testing cases, we can see that there is no such $b$ .

Subcase 3.3: $b>a=3$

$(30+b)^2=3+b!$

Testing cases, we can see that there is no such $b$ .

Subcase 3.4: $b>a=4$

$(40+b)^2=4+b!$

Testing cases, we can see that there is no such $b$ .

We see that $(a,b)=(7,1) \implies a+b=\boxed{008}$ . $\blacksquare$ ~Ddk001

Solution 2 (Official)

$\overline{ab}^2=a! +b!$ cleary square of a positive double-digit integer is either 3-digit or 4-digit as it ranges between $100(10^{2})$ and $9801(99^{2})$ .

This means both $a$ and $b$ are less than or equal to 7 as any $factorial$ greater than 7! exceeds 9999.

Now at least one of $a$ or $b$ must be greater than or equal to 5 or else the maximum possible value of LHS is 4!+4!=48<100 also, both $a$ and $b$ can't be greater than or equal to 5 as if they were so. The unit digit of LHS would be zero but the unit digit of RHS would be either 5,6 or 9.

Hence, one of $a$ and $b$ is greater than or equal to 5 and the other is less than 5. this means the unit digit of RHS is the unit digit of a factorial which is less than 5.

Hence unit digit of RHS is 0,1,2, 6 or 4. 0,2,4 and 6 are rejected as follows:-

1. 2 can't be the unit digit of a perfect square.

2. If 6 is the unit digit of OF LHS this means one of the numbers is 3( as only 3! has the unit digit 6) and also this would mean that $b$ is either 4 or 6. This directly means that 36 and 34 are the only cases to be tested. 34 can't be as both the digits are less than 5 and 36 clearly doesn't satisfy

3. If 0 is the unit digit of LHS then 50 60 70 are the only cases (as one of the digits is greater than or equal to 5) that don't satisfy

4. If 4 is the unit digit of LHS this means one of the numbers is 4( as only 4! has the unit digit 4) and also this would mean that $b$ is either 2 or 8. This directly means that 42 and 48 are the only cases to be tested. 42 can't be as both the digits are less than 5 and 48 can't also as one of the digits is 8

Hence, the unit digit of LHS must be 1 for which $b$ =1 or 9(rejected). This means 51 61 and 71 are the only cases to be tried now which is really very easy to calculate and get 71 as the only possible solution to the problem and

thus, our answer is 7+1= $\boxed{008}$ .~ SANSGANKRSNGUPTA

Solution 3 (Official and Fastest)

$\overline{ab}^2=a! +b!$ cleary square of a positive double-digit integer is either 3-digit or 4-digit as it ranges between $100(10^{2})$ and $9801(99^{2})$ .

This means both $a$ and $b$ are less than or equal to 7 as any $factorial$ greater than 7! exceeds 9999.

Now at least one of $a$ or $b$ must be greater than or equal to 5 or else the maximum possible value of LHS is 4!+4!=48<100 also, both $a$ and $b$ can't be greater than or equal to 5 as if they were so. The unit digit of LHS would be zero but the unit digit of RHS would be either 5,6 or 9.

Hence, one of $a$ and $b$ is greater than or equal to 5 and the other is less than 5. Also, RHS is a perfect square, so it must be 0 or 1 $(mod 4)$

Hence the number less than 5 can be either 1 or 4 because 2! and 3! are 2 $(mod 4)$

If it is 4, then the unit digit of RHS is 4 meaning $b$ is either 2 or 8 both of which aren't possible as 8>7 and the number less than 5 is 4 not 2.

If it is 1 then the unit digit of RHS is 1 implying $b$ is either 9(rejected 9>7) or 1. This means $b$ is 1, this means 51 61 and 71 are the only cases to be tried

which is very easy to calculate and get 71 as the only possible solution to the problem and

thus, our answer is 7+1= $\boxed{008}$ .~ SANSGANKRSNGUPTA

Problem 8

Suppose $f(x)$ is a $10000000010$ -degrees polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \dots, r_{10000000010}$ . Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$ . Also, suppose that

$(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!$

for an integer $m$ . If $p$ is the minimum possible positive integral value of

$(1+r_1)(1+r_2) \dots (1+r_{10000000010})$ .

Find the number of factors of the prime $999999937$ in $p$ .

Solution 1

Since all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$ , we have that all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)-n=0$ , so by the Factor Theorem,

$n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n$

$\implies (n+1)n \dots (n-10000000008)|f(n)-n$ .

$\implies f(n)=a(n+1)n \dots (n-10000000008)+n$

since $f(n)$ is a $10000000010$ -degrees polynomial, and we let $a$ to be the leading coefficient of $f(n)$ .

Also note that since $r_1, r_2, \dots, r_{10000000010}$ is the roots of $f(n)$ , $f(n)=a(n-r_1)(n-r_2) \dots (n-r_{10000000010})$

Now, notice that

$m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})$

$=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})$

$=\frac{f(-2)}{a}$

$=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}$

$=\frac{10000000010! \cdot a-2}{a}$

$=10000000010!-\frac{2}{a}$

Similarly, we have

$(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}$

To minimize this, we minimize $m$ . The minimum $m$ can get is when $m=10000000011$ , in which case

$-\frac{2}{a}=10000000011!-10000000010!$

$=10000000011 \cdot 10000000010!-10000000010!$

$=10000000010 \cdot 10000000010!$

$\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})$

$=-\frac{1}{a}$

$=\frac{10000000010 \cdot 10000000010!}{2}$

$=5000000005 \cdot 10000000010!$

, so there is $\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}$ factors of $999999937$ . $\square$ ~Ddk001

Problem 9

$\Delta ABC$ is an isosceles triangle where $CB=CA$ . Let the circumcircle of $\Delta ABC$ be $\Omega$ . Then, there is a point $E$ and a point $D$ on circle $\Omega$ such that $AD$ and $AB$ trisects $\angle CAE$ and $BE<AE$ , and point $D$ lies on minor arc $BC$ . Point $F$ is chosen on segment $AD$ such that $CF$ is one of the altitudes of $\Delta ACD$ . Ray $CF$ intersects $\Omega$ at point $G$ (not $C$ ) and is extended past $G$ to point $I$ , and $IG=AC$ . Point $H$ is also on $\Omega$ and $AH=GI<HB$ . Let the perpendicular bisector of $BC$ and $AC$ intersect at $O$ . Let $J$ be a point such that $OJ$ is both equal to $OA$ (in length) and is perpendicular to $IJ$ and $J$ is on the same side of $CI$ as $A$ . Let $O’$ be the reflection of point $O$ over line $IJ$ . There exist a circle $\Omega_1$ centered at $I$ and tangent to $\Omega$ at point $K$ . $IO’$ intersect $\Omega_1$ at $L$ . Now suppose $O’G$ intersects $\Omega$ at one distinct point, and $O’, G$ , and $K$ are collinear. If $IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2$ , then $\frac{EH}{BH}$ can be expressed in the form $\frac{\sqrt{b}}{a} (\sqrt{c} + d)$ , where $b$ and $c$ are not divisible by the squares of any prime. Find $a^2+b^2+c^2+d^2+abcd$ .

Someone mind making a diagram for this?

Solution 1

Line $IJ$ is tangent to $\Omega$ with point of tangency point $J$ because $OJ=OA \implies \text{J is on } \Omega$ and $IJ$ is perpendicular to $OJ$ so this is true by the definition of tangent lines. Both $G$ and $K$ are on $\Omega$ and line $O’G$ , so $O’G$ intersects $\Omega$ at both $G$ and $K$ , and since we’re given $O’G$ intersects $\Omega$ at one distinct point, $G$ and $K$ are not distinct, hence they are the same point.

Now, if the center of $2$ tangent circles are connected, the line segment will pass through the point of tangency. In this case, if we connect the center of $2$ tangent circles, $\Omega$ and $\Omega_1$ ( $O$ and $I$ respectively), it is going to pass through the point of tangency, namely, $K$ , which is the same point as $G$ , so $O$ , $I$ , and $G$ are collinear. Hence, $G$ and $I$ are on both lines $OI$ and $CI$ , so $CI$ passes through point $O$ , making $CG$ a diameter of $\Omega$ .

Now we state a few claims :

Claim 1: $\Delta O’IO$ is equilateral.

Proof:

$\frac{3}{4} (IK+O’L)^2$

$=\frac{3}{4} IK^2+\frac{3}{2} IK \cdot O’L+\frac{3}{4} O’L^2$

$=IG^2+IG \cdot GC$

$=IG \cdot (IG+GC)$

$=IG \cdot IC$

$=IJ^2$

where the last equality holds by the Power of a Point Theorem.

Taking the square root of each side yields $IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2$ .

Since, by the definition of point $L$ , $L$ is on $\Omega_1$ . Hence, $IK=IL$ , so

$IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2=\frac{\sqrt{3}}{2} (IL+O’L)^2=\frac{\sqrt{3}}{2} IO’^2$ , and since $O’$ is the reflection of point $O$ over line $IJ$ , $OJ=O’J=\frac{OO’}{2}$ , and since $IJ=\frac{\sqrt{3}}{2} IO’^2$ , by the Pythagorean Theorem we have

$JO’=\frac{IO’}{2} \implies \frac{OO’}{2}=\frac{IO’}{2} \implies OO’=IO’$

Since $IJ$ is the perpendicular bisector of $OO’$ , $IO’=IO$ and we have $IO=IO’=OO’$ hence $\Delta O’IO$ is equilateral. $\square$

With this in mind, we see that

$2OJ=OO’=OI=OK+KI=OJ+GI=OJ+AC \implies OA=OJ=AC$

Here, we state another claim :

Claim 2 : $BH$ is a diameter of $\Omega$

Proof: Since $OA=OC=AC$ , we have

$\angle AOC =60^\circ \implies \angle ABC=\frac{1}{2} \angle AOC=30^\circ \implies AB=\sqrt{3} AC$

and the same reasoning with $\Delta CAH$ gives $CH=\sqrt{3} AC$ since $AH=IG=AC$ .

Now, apply Ptolemy’s Theorem gives

$BH \cdot AC+BC \cdot AH=CH \cdot AB \implies BH \cdot AC+AC^2=3AC^2 \implies BH=2AC=2OA$

so $BH$ is a diameter. $\square$

From that, we see that $\angle BEH=90^\circ$ , so $\frac{EH}{BH}=\cos{BHE}$ . Now,

$\angle BHE=\angle BAE=\frac{1}{2} \angle CAB=15^\circ$

, so

$\frac{EH}{BH}=\cos{15}=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{2}}{4} (\sqrt{3}+1)$

, so

$a=4, b=2, c=3, d=1 \implies a^2+b^2+c^2+d^2+abcd=1+4+9+16+24=\boxed{054}$

, and we’re done. $\blacksquare$ ~Ddk001

Problem 10

Suppose $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]=\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution 1(Wordless endless bash)

$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}$

$=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn(m+n+2)}$

$=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{m(m+n+2)}$

$=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{n+2} (\frac{1}{m}-\frac{1}{m+n+2})$

$=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \sum_{m=1}^{\infty} (\frac{1}{m}-\frac{1}{m+n+2})$

$=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot [(1-\frac{1}{n+3})+(\frac{1}{2}-\frac{1}{n+4})+ \dots]$

$=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})$

$=\sum_{n=1}^{\infty} (\frac{\frac{1}{2}}{n}-\frac{\frac{1}{2}}{n+2}) \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})$

$=\frac{1}{2} [(1-\frac{1}{3})(1+\frac{1}{2}+\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+ \dots$

$=\frac{1}{2} [[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\dots](1+\frac{1}{2}+\frac{1}{3})+[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{6})+\dots](1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+\dots](\frac{1}{4}+\frac{1}{5})+\dots]$

$=\frac{1}{2} [(1+\frac{1}{2}+\frac{1}{3})+\frac{1}{2} (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+\frac{1}{3} (\frac{1}{4}+\frac{1}{5})+\dots]$

$=\frac{1}{2} [\frac{11}{6}+\frac{1}{2} \cdot \frac{25}{12}+(\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\dots)+(\frac{1}{3 \cdot 5}+\frac{1}{4 \cdot 6}+\dots)]$

$=\frac{1}{2} [\frac{69}{24}+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\dots ]+\frac{1}{2} [(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+\dots ]$

$=\frac{1}{2} [\frac{69}{24}+\frac{1}{3}+\frac{1}{6}+\frac{1}{8}]$

$=\frac{1}{2} \cdot \frac{84}{24}$

$=\frac{7}{4}$

$(1+\frac{1}{x})^x=e^{x \cdot \ln (1+\frac{1}{x})}$

$=e^{x \cdot [(\frac{1}{x})-\frac{(\frac{1}{x})^2}{2}+\frac{(\frac{1}{x})^3}{3}+\dots]}$

$=e^{1-\frac{1}{2} (\frac{1}{x})+\frac{1}{3} (\frac{1}{x})^2+\dots}$

$=e \cdot e^{-\frac{1}{2} (\frac{1}{x})} \cdot e^{\frac{1}{3} (\frac{1}{x})^2} \dots$

$=e \cdot [1-\frac{1}{2x}+\frac{1}{2!} (\frac{1}{2x})^2- \dots] \cdot [1+\frac{1}{3x^2}+\frac{1}{2!} (\frac{1}{3x^2})^2+ \dots]$

$=e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]$

$\implies \lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]$

$=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]}{e}-1]]$

$=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots-1)]$

$=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots)]$

$=\lim_{x\rightarrow \infty} (\frac{x}{2}-\frac{x}{2}+\frac{11}{24}+\dots)$

$=\frac{11}{24}$

$\implies \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]$

$=\frac{7}{4}+\frac{11}{24}$

$=\frac{53}{24}$

$\implies p=53,q=24$

$\implies p+q=\boxed{077}$ $\square$ ~Ddk001

Problem 11

In $\Delta ABC$ with $AB=AC$ , $D$ is the foot of the perpendicular from $A$ to $BC$ . $E$ is the foot of the perpendicular from $D$ to $AC$ . $F$ is the midpoint of $DE$ . Prove that $AF \perp BE$ .

Solution 1 (Analytic geo)

Let

$A=(0,0)$

$B=(4a,4b)$

$C=(4 \sqrt{a^2+b^2},0)$

We set it this way to simplify calculation when we calculate the coordinates of $E$ and $F$ (Notice to find $E$ , you just need to take the x coordinate of $D$ and let the y coordinate be $0$ ).

Obviously,

$D=(\frac{4a+4 \sqrt{a^2+b^2}}{2},\frac{4b+0}{2})=(2a+2 \sqrt{a^2+b^2},2b)$

$\implies E=(2a+2 \sqrt{a^2+b^2},0)$

$\implies F=(\frac{2a+2 \sqrt{a^2+b^2}+2a+2 \sqrt{a^2+b^2}}{2},\frac{2b+0}{2})=(2a+2 \sqrt{a^2+b^2},b)$

Now, we see that

$\text{Slope} _ {AF}=\frac{b}{2a+2 \sqrt{a^2+b^2}}$

$\text{Slope} _ {BE}=\frac{0-4b}{2a+2 \sqrt{a^2+b^2}-4a}=\frac{-2b}{\sqrt{a^2+b^2}-a}$

$\implies \text{Slope} _ {AF} \cdot \text{Slope} _ {BE}=\frac{b}{2a+2 \sqrt{a^2+b^2}} \cdot \frac{-2b}{a+ \sqrt{a^2+b^2}-2a}=\frac{-2b^2}{2(a+\sqrt{a^2+b^2})(\sqrt{a^2+b^2}-a)}=\frac{-2b^2}{2b^2}=-1$

, so $AF \perp BE$ , as desired. $\square$ ~Ddk001

Solution 2 (Hard vector bash)

Solution 2a (Hard)

$\overrightarrow{AF} \cdot \overrightarrow{BE}$

$=(\overrightarrow{AE}+\overrightarrow{EF}) \cdot (\overrightarrow{BD}+\overrightarrow{DE})$

$=\overrightarrow{AE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{DE}+\overrightarrow{AE} \cdot \overrightarrow{DE}$

$=\overrightarrow{AE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{DE}$

$=(\overrightarrow{AD}+\overrightarrow{DE}) \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD} + \overrightarrow{EF} \cdot \overrightarrow{DE}$

$=\overrightarrow{DE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD} + \overrightarrow{EF} \cdot \overrightarrow{DE}$

$=\overrightarrow{DE} \cdot \overrightarrow{DC}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{BD}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DE}$

$=\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DC}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DE}$

$=\overrightarrow{DE} \cdot (\frac{\overrightarrow{DC}-\overrightarrow{DE}}{2})$

$=\frac{\overrightarrow{DE} \cdot \overrightarrow{EC}}{2}$

$=0$

Hence, $AF \perp BE$ . $\square$ ~Ddk001

Solution 2b (Harder)

$\angle ACD=\angle ECD$

$\angle ADC=\angle DEC$

$\implies \Delta ADC \sim \Delta DEC$

$\implies \frac{EC}{DC}=\frac{DC}{AC}$

$\implies EC=\frac{DC^2}{AC}$

$\implies \overrightarrow{E}=\overrightarrow{C}+\overrightarrow{CE}$

$=\overrightarrow{C}+\frac{CE}{AC} \cdot \overrightarrow{CA}$

$=\overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{CA}$

$=\overrightarrow{C}+\frac{DC^2}{AC^2} (\overrightarrow{A}-\overrightarrow{C})$

$=\frac{AC^2-DC^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}$

$=\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}$

$\overrightarrow{D}=\frac{\overrightarrow{B}+\overrightarrow{C}}{2}$

Since $F$ is the midpoint of $DE$ ,

$\overrightarrow{F}=\frac{\overrightarrow{D}+\overrightarrow{E}}{2}$

$=\frac{\frac{\overrightarrow{B}+\overrightarrow{C}}{2}+\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}}{2}$

$=\frac{\overrightarrow{B}}{4}+\frac{AC^2+2AD^2}{4AC^2} \overrightarrow{C}+\frac{DC^2}{2AC^2} \overrightarrow{A}$

$\overrightarrow{AF}=\overrightarrow{F}-\overrightarrow{A}=\frac{\overrightarrow{B}}{4}+\frac{AC^2+2AD^2}{4AC^2} \overrightarrow{C}+\frac{DC^2-2AC^2}{2AC^2} \overrightarrow{A}$

$\overrightarrow{BE}=\overrightarrow{E}-\overrightarrow{B}=\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}-\overrightarrow{B}$

Now come the coordinates. Let

$A=(0,0)$

$B=(-a,-b)$

$C=(a,-b)$

so that

$\overrightarrow{A}=\langle 0,0 \rangle$

$\overrightarrow{B}=\langle -a,-b \rangle$

$\overrightarrow{C}=\langle a,-b \rangle$ .

Therefore,

$\overrightarrow{AF}=\langle \frac{-a}{4},\frac{-b}{4} \rangle + \frac{(a^2+b^2)+2b^2}{4(a^2+b^2)} \langle a,-b \rangle=\langle \frac{ab^2}{2(a^2+b^2)},\frac{-a^2b-2b^3}{2(a^2+b^2)} \rangle$

$\overrightarrow{BE}=\frac{b^2}{a^2+b^2} \langle a,-b \rangle-\langle -a,-b \rangle=\langle \frac{2ab^2+a^3}{a^2+b^2},\frac{a^2b}{a^2+b^2} \rangle$

$\implies \overrightarrow{AF} \cdot \overrightarrow{BE}=\langle \frac{ab^2}{2(a^2+b^2)},\frac{-a^2b-2b^3}{2(a^2+b^2)} \rangle \cdot \langle \frac{2ab^2+a^3}{a^2+b^2},\frac{a^2b}{a^2+b^2} \rangle$

$=\frac{1}{2(a^2+b^2)^2}[ab^2(a^3+2ab^2)-a^2b(a^2b+2b^3)]$

$=\frac{ab}{2(a^2+b^2)^2} (a^3b+2ab^3-a^3b-2ab^3)$

$=0$

Hence, we have that $AF$ is perpendicular to $BE$ . $\square$ ~Ddk001

Problem 12

Show that the series

$\sum_{n=2}^\infty \frac{1}{n^m (\log{n})^p}$

where p and m are real numbers converge if $m>1$ or $m=1$ but $p>1$ and diverge otherwise.

Solution 1 (Plain old chapter 3 analysis bash)

Before we even get started, let's state a few definitions first.

Definition 1: A set $X$ , whose elements we shall call points, is said to be a metric space if with any two points $p$ and $q$ of $X$ there is associated a real number $d(p,q)$ , called the distance from $p$ to $q$ , such that

$(a) d(p,q)>0$ if $p \ne q; d(p,p)=0;$

$(b) d(p,q)=d(q,p);$

$(c) d(p,q)\le d(p,r)+d(r,q)$ , for any $r \in X$ .

Any function with these properties is called a distance function or a metric

Definition 2: Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$ .

(a) A neighborhood of a point $p$ is a set $N_r (p)$ consisting of all points such that $d(p,q)<r$ . The number $r$ is said to be the radius of $N_r (p)$ .

(b) A point $p$ is a limit point of a set $E$ if every neighborhood of $p$ contains a point $q \ne p$ such that $q \in E$ .

(c) If $p \in E$ and $p$ is not a limit point of $E$ , then $p$ is called a isolated point of $E$ .

(d) $E$ is closed if every limit point of $E$ is a point of $E$ .

(e) A point $p$ is an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N \subseteq E$

(f) $E$ is open if every point of $E$ is a interior point of $E$ .

(g) The complement of $E$ (denoted by $E^c$ ) is the set of points $p$ such that $p \not\in E$ but $p \in X$ .

(h) $E$ is bounded if there is a ream number $M$ and a point $q \in X$ such that $d(p,q)<M$ for all $p \in E$ .

Definition 3: Let $X$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $X$ .

(a) By a open cover of a set $E$ in $X$ we mean a collection ${G_{\alpha}}$ of open subsets of $X$ such that $E \subseteq \cup _{\alpha} G_{\alpha}$ .

(b) A subset $K$ of $X$ is said to be compact if every open cover of $K$ contain a finite subcover.

Definition 4: Let $X$ be a metric space and $E \subseteq X$ . Let $E'$ denote the set of limit points of $E$ . The closure of $E$ is said to be $\bar{E} =E \cup E'$ .

Lemma 1: Closed subsets of compact sets are compact.

Proof: Suppose $F \subseteq K \subseteq X$ , $F$ is closed and $K$ is compact. Let $\{ V_{\alpha} \}$ be a open cover of $F$ . We wish to show that $\{ V_{\alpha} \}$ have no finite subcover of $F$ . Let $F^c$ be adjoined to $\{ V_{\alpha} \}$ . Then we obtain an open cover $\Omega$ of $K$ . (Note that complements of closed sets are open since if $E$ is closed, all the limit points of $E$ is a point of $E$ so if $p \in E'$ , $p\in E$ . Every neighborhood of $p$ contain a point of $E$ that is not $p$ itself so no neighborhood of $p$ is completely in $E^c$ so $p$ is not a interior point of $E^c$ , so $E^c$ is open. Thus $F^c$ is open.) Since $K$ is compact, there is a finite subcollection $\phi$ of $\Omega$ that cover $K$ , and hence $F$ . If $F^c \subseteq \phi$ , we can exclude it from $\phi$ and can thus still have a open cover of $F$ . The obtained open cover is thus a finite subcollection of $\{ V_{\alpha} \}$ that cover $F$ , so $F$ is compact. $\square$

Definition 5: A subset $E$ of $R^k$ is said to be a k-cell if $E$ is the set of $x$ such that $E= \{ x=(x_1,\dots , x_k)| a_1 \le x_1 \le b_1, \dots , a_k \le x_k \le b_k \}$ , where the $a_i$ 's and $b_i$ 's are real numbers.

Lemma 2: Every k-cell is compact.

Proof: Let $I$ be a k-cell, consisting of the points $x$ such that $x=(x_1,\dots , x_k), a_1 \le x_1 \le b_1, \dots , a_k \le x_k \le b_k$ . Put

$\delta = \{ \sum_{i=1} ^{k} (b_i-a_i)^2 \} ^{\frac{1}{2}}$

Thus $x \in I,y \in I$ would imply $|x-y| \le \delta$ .

Suppose, for the sake of contradiction, that there exist an open cover $\{ G_{\alpha} \}$ of $I$ which contain no finite subcover. Put $c_i=\frac{a_i+b_i}{2}$ . The intervals $[a_j,c_j]$ and $[c_j,b_j]$ then determine $2^k$ k-cells $Q_i$ whose union is $I$ . Thus at least one of these sets, call it $I_1$ , cannot be covered by any finite subcollection of $\{ G_{\alpha} \}$ (otherwise $I$ could be so covered). We divide $I_1$ like we did $I$ , and obtain $I_2$ as we did $I_1$ . As the process continue we thus obtain a sequence of $I_n$ 's such that

(a) $I \supseteq I_1 \supseteq I_2 \supseteq \dots$ ;

(b) $I_n$ is not covered by any subcollection of $\{ G_{\alpha} \}$ ;

(c) If $x \in I_n, y \in I_n$ , then $|x-y| \le 2^{-n} \delta$ .

We claim that there is a point $x*$ in every $I_n$ . In fact, we can even prove the stronger statement

"Let $k$ be a positive interger. If $\{ I_n \}$ is a sequence of k-cells such that $I \supseteq I_1 \supseteq I_2 \supseteq \dots$ , then $\cap _{1} ^{\infty} I_n \ne {\o}$ ."

To prove that, it suffices to prove it with $k=1$ . The rest follows easily from induction. The $k=1$ case can be rephrased as follows:

"If $\{ I_n \}$ is a sequence of intervals such that $I \supseteq I_1 \supseteq I_2 \supseteq \dots$ , then $\cap _{1} ^{\infty} I_n \ne {\o}$

To prove this, let $I_n=[a_n,b_n]$ . Then let $E$ denote the set of all the $a_i$ 's. Obviously $E$ is non-empty and is bounded above (by $b_1$ , for one). Then the least upper bound of $E$ exist, call it $x$ . If $m$ and $n$ are any positive integers, then

$a_n \le a_{m+n} \le b_{m+n} \le b_m$

so $x \le b_m$ for each $m$ . Since $a_m \le x$ , obviously $x \in \cap _{1} ^{\infty} I_n$ .

The result follow.

Thus there is a point $x*$ in every $I_n$ . Now, since $\{ G_{\alpha} \}$ covers $I$ it follows that $x* \in G_{\alpha}$ for some $\alpha$ . Since $G_{\alpha}$ is open there is a $r>0$ such that $|y-x*|<r$ implies $y \in G_{\alpha}$ . If n is so large that $2^{-n} \delta <r$ then (c) implies that $I_n \subseteq G_{\alpha}$ , which contradicts (b). A contradiction is obtained and the result must follow. $\square$

Corollary 1 (Heine-Borel Theorem): Suppose $E \subseteq R^k$ . Then the following two statements are equivalent:

(a) $E$ is closed and bounded

(b) $E$ is compact

Proof: It suffices to show that $(a) \Longleftrightarrow (b)$ . We shall only prove that $(a) \implies (b)$ since we will only use that part. However, a proof of the converse is given in a remark after the solution.

Since $E$ is closed and bounded, $E$ is a closed subset of a k-cell, a compact set. Thus $E$ is compact by Lemma 1. $\square$

Thus, in $R^k$ , every single bounded set have a compact closure.

Definition 6: Let $E$ be a subset of a metric space $X$ , and let $S$ be the set such that

$S= \{ d(p,q):|p,q \in E \}$ .

Then the smallest upper bound of $S$ is called the diameter of $E$ and is often written as $\text{diam} (E)$ .

Lemma 3: (a) Let $\bar{E}$ be the closure of $E$ in a metric space $X$ . Then

$\text{diam} (\bar{E})=\text{diam} (E)$

(b) If $K_n$ is a sequence of compact sets in $X$ such that $K_n \supseteq K_{n+1}, (n=1,2,3, \dots)$ and if

$\lim_{n \to \infty} \text{diam} (K_n)=0$

, then $\cap _{n=1} ^{\infty} K_n$ consist of exactly one point.

Proof: (a) Since $E \subseteq \bar{E}$ , it is obvious that

$\text{diam} (\bar{E})\supseteq \text{diam} (E)$

Thus it suffice to show that

$\text{diam} (\bar{E})\subseteq \text{diam} (E)$

To do so, choose $p,q \in \bar{E}$ . Then for each $\epsilon >0$ there exist points $p'$ and $q'$ such that $d(p,p')< \epsilon$ and that $d(q,q')<\epsilon$ . Hence,

$d(p,q) \le d(p,p')+d(p',q')+d(q',q) < 2\epsilon +d(p',q') \le 2\epsilon+ \text{diam} (E)$

It follows that

$\text{diam} (\bar{E}) \le 2 \epsilon + \text{diam} (E)$

. Since $\epsilon$ is ambitarily chosen, (a) is proved. $\square$

(b) Put $K=\cap _{n=1} ^{\infty} K_n$ . It suffice to show that $K$ has at most 1 point and that it exist. The former is easy. If $K$ have 1 or more points, $\text{diam} (K) >0$ which contradict the fact that $\lim_{n \to \infty} \text{diam} (K_n)=0$ .

It remains to show that $K$ is non-empty.

Sublemma 3.1: If $\{ K_{\alpha} \}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{ K_{\alpha} \}$ is non-empty, then $\cap K_{\alpha}$ is non-empty.

Proof: Fix a member $K_1$ of $\{ K_{\alpha} \}$ and let $G_{\alpha}= {K_{\alpha}}^{c}$ for each $\alpha$ . Suppose, for the sake of contradiction that no point of $K_1$ belong to every single member of $\{ K_{\alpha} \}$ . Note that every $G_{\alpha}$ is open by the parenthetical remark in the proof of lemma 1. Thus the sets $G_{\alpha}$ form a open cover of $K_1$ ; however, since $K_1$ is compact there exist finitely many indices $\alpha_{1}, \alpha_{2}, \alpha_{3}, \dots, \alpha_{n}$ such that

$K_1 \subseteq G_{\alpha_{1}} \cup G_{\alpha_{2}} \cup \dots \cup G_{\alpha_{n}}$

But this would mean

$K_1 \cap K_{\alpha_{1}} \cap K_{\alpha_{2}} \cap \dots \cap K_{\alpha_{n}}$

, contradictory to the hypothesis.

A contradiction is obtained so we are done. $\square$ .

The result follow from the sublemma. $\blacksquare$

Now, we show state another important lemma that will use the above lemma (lemma 3). Part (a) and (c) of the following lemma is called the Cauchy Criterion.

Before starting the proof, however, we have one more definition to make.

Definition 7: A sequence $\{ p_n \}$ of a metric space $X$ is said to be a Cauchy Sequence if for every $\epsilon >0$ there exist a integer $N$ such that $d(p_n,p_m)< \epsilon$ if $n,m \ge N$ .

Obviously, we can state definition 7 in terms of definition 6:

" A sequence $\{ p_n \}$ of a metric space $X$ is said to be a Cauchy Sequence if

$\lim_{n \to \infty} \text{diam} (E_n)=0$

where $E_n$ is the set consisting of the points $p_n, p_{n+1}, \dots$ . "

In fact, its converse is also true (and its proof is trivial):

"Let $\{ E_n \}$ be a sequence of sets such that

$\lim_{n \to \infty} \text{diam} (E_n)=0$ .

Then if a sequence $\{ p_n \}$ is chosen with $p_n \in E_n$

More to come

Problem 13

Let $x,y$ and $z$ be real numbers. Show that

$(x^2+z^2)^2+y^4 \ge 4xzy^2$

Solution 1 (Plain high school contest)

It might seem insane, but we are going to divide this problem into 2 cases:

$xy+yz \ge 0$

and

$xy+yz \le 0$

Case 1: $xy+yz \ge 0$

By the Trivial Inequality, we have

$(x-\frac{y}{\sqrt{2}})^2 \ge 0$ $(z-\frac{y}{\sqrt{2}})^2 \ge 0$

. Add these inequalities gives

$(x-\frac{y}{\sqrt{2}})^2+(z-\frac{y}{\sqrt{2}})^2 \ge 0$

and expanding yields

$x^2+y^2+z^2 \ge \sqrt{2} (xy+yz)$

Since $xy+yz \ge 0$ , we can square both sides and obtain

$(x^2+y^2+z^2)^2 \ge 2(xy+yz)^2$

so

$x^4+y^4+z^4+2x^2 y^2+2y^2z^2+2x^2z^2 \ge 2(x^2y^2+y^2z^2+2xzy^2)$ $\implies (x^2+z^2)^2+y^4=x^4+y^4+z^4+2x^2z^2 \ge 4xzy^2$

so the result follow if $xy+yz \ge 0$ . $\square$

Case 2: $xy+yz \le 0$

By the Trivial Inequality, we have

$(x+\frac{y}{\sqrt{2}})^2 \ge 0$ $(z+\frac{y}{\sqrt{2}})^2 \ge 0$

. Add these inequalities gives

$(x+\frac{y}{\sqrt{2}})^2+(z+\frac{y}{\sqrt{2}})^2 \ge 0$

and expanding yields

$x^2+y^2+z^2 \ge - \sqrt{2} (xy+yz)$

Since $-(xy+yz) \ge 0$ , we can square both sides and obtain

$(x^2+y^2+z^2)^2 \ge 2(xy+yz)^2$

and the rest proceeds as in case 1. $\square$

The result follows. $\blacksquare$ ~Ddk001

Solution 2 (Lagrange Multipliers)

Put $f(x,y,z)=(x^2+z^2)^2+y^4-4xzy^2$ and let

Under construction

Solution 3 (Finding relative extrema with calculus)

Let $f(x, y, z) = (x^2 + z^2)^2 + y^4 - 4xzy^2$ . We will prove that $f$ is never negative (equivalent to the problem statement) by finding the critical values of $f$ . (Note that $f$ is a polynomial of $x$ , $y$ , and $z$ . Therefore, we do not need to consider discontinuities.) We will begin by setting partial derivatives of $f$ to $0$ . (Note that if $x = 0$ , $y = 0$ , or $z = 0$ , the inequality in the problem statement reduces to $(x^2 + z^2)^2 + y^4 \ge 0$ , which is true by the Trivial Inequality. Here on, we assume $x, y, z \neq 0$ ).

$\frac{\partial f}{\partial x} = 4x(x^2 + z^2) - 4zy^2 = 0$

$\therefore x^3 + z^2x = zy^2 \textbf{(1)}$

$\frac{\partial f}{\partial z} = 4z(z^2 + x^2) - 4xy^2 = 0$ $\therefore z^3 + x^2z = xy^2 \textbf{(2)}$

$\frac{\partial f}{\partial y} = 4y^3 - 8xzy = 0$ $\therefore y^2 = 2xz \textbf{(3)}$

If $(x, y, z)$ is a critical point of $f$ , then it must satisfy equations (1), (2), and (3). Substituting (3) into (1), we find:

$x^3 + z^2x = 2xz^2$ $\therefore x^3 = xz^2$ $\therefore x^2 = z^2$ $\therefore |x| = |z|$

If $x \neq z$ , then $x = -z$ and $(x^2 + z^2)^2 + y^4 \ge 0 \ge 4xzy^2 = -4x^2y^2$ by the Trivial Inequality. We now only need to consider the case where $x = z$ . Let $g(x, y)$ be the function obtained when we substitute $x = z$ into function $f$ .

$f(x, y, z) = (x^2 + z^2)^2 + y^4 - 4xzy^2$ $\therefore g(x, y) = (2x^2)^2 + y^4 - 4x^2y^2$ .

We must now prove that $g$ is never negative (proving this statement finishes the last case of the proof). We similarly find the critical values of $g$ by taking partial derivatives of $g$ and setting them to $0$ .

$\frac{\partial g}{\partial x} = 16x^3 - 8y^2x = 0$ $\therefore 2x^2 = y^2$

$\frac{\partial g}{\partial y} = 4y^3 - 8x^2y = 0$ $\therefore 2x^2 = y^2$

If $(x, y)$ is a critical point of $g$ , then $2x^2 = y^2$ . If $2x^2 = y^2$ then $g(x, y) = (2x^2)^2 + y^4 - 4x^2y^2 = 4x^4 + 4x^4 - 8x^4 = 0$ . Therefore, $g$ is never negative. QED

Solution 4 (AM-GM bash)

Under construction (I am not very sure if this could work)

Problem 14

Common blood types are determined genetically by 3 alleles: $A$ , $B$ , and $O$ . A person whose blood type is $AB$ , $AO$ , or $BO$ is heterozygous while $BA$ , $OA$ , or $OB$ is homozygous. The Hardy-Weinberg Law states that the proportion P of heterozygous individuals in a certain population is given by

$P(p,q,r)=2pq+2pr+2qr$

where $p$ is the percent of allele $A$ , $q$ the percent of allele $B$ , and $r$ the percent allele $O$ . Find the maximum possible proportion of heterozygous individuals in a certain population.

Solution 1 (Lagrange Multipliers)

Note that allele $A$ , $B$ , and $O$ make up all the alleles, so therefore,

$p+q+r=1$

The problem reduces to determining the maximum value of $f(p,q,r)=2pq+2pr+2qr$ given that $g(p,q,r)=p+q+r=1$ . Now, by the Method of Lagrange multipliers, we find all values of $p$ , $q$ , and $r$ and $\lambda$ such that the gradient of $f$ is $\lambda$ times the gradient of $g$ , or

$\nabla f(p,q,r)= \lambda \nabla g(p,q,r)$

$\implies (2q+2r) \overrightarrow{i} + (2p+2r) \overrightarrow{j} + (2p+2q) \overrightarrow{k} = \lambda (\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})$

or

$2q+2r=\lambda$

$2p+2r=\lambda$

$2p+2q=\lambda$

Solving yields

$p= \frac{\lambda}{4}$

$q= \frac{\lambda}{4}$

$r= \frac{\lambda}{4}$

Now, since $p+q+r=1$ , we have

$\lambda =\frac{4}{3}$

so $p=\frac{1}{3},q=\frac{1}{3},r=\frac{1}{3}$ . Therefore, the maximum of $f$ given $g=1$ is

$f(\frac{1}{3} ,\frac{1}{3} ,\frac{1}{3} )=\boxed{\frac{2}{3}}$

$\blacksquare$ .

Solution 2 (Relative extrema by calculus)

Just as in solution 1, we see that $p+q+r=1$ . To turn this into a 2-D problem, substitute

Problem 15

Let $F_n$ denote the $n$ th Fibonacci number. Prove that if $n$ is odd, then all odd prime divisors of $F_n$ are $1 \mod{4}$ .

Solution 1 (short and elegant)

Let $n=2k+1$ for some nonnegative integer $k$ . Then $F_n=F_{2k+1}=(F_k)^2+(F_{k+1})^2$ . Since $F_k$ and $F_{k+1}$ are relatively prime, we are done.

Solution by Yiyj1 (talk) 23:48, 3 December 2024 (EST)

Contributors

Ddk001
Yiyj1

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Problem 12

Solution 1 (Plain old chapter 3 analysis bash)

Problem 13

Solution 1 (Plain high school contest)

Solution 2 (Lagrange Multipliers)

Solution 3 (Finding relative extrema with calculus)

Solution 4 (AM-GM bash)

Problem 14

Solution 1 (Lagrange Multipliers)

Solution 2 (Relative extrema by calculus)

Problem 15

Solution 1 (short and elegant)

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