Difference between revisions of "Problems Collection"

Latest revision as of 21:50, 29 June 2024

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

WARNING: This page is going to be a very very messy page and will be hard to clear out!

AMC styled

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 $q$ are 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 is 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 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 $a^2+b^2+c^2+d^2+abcd$ .

Someone mind making a diagram for this?

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 $p+q$ .

Proofs

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 $AF$ is perpendicular to $BE$ .

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

Problems I made

Aime styled

Introductory

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 $q$ are prime. Find that perfect square.

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

Intermediate

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 is 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 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. 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 $p$ .

Olympiad

8. (Much harder) $\Delta ABC$ 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?

9. 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 $p+q$ .

Proofs

10. 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 $AF$ is perpendicular to $BE$ .

I will leave a big gap below this sentence so you won't see the answers accidentally.

Solutions

$\textbf{I wrote a couple of solutions here. Hope it's okay :) - cxsmi (please feel free to delete this note and/or the solutions)}$

I like your solutions.~Ddk001

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, $(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}$ . $\square$

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,

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

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.

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

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 $\boxed{019}$ . $\square$

Solution 3

Expand the denominator. We now have $\frac{ab + bc + ac}{a^2 + b^2 + c^2 + 2ab + 2ac + 2bc}$ . Consider its reciprocal; if this expression takes values on the interval $(p,q]$ , then its reciprocal will take values on the interval $[\frac{1}{q},\frac{1}{p})$ . This is important because we can now write the reciprocal of the expression as $\frac{a^2 + b^2 + c^2}{ab + ac + bc} + 2$ . We attempt to maximize and minimize $\frac{a^2 + b^2 + c^2}{ab + ac + bc}$ . To maximize the expression, we consider the triangle inequality. From it, we find the following. $a + b > c$ $a + c > b$ $b + c > a$ We rewrite. $ac + bc > c^2$ $ab + bc > b^2$ $ab + ac > a^2$ Add all of the inequalities. We find the following. $2ab + 2ac + 2bc > a^2 + b^2 + c^2$ Considering the equality case and plugging into the expression, we find that the maximum value of the expression is $4$ . 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 $a$ , $b$ , and $c$ , we find the following. $\sqrt{\frac{a^2 + b^2 + c^2}{3}} \geq \frac{a + b + c}{3}$ Square both sides. $\frac{a^2 + b^2 + c^2}{3} \geq \frac{a^2 + b^2 + c^2 + 2ab + 2ac + 2bc}{9}$ Rewrite as follows. $3(a^2 + b^2 + c^2) \geq a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$ $2a^2 + 2b^2 + 2c^2 \geq 2ab + 2ac + 2bc$ $a^2 + b^2 + c^2 \geq ab + ac + bc$ Considering the equality case and plugging into the expression, we find that the minimum value of the expression is $3$ . Since the expression (which we said was the reciprocal of the original expression) takes values on the interval $[3, 4)$ , the original expression must take values on the interval $(\frac{1}{4}, \frac{1}{3}]$ . Then $p + q = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}$ , so our final answer is $7 + 12 = \boxed{019}$ .

~ cxsmi

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.

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. $\square$

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, $x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}$ . $\square$

Problem 7

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 $999999937$ . $\square$

Problem 8

$\Delta ABC$ 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. $\blacksquare$