User:Ddk001

Introduction

I am a 5th grader who likes making and doing problems, doing math, and redirecting pages (see Principle of Insufficient Reasons). I like geometry and don't like counting and probability. My number theory skill are also not bad.

User Counts

If this is you first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)

$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{1}}}}}}}}}}}}}}}}$

(Please don't mess with the user count)

Doesn't that look like a number on a pyramid?

Cool asyptote graphs

Asymptote is fun! $[asy]draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));[/asy]$

$[asy]draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));[/asy]$

Problems Sharing Contest

Here, you can post all the math problem that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start:

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

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

where $p$ and $q$ are prime. Find that perfect square. (DO NOT LOOK AT MY SOLUTIONS YET)

Contributions

2005 AMC 8 Problems/Problem 21 Solution 2

2022 AMC 12B Problems/Problem 25 Solution 5 (Now it's solution 6)

2023 AMC 12B Problems/Problem 20 Solution 3

2016 AIME I Problems/Problem 10 Solution 3

2017 AIME I Problems/Problem 14 Solution 2

2019 AIME I Problems/Problem 15 Solution 6

2022 AIME II Problems/Problem 3 Solution 3

Restored diagram for 1994 AIME Problems/Problem 7

Divergence Theorem

Stokes' Theorem

Principle of Insufficient Reasons

Problems I made

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

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

dsf

fsd

Answer key

1. 049

2. 019

3. 092

4. 170

5. 736

6. 895

7. 011

8. 054

Solutions

Note: All the solutions so far have been made by me :)

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

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

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

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

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

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

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

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