User:Ddk001

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$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{68}}}}}}$

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 I made

See if you can solve these:

1. (Much easier) 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. 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 $x_{1}^3+x_{2}^3+x_{2}^3$ .

3. 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.

4. 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 $p$ .

5. (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?

Answer key & solution to the problems

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

dsf

fsd

Answer key

1. 049

2. 170

3. 736

4. 011

5. 054

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

Problem 2

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

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

As we have done many times before, 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 3

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

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

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 4

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 5

$\Delta ABC$ 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?

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User:Ddk001

Contents

User Counts

Cool asyptote graphs

Problems I made

Answer key & solution to the problems

Answer key

Solutions

Problem 1

Solution 1

Problem 2

Solution 1

Problem 3

Solution 1 (Euler's Totient Theorem)

Problem 4

Solution 1

Problem 5

Solution 1