Difference between revisions of "User:Ddk001"

Revision as of 19:03, 1 January 2024

Problems

See if you can solve these:

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. 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 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?

User Counts

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

$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{0}}}}}}$

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

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

Here:

1. 049

2. 170

3. 736

4. 011

5. 054

Solutions:

Problem 1

There is one and only one perfect square in the form

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

Find that perfect square.

Solution

$(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2 \cdot q^2 +p^2+q^2+1-p^2 \cdot q^2 +pq-1=p^2+q^2+pq$ . 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 $2p+2q=pq+1 \implies (p-2)(q-2)=3 \implies (p,q)=(3,5)$ in some order. Hence, $(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}$ .

@@ Line 138: / Line 138: @@
 Find that perfect square.
 ==Solution==
-<math>(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</math>
+<math>(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</math>.
 Suppose <math>n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)</math>.
-Then, <math>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)</math>
+Then, <math>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)</math>, so since <math>n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}</math>, <math>n>p,n>q</math> so <math>p+q-n</math> is less than both <math>p</math> and <math>q</math> and thus we have <math>p+q-n=1</math> and <math>p+q+n=pq</math>. Adding them gives <math>2p+2q=pq+1 \implies (p-2)(q-2)=3 \implies (p,q)=(3,5)</math> in some order. Hence, <math>(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}</math>.

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