Algebra Problems

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Algebra Problems

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilikemath247365

267 posts

ilikemath247365

#1 h Apr 14, 2025, 4:52 PM

Y by

Find all real $(a, b)$ with $a + b = 1$ such that

$(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} = \frac{25}{2}$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2506 posts

#2 h Apr 14, 2025, 4:57 PM

Y by

xiooix !
a^2+b^2+1/a^2+1/b^2=17/2
let x=a^2 and y=b^2

thus x+y+1/x+1/y=17/2 so (x+y)(1+1/xy)=17/2

thus xy=2/15 and x+y=1, and the rest is computation.
oops wrong

This post has been edited 1 time. Last edited by fruitmonster97, Apr 14, 2025, 5:25 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilikemath247365

267 posts

ilikemath247365

#3 h Apr 14, 2025, 5:15 PM

Y by

I think that is incorrect. My solution:

Click to reveal hidden text

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilikemath247365

267 posts

ilikemath247365

#5 h Apr 14, 2025, 5:22 PM

Y by

Wait sorry never mind I didn't see your substution of x = a^2 and y = b^2.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2506 posts

#6 h Apr 14, 2025, 5:24 PM

Y by

ilikemath247365 wrote:

I think that is incorrect. My solution:

Click to reveal hidden text

oops im sobad we have a+b=1 not a^2+b^2=1

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilikemath247365

267 posts

ilikemath247365

#7 h Apr 14, 2025, 5:25 PM

Y by

Oh yeah.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

483 posts

#8 h Apr 14, 2025, 6:27 PM

Y by

$2\left( \left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \right) \geq \left( a + \frac{1}{a} + b + \frac{1}{b} \right)^2$ $RHS = a + \frac{a+b}{a} + b + \frac{a + b}{b} = 3 + \frac{a}{b} + \frac{b}{a} \geq 5,$ $QED.$

edit: nvm i didnt see this wasnt a proof. oh well, the rest is easy anyway. equality must occur at $a + \frac{1}{a} = b +\frac{1}{b},$ or $a=b.$

This post has been edited 1 time. Last edited by anduran, Apr 14, 2025, 6:29 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

10640 posts

#9 h Apr 14, 2025, 9:56 PM

Y by

We have $a^2+b^2+{{a^2+b^2}\over {a^2b^2}}=17/2$ and with $p:=ab$ we get $1-2p+{{1-2p}\over {p^2}}=17/2$ hence $4p^3+15p^2+4p-2=0$ . Factoring gives $(4p-1)(p^2+4p+2)=0$ .
Now $p=a(1-a)\le 1/4$ hence each value from $p=1/4, -2\pm \sqrt{2}$ gives a solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

51 posts

lgx57

#10 h Apr 18, 2025, 1:28 PM

Y by

ilikemath247365 wrote:

Find all real $(a, b)$ with $a + b = 1$ such that

$(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} = \frac{25}{2}$ .

Only for $a,b>0$ .

$(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} = \frac{25}{2} \Leftrightarrow (a^2+b^2)+(\frac{1}{a^2}+\frac{1}{b^2})=\frac{17}{2}$

$(a^2+b^2)+(\frac{1}{a^2}+\frac{1}{b^2}) \ge 2ab+\frac{2}{ab}$ . Because $a,b>0$ , $ab \in[0,\frac{1}{4}]$ .

Obviously , $f(x)=2x+\frac{2}{x}$ is monotonic decline in $[0,\frac{1}{4}]$ .

So $2ab+\frac{2}{ab} \ge f(\frac{1}{4})=\frac{17}{4}$

Then $(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} \ge \frac{17}{4} +2 = \frac{25}{2}$

The equality holds iff $a=b$ and $ab=\frac{1}{4}$ .

So the answer is $(a,b)=(\frac{1}{2},\frac{1}{2})$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

SomeonecoolLovesMaths

3309 posts

SomeonecoolLovesMaths

#11 h Apr 18, 2025, 2:14 PM • 1 Y

Y by EntangledElectron99

If $\mid z -2 + i \mid \leq 2$ , find the greatest and least values of $\mid z \mid$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

51 posts

lgx57

#12 h Apr 19, 2025, 4:25 AM

Y by

SomeonecoolLovesMaths wrote:

If $\mid z -2 + i \mid \leq 2$ , find the greatest and least values of $\mid z \mid$ .

You can solve this problem in complex plane.

$\mid z-2+i \mid$ means distance from $(2,-1)$ to $z$ , so $\mid z-i+i \mid \leq 2$ stands for a circle with a center at $(2,-1)$ and a radius of $2$ .

So $\mid z \mid \in [\sqrt{5}-2,\sqrt{5}+2]$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a