square root problem that involves geometry

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

B

No tags match your search

M

square root problem that involves geometry

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.

42 posts

#1 h Apr 5, 2025, 3:56 AM • 1 Y

Y by PikaPika999

If x is a nonnegative real number , find the minimum value of √x^2+4 + √x^2 -24x +153

Z K Y

The post below has been deleted. Click to close.

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

42 posts

#2 h Apr 5, 2025, 3:57 AM • 1 Y

Y by PikaPika999

Here’s a hint: Try completing the square

This post has been edited 1 time. Last edited by kjhgyuio, Apr 5, 2025, 3:57 AM
Reason: nil

Z K Y

The post below has been deleted. Click to close.

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

44 posts

ND_

#3 h Apr 5, 2025, 5:14 AM • 1 Y

Y by PikaPika999

kjhgyuio wrote:

If x is a nonnegative real number , find the minimum value of $\sqrt{x^2+4} + \sqrt{x^2 -24x +153}$

Let $P$ and $Q$ be $(0,2)$ and $(12,3)$ and $X$ be a point on the x axis, then our required expression is $XP + XQ$ , which is minimized when $X$ lies on $P'Q$ , where $P' = (0,-2)$ . So the minimum value is $\sqrt{12^2 + (3 - (-2))^2} = 13$ .

Z K Y

The post below has been deleted. Click to close.

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

42 posts

#4 h Apr 5, 2025, 9:07 AM

Y by

ND_ wrote:

kjhgyuio wrote:

If x is a nonnegative real number , find the minimum value of $\sqrt{x^2+4} + \sqrt{x^2 -24x +153}$

Let $P$ and $Q$ be $(0,2)$ and $(12,3)$ and $X$ be a point on the x axis, then our required expression is $XP + XQ$ , which is minimized when $X$ lies on $P'Q$ , where $P' = (0,-2)$ . So the minimum value is $\sqrt{12^2 + (3 - (-2))^2} = 13$ .

Another method : complete the square to find √x^2-24x+153 =√(x-12)^2+9 let 2 and x be the legs of a right angles triangle .then ,based on pythagoras.c^2=√x^2+4 similarly for √x^2-24x+153 let (x-2) and 3 be the legs of a right angled triangle C^2 =√(x-2)^2+9 find that the smallest configuration of x leads to making a big triangle with hypotenuse √x^2+4 + √x^2 -24x +153 and the two legs have lengths 5 and 12.Using pythagoras again 5^2+12^2=169 √169 =13

Z K Y

The post below has been deleted. Click to close.

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

810 posts

#5 h Apr 5, 2025, 9:22 AM

Y by

Released!! That Minkoski problem being shared per a week have been released.
$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$

Z K Y

The post below has been deleted. Click to close.

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

42 posts

#6 h Apr 5, 2025, 9:46 AM

Y by

ehuseyinyigit wrote:

Released!! That Minkoski problem being shared per a week have been released.
$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$

?

Z K Y

The post below has been deleted. Click to close.

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

Primeniyazidayi

71 posts

Primeniyazidayi

#8 h Apr 5, 2025, 11:39 AM

Y by

kjhgyuio wrote:

ehuseyinyigit wrote:

Released!! That Minkoski problem being shared per a week have been released.
$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$

?

Here is the Minkowski inequality:
If $x_1,\cdots,x_n,y_1,\cdots,y_n$ are real numbers and $p$ is a random positive real number such that $1 < p < \infty$ ,then $\sqrt[p]{\sum_{i=1}^n \left| x_i + y_i \right|^p} \le \sqrt[p]{\left| \sum_{i=1}^n x_i \right|^p} + \sqrt[p]{\left| \sum_{i=1}^n y_i \right|^p}$

This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 5, 2025, 12:44 PM

Z K Y

The post below has been deleted. Click to close.

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

42 posts

#9 h Apr 6, 2025, 12:23 AM

Y by

Primeniyazidayi wrote:

kjhgyuio wrote:

ehuseyinyigit wrote:

Released!! That Minkoski problem being shared per a week have been released.
$\sqrt{x^2+4}+\sqrt{(x-12)^2+9}\geq \sqrt{12^2+5^2}=13$

?

Here is the Minkowski inequality:
If $x_1,\cdots,x_n,y_1,\cdots,y_n$ are real numbers and $p$ is a random positive real number such that $1 < p < \infty$ ,then $\sqrt[p]{\sum_{i=1}^n \left| x_i + y_i \right|^p} \le \sqrt[p]{\left| \sum_{i=1}^n x_i \right|^p} + \sqrt[p]{\left| \sum_{i=1}^n y_i \right|^p}$

ok thanks

Z K Y

The post below has been deleted. Click to close.

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

mathprodigy2011

313 posts

mathprodigy2011

#10 h Apr 6, 2025, 1:21 AM

Y by

taking the derivative is probably much easier if you don't see the geometric intuition

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a