circle tangent to circle, semicircle (2009 All-Siberian Correspondence 10.2)

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0 replies

jlacosta

Apr 2, 2025

0 replies

Polynomials

CuriousBabu 5

N 17 minutes ago by wh0nix

$\frac{(x+y+z)^5 - x^5 - y^5 - z^5}{(x+y)(y+z)(z+x)} = 0$
Find the number of real solutions.

5 replies

CuriousBabu

Apr 14, 2025

wh0nix

17 minutes ago

Inequalities

sqing 9

N 23 minutes ago by sqing

Let $a,b$ be reals such that $a^2+ab+b^2 =3$ . Prove that
$\frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$ $\frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$ $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$ $\frac{19}{ 6}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$ Let $a,b$ be reals such that $a^2-ab+b^2 =1$ . Prove that
$\frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$ $\frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$ $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$ $\frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$

9 replies

sqing

Apr 16, 2025

sqing

23 minutes ago

Geometry

AlexCenteno2007 4

N 37 minutes ago by sunken rock

Let ABC be an isosceles triangle with AB = AC and M the midpoint of BC. Consider a point E outside the triangle such that BE = BM and CE perpendicular to AB. The point of intersection of the perpendicular bisector of segment EB with the circumcircle of triangle AMB, which is on the same side as A with respect to BE, is point F. Show that angle FME = 90°

4 replies

AlexCenteno2007

Yesterday at 3:49 AM

sunken rock

37 minutes ago

JEE Related ig?

mikkymini2 16

N 2 hours ago by mikkymini2

Hey everyone,

Just wanted to see if there are any other JEE aspirants on this forum currently prepping for it[mention year if you can]

I am actually entering 10th this year and have decided to try for it...So this year is just going to go in me strengthening my math (IOQM level (heard its enough till Mains part, so will start from there) for the problem solving part, and learn some topics from 11th and 12th as well)

It would be great to connect with others who are going through the same thing - share study strategies, tips, resources, discuss, and maybe even form study groups(not sure how to tho :maybe: ) and motivate each other ig?. :D
So yea, cya later

16 replies

mikkymini2

Apr 10, 2025

mikkymini2

2 hours ago

No more topics!

circle tangent to circle, semicircle (2009 All-Siberian Correspondence 10.2)

parmenides51 3

N May 5, 2021 by franzliszt

A semicircle of radius $2$ is inscribed with a circle $\omega$ of radius $1$ , touching a semicircle in the midpoint $O$ of its diameter $AB$ . Find the radius of the circle tangent to $\omega$ and the semicircle as shown in the image.
IMAGE

3 replies

parmenides51

May 4, 2021

franzliszt

May 5, 2021

circle tangent to circle, semicircle (2009 All-Siberian Correspondence 10.2)

G H J

Reveal topic tags

geometry

tangent circles

semicircle

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parmenides51

30629 posts

parmenides51

#1 h May 4, 2021, 9:00 AM • 3 Y

Y by Mango247, Mango247, Mango247

https://cdn.artofproblemsolving.com/attachments/0/e/9955af9e06ce755befe7b0b6f3b5d6b9b8f884.png

This post has been edited 1 time. Last edited by parmenides51, May 4, 2021, 9:01 AM

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parmenides51

30629 posts

parmenides51

#2 h May 4, 2021, 9:00 AM • 3 Y

Y by Mango247, Mango247, Mango247

posted for the image link

Attachments:

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natmath

8219 posts

natmath

#3 h May 4, 2021, 2:27 PM • 1 Y

Y by Mango247

solution

If $r$ is the desired radius, then the horizontal distance of the small circle from the center is $2\sqrt{r}$
We require that
$2=r+\sqrt{r^2+4r}$ $4=(r+2)+\sqrt{(r+2)^2-4}$ Define $u=r+2$ ,
$4-u=\sqrt{u^2-4}$ $u^2-8u+16=u^2-4$ $u=2.5$ $r=.5$

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franzliszt

23531 posts

franzliszt

#4 h May 5, 2021, 5:15 PM • 1 Y

Y by Mango247

Any nice solutions with inversion or Descartes' circle formula (I don't think DCT will work but maybe there is an extra clever use here somewhere...)

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