# how to inscribed tow circle in triangle [closed]

I have problem how can I understand that I can inscribed tow circle in one triangle and we have just 2 side of triangle and both circle radii. for example radii 1,1 side:5 4 in this case we can but what is the formula???

-

## closed as unclear what you're asking by dfeuer, Daniel Rust, Old John, Thomas Andrews, egregDec 9 '13 at 21:30

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

What is a "tow circle"? –  Henning Makholm Dec 9 '13 at 18:02
@HenningMakholm, I think it's "two circles". –  dfeuer Dec 9 '13 at 18:02
@HenningMakholm: I would guess it is supposed to be "two circles" Danial: Please draw a picture. We can't see what you are asking. –  Ross Millikan Dec 9 '13 at 18:04
I cant but that is the real question: two circles and a rectangle, your task is to judge wheather the two circles can be put into the rectangle with no part of circles outside the retangle.and we have tow side of rectangle and both circles radii –  Danial Dec 9 '13 at 18:09
You need to edit the question, and correct it. –  ja72 Dec 9 '13 at 18:45

Here is a guess of what is being asked. We are given a rectangle $A \times B$ and two numbers $R_1,R_2$. We are asked if two circles, one of radius $R_1$ and one of radius $R_2$ will fit inside the rectangle without overlapping. We assume $A \ge B,$, otherwise switch them. If either $2R_1\gt B$ or $2R_2 \gt B$ we fail immediately. Otherwise, stack them as shown in the figure and they fit as long as the total width is less than $A$. To get the total width, note that the vertical distance between the centers is $B-R_1-R_2$. The total distance between the centers is $R_1+R_2$, so the horizontal distance between the centers is $\sqrt{(R_1+R_2)^2-(B-R_1-R_2)^2}$ The total width is then $R_1+R_2+\sqrt{(R_1+R_2)^2-(B-R_1-R_2)^2}$, which must be less than or equal to $A$
@Danial: the formula is the last line. The expression given has to be less than or equal to $A$. –  Ross Millikan Dec 9 '13 at 20:13
It seems clear that if you can do it at all, you can do it with one circle touching two neighboring sides, and the other circle touching the two other sides. If we put the rectangle into a coordinate system, this places the centers of the circles at $(1,1)$ and $(4,3)$. They will overlap if the distance between those centers is less than the sum of the radii.