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Two half circles and a full circle fit inside a larger quarter circle as shown in the diagram. The centers of the two half circles are on the two sides of the quarter circle, respectively.

enter image description here

Prove that the triangle formed by the centers of three smaller circles, $\triangle O_1O_2O_3$, is a right triangle.

I was able to apply the Pythagorean formula to a few triangles involving the radii of the inscribed circles and proved that the ratios of the three radii are 1:2:3. Then, the centers of the three circles form a triangle with side-length ratios 3:4:5, hence, a right triangle.

On the other hand, I feel the proof may be an overkill, and evaluating the three radii explicitly may be unnecessary. There ought to be clean geometric solutions to show directly that the vertex $O_3$ is of a right angle, which I am not sure how to figure out.

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    $\begingroup$ Suppose $OO_1$ and $OO_2$ are x and y axis. If you complete half and quarter circles, you will have a big circle containing two big circles, two medium and four little circles. the center of little circles are symmetric about x and y axis. since x and y axis are perpendicular then the lines connecting the centers of little circles are perpendicular that is triangle $O_1O_2O_3$ is right angle at $O_3$. $\endgroup$ – sirous Sep 14 at 21:01
  • $\begingroup$ Insightful. A good case of can’t-see-forest-staring at trees. Thanks $\endgroup$ – Quanto Sep 14 at 21:10
  • $\begingroup$ @sirous I finally read your comment. I think it would be clearer if you "complete" the half and quarter circles by reflecting the figure over one axis and then the other. But your construction is much more elegant than mine and would make a fine answer. $\endgroup$ – David K Sep 15 at 0:30
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    $\begingroup$ @sirous: Certainly, "the lines connecting the centers of the little circles are perpendicular", but how do you know that those lines contain $O_1$ and $O_2$? $\endgroup$ – Blue Sep 15 at 3:51
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    $\begingroup$ @Blue - critical observation. guess a simple geometric proof remains elusive $\endgroup$ – Quanto Sep 15 at 14:42
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Given the quarter circle with $O$ as the center of the arc, and given semicircles with centers $O_1$ and $O_2$ on the two straight sides of the quarter circle such that $O_1$ and $O_2$ both are tangent to the arc of the quarter circle and such that $O_1$ and $O_2$ are tangent to each other.

This is based on the figure in the question, except that we do not assume that $O_1$ is the midpoint of one side of the quarter circle.

Let the radii of the quarter circle, the semicircle about $O_1$, and the semicircle about $O_2$ be $r,$ $r_1,$ and $r_2$ respectively.

Let $P$ be the fourth vertex of the rectangle that has three of its vertices at $O,$ $O_1,$ and $O_2.$ The sides of this rectangle are $OO_1 = r - r_1$ and $OO_2 = r - r_2.$ The diagonals are $OP = O_1O_2 = r_1 + r_2.$

Let $P_1$ be the point of intersection of the semicircle about $O_1$ with $O_1P$ and $P_2$ be the point of intersection of the semicircle about $O_2$ with $O_2P.$ Then $PP_1 = r - r_1 - r_2 = PP_2.$ Moreover, the distance from $P$ to the arc of the quarter circle is $r - (r_1 + r_2) = r - r_1 - r_2.$

Then a circle of radius $r - r_1 - r_2$ about $P$ is tangent to the quarter circle and to the semicircles about $O_1$ and $O_2.$ Therefore $P$ is $O_3$ in the figure, the center of the circle inscribed between the two semicircles and the quarter circle.

Since $O_3$ is a vertex of the rectangle $OO_1O_3O_2,$ the triangle $\triangle O_1O_3O_2$ is a right triangle. $\square$

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  • $\begingroup$ +1 a nice generalization. $\endgroup$ – achille hui Sep 15 at 16:27
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Let $R=6a$ then $r_1=3a$

$$\triangle OO_1O_2: \;\;\;\; (R-r_2)^2+r_1^2 =(r_1+r_2)^2$$

So $$ \boxed{r_2= 2a}$$

So $\triangle O_3O_1O_2$ is right iff $$ (r_1+r_2)^2 = (r_1+r_3)^2+(r_2+r_3)^2$$

or $$r_1r_2 = r_3(r_1+r_2+r_3)$$

or $$6a^2 = r_3(5a+r_3)\iff r_3^2+5ar_3-6a^2 =0$$

So we have to prove: $$\boxed{\color{red}{r_3=a}}$$

If we put everything in coordinate system then we have: $O(0,0)$, $O_1(3a,0)$, $O_2(0,4a)$ and let $O_3(m,n)$, so \begin{eqnarray} m^2+n^2 &=& (6a-r_3)^2\;\;\;\;\;\; (OO_3 = R-r_3)\\ (m-3a)^2+n^2 &=& (3a+r_3)^2\;\;\;\;\;\; (O_1O_3 = r_1+r_3)\\ m^2+(n-4a)^2 &=& (2a+r_3)^2\;\;\;\;\;\; (O_2O_3 = r_2+r_3) \end{eqnarray}

If we substract 1.st and 2.nd equation we get $$m = 6a-3r_3$$ and if we substract 1.st and 3.rd equation we get $$n=6a-2r_3$$

Now put this again in 1.st equation and we get $\boxed{\color{green}{r_3=a}}$

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Following is an answer based on circle inversion w/o computing the radius of circle centered at $O_3$.

Label the contact points as illustrated below. We are going to show $$\angle OYD = \angle OXC = 45^\circ$$ Form these, it is easy to deduce $\begin{cases} O_2O_3 \parallel O_2D \parallel OX,\\ O_1O_3 \parallel O_1C \parallel OY\end{cases}$ and hence $\angle O_1O_3O_2 = 90^\circ$

Original configuration

Choose a coordinate system so that $O$ is the origin and $X = (1,0)$, $Y = (0,1)$. The arc $XEY$ will be part of the unit circle. Under circle inversion with respect to the unit circle centered at $O$, we get what illustrated below:

circle inversion wrt O

  • The red semicircle centered at $O_1$ get mapped to a ray $y = 1$ in the upper half plane.
  • The green semicircle centered at $O_2$ get mapped to another semicircle centered at some point of $y$-axis and touching the red line. This means this inverted green semicircle has radius $1$.
    As a result, the inverted image of point $A$ is $A' = (0,3) \implies A = (0,\frac13)$.

  • The blue circle get mapped to a circle placed symmetrically between the inverted green semicircle and the arc $XEY$. As a result, the inverted image of point $C$ is $C' = (1,1)$.
    This implies $\angle OXC = \angle OXC' = 45^\circ$.

For the other angle, translate the coordinate axis so that $Y$ is the new origin. The new coordinate of $A$ is now $(0,-\frac23)$. Under circle inversion with respect to the unit circle centered at $Y$, it now looks like:

Circle inversion wrt Y

  • The arc $XEY$ get mapped to a ray on the line $y = -\frac12$.
  • The green semicircle centered at $O_2$ get mapped to a ray on the line $y = -\frac32$.
  • The red semicircle get mapped to an arc on a circle sandwiched between above two rays touching the $y$-axis. So its radius is $\frac12$.
  • The blue circle get mapped to an circle also sandwiched between above two rays. This means the inverted image of $D$ is located at $D'' = (\frac32,-\frac32)$. As a result, $\angle OYD = \angle OYD' = 45^\circ$.
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Let $OO_1$ and $OO_2$ are x and y axis respectively. Mirror the figure about y axis, then mirror the resulted figure about x axis. You get two big, two medium and four small circles indescribable in a large circle. If you connect the centers of four small circles you get a square. So angle at $O_3$ is $90^o$. Now let the radius of small circle is $r_1=1$ and that of large circle is $r_2=6$. Suppose the location of center of small circle is $O_3(3, 4)$ which satisfies:

$3^2+4^2= [OO_3=6-1=5]^2$

Then we can claim that the radius of medium circle is $r_3=3-1=2$ and those of big and large circles are $r_4=4-1=3$ and $r_2=2\times 3=6$ respectively that is all involved circle are mutually tangent.If radius of medium circle is $r_3=2$ that means that it's center locates on a line passing $O_3$ which has a distance=4 from x axis. similarly if the radius of big circle is $r_4=3$ that would mean that it's center locates on a line passing $O_3$ which has distance =3 from y axis.That is triangle $O_1O_2O_3$ is right triangle at $O_3$

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