Showing that the centers of two semicircles and a circle inscribed in a quarter circle form a right triangle

Question

The challenge in this image is to determine the radii of the two semicircles and the full circle.

Determining the radii of the two semicircles is straightforward; if the radius of the quarter circle is $6$, the radius of the larger semicircle is $3$, and of the smaller semicircle is $2$. Solving for the radius of the full circle is also easy if you assume the angle in the red triangle is a right angle - the radius is $1$. My question is how do we know that the marked angle is actually a right angle?

Answer 1

Let $r_1, r_2, r_3, r_4$ be the radii of the circular arcs in decreasing order. So $r_1 = 6$, and we obviously must have $r_2 = r_1/2 = 3$, since the larger semicircle shares its diameter with the radius of the quarter circle.

Since the angle at the quarter circle is by definition a right angle, it follows that the triangle joining the centers of the three largest circular arcs is a right triangle, hence $r_3$ satisfies

$$r_2^2 + (r_1 - r_3)^2 = (r_2 + r_3)^2.$$

Substituting the known values for $r_1$ and $r_2$ and solving for $r_3$ yields $r_3 = 2$, which also implies that the right triangle is a $3$-$4$-$5$ right triangle.

At this point, we still do not know if that red triangle is actually right. But we can proceed to solve for $r_4$ by placing the figure in the first quadrant of a Cartesian coordinate plane, and letting the center of the smallest circle be located at $(x_4, y_4)$. Then the relationships

$$\begin{align} \sqrt{x_4^2 + y_4^2} + r_4 &= r_1, \\ \sqrt{(x_4 - 4)^2 + y_4^2} &= r_3 + r_4, \\ \sqrt{x_4^2 + (y_4 - 3)^2} &= r_2 + r_4 \end{align}$$

must simultaneously hold. This is a system of $3$ equations in $3$ unknowns $(x_4, y_4, r_4)$, which we can solve with relative ease. We obtain two solutions, only one of which has $(x_4, y_4)$ in the first quadrant. I leave the computation to you as an exercise.

Answer 2

Let us put some letters on the involved points, so that propositions can simpler address them.

The picture was constructed in the following steps. Draw the axes, take $A=(0,0)$ as the origin. Let $B,C$ be on the positive part of the axes at distance six from $A$, so using cartesian coordinates $B=(6,0)$, $C=(0,6)$. Let $D$ be the mid point of $AC$, and we draw the half-circle $\overset\frown{AEC}$ with $DE$ parallel to $AB$. So $D=(0,3)$ and $E=(3,3)$. Let $H,F$ be on $AB$ so that $AH=HF=FB$. In cartesian coordinates $F=(4,0)$. The triangle $\Delta AFD$ has sides $3,4,5$, and so, when we construct $G$ the intersection of the semicircle $\overset\frown{AEC}$ with $DF$ we have $FG=FH=FB=2$. It turns out that the semicircles $\overset\frown{AEC}$ and $\overset\frown{BGH}$ are tangent in $G$. (Which is a point of intersection on the line $FD$ connecting the centers.)

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We can now address the asked question about the center of the third circle. Where do we know where it is?

We can try to answer this questions from two perspectives: $(A)$ Well, we know where it is, and use this information to conclude in seconds. $(B)$ We simulate we do not have this information and get it in a "fair manner".

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$(A)$ Let $I=(4,3)$ be the point making $IDAF$ a rectangle. The distance from $I$ to the semicircle $\overset\frown{AEC}$ is $DI-DE=4-3=1$. The distance from $I$ to the semicircle $\overset\frown{BGH}$ is also a difference, $IF$ minus the radius $FB=FG=FH$ for the semicircle, so it is $3-2=1$. The distance from $I$ to the quarter circle $\overset\frown{BC}$ is also a difference, radius $AB=AC=6$ minus the distance $AI$, so it is $6-5=1$.

This means that the point $I$ is the center of a circle (of radius one) tangent to the drawn three arcs.

$\square$

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$(B)$ Let us use now a "fair method" to locate the "full circle" from the given picture. For this, we use the inversion - denoted by a star - centered in $B$ with power $24$. Then the three arcs, their corresponding circles, transform as in the picture in the orange curves:

• The big circle centered in $A$ (with radius six), that goes through $C=(0,6)$ is transformed in the line perpendicular on $AB$ that goes through $C^*=J=(4,2)$, because $BJ\cdot BC=2\sqrt2\cdot 6\sqrt 2=24$. This line is thus $JF$.

• The small circle centered in $F$ (with radius two), that goes through $H$ is transformed in the line perpendicular on $AF$ that goes through $H^*=A$, because $BH\cdot BA=4\cdot 6=24$. This line is thus $ADC$. (And $A=H^*$, $C=J^*$.)

• The third circle, the one centered in $D$, tangent to the two above circles, and to $BA$, transforms in a circle. Which passes through $J=C^*$ and $H=A^*$, and is tangent to $BA^*$ in $A^*=H$, and to $FJ$ in $J=C^*$. So this is the (lower) circle centered in $(2,2)$ with radius two in orange in the picture.

It remains to find an other circle tangent to the two orange parallel lines $AC\|FJ$, and to the above (lower) orange circle. Such a circle has two changes only inside the strip delimited by the parallels, and we take the upper orange circle. It is centered in $(2,6)$. And passes through $C=J^*$. And through $(2,4)$.

To get the original full circle, we transform this circle back. It passes through $J$. And through $(2,4)^*=E$. So we know its points of tangency with two circles, so its center is the intersection $FJ\cap DE$. Yes, this is the point $I$, on the line through $B$ and the center of the upper orange circle.

This is exactly the wanted information about the right angle between the lines $DEI$ and $FJI$ from the question.

$\square$

Answer 3

Descartes' kissing circles theorem states that when four circles are pairwise tangent at distinct points, $(k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2)$, where $k_i$ is the signed curvature of circle $i$: the reciprocal of its radius, but positive for external tangencies, and negative for external ones. (For each circle, all three tangencies will have the same type.)

In the diagram, three of the curvatures are known: they are $-\frac16$ (for the quarter circle), $\frac13$ (for one semicircle), and $\frac12$ (for the other semicircle). Solving $$ \left(-\frac16+\frac13+\frac12+k_4\right)^2 = 2\left(\frac1{36} + \frac19 + \frac14 + k_4^2\right) \iff \left(k_4 + \frac23\right)^2 = 2 k_4^2 + \frac79 $$ for $k_4$, we get the quadratic equation $k_4^2 - \frac43 k_4 + \frac13 = 0$, which has two roots: $1$ and $\frac13$. These correspond to circles of radius $1$ and $3$.

The theorem is right: there are actually two solutions if we extend the quarter-circle and semicircles to full circles, as shown in the diagram below.

The larger solution - a circle of radius $3$ - is placed directly below the other circle of radius $3$, and all four circles are pairwise tangent due to symmetry. It is not the solution we want. So we conclude that the solution we do want has radius $1$.

Answer 4

Complete the semi-circles and the quarter-circle. Let $a, b, c$ and $R$ be the radii of circles with centers $A, B, C$, and $O$ respectively such that $b<a,c<R$ as shown in the Figure below.

The radius $R$ of the smallest circle internally touching three external tangent circles with radii $a,b$ and $c$ (as shown in the Figure above) is given by the following generalized formula $$\boxed{R=\dfrac{abc}{2\sqrt{abc(a+b+c)}-(ab+bc+ca)}}$$ Now, substituting the corresponding values of the radii $a=3, c=2$ and $R=6$ in the above generalized formula, one should obtain

$$6=\dfrac{3\cdot b\cdot 2}{2\sqrt{3\cdot b\cdot 2(3+b+2)}-(3\cdot b+b\cdot 2+2\cdot 3)}$$ $$b^2-4b+3=0$$$$ b=1,3$$

$$\implies b=1\quad (\because b<c=2)$$ Now, from the above Figure, $$ AB=a+b=3+1=4$$ $$ BC=b+c=1+2=3$$ $$ AC=a+c=3+2=5$$ From the Pythagorean theorem, $\triangle ABC$ having sides $3, 4,$ and $5$ is a right triangle which implies that $\angle B $ (as marked in the above Figure) is a right angle.

Answer 5

Since the radius of the big quarter circle is $6$, then the radius of the semi-circle on the left is $3$. Now, for the semi-circle with diameter on the $x$-axis, we can take its center $(x, 0)$, and its radius $r_3$, then from the right extremity, we know that $ x + r_3 = 6 $ and from tangency with the semi-circle on the left, we know that $ x^2 + 3^2 = (r_3 + 3 )^2 $. And this implies that,

$ (6 - r_3)^2 + 3^2 = (r_3 + 3)^2 $

which has a single solution $r_3 = 2 $ and hence, $x = 4 $

Now, we can apply Descartes' kissing circle theorem, to determine the radius of the small circle. It is tangent externally to the two semi-circles and internally to the quarter circle. Hence, its radius $r_4$ satisfies

$ \dfrac{1}{r_4} = \dfrac{1}{r_2} + \dfrac{1}{r_3} - \dfrac{1}{r_1} + 2 \sqrt{ \dfrac{1 }{r_2 r_3} - \dfrac{1}{r_1 r_2} - \dfrac{1}{r_1 r_3} } $

$ \dfrac{1}{r_4} = \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{6} + 2 \sqrt{ \dfrac{1}{6} - \dfrac{1}{18} - \dfrac{1}{12} } = 1 $

So the small circle is of radius $1$ but we need to find the location of its center. We know that the center $C_4 = (x_4 , y_4)$ lies on the following arcs

$ (x_4 , y_4) = (0, 3) + 4 (\cos t ,\sin t ) = (4, 0) + 3 (\cos s , \sin s) = (0,0) + 5 (\cos r , \sin r ) $

From $(0, 3) + 4 (\cos t , \sin t ) = (4, 0) + 3 (\cos s, \sin s ) $

We get the following two trigonometric equations in $t$ and $s$

$4 \cos t = 4 + 3 \cos s $

$ 3 + 4 \sin t = 3 \sin s $

Re-arranging the second equation:

$ 4 \sin t = 3 (\sin s - 1) $

Squaring both equations and adding, we get

$ 16 = 16 + 24 \cos s + 9 \cos^2 s + 9 \sin^2 s - 18 \sin s + 9 $

So that,

$ 0 = 18 + 24 \cos s - 18 \sin s $

By inspection, $s = \dfrac{\pi}{2} $ is a solution. The other solution can found by the standard trick to manipulate a weighted sum of a $\cos$ and a $\sin$, which is as follows

$ 18 + 24 \cos s - 18 \sin s = 18 + 30 \cos(s - \phi) $

where $\phi = \text{atan2} ( 24, -18 ) = - \tan^{-1}\left(\dfrac{18}{24}\right) = - \tan^{-1} \left( \dfrac{3}{4} \right) $

Hence,

$ s = \phi \pm \cos^{-1} \left( - \dfrac{18}{30} \right) = \phi \pm \cos^{-1} \left( - \dfrac{3}{5} \right) $

But $ \cos^{-1} (- x) = \pi - \cos^{-1}(x) $

Therefore, the two solutions for $s$ are

$ s = - \tan^{-1} \left( \dfrac{3}{4} \right) \pm \left(\pi - \cos^{-1} \left( \dfrac{3}{5} \right) \right) = 90^\circ , -163.74^\circ $

Clearly the correct value for $s$ is $90^\circ$ i.e $\dfrac{\pi}{2}$ while the other value is extraneous. It corresponds to the other intersection point of the two arcs. So for the purposes of this problem we can safely assume that $s = 90^\circ$. The corresponding value for $t$ is $0^\circ$.

This proves that the center of the small circle is directly above the lower semi-circle and directly to the right of the left semi-circle.

Answer 6

Call radius of big quarter circle $R$. Then big semi-circle has radius $R/2$ since its diameter is the radius of the quarter-circle.

Let $(z,0)$ be center of smaller semi-circle and its radius is $R-z$

The hypotenuse is the sum of the two circle's radii.

Pythagorean theorem: $R^2/4+z^2=(R/2+R-z)^2=(3R/2-z)^2$

$R^2/4+z^2=9R^2/4-3Rz+z^2$

$3Rz=2R^2\implies z=2R/3$

So the smaller semi-circle's radius is $r=R/3. $

Looks like the center of the smaller whole circle is $(z,R/2)$ having radius $R/2-R/3=R/6$

I don't think I used that that was a right angle.

Answer 7

Using the equations and symbols in heropup's answer, we begin by assuming that the angle in the red triangle is right. Then the coordinates $(x_4,y_4)$ of the full circle's center must be $(4,3)$.

The red triangle is completed by its twin to form a rectangle with one diagonal joining the full circle's center $(4,3)$ and the origin. Since the length of this diagonal is $5$, $r_4$ must be $1$. Or, more formally, substitute the known values for $x_4$ and $y_4$, along with $r_1=6$, into the first of heropup's three simultaneous equations $$\sqrt{x_4^2 + y_4^2} + r_4 = r_1$$ to solve for $r_4$.

Finally, we substitute these values along with those that heropup found, $r_2=3$ and $r_3=2$, into the remaining two of his simultaneous equations $$\begin{align} \sqrt{(x_4 - 4)^2 + y_4^2} &= r_3 + r_4, \\ \sqrt{x_4^2 + (y_4 - 3)^2} &= r_2 + r_4 \end{align}$$ and verify that they are satisfied. Since they are we know that our initial assumption of a right angle was correct. While along the way we have also solved the problem.