MathFun4Kids

In this 10th post of this series, consider the following figure with two co-tangent semi-circles drawn, centered at point $E$ on $\overline{AB}$ and $F$ on $\overline{BC}$. A full circle centered at point G, tangent with both semi-circles, line $\overline{AD}$ and $\overline{CD}$ in a unit square $ABCD$. Find the radii of the semi-circles and the full circle? Click here to show the solution.

Assume the radius of the two congruent semi-circle is $r$, the radius of the full circle is $s$. First, we need to find the value of $r$, as as the semi-circles are drawn first in the set up. Let’s draw line $\overline{EF}$, connecting the centers of the semi-circles, as shown below:

Obviously, $\triangle{BEF}$ is a right triangle. We have $$\overline{EF}^2=\overline{BE}^2+\overline{BF}^2$$ Since $\overline{AE}=\overline{CF}=r$, $\overline{BE}=\overline{BF}=\overline{AB}-\overline{AE}=1-r$, we have: $$(2r)^2=(1-r)^2+(1-r)^2$$ The solution of the above equation, we have $r=\pm\sqrt{2}-1$. Ignoring the negative $r$ value, we have the radius of the semi-circles as $$r=\sqrt{2}-1$$ Next, we consider the full circle center at point $G$. Connect the center of one semi-circle and the center point of the full circle, we have $\overline{EG}$ intersecting the semi and full circles at point $I$. We have $$\overline{EG}=\overline{EI}+\overline{GI}=\sqrt{2}-1+s$$ Draw line $\overline{KL}$ passing thru $G$ and parallel to $\overline{CD}$, and line $\overline{EM}$ perpendicular to $\overline{CD}$ and intersecting with line $\overline{KL}$ at point $N$. As $\triangle{EGN}$ is a right triangle, we have $$\overline{EG}^2=\overline{EN}^2+\overline{GN}^2$$ Since $$\overline{EN}=\overline{EM}-\overline{NM}=1-s$$ $$\overline{GN}=\overline{KN}-\overline{GK}=\overline{AE}-s=\sqrt{2}-1-s$$ Therefore $$(\sqrt{2}−1+s)^2=(1−s)^2+(\sqrt{2}−1−s)^2$$ Solve the above equation, we have $$s=-1+2\sqrt{2}\pm2\sqrt{2-\sqrt{2}}$$ Ignoring the $s$ value greater than 1, we have the radius of the full circle as $$s=-1+2\sqrt{2}-\sqrt{2-\sqrt{2}} \approx 0.2976933952$$

To be continued…

Continue the topic of the previous post in this series, if we construct a semi-circle first, then a quarter circle, and finally a full circle, as the following, what the radius of the full circle?

The problem is harder than the previous one, but it is still solvable by following the same steps.

Obviously, the radius of the semi-circle is $\dfrac{1}{2}$. Assume radius of the quarter circle is $q$. According to Pythagoras theorem, $$(\overline{BF}+\overline{EF})^2=\overline{BK}^2+\overline{EK}^2$$ We have $$(q+\dfrac{1}{2})^2=(\dfrac{1}{2})^2+1^2$$ The above equation can be solved as $$q=\dfrac{\pm\sqrt{5}-1}{2}$$ Ignoring the negative $q$ value, we have $$q=\dfrac{\sqrt{5}-1}{2}$$ Let’s denote the radius of the full circle as $r$, and the length of line $\overline{IL}$ and $\overline{MK}$ as $t$. Since $\triangle{BIL}$ and $\triangle{EIM}$ are right triangles, we have $$(q+r)^2 = t^2 + (1-r)^2$$ $$(\dfrac{1}{2}+r)^2=(1-t)^2+(\dfrac{1}{2}-r)^2$$ The above equations can be solved as $$r=\dfrac{3-2\sqrt{2}}{3-\sqrt{5}}, \ \ \ t=\dfrac{2+\sqrt{2}-\sqrt{10}}{3-\sqrt{5}}$$ Simplifying the above values, we have the radius of the full circle as $$ r = \dfrac{9-6\sqrt{2}+3\sqrt{5}-2\sqrt{10}}{4} \approx 0.2245918095$$

To be continued…

Continue the same topic of the previous post in this series, a line is drawn between point L and the tangent point M between the full circle and the quarter circle, as shown in the following figure. Prove $\triangle{HLM}$ is a right triangle.

Based on the result of the previous post, the radius of the full circle is $\dfrac{1}{9}$, the radius of the semi-circle is $\dfrac{1}{4}$, we have: $$\dfrac{\overline{HM}}{\overline{HL}}=\dfrac{\overline{HM}}{\overline{KL}-\overline{HK}}=\dfrac{\dfrac{1}{9}}{\dfrac{1}{4}-\dfrac{1}{9}}=\dfrac{4}{5}$$ $$\dfrac{\overline{HJ}}{\overline{HC}}=\dfrac{\overline{KJ}-\overline{HK}}{\overline{HM}+\overline{MC}}=\dfrac{1-\dfrac{1}{9}}{1+\dfrac{1}{9}}=\dfrac{4}{5}$$ Therefore $$\dfrac{\overline{HM}}{\overline{HL}}=\dfrac{\overline{HJ}}{\overline{HC}}$$ Since $\triangle{HLM}$ and $\triangle{HCJ}$ share the same angle at point $H$, $\triangle{HLM}$ and $\triangle{HCJ}$ are similar. Because $\triangle{HCJ}$ is a right triangle, $\triangle{HLM}$ is also a right triangle.

Can you prove it without calculating the radius of the full circle? To be continued…

Continue the topic in the previous post of this series, we add another circle inscribed in the area bounded by one side of the unit squares, the simi-circle and the quarter-circle, as the following. What is the radius of the circle?

Let’s connect several lines as the following, with line $\overline{GI}$ and $\overline{JK}$ perpendicular to each other and intersecting at point $L$, and point $H$ as the center of the full circle:

Therefore, both $\triangle{GLH}$ and $\triangle{CJH}$ are right triangles. According to Pythagoras theorem, we have: $$\overline{GL}^2+\overline{HL}^2=\overline{GH}^2,\ \ \ \overline{CJ}^2+\overline{HJ}^2=\overline{CH}^2$$ Based on the result of the previous post in this series, the radius of the semi-circle is $\dfrac{1}{4}$. Additionally, since point $C$ is the center of the quarter circle and point $G$ is the center of the semi-circle, both of them are tangent with the full circle centered at point $H$, by assuming the radius of the full circle is $r$ and $\overline{GL} = t$, we have: $$\overline{GH}=\dfrac{1}{4}+r,\ \ \ \overline{HL}=\dfrac{1}{4}-r$$ $$\overline{CH}=1+r, \ \ \ \overline{CJ}=1-t. \ \ \ \overline{HJ}=1-r$$ Therefore, we have the following equations: $$t^2+(\dfrac{1}{4})^2=(\dfrac{1}{4}+r)^2$$ $$(1-t)^2+(1-r)^2=(1+r)^2$$ Simplify the above equations, we have: $$t^2=r, \ \ \ (1-t)^2=4r$$ We have two solutions $$t=\dfrac{1}{3}, \ \ \ r=\dfrac{1}{9}$$ $$t=−1, \ \ \ r=1$$ By ignoring invalid answer $r=1$, we have the answer $r=\dfrac{1}{9}$.

To be continued…