This question was uploaded on 27/08/21 on social media accounts.
A circle passing through the center of a semicircle such that their common is the perpendicular bisector of the line joining the center of the circle and semicircle. Find the radius of the semicircle if the radius of the small circle is 1.
Solution 1
Step 1:
Draw lines between the center and common point for both circles.
Join the end of the radius of the semicircle and common point of both circles.
Give names to centers and intersection points.
Let the radius of the semicircle is 'r'.
Step 2:
Let angle between AC and AD = θ.
OBC and DBC are similar triangles.
⇒ OC = DC;
AO = OC = DC = 1;
AC = AD = r;
Step 3:
Use cosine rule at the angle CAD for triangle AOC and ADC:
\[\Rightarrow\frac{1^2+r^2-1^2}{2r}=\frac{r^2+r^2-1^2}{2r^2}\]
\[\Rightarrow r^3-2r^2+1=0\]
\[\Rightarrow r=\frac{1-\sqrt{5}}{2},1,\frac{1+\sqrt{5}}{2}\]
Here, only third answer is possible which is also known as Golden Ratio φ.
\[\Rightarrow r=\frac{1+\sqrt{5}}{2}=φ\]
Image Solution:
Solution 2
Step 1:
Join both centers and a common point.
Gove names to centers and intersection of lines.
Let the radius of the semicircle is 'r'.
Step 2:
AO = OC = 1;
AC = r;
OB = BD = x = (r - 1)/2;
Step 3:
Use Pythagoras theorem at triangles ABC and OBC.
For OBC:
\[BC^2=1-\left(\frac{r-1}{2}\right)^2----(1)\]
For ABC:
\[BC^2=r^2-\left(1+\frac{r-1}{2}\right)^2----(2)\]
From (1) and (2):
\[1-\left(\frac{r-1}{2}\right)^2=r^2-\left(1+\frac{r-1}{2}\right)^2\]
\[\Rightarrow r^2-r-1=0\]
\[\Rightarrow r=\frac{1\pm\sqrt{5}}{2}\]
Here, only second answer is possible which is also known as Golden Ratio φ.
\[\Rightarrow r=\frac{1+\sqrt{5}}{2}=φ\]
Image Solution:
Solution 3
Step 1:
Join Center of the semicircle and a common point for both circles.
Extend line passing through the center of both circles.
Give names to both centers and Intersection points.
Let the radius of the semicircle is 'r'.
Step 2:
AO = 1,
OB = BD = x,
DE = 1-2x,
BC = y
x = (r - 1)/2
Step 3:
Use chord theorem for both circles at point B.
For Large circle:
\[y^2=x(2r-x)----(1)\]
For Small circle:
\[y^2=(1+x)(1-x)----(2)\]
From (1) and (2):
\[\Rightarrow 2rx-x^2=1-x^2\]
\[\Rightarrow 2rx=1\]
But x = (r - 1)/2,
\[\Rightarrow r^2-r-1=0\]
\[\Rightarrow r=\frac{1\pm\sqrt{5}}{2}\]
Here, only second answer is possible which is also known as Golden Ratio φ.
\[\Rightarrow r=\frac{1+\sqrt{5}}{2}=φ\]
Image Solution:
Puzzle related to Geometry, Circle, Length
