44 of 100: Semitricky Semicircles

If the radii of the circles are $r$ and $R$ , then the cord is $r$ above the shared diameter

Pythagorean theorem in triangle $ABC$ gives us $R^2=r^2+12^2$ or $R^2-r^2=144$

Pink area $=\frac{\pi}2R^2-\frac{\pi}2r^2=\frac{\pi}2(R^2-r^2)=\frac{\pi}2\times144=\boxed{72\pi}$

The beauty of the problem lies in the fact that the pink area does not depend on $R$ directly. For example, the following pink areas are all the same size.

Very simple solution without the depth of proof.

Note 1: We aren't given any information about the size of the smaller circle. Note 2: There isn't an option of "Not enough information".

Conclusion: Area of pink region must be independent of size of small circle.

Let the small circle have radius of zero.

Area of the pink region is simply the area of a semi-circle with radius 12: 72pi.

I also created a graph on desmos to justify that the area is independent.

Denote the radius of the small semicircle r and the radius of the large semicircle R . Create a diagram similar to this:

Set the equation up as $r^2+12^2=R^2$

The area of the pink region is the area of the large semicircle subtracted by the area of the smaller smeicircle. In other words, $\frac{πR^2}{2}$ - $\frac{πr^2}{2}$ or $\frac{π(R^2-r^2)}{2}$ .

Using the first equation, $R^2-r^2=12^2=144$

The area of the pink region is $\frac{π(144)}{2}$ = 72π

Edit: I apologize for the hand drawn figure, I don't know how to make good images like those in the solutions above.

Desmos demo A demonstration in desmos that the area of the shaded region doesn't change as the size of the semicircles change. Click on link. Drag purple dot.

Lazy solution

Apparently, the answer does not depend on how big or small the white semicircle is.

Let's shrink it down to size zero. Then the chord of length 24 has the same length as the diameter of the pink circle. Thus its radius is 12, and the pink area is half the area of a circle, $A = \tfrac12\cdot \pi (12)^2 = 72\pi.$

Coordinate solution

Let the center of the pink semicircle by at $(0,0)$ and its radius $R$ ; the white semicircle has radius $r$ . The chord connects the points $(\pm 12, r)$ . Since these points lie on the pink semicircle, $12^2 + r^2 = R^2\ \ \ \ \therefore\ \ \ \ R^2 - r^2 = 12^2.$ Now the pink area is $\tfrac12 \pi R^2 - \tfrac12 \pi r^2 = \tfrac12\pi(R^2 - r^2) = \tfrac12\pi(12^2) = 72\pi.$

Let, radius of small circle is $r$ and radius of large circle is $c$ , So, area of white portion is $\frac{\pi r^2}{2}$ , From the triangle we can write $c^2=12^2 + r^2$ , Now, Area of pink portion is $= \frac{\pi c^2}{2} - \frac{\pi r^2}{2} = \frac{144 \pi}{2}+ \frac{\pi r^2}{2} - \frac{\pi r^2}{2} = 72\pi$

Compare to an earlier problem.

I have cheated slightly. :) Note that the configuration isn't unique - if we increase the radius of the big circle and move the horizontal line up so it stays at 24 units, it should still give the same answer. So consider the limit as the small circle's radius tends to zero. Then the diameter of the big circle would be 24 as the chord would be the same as the diameter. So the answer would be $\frac{\pi\times 12^2}{2}$ .

It reminds me of a similar Gardner problem about a sphere with a hole.

Assuming that there is an exact solution, it does not depend on the radiuses of the two circles. So, if the smaller circle is infinetely small, the solution is the area of a semicircle with radius 12.

Using logic:

Let the radius of the small and large semi-circles be r and R respectively.

The problem statement implies that the value of r/R does not affect the pink area.

As r tends to zero, R tends to 12, so the pink area tends to $\frac{1}{2}\times \pi R^2$ which equals $72\pi$

suppose, radius of the bigger semicircle is R and the radius of the smaller semicircle is r.

so, It can be observed that R^2=12x12-r^2. => R^2=144-r^2.

So, pink area=(1/2)pi(R^2-r^2)=(1/2)pie144=72*pie

Solution as per intersecting chord theorem.
Let the radii of large and small circles be R and r. Draw a diameter in the large circle, perpendicular to the chord of length 24, bisecting it in two segments of lengths 12 & 12.
The intersecting chord would cut the diameter into two segments of lengths (R+r) and (R-r).
As per intersecting chord theorem, the product of the segments of intersecting chords are equal.
So we have, 12×12 = (R+r)(R-r)=R^2 - r^2,.
Or, π(R^2 - r^2)/2 = π(12×12)/2= 72π.
Pink area = π(R^2 - r^2)/2 = 72π.

Let the radius of the bigger semicirce be R and the radius of the smaller semicircle be r. It can be observed that R^2=12x12-r^2. => R^2=144-r^2. So, pink area=(1/2) pi (R^2-r^2)=(1/2) pi 144=72*pi.

Area of Pink Region = $\frac{A}{2}$ - $\frac{a}{2}$

= $\frac{\pi(R^{2}) - \pi(r^{2})}{2}$

= $\frac{\pi(R^{2} - r^{2})}{2}$

= $\frac{\pi(12^{2})}{2}$ [Look in the fig, for this step]

= 72 $\pi$

LaTeX: $Let \; A_C, \text {the area of the large semi-circle} \; A_c, \text {the area of the little semi-circle and} \;A_p \; \text {the pink area}$

LaTex: $\Large \text {As you can see, } \; R = 3r \; , \; A_C = \frac{\pi R^2}{2} \text {and} \; A_c = \frac{\pi r^2}{2}$

LaTeX: $\Large \text {The area we are looking for is : } \; A_C - A_c = A_p \iff A_p = \frac{\pi R^2}{2} - \frac{\pi r^2}{2}$

LaTeX: $\Large A_p = \frac{\pi R^2}{2} - \frac{\pi r^2}{2} \iff A_p = \frac{\pi {(3r)}^2}{2} - \frac{\pi r^2}{2} \iff A_p = \frac{8\pi r^2}{2} \iff A_p = 4\pi r^2$

LaTex: $\large \text {As we have a cord [EF], we need half that cord : } \; [EF] = 24 \iff \frac{[EF]}{2} = 12$

LaTex: $\large \text {the radii r and R and we get a right angled triangle. Applying the pithagorean theorem,}$

LaTex: $\large \text {we find the values of those 2 radii.}$

LaTeX: $\large \text {then we get : } \; r^2 + 12^2 = R^2 \iff r^2 + 12^2 = (3r)^2 \iff r^2 + 12^2 = 9r^2 \iff r^2 = \frac{3^2 \times 4^2}{2 \times 4}$

LaTex: $\Large \text {Finally, the radii are :} \; r^2 = 3^2 \times 2 \iff (r = 3 \sqrt2) \; \text {and} \; (R =9 \sqrt2)$

LaTex: $\Large \text {The pink area is :} \; A_p = 4\pi r^2 = 4\pi (3 \sqrt2) ^2 = 4 \pi \times 9 \times 2 = \color{#D61F06}{\boxed {72 \pi}}$

OC is the radius of the larger semicircle; so OC=R

We can apply the Pythagorean theorem in triangle OCE; $OC=\sqrt { (OE)^{ 2 }+({ CE) }^{ 2 } }$

OE=r

CE=12 (half of CD)

$OC=\sqrt { { r }^{ 2 }+144 }$

So, $\sqrt { { r }^{ 2 }+144 } =R$

${ R }^{ 2 }={ r }^{ 2 }+144$

The pink area is the difference between the areas of the semicircles: $A=\frac { 1 }{ 2 } \pi { R }^{ 2 }-\frac { 1 }{ 2 } \pi { r }^{ 2 }=\frac { 1 }{ 2 } \pi ({ r }^{ 2 }+144)-\frac { 1 }{ 2 } \pi { r }^{ 2 }=72\pi$

Let the radiuses of big and small hemi-circles be r and r’. Forming a right tringle by connecting the center of the big hemi-circle to the right side of 24 unit chord and drawing a perpendicular from the center of big hemi-circle to this chord we have a right triangle and so 144 + r’2 = r2, or r2 – r’2 = 144. The area A of the pink part is: A = pi/2(r2 - r’2) = pi/2(r2 – r’2) = pi/2(144) = 72pi.

Let's call the radius of the small circle r and the radius of the large circle R. r is also the distance between the diameter and the chord, so you can draw a right triangle from the center to the edge along the chord (side lengths are 12, r and R). By Pythagorean Theorem, R^2=144+r^2. The pink area is (R^2-r^2) pi/2, which is 144 pi/2 or 72*pi.

We're given no information about the smaller diameter. Turns out, the pink area is constant for all such circles. That is, as the white area grows, so must the pink area in order to accommodate the chord constraint.

So, solve for the case when the diameter of the small circle approaches zero. The diameter of the large circle approaches 24. And the area of a half circle of diameter 24 is 72 π.

By Pythagoras theorem, R^2=r^2+(24/2)^2. Therefore, R^2-r^2=144. Finally, pinkArea=0.5 pi (R^2-r^2)=72*pi.