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Four externally tangent circles with radius $ 2 $ are internally tangent to a larger circle and externally tangent to a smaller circle. A square is drawn by connecting the points of tangency of the four medium- sized circles to the larger circle. The area of the shaded region in the diagram can be written as $a+(b+cπ)\sqrt2+dπ$, where $a, b, c,$ and $d $ are integers. Find $a + b + c + d.​$

One possible way to solve this problem is to use the Pythagorean Theorem to find the length of the diagonal of the square, and then use the formula for the area of a square to find its area. Then, subtract the areas of the four right triangles in the corners of the square and the area of the small circle to get the area of the shaded region. By solving with this method I got $26$ as the answer but the original answer is $12$.

Where did I go wrong? Is there any other method to solve this problem?

  • 1
    $\begingroup$ We don't know where you went wrong, unless you show your steps in detail. Also seemingly some steps are missing in your approach. $\endgroup$
    – ACB
    May 21, 2023 at 6:56

2 Answers 2


To tackle this, I first calculated the radius of the large circle. 🙂

If we draw a square from the center of the 4 smaller circles, we can see the square's sides are of length $4$ and its diagonal is $4\sqrt2$.

Let's draw that (apologies for the cluttery mess):


From above, we can see that $C_4C_1=4\sqrt2$, while $CC_4$ and $C_1D$ are each of length $2$.

So, the diameter of the large circle is $4+4\sqrt2$ and its radius is $2+2\sqrt2$. Yeay!

Now, let's draw a big square (like depicted in the problem):


Since $AC=AE=2+2\sqrt2$, $\triangle ACE$ is an isosceles right triangle. That means: $$CE=AC\sqrt2$$ $$CE=2\sqrt2+4$$

That means the square's sides are of length $4+2\sqrt2$. Now, let's zoom in on a smaller circle:


In the problem you presented, we need to find the area of $\triangle CGH$ and half of the circle.

As we can see, $\triangle CGH$ is yet another isosceles right triangle! So, $CG=CH=\frac{GH}{\sqrt2}$. Since $GH=4$, $CG=CH=2\sqrt2$ and $[\triangle CGH]=4$.

Furthermore, the area of the half-circle is $\frac12\pi\cdot2^2=2\pi$. So, the area of the circle inside the square is $4+2\pi$.

Now, let's talk about the small circle in the middle. By looking for a bit, we can find that its diameter is the large circle's diameter, minus $8$: $$d=(4+4\sqrt2)-8$$ $$d=4\sqrt2-4$$ $$r=\frac12d=2\sqrt2-2$$

Which means its area is... $$A=r^2\cdot\pi$$ $$A=(2\sqrt2-2)^2\cdot\pi$$ $$A=(12-8\sqrt2)\pi$$


We now have enough information to get the answer. So:

$$A_{shaded}=A_{square}-A_{small\;circle}-A_{circle\;inside\;the\;square}$$ $$A_{shaded}=(4+2\sqrt2)^2-(12-8\sqrt2)\pi-4(4+2\pi)$$ $$A_{shaded}=(24+16\sqrt2)-(12-8\sqrt2)\pi-(16+8\pi)$$ $$A_{shaded}=\color{#880000}{24}+\color{#AA8800}{16}\sqrt2\color{#AAAA00}{-12}\pi+\color{#88AA00}{8}\sqrt2\pi\color{#880000}{-16}\color{#AAAA00}{-8}\pi$$ $$A_{shaded}=\color{#880000}{8}+(\color{#AA8800}{16}+\color{#88AA00}{8}\pi)\sqrt2\color{#AAAA00}{-20}\pi$$


...and $a+b+c+d=12$. That was long haha


HINT…rather than write out the full solution for you, I will tell you what you should get for each stage of the calculation. You can then piece together the solution and get the correct answer.

  1. The side of the square is $4+2\sqrt{2}$
  2. The diagonal of the square is $4+4\sqrt{2}$
  3. The diameter of the innermost circle is $4\sqrt{2}-4$
  4. The area of each of the eight white segments is $\pi-2$
  5. The area of each of the white regions inside the four circles and inside the square is $2\pi+4$

I hope this helps.


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