Dynamic Geometry: P59

Geometry Level 4

The diagram shows an orange semicircle with radius $1$ . Two cyan semicircles are growing and shrinking symmetrically so they are internally tangent to the orange semicircle. Each pink circle is internally tangent to the orange semicircle and tangent to a cyan semicircle. One cyan semicircle and its tangent pink circle share the same $x$ coordinate. Using the \four centers, we draw a black rectangle. When the area of the black rectangle is maximum , its perimeter can be expressed as $p-q\sqrt{m}$ , where $q$ , $p$ and $m$ are positive integers and $m$ is square-free. Find $\sqrt{p+q-m}$ .

The answer is 4.

1 solution

Chew-Seong Cheong
Mar 4, 2021

Let the center of the large semicircle be $O$ , radii of the cyan semicircle and pink circle at any instant be $r_1$ and $r_2$ respectively. By Pythagorean theorem ,

$\begin{aligned} (r_1+r_2)^2 + (1-r_1)^2 & = (1-r_2)^2 \\ \implies r_2 & = \frac {r_1-r_1^2}{1+r_1} \end{aligned}$

The area of the rectangle is given by:

$\begin{aligned} A & = 2(1-r_1)(r_1+r_2) = \frac {4r_1(1-r_1)}{1+r_1} \\ \frac {dA}{dr_1} & = \frac {4(1-2r_1)(1+r_1) - 4r_1(1-r_1)}{(1+r_1)^2} \\ & = \frac {1-2r_1+r_1^2}{(1+r_1)^2} & \small \blue{\text{Putting }\frac {dA}{dr_1}=0} \\ 1 - 2r_1 - r_1^2 & = 0 \\ \implies r_1 & = \sqrt 2 - 1 & \small \blue{\text{when }A \text{ is maximum.}} \\ \implies r_2 & = \frac {r_1-r_1^2}{1+r_1} = 3-2\sqrt 2 \end{aligned}$

Then the perimeter of the rectangle is $4(1-r_1) + 2(r_1 + r_2) = 12 - 6\sqrt 2$ . And $\sqrt{p+q-m} = \sqrt{12+6-2} = \boxed 4$ .

Nice solution! It's interesting that $A = 4r_2$ .

David Vreken - 3 months, 1 week ago

Dynamic Geometry: P59

The answer is 4.

1 solution

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