Dynamic Geometry: P51

Geometry Level pending

The diagram shows a black semicircle with radius 1 1 . A red point is moving along its arc. We use this point and the semicircle's diameter to draw a blue triangle. The centroid of the triangle (green point) traces a locus (orange curve). The area bounded by the orange curve and the cyan segment can be expressed as π a \dfrac{\pi }{a} , where a a is an integer. What is a a ?


The answer is 18.

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1 solution

Let the center of the semicircle be O ( 0 , 0 ) O(0,0) , the origin of the x y xy -plane, the triangle be A B C ABC , where A B AB is the diameter of the semicircle and C ( u , v ) C(u,v) , the red point, and P ( x , y ) P(x,y) , the centroid. Since P P is the centroid of A B C \triangle ABC , its coordinates are given by:

{ x = x A + x B + x C 3 = 1 + 1 + u 3 = u 3 y = y A + y B + y C 3 = 0 + 0 + v 3 = v 3 \begin{cases} x = \dfrac {x_A+x_B+x_C}3 = \dfrac {-1+1+u}3 & = \dfrac u3 \\ y = \dfrac {y_A+y_B+y_C}3 = \dfrac {0+0+v}3 & = \dfrac v3 \end{cases}

Since the locus of C ( u , v ) C(u,v) is a semicircle with radius 1 1 , the locus P ( x , y ) P(x,y) is also a semicircle with a smaller radius of 1 3 \dfrac 13 . Therefore the area bounded by the locus and the cyan segment is A = 1 2 π ( 1 3 ) 2 = π 18 A = \dfrac 12 \cdot \pi \left(\dfrac 13 \right)^2 = \dfrac \pi{18} . Therefore a = 18 a = \boxed{18} .

For once we share the same solution, I even ate a fruit to celebrate. Thank you for posting.

Valentin Duringer - 3 months, 1 week ago

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You are welcome

Chew-Seong Cheong - 3 months, 1 week ago

Cool, even I solved it the same way

Omek K - 3 months, 1 week ago

So cool :) !

Valentin Duringer - 3 months, 1 week ago

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