Dynamic Geometry: P5

Geometry Level 5

The diagram shows a black parabola y = x 2 y=x^{2} and a blue square moving so that it's tangent to the parabola and to the line y = 0 y=0 . As the square is moving, its center (cyan) is following the path of a pink curve. The center bounces between an orange vertical segment and a green vertical segment. When the cyan point touches the green segment, the area of the square is equal to 1 1 . When the cyan point touches the orange segment, the area of the square is equal to 16 16 . The area bounded by the pink curve, the green segment, the orange segment and the line y = 0 y=0 can be expressed as a b \dfrac{a}{b} where a a , and b b are coprime positive integers. Find a + b a+b .


The answer is 97.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

Karan Chatrath
Feb 2, 2021

Let any point on the parabola be ( h , h 2 ) (h,h^2) . The square, one edge of which touches the parabola and the opposite side lying on the X-axis has a side of h 2 h^2 . Therefore, the area of the square is A s = h 4 A_s = h^4 .

The coordinates of the center of the circle are therefore:

x = h + h 2 2 x = h + \frac{h^2}{2} y = h 2 2 y = \frac{h^2}{2}

When the area of the square is unity, then h = 1 h=1 and when A s = 16 A_s=16 then h = 2 h = 2 . This means that as the center moves along the pink curve, 1 ≤ h ≤ 2 1 \le h \le 2 . The area of this pink curve in the given interval is:

A p = ∫ h = 1 h = 2 y d x A_p = \int_{h=1}^{h=2} y \ dx ⟹ A p = ∫ 1 2 h 2 2 ( 1 + h ) d h \implies A_p = \int_{1}^{2} \frac{h^2}{2} \left(1 + h\right) \ dh ⟹ A p = 73 24 \implies A_p = \frac{73}{24}

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Thanks for posting -)

Valentin Duringer - 4 months, 1 week ago

Let the point where the square touches the parabola be P = ( x ; x 2 ) P=(x; x^2) The square has side y = x 2 y=x^2 , so the area of the square is A = y 2 = x 4 A=y^2=x^4 Hence P m i n = ( 1 , 1 ) P_{min}=(1,1) and P m a x = ( 2 , 4 ) P_{max}=(2,4) . The cyan point at the centre of the square is at C = ( x + 1 2 y , 1 2 y ) = ( f ( x ) ; g ( x ) ) where f ( x ) = x + 1 2 x 2 and g ( x ) = 1 2 x 2 C=(x+\frac12 y, \frac12 y)= (f(x); g(x))\text{ where }f(x)=x+\frac12 x^2 \text{ and } g(x)= \frac12 x^2

Now the requested area is given by ∫ x = 1 2 g ( x ) d f ( x ) = ∫ 1 2 g d f d x d x = ∫ 1 2 1 2 x 2 ( 1 + x ) d x = 1 2 ( 1 3 x 3 + 1 4 x 4 ) ∣ 0 1 = 1 2 ( 7 3 + 15 4 ) = 28 + 45 24 = 73 24 \int_{x=1}^{2} g(x)df(x) = \int_{1}^{2} g \frac{d f}{d x} dx=\int_{1}^{2} \frac12 x^2 (1+x) dx= \frac12(\frac13 x^3+ \frac14 x^4)\biggr\rvert_0^1= \frac12(\frac{7}{3}+\frac{15}{4})=\frac{28+45}{24} =\frac{73}{24}

The answer is 73+24=97

K T - 4 months, 1 week ago

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