The minimum distance between the figures is not 1.

Calculus Level 5

Since the figures are symmetrical in all four quadrants, the problem is to be worked in the first quadrant only, where x ≥ 0 x\geq 0 and y ≥ 0 y\geq 0 .

The two figures, respectively, have the equations: ( 2 x 1 ) 2 + ( 3 y 1 ) 2 = ( 2 × 3 ) 2 \left(2 x_1\right){}^2+\left(3 y_1\right){}^2=(2\times 3)^2 and ( 3 x 2 ) 2 + ( 4 y 2 ) 2 = ( 3 × 4 ) 2 \left(3 x_2\right){}^2+\left(4 y_2\right){}^2=(3\times 4)^2 .

The problem is to find the minimum distance between the figures. The answer will have the form of a relatively prime fraction possibly under a square root. The answer will be the concatenation (i.e., list altogether) of a 1 if there is no square root is required and a 2 if a square root is required, the numerator of the fraction, and the denominator of the fraction. If the answer were 47 51 \sqrt{\frac{47}{51}} and this is not the answer, then the number to be entered would be 24751. If the answer were 1 1 and this also is not the answer, then the number would be 111 because a fraction is required and in that case would be written as the answer over 1. If the answer were 19 20 \frac{19}{20} and again this is not the answer, then the number to be entered would be 11920, Last example, 999 1001 \sqrt{\frac{999}{1001}} gives 29991001.


The answer is 23435.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

( 2 x 1 ) 2 + ( 3 y 1 ) 2 = ( 2 3 ) 2 ∧ x 1 ≥ 0 ∧ y 1 ≥ 0 → 0 ≤ x 1 ≤ 3 ∧ y 1 = 2 3 9 − x 1 2 \left(2 x_1\right){}^2+\left(3 y_1\right){}^2=(2\ 3)^2\land x_1\geq 0\land y_1\geq 0\to 0\leq x_1\leq 3\land y_1=\frac{2}{3} \sqrt{9-x_1^2}

( 3 x 2 ) 2 + ( 4 y 2 ) 2 = ( 3 4 ) 2 ∧ x 2 ≥ 0 ∧ y 2 ≥ 0 → 0 ≤ x 2 ≤ 4 ∧ y 2 = 3 4 16 − x 2 2 \left(3 x_2\right){}^2+\left(4 y_2\right){}^2=(3\ 4)^2\land x_2\geq 0\land y_2\geq 0\to 0\leq x_2\leq 4\land y_2=\frac{3}{4} \sqrt{16-x_2^2}

The expression for the Euclidean distance is 5 x 1 2 9 − 2 x 2 x 1 + 7 x 2 2 16 − 9 − x 1 2 16 − x 2 2 + 13 \sqrt{\frac{5 x_1^2}{9}-2 x_2 x_1+\frac{7 x_2^2}{16}-\sqrt{9-x_1^2} \sqrt{16-x_2^2}+13} .

The minimum distance is 34 35 \sqrt{\frac{34}{35}} for which the number to be entered as 23435.

Using the square of the Euclidean distance simplifies the computations and does not change the answer.

∂ ( 5 x 1 2 9 − 2 x 2 x 1 + 7 x 2 2 16 − 9 − x 1 2 16 − x 2 2 + 13 ) ∂ x 1 → 16 − x 2 2 x 1 9 − x 1 2 + 10 x 1 9 − 2 x 2 \frac{\partial \left(\frac{5 x_1^2}{9}-2 x_2 x_1+\frac{7 x_2^2}{16}-\sqrt{9-x_1^2} \sqrt{16-x_2^2}+13\right)}{\partial x_1}\to \frac{\sqrt{16-x_2^2} x_1}{\sqrt{9-x_1^2}}+\frac{10 x_1}{9}-2 x_2

∂ ( 5 x 1 2 9 − 2 x 2 x 1 + 7 x 2 2 16 − 9 − x 1 2 16 − x 2 2 + 13 ) ∂ x 2 → − 2 x 1 + 7 x 2 8 + 9 − x 1 2 x 2 16 − x 2 2 \frac{\partial \left(\frac{5 x_1^2}{9}-2 x_2 x_1+\frac{7 x_2^2}{16}-\sqrt{9-x_1^2} \sqrt{16-x_2^2}+13\right)}{\partial x_2}\to -2 x_1+\frac{7 x_2}{8}+\frac{\sqrt{9-x_1^2} x_2}{\sqrt{16-x_2^2}}

Setting the expression as being equal to 0 0 and solving the simultaneous equations with additional constraints that x 1 ≥ 0 x_1\geq 0 and x 2 ≥ 0 x_2\geq 0 gives { x 1 → 0 , x 2 → 0 } , { x 1 → 12 5 17 , x 2 → 12 5 17 } , { x 1 → 9 29 17 5 , x 2 → 16 29 17 7 } \left\{x_1\to 0,x_2\to 0\right\},\left\{x_1\to 12 \sqrt{\frac{5}{17}},x_2\to 12 \sqrt{\frac{5}{17}}\right\},\left\{x_1\to \frac{9 \sqrt{\frac{29}{17}}}{5},x_2\to \frac{16 \sqrt{\frac{29}{17}}}{7}\right\} .

Now the second derivative test is needed to determine whether any of these situations qualify.

∂ 2 ∂ x 1 ∂ x 1 ( 13 + 5 x 1 2 9 − 2 x 1 x 2 + 7 x 2 2 16 − 9 − x 1 2 16 − x 2 2 ) 16 − x 2 2 x 1 2 ( 9 − x 1 2 ) 3 / 2 + 16 − x 2 2 9 − x 1 2 + 10 9 ∂ 2 ∂ x 2 ∂ x 2 ( 13 + 5 x 1 2 9 − 2 x 1 x 2 + 7 x 2 2 16 − 9 − x 1 2 16 − x 2 2 ) 9 − x 1 2 x 2 2 ( 16 − x 2 2 ) 3 / 2 + 9 − x 1 2 16 − x 2 2 + 7 8 ∂ 2 ∂ x 1 ∂ x 2 ( 13 + 5 x 1 2 9 − 2 x 1 x 2 + 7 x 2 2 16 − 9 − x 1 2 16 − x 2 2 ) − x 1 x 2 9 − x 1 2 16 − x 2 2 − 2 \begin{array}{|c|c|} \hline \frac{\partial ^2}{\partial x_1\, \partial x_1}\left(13+\frac{5 x_1^2}{9}-2 x_1 x_2+\frac{7 x_2^2}{16}-\sqrt{9-x_1^2} \sqrt{16-x_2^2}\right) & \frac{\sqrt{16-x_2^2} x_1^2}{\left(9-x_1^2\right){}^{3/2}}+\frac{\sqrt{16-x_2^2}}{\sqrt{9-x_1^2}}+\frac{10}{9} \\ \frac{\partial ^2}{\partial x_2\, \partial x_2}\left(13+\frac{5 x_1^2}{9}-2 x_1 x_2+\frac{7 x_2^2}{16}-\sqrt{9-x_1^2} \sqrt{16-x_2^2}\right) & \frac{\sqrt{9-x_1^2} x_2^2}{\left(16-x_2^2\right){}^{3/2}}+\frac{\sqrt{9-x_1^2}}{\sqrt{16-x_2^2}}+\frac{7}{8} \\ \frac{\partial ^2}{\partial x_1\, \partial x_2}\left(13+\frac{5 x_1^2}{9}-2 x_1 x_2+\frac{7 x_2^2}{16}-\sqrt{9-x_1^2} \sqrt{16-x_2^2}\right) & -\frac{x_1 x_2}{\sqrt{9-x_1^2} \sqrt{16-x_2^2}}-2 \\ \hline \end{array}

The discriminant is ( 9 − x 1 2 x 2 2 ( 16 − x 2 2 ) 3 / 2 + 9 − x 1 2 16 − x 2 2 + 7 8 ) ( 16 − x 2 2 x 1 2 ( 9 − x 1 2 ) 3 / 2 + 16 − x 2 2 9 − x 1 2 + 10 9 ) − ( − x 1 x 2 9 − x 1 2 16 − x 2 2 − 2 ) 2 \left(\frac{\sqrt{9-x_1^2} x_2^2}{\left(16-x_2^2\right){}^{3/2}}+\frac{\sqrt{9-x_1^2}}{\sqrt{16-x_2^2}}+\frac{7}{8}\right) \left(\frac{\sqrt{16-x_2^2} x_1^2}{\left(9-x_1^2\right){}^{3/2}}+\frac{\sqrt{16-x_2^2}}{\sqrt{9-x_1^2}}+\frac{10}{9}\right)-\left(-\frac{x_1 x_2}{\sqrt{9-x_1^2} \sqrt{16-x_2^2}}-2\right){}^2 .

Evaluating the discriminant for each of the three points gives { − 1 36 , − 1445 9072 , 8381 53136 } \left\{-\frac{1}{36},-\frac{1445}{9072},\frac{8381}{53136}\right\} .

Only the last point evaluated to a positive, which is what is required for a minimization. Therefore, the minimum occurs at x 1 → 9 29 17 5 , x 2 → 16 29 17 7 x_1\to \frac{9 \sqrt{\frac{29}{17}}}{5},x_2\to \frac{16 \sqrt{\frac{29}{17}}}{7} . The cooresponding y y s are: { 4 41 17 5 , 9 41 17 7 } \left\{\frac{4 \sqrt{\frac{41}{17}}}{5},\frac{9 \sqrt{\frac{41}{17}}}{7}\right\} .

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Uahnbu Tran
Oct 15, 2020

For x 1 x_1 of f ( x ) f(x) and x 2 x_2 of g ( x ) g(x) which yield the minimum distance, there shall be 2 equations:

( 1 ) f ′ ( x 1 ) = g ′ ( x 2 ) (1) f'(x_1)=g'(x_2) since slopes at intersections should be equal

( 2 ) f ( x 1 ) + x 1 f ′ ( x 1 ) = g ( x 2 ) + x 2 g ′ ( x 2 ) (2) f(x_1)+\frac{x_1}{f'(x_1)}=g(x_2)+\frac{x_2}{g'(x_2)} since y intercepts for the lines should be equal

For this problem, (1) means − 2 x 1 3 9 − x 1 2 = − 3 x 2 4 16 − x 2 2 \frac{-2x_1}{3\sqrt{9-x_1^2}}=\frac{-3x_2}{4\sqrt{16-x_2^2}}

and (2) is 7 12 16 − x 1 2 = 5 6 9 − x 2 2 \frac7{12}\sqrt{16-x_1^2}=\frac56\sqrt{9-x_2^2}

which both combined shall reduce to a quadratic equation in terms of x 1 2 x_1^2 .

The value is 9 5 29 17 \frac95\sqrt\frac{29}{17} and 16 7 29 17 \frac{16}7\sqrt\frac{29}{17} for x 1 x_1 and x 2 x_2 respectively.

Plug them into ( x 1 − x 2 ) 2 + ( 2 3 9 − x 1 2 − 3 4 16 − x 2 2 ) 2 \sqrt{\left(x_1-x_2\right)^2+\left(\frac23\sqrt{9-x_1^2}-\frac34\sqrt{16-x_2^2}\right)^2} to obtain the distance.

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