Rhombus Rhumble

Geometry Level pending

A B C D ABCD is a rhombus positioned on the coordinate plane with A A on the origin, as shown in the figure. Its inferior angle is 6 0 ∘ 60^\circ and the length of each side is 4 4 . E E is positioned on B C BC such that the radius of the larger circle is twice the radius of the smaller circle. Find the value of the x x -coordinate of E E . If it is a root of f ( x ) = a x 3 + b x 2 + c x + d + e f(x) = ax^3 + bx^2 + cx + d + e , where a > 0 a>0 , and gcd ( a , b , c , d , e ) = 1 (a,b,c,d,e)=1 , submit f ( 6 ) f(6) .


The answer is 13.

Fletcher Mattox by Fletcher Mattox , 72, USA

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1 solution

David Vreken
May 22, 2021

Let the x x -coordinate of E E be x x .

Since A B = 4 AB = 4 and the exterior angle of ∠ A B E \angle ABE is 60 ° 60° , B E = 2 ( x − 4 ) = 2 x − 8 BE = 2(x - 4) = 2x - 8 , which means C E = B C − B E = 4 − ( 2 x − 8 ) = 12 − 2 x CE = BC - BE = 4 - (2x - 8) = 12 - 2x .

By the law of cosines on △ A B E \triangle ABE , A E = A B 2 + B E 2 − 2 ⋅ A B ⋅ B E ⋅ cos ⁡ A B E = 4 2 + ( 2 x − 8 ) 2 − 2 ⋅ 4 ⋅ ( 2 x − 8 ) ⋅ cos ⁡ 120 ° = 2 x 2 − 6 x + 12 AE = \sqrt{AB^2 + BE^2 - 2 \cdot AB \cdot BE \cdot \cos ABE} = \sqrt{4^2 + (2x - 8)^2 - 2 \cdot 4 \cdot (2x - 8) \cdot \cos 120°} = 2\sqrt{x^2 - 6x + 12} .

By the law of cosines on △ D C E \triangle DCE , D E = C D 2 + C E 2 − 2 ⋅ C D ⋅ C E ⋅ cos ⁡ D C E = 4 2 + ( 12 − 2 x ) 2 − 2 ⋅ 4 ⋅ ( 12 − 2 x ) ⋅ cos ⁡ 60 ° = 2 x 2 − 10 x + 28 DE = \sqrt{CD^2 + CE^2 - 2 \cdot CD \cdot CE \cdot \cos DCE} = \sqrt{4^2 + (12 - 2x)^2 - 2 \cdot 4 \cdot (12 - 2x) \cdot \cos 60°} = 2\sqrt{x^2 - 10x + 28} .

The area of △ A D E \triangle ADE is A △ A D E = 1 2 ⋅ b ⋅ h = 1 2 ⋅ 4 ⋅ 2 3 = 4 3 A_{\triangle ADE} = \frac{1}{2} \cdot b \cdot h = \frac{1}{2} \cdot 4 \cdot 2\sqrt{3} = 4\sqrt{3} .

The area of △ D C E \triangle DCE is A △ D C E = 1 2 ⋅ C D ⋅ C E ⋅ sin ⁡ D C E = 1 2 ⋅ 4 ⋅ ( 12 − 2 x ) ⋅ sin ⁡ 60 ° = 2 3 ( 6 − x ) A_{\triangle DCE} = \frac{1}{2} \cdot CD \cdot CE \cdot \sin DCE = \frac{1}{2} \cdot 4 \cdot (12 - 2x) \cdot \sin 60° = 2\sqrt{3}(6 - x) .

The inradius of △ A D E \triangle ADE is R = 2 A △ A D E A D + D E + A E = 2 ⋅ 4 3 4 + 2 x 2 − 10 x + 28 + 2 x 2 − 6 x + 12 = 4 3 2 + x 2 − 10 x + 28 + x 2 − 6 x + 12 R = \cfrac{2A_{\triangle ADE}}{AD + DE + AE} = \cfrac{2 \cdot 4\sqrt{3}}{4 + 2\sqrt{x^2 - 10x + 28} + 2\sqrt{x^2 - 6x + 12}} = \cfrac{4\sqrt{3}}{2 + \sqrt{x^2 - 10x + 28} + \sqrt{x^2 - 6x + 12}} .

The inradius of △ D C E \triangle DCE is r = 2 A △ D C E C D + C E + D E = 2 ⋅ 2 3 ( 6 − x ) 4 + ( 12 − 2 x ) + 2 x 2 − 10 x + 28 = 2 3 ( 6 − x ) 8 − x + x 2 − 10 x + 28 r = \cfrac{2A_{\triangle DCE}}{CD + CE + DE} = \cfrac{2 \cdot 2\sqrt{3}(6 - x)}{4 + (12 - 2x) + 2\sqrt{x^2 - 10x + 28}} = \cfrac{2\sqrt{3}(6 - x)}{8 - x + \sqrt{x^2 - 10x + 28}} .

Since the radius of the larger circle is twice the radius of the smaller circle, R = 2 r R = 2r , and this simplifies to:

4 3 2 + x 2 − 10 x + 28 + x 2 − 6 x + 12 = 2 ⋅ 2 3 ( 6 − x ) 8 − x + x 2 − 10 x + 28 \cfrac{4\sqrt{3}}{2 + \sqrt{x^2 - 10x + 28} + \sqrt{x^2 - 6x + 12}} = 2 \cdot \cfrac{2\sqrt{3}(6 - x)}{8 - x + \sqrt{x^2 - 10x + 28}}

8 − x + x 2 − 10 x + 28 = ( 6 − x ) ( 2 + x 2 − 10 x + 28 + x 2 − 6 x + 12 ) 8 - x + \sqrt{x^2 - 10x + 28} = (6 - x)(2 + \sqrt{x^2 - 10x + 28} + \sqrt{x^2 - 6x + 12}) (cross-multiply)

8 − x + x 2 − 10 x + 28 = 2 ( 6 − x ) + ( 6 − x ) x 2 − 10 x + 28 + ( 6 − x ) x 2 − 6 x + 12 8 - x + \sqrt{x^2 - 10x + 28} = 2(6 - x) + (6 - x)\sqrt{x^2 - 10x + 28} + (6 - x)\sqrt{x^2 - 6x + 12} (distribute)

( x − 5 ) x 2 − 10 x + 28 = 4 − x + ( 6 − x ) x 2 − 6 x + 12 (x - 5)\sqrt{x^2 - 10x + 28} = 4 - x + (6 - x)\sqrt{x^2 - 6x + 12} (rearrange)

( ( x − 5 ) x 2 − 10 x + 28 ) 2 = ( 4 − x + ( 6 − x ) x 2 − 6 x + 12 ) 2 ((x - 5)\sqrt{x^2 - 10x + 28})^2 = (4 - x + (6 - x)\sqrt{x^2 - 6x + 12})^2 (square both sides)

x 4 − 20 x 3 + 153 x 2 − 530 x + 700 = x 4 − 18 x 3 + 121 x 2 − 368 x + 448 + ( 2 x 2 − 20 x + 48 ) x 2 − 6 x + 12 x^4 - 20x^3 + 153x^2 - 530x + 700 = x^4 -18x^3 + 121x^2 - 368x + 448 + (2x^2 - 20x + 48)\sqrt{x^2 - 6x + 12} (expand)

− 2 x 3 + 32 x 2 − 162 x + 252 = ( 2 x 2 − 20 x + 48 ) x 2 − 6 x + 12 -2x^3 + 32x^2 - 162x + 252 = (2x^2 - 20x + 48)\sqrt{x^2 - 6x + 12} (simplify terms)

− 2 ( x − 3 ) ( x − 6 ) ( x − 7 ) = 2 ( x − 4 ) ( x − 6 ) x 2 − 6 x + 12 -2(x - 3)(x - 6)(x - 7) = 2(x - 4)(x - 6)\sqrt{x^2 - 6x + 12} (factor)

− ( x − 3 ) ( x − 7 ) = ( x − 4 ) x 2 − 6 x + 12 -(x - 3)(x - 7) = (x - 4)\sqrt{x^2 - 6x + 12} (cancel 2 ( x − 6 ) 2(x - 6) )

( − ( x − 3 ) ( x − 7 ) ) 2 = ( ( x − 4 ) x 2 − 6 x + 12 ) 2 (-(x - 3)(x - 7))^2 = ((x - 4)\sqrt{x^2 - 6x + 12})^2 (square both sides)

x 4 − 20 x 3 + 142 x 2 − 420 x + 441 = x 4 − 14 x 3 + 76 x 2 − 192 x + 192 x^4 - 20x^3 + 142x^2 - 420x + 441 = x^4 - 14x^3 + 76x^2 - 192x + 192 (expand)

6 x 3 − 66 x 2 + 228 x − 249 = 0 6x^3 - 66x^2 + 228x - 249 = 0 (simplify terms)

2 x 3 − 22 x 2 + 76 x − 83 = 0 2x^3 - 22x^2 + 76x - 83 = 0 (divide by 3 3 )

Therefore, f ( 6 ) = 2 ( 6 ) 3 − 22 ( 6 ) 2 + 76 ( 6 ) − 83 = 13 f(6) = 2(6)^3 - 22(6)^2 + 76(6) - 83 = \boxed{13} .

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Thank you for slogging through all the algebra. Good job.

Fletcher Mattox - 3 weeks ago

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Thanks, and that was another fun question!

David Vreken - 3 weeks ago

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