Geometric mean in geometry

Geometry Level 5

Large version of the image
For A B C \triangle ABC , I I is the midpoint of B C BC , point D D is in line segment A C AC such that C D = 3 A D CD=3AD and point E E is in line segment A B AB such that [ B I E ] = [ C I D ] × [ A D E ] [BIE]=\sqrt{[CID] \times [ADE]} .

If the ratio A E E B \dfrac{AE}{EB} can be expressed as a b c \dfrac{\sqrt{a}-b}{c} , then find a + b + c . a + b + c.

(The figure is not drawn to scale)

Note: [ X Y Z ] [XYZ] denotes the area of triangle X Y Z XYZ .


The answer is 114.

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1 solution

L e t A E = X , a n d E B = 1 L e t a r e a Δ A B C = S A B C S B I E = 1 2 1 1 + X S A B C . S C I D = 1 2 3 4 S A B C . S A D E = 1 4 X 1 + X S A B C . ( 1 2 1 1 + X S A B C . ) 2 = ( 1 2 3 4 S A B C . ) ( 1 4 X 1 + X S A B C ) . 1 1 + X = 3 8 X 3 X 2 + 3 X 8 = 0. S o l v i n g t h e q u a d r a t i c X = 3 + 9 + 96 6 = 3 + 105 6 . B u t A E E B = X . a + b + c = 114 Let\ AE=X,\ \ and\ \ EB=1\ \ \ \ \ \ Let\ area \ \Delta\ ABC=S_{ABC}\\ S_{BIE} = \dfrac 1 2 *\dfrac 1 {1+X}*S_{ABC}.\\ S_{CID} = \dfrac 1 2 *\dfrac 3 4*S_{ABC}.\\ S_{ADE} = \dfrac 1 4 *\dfrac X {1+X}*S_{ABC}.\\ \therefore\ ( \dfrac 1 2 *\dfrac 1 {1+X}*S_{ABC}.)^2\ =\ ( \dfrac 1 2 *\dfrac 3 4*S_{ABC}.)*( \dfrac 1 4 *\dfrac X {1+X}*S_{ABC}). \\ \implies\ \dfrac 1 {1+X}\ =\ \dfrac 3 8 * X\\ \therefore\ 3X^2 \ +\ 3X - 8 \ =\ 0.\\ Solving\ the\ quadratic\ \ X\ =\ \dfrac{-3 +\ \sqrt{9+96} } 6 =\ \dfrac{-3 +\ \sqrt{105} } 6.\\ But\ \ \dfrac {AE}{EB} \ =\ X. \ \ \ \therefore \ a+b+c=\Large \ \ \ \ \ \color{#D61F06}{114} .

Same solution . Nice a good presentation .

vishwash kumar - 4 years, 7 months ago

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Thank you.

Niranjan Khanderia - 4 years, 1 month ago

Same solution.Hats off to your great presentation.

rajdeep brahma - 4 years, 2 months ago

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Thank you.

Niranjan Khanderia - 4 years, 1 month ago

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