Conversion of Algebra To Geometry is an art

Geometry Level 5

Let a , b R a,b \in R such that E E is minimum where E = a 2 + b 2 + ( a 12 ) 2 + b 2 + a 2 + ( b 5 ) 2 E=\sqrt { { a }^{ 2 }+{ b }^{ 2 } } +\sqrt { { (a-12) }^{ 2 }+{ b }^{ 2 } } +\sqrt { { a }^{ 2 }+{ (b-5) }^{ 2 } } Let E min 2 = m + n l { E }_{ \text{min} }^{ 2 }=m+n\sqrt { l } where m , n m,n are co-prime integers and l l is a square free integer. Then, find the value of m + n + l m+n+l .

Hint:

Don't Use Calculus since here it is Poisonous for you.

This is Part of set Click here .


The answer is 232.

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2 solutions

Deepanshu Gupta
Sep 13, 2014

I a m w a i t i n g f o r o t h e r s t o p o s t a f u l l s o l u t i o n f o r t h i s c o o l p r o b l e m . I c a n g i v e y o u h i n t t h a t a s s u m e a n r i g h t a n g l e d T r i a n g l e A B C i n X Y p l a n e a n d l e t A ( 0 , 5 ) & B ( 12 , 0 ) & C ( 0 , 0 ) . N o w l e t a p o i n t P ( a , b ) T h u s w e h a v e t o c a l c u l a t e m i n i m u m v a l u e o f E = P A + P B + P C B y L o g i c a l l y t h i n k i n g w e c o n c l u d e t h a t P m u s t l i e i n s i d e t h e t r i a n g l e a n d m a k e e q u a l A n g l e s w i t h r e s p e c t i v e s i d e s i . e a n g l e = 120 d e g r e e N o w u s e c o s i n e l a w i n r e s p e c t i v e 3 t r i a n g l e s . A n d D o S o m e m a t h a m a t i c a l C a l c u l a t i o n a n d c o m p u t e t h e v a l u e o f E m i n 2 = 169 + 60 3 l = 169 m = 60 n = 3 l + m + n = 232 I'am\quad waiting\quad for\quad others\quad to\quad post\quad a\quad full\quad solution\quad for\\ this\quad cool\quad problem.\\ I\quad can\quad give\quad you\quad hint\quad that\quad assume\quad an\quad right\quad angled\\ Triangle\quad ABC\quad in\quad X-Y\quad plane\quad and\quad let\quad A(0,5)\\ \& \quad B(12,0)\quad \& \quad C(0,0).\\ Now\quad let\quad a\quad point\quad P(a,b)\\ Thus\quad we\quad have\quad to\quad calculate\quad minimum\\ value\quad of\quad E=\quad PA+PB+PC\\ By\quad Logically\quad thinking\quad we\quad conclude\quad that\\ 'P'\quad must\quad lie\quad inside\quad the\quad triangle\quad and\quad make\quad equal\quad \\ Angles\quad with\quad respective\quad sides\quad i.e\quad angle=120\quad degree\\ Now\quad use\quad cosine\quad law\quad in\quad respective\quad 3\quad triangles.And\quad Do\quad \\ Some\quad mathamatical\quad Calculation\quad and\quad compute\quad the\quad value\quad of\\ { E }_{ min }^{ 2 }=\quad 169\quad +\quad 60\sqrt { 3 } \\ l=169\\ m=60\\ n=3\\ \Longrightarrow l+m+n=232 .

I think you meant B(12,0). right???

Chirag Singapore - 6 years, 3 months ago

can you explain how will p make 120 degrees .Actually , I am unable to make the figure.

Raven Herd - 6 years, 6 months ago

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It's because in an acute angled triangle the F e r m a t s P o i n t Fermat's Point lies on the pint that makes an angle of 12 0 o 120^o with the lines joining the vertices of the triangle to it. For more info. you can google ,"Fermat's Point"

Kunal Gupta - 6 years, 3 months ago

Could you elaborate your calculations? I seem to miss a factor somewhere

Suhas Sheikh - 3 years ago

Does that point lies on the centroid of this triangle for which the value would be minimum

Avi Ghanshani - 5 years, 10 months ago

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No the centroid is the point that minimises the sum of the square of the distances.

Spandan Senapati - 4 years, 1 month ago
Nguyễn Phát
Sep 29, 2014
  • With D(-(5 square root of(3)/2, 5/2)
  • Using theory of 'Torricelli/Fermat Point Triangle' we have Emin=BD then Emin^2= BD^2 =....
  • Actually, B(12,0)

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