Triangle defined by one point II

Geometry Level 5

Point O ( 1 , 2 , 3 ) O (1,2,3) is a circumcentre of a triangle. Vertices of a triangle are on the axes of the coordinate system. R R is circumradius of the triangle.

Find ⌊ R × 1000 ⌋ \lfloor R \times 1000 \rfloor


The answer is 4844.

Maria Kozlowska by Maria Kozlowska , 57, Canada

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Maria Kozlowska
Aug 4, 2015

Let A ( a , 0 , 0 ) , B ( 0 , b , 0 ) , C ( 0 , 0 , c ) A (a,0,0) , B (0,b,0), C(0,0,c) denote vertices of the triangle. ∣ O A ⃗ ∣ = ∣ O B ⃗ ∣ = ∣ O C ⃗ ∣ = R ⇒ |\vec{OA} |=|\vec{OB} |=|\vec{OC}| =R \Rightarrow

a ² − 2 a + 14 = R 2 a² - 2a + 14 = R^2

b ² − 4 b + 14 = R 2 b² - 4b + 14 = R^2

c ² − 6 c + 14 = R 2 c² - 6c + 14 = R^2

Three vectors need to be coplanar ⇒ O A ⃗ ∗ ( O B ⃗ × O C ⃗ ) = 0 ⇒ a b c − 3 a b − 2 a c − b c = 0 \Rightarrow \vec{OA} * (\vec{OB} \times \vec {OC})=0 \Rightarrow abc-3ab-2ac-bc=0

Geogebra gives the following as solutions for the set of the 4 equations above: ( a = 0 , b = 0 , c = 0 , r = − 14 ) , ( a = 0 , b = 0 , c = 0 , r = 14 ) , ( a = 0 , b = 0 , c = 6 , r = − 14 ) , ( a = 0 , b = 0 , c = 6 , r = 14 ) , ( a = 0 , b = 4 , c = 0 , r = − 14 ) , ( a = 0 , b = 4 , c = 0 , r = 14 ) , ( a = 2 , b = 0 , c = 0 , r = − 14 ) , ( a = 2 , b = 0 , c = 0 , r = 14 ) , ( a = 4.235088532729 , b = 5.669577334592 , c = 7.297184870884 , r = − 4.844150886878 ) , ( a = 4.235088546276 , b = 5.669577381281 , c = 7.297184929667 , r = 4.844151037412 ) {(a = 0, b = 0, c = 0, r = -\sqrt{14}), (a = 0, b = 0, c = 0, r = \sqrt{14}), (a = 0, b = 0, c = 6, r = -\sqrt{14}),(a = 0, b = 0, c = 6, r = \sqrt{14}),(a = 0, b = 4, c = 0, r = -\sqrt{14}),(a = 0, b = 4, c = 0, r = \sqrt{14}),(a = 2, b = 0, c = 0, r = -\sqrt{14}),(a = 2, b = 0, c = 0, r = \sqrt{14}),(a = 4.235088532729, b = 5.669577334592, c = 7.297184870884, r = -4.844150886878),(a = 4.235088546276, b = 5.669577381281, c = 7.297184929667, r = 4.844151037412)} The only acceptable solution is: 4.844150886851 4.844150886851

Alternatively one can try to solve the following: ( R ² − 5 ) ( R ² − 10 ) ( R ² − 13 ) − 2 R ² − 5 − 3 R ² − 10 − 6 R ² − 13 − 12 = 0 \sqrt{(R² - 5)(R² - 10)(R² - 13)} - 2\sqrt{R² - 5} - 3\sqrt{R² - 10} - 6\sqrt{R² - 13} - 12 = 0

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Hm, this seems incomplete to me.

Calvin Lin Staff - 5 years, 10 months ago

Log in to reply

I do get the same answer 4844, but I had to resort to pretty messy numerical approximations. Let me compare my method with hers later on.

Edit: Here's how I did this (after cleaning up a bit)

Let ( x , 0 , 0 ) , ( 0 , y , 0 ) , ( 0 , 0 , z ) \left( x,0,0 \right) , \left( 0,y,0 \right) ,\left( 0,0,z \right) be the coordinates of the vertices of the triangle, and r r be the radius of the circumcircle of the triangle, the center being at ( 1 , 2 , 3 ) \left(1, 2, 3\right) . Then we know that

( x − 1 ) 2 + ( 0 − 2 ) 2 + ( 0 − 3 ) 2 = r 2 { \left( x-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 }+{ \left( 0-3 \right) }^{ 2 }={ r }^{ 2 }
( 0 − 1 ) 2 + ( y − 2 ) 2 + ( 0 − 3 ) 2 = r 2 { \left( 0-1 \right) }^{ 2 }+{ \left( y-2 \right) }^{ 2 }+{ \left( 0-3 \right) }^{ 2 }={ r }^{ 2 }
( 0 − 1 ) 2 + ( 0 − 2 ) 2 + ( z − 3 ) 2 = r 2 { \left( 0-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 }+{ \left( z-3 \right) }^{ 2 }={ r }^{ 2 }

and, because the plane that passes through the vertices passes through ( 1 , 2 , 3 ) \left(1, 2, 3\right)

1 x + 2 y + 3 z = 1 \dfrac { 1 }{ x } +\dfrac { 2 }{ y } +\dfrac { 3 }{ z } =1

we end up with the equation to solve for r r directly, by numerical means

1 1 + r 2 − 13 + 2 2 + r 2 − 10 + 3 3 + r 2 − 5 = 1 \dfrac { 1 }{ 1+\sqrt { { r }^{ 2 }-13 } } +\dfrac { 2 }{ 2+\sqrt { { r }^{ 2 }-10 } } +\dfrac { 3 }{ 3+\sqrt { { r }^{ 2 }-5 } } =1

resulting in r = 4.84415088685 … r=4.84415088685…

Using this approach gets a clean plot with only one positive real solution for r r

Michael Mendrin - 5 years, 10 months ago

Log in to reply

My confusion is over how he went from the 2nd last line to the last line. I understand that a − 1 = R 2 − 13 a - 1 = \sqrt{ R^2 - 13 } . However, I'm not sure how all of that substitution works into the final result that he stated.

Calvin Lin Staff - 5 years, 10 months ago

Log in to reply

@Calvin Lin – Calvin

I looked for simple solution but so far I came short. You are right. Those equations get so huge and proper analysis should account for all the cases including ( a < 1 , b < 2 , c < 3 (a<1, b<2, c<3 etc. I did it using software and the value given is the only non-generate case. I leave the rest of the work to some algebra pros :).

Maria Kozlowska - 5 years, 10 months ago

Log in to reply

@Maria Kozlowska – Actually, without proof, I do believe that given any arbitrary point (at least in a solid which extends into infinity) not on any of the axes, there can only be one such circumcircle and radius. The rest of the solutions are imaginary and clutter up the landscape. I can see that proving this could be interesting.

Let me get back to you on this subject, on how extensive this solid is.

Edit; Yes, any point that is not on any of the axes has an unique real solution for the circumcircle and radius.

Michael Mendrin - 5 years, 10 months ago

Log in to reply

@Michael Mendrin – Thanks for checking this. What I really tried to do is the question on locus of all circumcentres for circumradius = 1. The equation I got did not look encouraging so I gave up on the idea and posted this question instead as a simpler version.

The equation is this: − 2 x y z − x y − x ² − y ² + 1 − x z − x ² − z ² + 1 − y z − y ² − z ² + 1 + − x ² − y ² + 1 − x ² − z ² + 1 − y ² − z ² + 1 = 0 -2 x y z - x y \sqrt{-x² - y² + 1} - x z \sqrt{-x² - z² + 1} - y z \sqrt{-y² - z² + 1} + \sqrt{-x² - y² + 1} \sqrt{-x² - z² + 1} \sqrt{-y² - z² + 1} = 0

I can not even plot it. I was wondering what it looks like and how is it related to the ellipsoid you got.

Maria Kozlowska - 5 years, 10 months ago

Log in to reply

@Maria Kozlowska – The locus of all the circumcenters for circumradius = 1 is related to the problem of the Trirectangular Corner Locus. Is that going to be a new problem coming up?

Michael Mendrin - 5 years, 10 months ago

Log in to reply

@Michael Mendrin – Once I find an answer to it.

Maria Kozlowska - 5 years, 10 months ago

Thank you. These look cleaner than the ones I posted.

Maria Kozlowska - 5 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...