Dilated Steiner Inellipse

Geometry Level 5

Triangle has vertices A = ( 0 , 0 ) , B = ( 9 , 0 ) , C = ( 3 , 6 ) A=(0,0), B=(9,0), C=(3,6) . Its Steiner inellipse is dilated from triangle centroid. New ellipse trisects all three sides of the triangle A B C ABC .

Given that the dilation factor can be expressed as a b \dfrac{a}{\sqrt{b}} , where a a and b b are positive integers with b b square-free.

Find a + b a+b .


The problem is original


The answer is 5.

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2 solutions

Maria Kozlowska
Feb 2, 2016

It is a factor of 2 3 \frac{2}{\sqrt{3}} . It is the same as a factor between trisecting circle and incircle radii for an equilateral triangle. It is independent of a triangle, in other words, Steiner inellipse dilated by a factor of 2 3 \frac{2}{\sqrt{3}} from centroid will trisect sides of any triangle.

Nice problem! I love problems that require very little computation when "done right".

May I steal this problem for the upcoming sixth edition of my text "Linear Algebra with Applications"? (with citation, of course)

Otto Bretscher - 5 years, 4 months ago

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Reminder: Send me the amazon link when it's published.

Pi Han Goh - 5 years, 4 months ago

Yes, you can use it.

Maria Kozlowska - 5 years, 4 months ago

Fine. Very simple.

Yuriy Kazakov - 9 months, 1 week ago
Yuriy Kazakov
Sep 7, 2020

Equation for big ellipse 4 x 2 + 2 x y + 7 y 2 36 x 36 y + 72 = 0 4x^2 + 2xy + 7y^2 -36x -36y + 72 = 0 and

a 2 = 72 11 13 a^2=\frac{72}{11-\sqrt{13}}

b 2 = 72 11 + 13 b^2=\frac{72}{11+\sqrt{13}}

c b 2 + b 2 = a 2 c_b^2+b^2=a^2

c b = 2 13 3 c_b=\frac {2 \sqrt{\sqrt{13}}} {\sqrt{3}}

From https://brilliant.org/problems/complex-ellipse/?ref_id=1595309 find for little ellipse c l = 13 c_l= \sqrt{\sqrt{13}}

c b c l = 2 3 \frac{c_b}{c_l}=\frac{2}{\sqrt{3}}

Answer 2 + 3 = 5 2+3=5 .

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