Feuerbachs problem

Geometry Level 3

if the inradius of a triangle is half the radius of the circumcircle then the triangle must be $\text{\_\_\_\_\_\_\_\_\_\_\_}$ .

equilateral isosceles right angled not necessarily any of the foregoing types

1 solution

A Former Brilliant Member
Sep 9, 2016

by Feuerbach's theorem , the incircle touches the nine-point circle and the distance between the centre of the nine-point circle and the incenter(centre of incircle) is given by R/2-r where R is the radius of the circumcircle and r is the inradius. since r=R/2, the incircle and the NPC are one and the same. the NPC passes through the feet of the altitudes and the midpts of the 3 sides of a triangle . however , the incircle only touches the three sides ie. it passes through one point on each side. therefore , the foot of the altitude and the the midpt of the side are one and the same so the median to any side will be perpendicular to the side . the medians are perpendicular bisectors of the three sides of the triangle. since these perpendicular bisectors pass through the three vertices of the triangle, the triangle is equilateral

Just use Euler's theorem of geometry:

The distance $d$ between the circumradius $R$ and the incenter $r$ satisfy $d^2 =R(R - 2r)$ , since $r = \dfrac12 R$ , then $d = 0$ . This means that the circumradius and the incenter of this triangle lies on the same point. This only happens when the triangle is equilateral.

Pi Han Goh - 4 years, 9 months ago

yes, this is the solution I just explained it in detail so that it would be clear . its a 2 liner solution you are correct. actually ,you don't need Euler's theorem Feuerbach's theorem is enough to prove it in a single line ::::::: by Feuerbach's theorem the circumradius and the incenter of this triangle lie on the same point. therefore the triangle is equilateral done! its an MCQ , you have to think fast!!!

A Former Brilliant Member - 4 years, 9 months ago

Feuerbachs problem

1 solution

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