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More Incenter Properties (Inspired by Alan Yan)

I have noticed that our fellow member has been posting a few proof challenges regarding the incircle and excircle configurations, and I thought why not join him and spread more awesome properties as challenges. Perhaps we could eventually compile them into a wiki?

Let the incircle of \(ABC\) touch \(BC,AC,AB\) at \(D,E,F\), \(M\) is the midpoint of \(BC\). If \(AM\cap EF=K\), prove that \(KD\perp BC\).

Note by Xuming Liang
1 year ago

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@Alan Yan There are proofs of this property that involve harmonic division+poles and polar, as well as law of sines. Here I present a fairly simple synthetic solution:

Let \(I\) be the incenter and define \(DI\cap EF=K'\), since \(ID\perp BC\), it suffices to prove \(K=K'\).

Through \(K'\) construct the line perpendicular to \(DK'\) that intersects \(AB,AC\) at \(Y,X\). Since \(IF\perp AB, IE\perp AC\), \(IK'YF, IK'EX\) are cyclic quadrilateral. Hence \(\angle K'YI=\angle EFI=\angle FEI=\angle K'XI\), which means \(IX=IY\). Since \(IK'\perp XY\), so \(K'\) is the midpoint of \(XY\). Because \(XY\parallel BC\), \(A,K',M\) are collinear and thus \(K=AM\cap EF=K'\). Q.E.D Xuming Liang · 11 months, 4 weeks ago

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This solution is "not-so-elegant". I will try to find one when I have time.

The "easy to find points" are practically crying out barycentric coordinates. Letting \(A = (1, 0, 0) , B = (0,1,0), C=(0,0,1)\), we have \[\begin{align*} D & = (0 : s-c: s-b)\\ E & = (s-c:0:s-a) \\ F & =(s-b : s-a : 0)\\ M & = (0:1:1) \\ \end{align*}\] We can find easily the line equations for \(AM\) and \(FE\). \[ \begin{align*} AM&: y = z \\ FE&: (a-s)x + (s-b)y + (s-c)z = 0 \\ \end{align*} \] Thus we solve these two equations to get that \(K = (a : s-a : s-a) \). It suffices to prove that \(K\) lies on the line equation of \(ID\) where \(I\) is the incenter. This is because \(ID \perp BC\). Since \(I = (a:b:c)\) we find that the line equation for \(ID\) is \[\frac{(b-c)(s-a)}{a}x + (s-b)y + (c-s)z = 0.\]

Plugging in \(K\) we get \[ (b-c)(s-a) + (s-b)(s-a) + (c-s)(s-a) = (s-a)[(b-c) + (s-b) + (c-s)] = 0 \] which is indeed zero. Thus, because Geometry is Algebra :P, we are done. Alan Yan · 1 year ago

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