contest-matheuclidean-geometry
I simplified this from a problem in IMO Shortlist:
The incircle β¦ of the acute-angled triangle π΄π΅πΆ is tangent to π΅πΆ at πΎ. Let π΄π· be an altitude of triangle π΄π΅πΆ and let π be the midpoint of π΄π·. If π is the other common point of β¦ and πΎπ, prove that NK is the bisector of angle BNC.
I have no idea on this. I tried by analytic and trigonometric way but the incircle radius is too complicated. Are there any euclidean tricks to solve it?
This link may be useful.There are quite a few solutions there.
In fact,this was the toughest problem at the IMO 2011.It was G8 on the Shortlist,meaning a hard problem. The IMO committee actually ended up misjudging the difficulty of the problems,as evident from the way they were numbered on the Shortlist.
Let us consider the triangle $ABC$. Since the triangle is acute, the altitude from $B$ necessarily intersects $AC$ internally, i.e. $E$ is between $A$ and $C$.
Let the lengths and angles of the triangle be labelled according to the figure above.
From the angular bisector theorem, we have $$\frac{c}{b} = \frac{m}{n}$$ where $m = |BD|$ and $n = |CD|$. Likewise, from the generalized angular bisector theorem, we have $$\frac{h\sin\beta}{k\sin\alpha} = \frac{m}{n}$$ where $h$ is the length of the altitude from $B$ and $k=|CE|$. Note that the claim $\alpha > 45^\circ$ is equivalent to $\alpha > \beta$ since $\alpha + \beta = 90^\circ$. Therefore the claim is true if and only if $\alpha > \beta$ if and only if $\sin\alpha > \sin\beta$. Therefore the problem is equivalent to proving $$\frac{c}{b} = \frac{h\sin\beta}{k\sin\alpha} < \frac{h}{k}$$ Let $hb = \Delta$ denote the area of the triangle. Then equivalently $$k < \frac{\Delta}{c} = h_c$$ where $h_c$ is the length of the altitude from $c$. The above inequality is easily seen to be true. Consider swinging the length $CE$ around $C$ to form a circle of radius $k$ centered at $C$. Then the line $BE$ is tangent to the circle at $E$ and clearly separates the circle and the side $AC$. The claim follows.
Best Answer
I will try to give two solutions, one solution is purely geometric, one is (very) analytic in its nature and its computational aspect. If the geometric solution is known, who needs an other one? Well, i am writing a book on computer algebraic support for proofs of results of all kind, and this is a wonderful example. With a computational engine the proof is easy and algorithmic. (Although it gives no geometric insight and destroys the beauty of the problem.)
The geometric solution.
Let $a,b,c$ be as usual the lengths of the sides in $\Delta ABC$, and let $s=\frac 12(a+b+c)$ be the semiperimeter. Then $BK=s-c$ and $CK=s-b$. We have to show that $NK$ is the angle bisector in $\Delta NBC$. To get the idea for the (construction involved in the) proof, let us step back and ask for the geometric locus of all points $P$ having "this property", i.e. the locus of $P$ such that $PK$ is the angle bisector in $\Delta PBC$. By cartesian considerations, letting $P=P(x,y)$ have the coordinates $x,y$, we obtain the equation (with obvious notations of the coordinates for $B,C$...) $$ \begin{aligned} \frac{PB^2}{PC^2} &= \frac{KB^2}{KC^2}\ ,\qquad\text{i.e.}\\ \frac {(x-x_B)^2+(y-y_B)^2} {(x-x_C)^2+(y-y_C)^2} &= \frac{(s-b)^2}{(s-c)^2} \end{aligned} $$ In case of $b=c$ the triangle is isosceles, the lines $AK$ and $AD$ coincide with the side bisector of $BC$, so $N$ is on this side bisector, and the result is easy. Else, $b\ne c$, and we may and do assume $b>c$, so that we are in the situation in the picture. Then the above equation in $(x,y)$ has degree two, the term $xy$ does not appear, and the terms in $x^2$ and $y^2$ have the same non-vanishing coefficient. The geometric locus of $P$ is thus a circle. By the reflection symmetry w.r.t. the line $BC$ for the property of $P$, this circle is symmetric w.r.t. $BC$, so its center $\Omega$ is on $BC$. Here is the picture so far:
The point $K\in BC$ is one point on this circle. Which is the other point $K'\in BC$? It is a point satisfying $$ \frac{K'B}{KB} = \frac{K'C}{KC} = \frac{K'C-K'B}{KC-KB} = \frac{a}{(s-c)-(s-b)} = \frac{a}{b-c}\ . $$ This implies $$ \frac{s-b}{b-c} =\frac{BK}{b-c} =\frac{K'B}a =\frac{K'B+BK}{a+b-c} =\frac{K'K}{2(s-c)}\ , $$ so the radius of the circle $(\Omega)$ is $$ \Omega K =\frac 12 K'K=\frac{(s-b)(s-c)}{b-c}\ . $$ Let us now address the point $N$. The given problem wants us to show that $N$ is on the circle $(\Omega)$. It turns out (and this can be immediately observed in a faithfull picture), that $\Omega K$ and $\Omega N$ are two equal segments, and (their lines) are tangent to the incircle $(I)$ in $K$, respectively $N$. Let us show this. This property is equivalent to $$ \Omega I\perp KMN\ , $$ and in this version, $\Omega I\perp KM$ we have eliminated $N$ from the discussion, and need to show a property in terms of the less involved point $M$. This is equivalent to $\angle MKD=\angle \Omega IK$, so let us compute using standard notations in a triangle $$ \begin{aligned} \cot\widehat{MKD} & =\frac{DK}{DM} =\frac{BK-DK}{AD/2} =\frac{a((s-b)-c\cos B)}{a\cdot AD/2} \\ &=\frac1{[ABC]} \cdot \frac 12\left(\ a(a+c-b)-(a^2+c^2-b^2)\ \right) \\ &=\frac1{[ABC]} \cdot \frac 12(b+c-a)(b-c) =\frac{(s-a)}{[ABC]} \cdot(b-c)\ . \\ \cot\widehat{\Omega IK} &=\frac{IK}{\Omega K} =\frac{r(b-c)}{(s-b)(s-c)} =\frac{[ABC]}{s(s-b)(s-c)}\cdot(b-c) \ . \end{aligned} $$ The two cotangents are equal because of the formula for the squared area (Heron): $$ [ABC]^2 = s(s-a)(s-b)(s-c)\ . $$ $\square$
We are done, but... The OP comes with an effort to solve the problem by using analytic tools, however the complexity is overwhelming. For this and only this reason, let us pass the maths & programmers Olympiad, so here is a solution using barycentric coordinates, b.c. for short in the sequel.
Take a look at bary-short.pdf for a quick, excellent introduction. Here are some words on b.c. - the reader may please skip the passage if this is no issue at all.
The normed b.c. of a point $P$ is a triple $(x,y,z)$ so that (more or less formally) $P=xA+yB+zC$. (E.g. by passing to complex numbers, or using this as a short cut for the vectorial writing $OP=xOA+yOB+zOC$, for some/any $O$.) Here, normed means $x+y+z=1$. Sometimes we relax this condition, denote the coresponding (possibly unnormed, projective) b.c. by $[x:y:z]$, and there is a simple passage to the normed b.c. by dividing the components $x,y,z$ by $(x+y+z)$, so this makes sense only in case $x+y+z\ne 0$. Using projective coordinates makes sense, since all geometric objects (like lines, circles, etc.) have a corresponding homogeneous equation in the b.c. unknowns $x,y,z$. OK, let's start.
Let $a,b,c$ be the sides of the given triangle $\Delta ABC$. Let $s=\frac 12(a+b+c)$ be the semiperimeter. Then the incircle has the equation
$$ \tag{$\diamondsuit$} a^2yz + b^2zx + c^2xy = (x+y+z)\left(\ (s-a)^2x + (s-b)^2y + (s-c)^2z \ \right)\ . $$
The needed points are:
$$ \begin{aligned} A &= (1,0,0)\ ,\\ B &= (0,1,0)\ ,\\ C &= (0,0,1)\ ,\\ K &= [0:s-c:s-b]\ ,\qquad\text{ and it verifies $(\diamondsuit)$ ,}\\ D &= [0:\tan B:\tan C]\\ &=\left[0\ :\ \frac{\sin B}{\cos B}\ :\ \frac{\sin C}{\cos C}\ \right]\\ &=\left[0\ :\ \frac{b}{(a^2+c^2 -b^2)/(2ac)}\ :\ \frac{c}{(a^2+b^2-c^2)/(2ab)}\ \right]\\ &=[0\ :\ a^2 +b^2-c^2\ :\ a^2-b^2 +c^2\ ]\ ,\\ &=\left(0\ ,\ \frac 1{2a^2}(a^2 +b^2-c^2)\ ,\ \frac 1{2a^2}(a^2-b^2 +c^2)\ \right)\ ,\\ M &= \frac 12(A+D)\\ &=\left(\frac 12\ ,\ \frac 1{4a^2}(a^2 +b^2-c^2)\ ,\ \frac 1{4a^2}(a^2-b^2 +c^2)\ \right)\ ,\\ &=[2a^2\ :\ a^2+b^2-c^2\ :\ a^2-b^2+c^2]\\ &=\left[a^2\ :\ \frac 12(a^2+b^2-c^2)\ :\ \frac12(a^2-b^2+c^2)\right]\\ &=[a^2\ :\ S_C\ :\ S_B]\ . \end{aligned} $$
The last line comes with the Conway notations ($S_B$ and $S_C$, defined as used). Now after solving the equations $(\diamondsuit)$ and the equation of the line $KM$
$$ \tag{$\dagger$} 0 = \begin{vmatrix} x&y&z\\ 0&s-c&s-b\\ a^2 & S_C & S_B \end{vmatrix} $$
it turns out that $N$ has (against my expectation) a rather "simple" expression (in homogeneous, unnormed b.c.): $$ N = [x_N:y_N:z_N] = [ a^2(s-b)(s-c)\ :\ (s-a)(s-c)^3\ :\ (s-a)(s-b)^3]\ . $$ Let us explicitly check $(\dagger)$ for $N$. The relation $\sim$ used below means "up to a (known) factor". $$ \begin{aligned} & \begin{vmatrix} x_N&y_N&z_N\\ 0&s-c&s-b\\ a^2 & S_C & S_B \end{vmatrix} = \begin{vmatrix} a^2(s-b)(s-c) & (s-a)(s-c)^3 & (s-a)(s-b)^3 \\ 0&s-c&s-b\\ a^2 & S_C & S_B \end{vmatrix} \\ &\qquad \sim \begin{vmatrix} (s-b)(s-c) & (s-a)(s-c)^3 & (s-a)(s-b)^3 \\ 0&s-c&s-b\\ 1 & S_C & S_B \end{vmatrix} \\ &\qquad \sim \begin{vmatrix} 0 & (s-a)(s-c)^3 - (s-b)(s-c)S_C & (s-a)(s-b)^3 - (s-b)(s-c) S_B\\ 0&s-c&s-b\\ 1 & * & * \end{vmatrix} \\ &\qquad \sim \begin{vmatrix} (s-a)(s-c)^2 - (s-b)S_C & (s-a)(s-b)^2 - (s-c) S_B\\ 1 & 1\\ \end{vmatrix} \\ &\qquad \sim (s-a)((s-c)^2-(s-b)^2) -s(S_C-S_B) + (bS_C-cS_B) \\ &\qquad \sim (s-a)a(b-c) -s(b-c)(b+c) + \frac 12(a^2+b^2+c^2+2bc)(b-c) \\ &\qquad \sim a(s-a) - \frac 12(a+(b+c))(b+c) + \frac 12(a^2+b^2+c^2+2bc) \\ &\qquad \sim a(s-a) - \frac 12a(b+c) + \frac 12a^2 \\ &\qquad \sim (s-a) - \frac 12(2s-a) + \frac 12a \\ &\qquad =0\ . % % % \\[3mm] &y_N + z_N \\ &\qquad =(s-a)[(s-c)^3+(s-b)^3]\\ &\qquad =(s-a)(2s-b-c)[(s-b)^2 -(s-b)(s-c) + (s-c)^2]\\ &\qquad =a(s-a)\cdot \frac 14[ a^2 + 3(b-c)^2] \\[3mm] &x_N + y_N + z_N \\ &\qquad = a(s-a)\cdot \frac 14[ a^2 + 3(b-c)^2] + a^2(s-b)(s-c)\ , \\[3mm] &(s-a)^2x_N + (s-b)^2y_N + (s-c)^2z_N \\ &\qquad = -\frac 14 a(s-a)(s - b)(s - c) (a^2 + b^2 + c^2 - 2ab - 2bc - 2ca) \\[3mm] &a^2y_Nz_N + b^2z_Nx_N + c^2x_Ny_N \\ &\qquad =a^2(s-a)^2(s-b)^3(s-c)^3 + a^2b^2(s-a)(s-b)^4(s-c) + a^2c^2(s-a)(s-b)(s-c)^4\\ &\qquad =a^2(s-a)(s-b)(s-c)\Big(\ (s-a)(s-b)^2(s-c)^2 + b^2(s-b)^3 +c^2(s-c)^3\ \Big)\ , \end{aligned} $$ and it turns out that in the last parenthesis one can factor $(a^2 + b^2 + c^2 - 2ab - 2bc - 2ca)$, and finally $(\diamondsuit)$ is satisfied.
It remains to compute the squared distances $NB^2$ and $NC^2$, and to show the corresponding proportionality $NB^2:NC^2=KB^2:KC^2=(s-b)^2:(s-c)^2$. Let us compute: $$ \begin{aligned} N &= [x_N:y_N:z_N] =\left(\ \frac{x_N}{x_N+y_N+z_N}\ ,\ \frac{y_N}{x_N+y_N+z_N}\ ,\ \frac{z_N}{x_N+y_N+z_N}\ \right)\ , \\ B &=(0,1,0)\ , \\ \overrightarrow{BN} &= N-B\\ &=\left(\ \frac{x_N}{x_N+y_N+z_N}\ ,\ \frac{y_N}{x_N+y_N+z_N}\ ,\ \frac{z_N}{x_N+y_N+z_N}\ \right)-(0,1,0)\\ &=\left(\ \frac{x_N}{x_N+y_N+z_N}\ ,\ \frac{-x_N-z_N}{x_N+y_N+z_N}\ ,\ \frac{z_N}{x_N+y_N+z_N}\ \right) \ , \\ NB^2 &=-\frac {a^2(-x_N-z_N)z_N + b^2z_Nx_N + c^2x_N(-x_N-z_N) }{(x_N+y_N+z_N)^2} \\ \overrightarrow{CN} &= N-C\\ &=\left(\ \frac{x_N}{x_N+y_N+z_N}\ ,\ \frac{y_N}{x_N+y_N+z_N}\ ,\ \frac{z_N}{x_N+y_N+z_N}\ \right)-(0,0,1)\\ &=\left(\ \frac{x_N}{x_N+y_N+z_N}\ ,\ \frac{y_N}{x_N+y_N+z_N}\ ,\ \frac{-x_N-y_N}{x_N+y_N+z_N}\ \right) \ , \\ NC^2 &=-\frac {a^2y_N(-x_N-y_N) + b^2(-x_N-y_N)x_N + c^2x_Ny_N }{(x_N+y_N+z_N)^2} \\ \end{aligned} $$ So we have to show the equality: $$ \ (s-c)^2\Big(\ a^2(x_N+z_N)z_N - b^2x_Nz_N + c^2(x_N+z_N)x_N\ ) \\ = (s-b)^2\Big(\ a^2(x_N+y_N)y_N + b^2(x_N+y_N)x_N -c^2x_Ny_N\ ) \ . $$ This can be shown by computing the above as a polynomial in $x_N$...
Let us give the short computer proof on the above lines.
The above sage code delivers:
It is also easy to check distances proportion.
And we get:
$\square$