Prove IZ is perpendicular to AN

Question

Given $\Delta ABC$. A circle pass through $B,C$ intersects $AC,AB$ at $E.F$. Let $N$ be the midpoint of $EF$. $(ANE) \cap AB=X, (ANF) \cap AB=Y$. A line through $A$ and parallel to $BC$ intersects $EF$ at $Z$. Let $I$ be the center of $(AXY)$. Prove $IZ \perp AN$.

Answer 1

First of all, let us restate the problem, so that all "simple" points are removed from the picture, and there is slightly more air on the paper to have a deep breath before going into matter. Consider for this the following picture, which adds the measures $\alpha$, $\beta$, $\gamma$ of the angles in -respectively- $A,B,C$ in the triangle $\Delta ABC$, and finds them again in the one or other place.

Above, we have $FBCE$ "inscriptible" (a cyclic / inscribed quadrilateral), so

• its exterior angle in $F$ is equal (in measure, congruent) to the one in its interior angle in $C$, $\gamma$,
• its exterior angle in $E$ is equal to the one in its interior angle in $B$, $\beta$.

The same argument in the "inscriptibles" $AENX$ and $AFNY$ shows the depicted angles in the two triangles $\Delta NFX$ and $\Delta NEY$, so we have the similitude (with exact corresponding order of the vertices) $$ \Delta NFX \sim \Delta NYE\ ,\qquad \text{ so } \frac {NF}{NY} = \frac {NX}{NE} \ , \qquad \text{ so } NE^2=NF^2=NX\cdot NY\ . $$

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Parenthesis, please skip it at first reading.

We prepare the terrain to apply an inversion $\mathcal I = \mathcal(N,NE^2)$ with center in $N$ and above factor, the common value of the two squares $NE^2=NF^2$ and of the product $NX\cdot NY$. Psychologically, if such a problem sparks and scatters in an Olympiad contest, there are only two viable choices of the center of inversion, $A$ and $N$, the only points with many circles through them and with a "symmetrical rôle". In such case one should always choose the "simpler point" first. The "simpler point" is the point with a lower level of complexity, as introduced in the problem, it would be $A$, since we introduce in steps separated by semicolons the objects $A,B,C$; circle trough $B,C$; $E,F$; $N$ on the one side and $Z$ in a different way; $X,Y$; $I$. But after we restate, the order is $E,F$ free; $N$; $A$ free (here or in front); $X,Y$; $I$, so here there is a reason to consider $N$ as a possible center for a potential inversion solving the problem.

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We are in position to restate:

Problem 2845259:

Fix a triangle $\Delta AEF$ with angles in $A,E,F$ respectively $\alpha,\beta,\gamma$, and assume $\beta>\gamma$. Let $N$ be the mid point of the segment $EF$. Let $X,Y$ be on $AF$, $AE$ so that we have the similarities $$\Delta AEF \sim \Delta NEY\sim NXF\ ,$$ where the vertices come in corresponding order. Let $I$ be the center of the circle $(AXY)$. Let $Z$ be constructed on the ray $[FE$ beyond $E$ (related to $\beta>\gamma$)
so that $\angle ZAE=\gamma=\hat F$. Then $$ZI\perp AN\ .$$

The notations correspond to the one in the above picture, the circle $(ABEF)$ is no longer needed. The points $B,C$ can also be removed from it.

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Let us further restate. The problem is solved, if the following Lemma is proven.

Lemma:

Let $\Delta AEF$ be a triangle with angles in $A,E,F$ respectively $\alpha,\beta,\gamma$. (We may and do also assume $\beta>\gamma$.) Let $AN$ and $AS$ be the median and the symmedian from $A$ in $\Delta AEF$, i.e. $N$ is the mid point of $EF$ and $\angle FAN=\angle SAE=\alpha_1$,$N,S\in EF$. Let $Z$ be on line $EF$ so that $\gamma = \hat F = \angle EAZ$.

Let $A'$ be on $AN$ so that the triangle $\Delta AZA'$ is isosceles. (Equivalently, $\angle A'ZN=\alpha_2-\alpha_1$.) In particular, $Z$ is a point on the perpendicular bisector of the segment $AA'$.

Then we have the following relations:

$\boxed 1$ $$ \begin{aligned} ZA^2 &= ZE\cdot ZF = ZN\cdot ZS\ ,\\ NE^2 &= NS\cdot NZ \\ &= NA'\cdot NA\ . \end{aligned} $$ $\boxed 2$ Let $\mathcal I$ be the inversion in $N$ with power factor $NE^2$, also denoted by $W\to W'$ on points. Then $A'=\mathcal I A$, compatible with notation, $S'=Z$, $Z'=S$, and the quadrilateral $SA'AZ$ is "inscriptible".

$\boxed 3$ Let $X,Y$ be points on $AF$, respectively $AE$, so that $NX$, $NY$ sit in an angle $\alpha$ w.r.t the line $(FEN)$. Then $A$, $A'$, $X$, $Y$ lie on a circle.

It is clear, that the above solves the problem. So let us give the...

Proof:

$\boxed 1$

The triangles $\Delta ZAE$ and $\Delta ZFA$ have the same angles, (a common angle in $Z$ and a $\gamma$ angle in $\angle ZAE$ and $\angle ZFA$,) so write the proportionality $\frac{ZA}{ZF} =\frac{ZE}{ZA} =\frac{AE}{FA} $ and obtain (from the first equality) $ZA^2=ZE\cdot ZF$.

With the same argument, observing that the triangles $\Delta ZAS$ and $\Delta ZNA$ have the same angles, (a common angle in $Z$, and a $\gamma+\alpha_1$ angle in $\angle ZAS$ and in $\angle ZNA$,) we get $\frac{ZA}{ZN} =\frac{ZS}{ZA} =\frac{AS}{NA} $ and obtain (from the first equality) $ZA^2=ZS\cdot ZN$.

For the other relations we need to use the point $A'$. Let us write down some angles, obtained consecutively: $$ \begin{aligned} \angle A'AZ &= \gamma +\alpha_2\ ,\\ \angle AA'Z &= \gamma +\alpha_2\ ,\ \text{ since $\Delta AA'Z$ isosceles,}\\ \angle AZA' &= \pi-2(\gamma +\alpha_2)\\ &=(\alpha_1+\alpha_2+\beta+\gamma)-2(\gamma +\alpha_2)\\ &=(\beta-\gamma)-(\alpha_2 -\alpha_1)\ ,\\ \angle A'ZN &=\angle AZN -\angle AZA'\\ &=\alpha_2-\alpha_1\\ &=\angle A'AS\ , \end{aligned} $$ so the quadrilateral $(AA'SZ)$ is inscriptible. The power of $N$ w.r.t. the circle $(AA'SZ)$ is then $$ NS\cdot NZ=NA\cdot NA'\ .$$

It remains to connect these products with $NE^2$.

For this, we will first show the listed trigonometric realtions which are breaking the beauty. (I decided to do so, in order to show which is the type of metric relation we need, the complexity of this realtion also mirrors the complexity of the problem.) The first equality is simple, using the sine theorem, which claims that the proportion of two sides in a riangle is to the proportion of the sines of the corresponding opposite angles. So $$ \frac{\sin\alpha_1}{\sin\gamma} = \frac{FN}{AN} = \frac{EN}{AN} = \frac{\sin\alpha_2}{\sin\beta}\ . $$ For the second one, let us use the following notations in the triangle $\Delta AFE$. The sides in it opposite to the vertices $A,F,E$ will be denoted by $a,f,e$. Let $m=AN$ be the length of the $A$-median. We will try to rewrite the second trigonometric equality equivalently by using only $a,e,f$, and at the beginning also $m$.

(There is also a formula for $m$, obtained by applying the generalized formula of Pythagoras in $\Delta AFN$ and $\Delta AEN$, explicitly $m^2+(a/2)^2 - f^2 = +2m(a/2)\cos(\dots)$, and $m^2+(a/2)^2 - e^2 = -2m(a/2)\cos(\dots)$, lead to $2m^2=e^2+f^2-\frac 12 a^2$. But we do not need this.)

So let us compute, using only these simple ideas: $$ \begin{aligned} \sin(\alpha_2-\alpha_1) &=\sin\alpha_2\cos\alpha_1-\sin\alpha_1\cos\alpha_2 \\ &= \sin\beta\frac{a/2}m\cdot\frac{e^2+m^2-a^2/4}{2em} - \sin\gamma\frac{a/2}m\cdot\frac{f^2+m^2-a^2/4}{2fm} \\ &= \sin\alpha \cdot\frac{e^2+m^2-a^2/4}{4m^2} - \sin\alpha \cdot\frac{f^2+m^2-a^2/4}{4m^2} \\ &= \sin\alpha \cdot\frac{e^2-f^2}{4m^2}\ , \\[2mm] \sin(\beta-\gamma) &= \sin\beta\cos\gamma-\sin\gamma\cos\beta \\ &= \sin\alpha\frac ea\cdot\frac 1{2ae}(e^2+a^2-f^2) - \sin\alpha\frac fa\cdot\frac 1{2af}(f^2+a^2-e^2) \\ &= \sin\alpha\cdot\frac{e^2-f^2}{a^2}\ , \\[2mm] \frac{\sin(\alpha_2-\alpha_1)}{\sin(\beta-\gamma)} &= \frac{a^2}{4m^2} = \frac {(a/2)^2}{m^2} \\ &= \frac{NE^2}{AN^2} = \frac{NE\cdot NF}{AN^2} = \frac{NF^2}{AN^2} \\ &= \frac{\sin ^2\alpha_1}{\sin^2\gamma} = \frac{\sin\alpha_1\sin\alpha_2}{\sin\gamma\sin\beta} = \frac{\sin ^2\alpha_2}{\sin^2\beta} \ . \end{aligned} $$ The above is routine, no geometrical idea was involved. This is needed to show now $NA\cdot NA'=NE^2$. So let us compute it: $$ \begin{aligned} \frac{NA}{NE} \cdot \frac{NA'}{NE} &= \frac{\sin\beta}{\sin\alpha_2} \cdot \frac{NA'}{ZA'}\cdot \frac{ZA'}{ZA}\cdot \frac{ZA}{AE}\cdot \frac{AE}{NE}\\ &= \frac{\sin\beta}{\sin\alpha_2} \cdot \frac{\sin(\alpha_2-\alpha_1)}{\sin(\gamma+\alpha_1)}\cdot \frac 11 \cdot \frac{\sin\beta}{\sin(\beta-\gamma)} \cdot \frac{\sin(\gamma+\alpha_1)}{\sin\alpha_2} \\ &=\frac{\sin(\alpha_2-\alpha_1)}{\sin(\beta-\gamma)} \cdot \frac{\sin^2\beta}{\sin^2\alpha_2} \\ &=1\ . \end{aligned} $$

$\boxed 2$

This follows from $\boxed 1$, since the points $A,A'$ correspond each other via $\mathcal I$ because of $NA\cdot NA'=NE^2$. The relation $NE^2=NS\cdot NZ$ shows that $S,Z$ are interchanged by $\mathcal I$.

$\boxed 3$

Consider $(s)\perp ENF$, $N\in(s)$. Let $Y'$ be the symmetrical of $X$ with respect to $(s)$. Since the rays $NX$, $NY$ make the same angle $\alpha$ with $EF$, $Y'$ is on $NY$, and $\angle Y'EN=\gamma=\angle NYE$, $\angle YEN=\beta=\angle NY'E$, so we have the similarity $$ \Delta NYE\sim \Delta NEY'\ , $$ which implies $$NY\cdot NY'=NE^2\ ,$$ so the notation $Y'$ is consistent with the notation for the image of $Y$ via $\mathcal I$, and from $NA\cdot NA'=NY\cdot NY'$ it follows $A, A', Y, Y'$ are on a circle. Analogously, let $X'$ be the mirror of $Y$ w.r.t $(s)$, then $NA\cdot NA'=NX\cdot NX'$.

The point $A$ is on the circle circumscribed to the the trapez $XY'YX'$, since $\angle XAY=\alpha=\angle XNF=\angle XX'Y$, so $A,X',X,Y$ are on a circle.

The points $A, A';X,X'; Y,Y'$ are thus on the same circle.

$\square$

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Postlude:

The above was the straight line to the solution. The problem is already solved, but since the landscape of the problem is very rich, here are some further thoughts.

We need a new picture. The inversion $\mathcal I=\mathcal I(N,NE^2)$ is a central vehicle in the following. For a general point $W$ let $W'$ be the tacitly chosen notation for its image $\mathcal IW=:W'$ through this inversion. Then $E'=E$ and $F'=F$ are already in the picture. Recall the situation from the problem, as restated.

We need first $X'$. It is the point on the ray $[NX$ so that $NX'=NY$. "Same" for $Y'$. As constructed, $X', Y'$ are the symmetric points to $Y,X$ w.r.t. the perpendicular in $N$ on $EF$! Let $\mathcal S$ be the name of this transformation, and let $W\to W^*=\mathcal SW$ be the short notation for the image of $W$ throuhg it.

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So let us restate again and redraw.

Let $EF$ be a segment. Let $N$ be its mid point. Let $(s)$ be the perpendicular bisector of the segment $EF$, i.e. $N\in(s)$, $EF\perp (s)$. Let $A$ be a point not on $EF$. Let $A^*$ be the symmetrical point of $A$ w.r.t. $(s)$. We generally denote by a $*$ superscript this mapping. Let $\alpha$ be $\angle EAF$. We construct starting in $N$ in the same half-plane as $A,A^ *$ the two reciprocally $(s)$-reflected rays forming the angle $\alpha$ with $EF$, and let $Y,X$ be the intersections of the rays with $AE$, respectively $AF$. We introduce

$X'=Y^*=(NY)^*\cap(AE)^*=(NX)\cap(A^*F)$, and

$Y'=X^*=(NX)^*\cap(AF)^*=(NY)\cap(A^*E)$.

Then we have with notations from the above picture:

$\bullet$ (1) By construction the triangles $\Delta AEF$, $\Delta NEY$, $\Delta NX'E$, $\Delta NFY'$, $\Delta NXF$ are similar.

$\bullet$ (2) From the similarities we get $NE^2=NF^2=NX\cdot NX'=NY\cdot NY'$.

$\bullet$ (3) Let $$\mathcal I$$ be the inversion w.r.t. the center $N$ and the power $NE^2=NF^2$. Then $\mathcal I$ sends $X\to X'$ and $Y\to Y'$. We will generally denote this transformation by $W\to W'$.

$\bullet$ (4) The point $A'\in NA$ is the intersection $$A'=FY'\cap EX' \ .$$

$\bullet$ (5) The eight points $$ A,\ A';\ A^*,\ {A^*}';\ X,\ X';\ Y,\ Y'$$ are on a circle, let $I$ be its center. This center lies on $(s)$, on the side perpendicular bisector of the chord $AA'$, and these two conditions determine $I$.

$\bullet$ (6) The three circles $(EYY')$, and by symmetry $(FXX')$, and $(EAA')$, are tangent to $EF$ in $E$ and/or $F$. (We have used this in the drawing.)

$\bullet$ (7) It is convenient to introduce short hand notations for the following arcs in the eight points circle $(I)$: $$ \begin{aligned} 2y &= \overset \frown {A'Y'} = \overset \frown {X{A^*}'} \ ,\\ 2z &= \overset \frown {Y'Y} = \overset \frown {X'X} \ . \end{aligned} $$ Let $T,W$ be further points on the eight point circle $(I)$ such that we have (as oriented arcs of common measure $2z$) $$ \begin{aligned} \overset \frown {AT} &= 2z = \overset \frown {Y'Y}\ ,\\ \overset \frown {WA} &= 2(y+z)=\overset \frown {A'Y}\ . \end{aligned} $$ Then $AY\|TY'$, $AA'\|WY$, $AYY'T$ is an isosceles trapez, and the points $T,A,Z$ are colinear.

$\bullet$ (8) The points $N,A',Y',E$ lie on a circle $(O_E)$.

$\bullet$ (9) Let $K$ be the intersection $$K=A'W\cap AYE\ .$$ Then the triangle $\Delta AA'K$ is isosceles in $K$.

The line $A'KW$ is tangent in $A'$ to the circle $(O_E)$.

$\bullet$ (10) Let $S=Z'$ be the point on $NZ$ such that $$ \angle NA'Z=\beta-\gamma = \angle NZA\ . $$ Then $Z'$ corresponds through the inversion $\mathcal I$ to $Z$, consistent with the notation. $AS=AZ'$ is the symmedian w.r.t. $A$ in the triangle $\Delta AEF$.

$\bullet$ (11) Let $T^*$ be the reflection of $T$ in $(s)$, and let $T'$ be the inversion of $T$ w.r.t. $\mathcal I$. Then $(Z'A'T^*)$ and $(Z'T'A^*)$ are colinearities (i.e. lines in the same plane).

$\bullet$ (12) $(NZ'A'T')$ is inscriptible and $$\widehat{ZA'W}=\gamma\ .$$ So $ZA'$ is tangent to the circle $(NZ'A'T')$ in $A'$.

$\bullet$ (13) $ZA=ZA'$ and the triangle $\Delta ZAA'$ is isosceles.

$\bullet$ (14) $ZI$ is perpendicular on the direction $AA'\| WY\|TX$.

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Proof:

The full picture is now as follows:

$\bullet$ The bullets (1), (2), (3) are clear.

$\bullet$ (4) Let us show the location of $A'$ as the above intersection.

The circle $AENX$ is transformed by inversion in $N$ into a (projective) line. This line must be $A'E'\infty X'=E'X'=EX'$. So $A'$ is on $EX'$.

The circle $AFNY$ is trasformed by inversion in $N$ into a (projective) line. This line must be $A'F'\infty Y'=F'X'=FY'$. So $A'$ is on $FY'$.

We're done.

$\bullet$ (5) Eight points on a circle.
By construction $NA\cdot NA'=NY\cdot NY'$. So the four points $A,A', Y, Y'$ are on a circle. (Reciprocal to the power of a point w.r.t. a circle.)

We know further $\angle Y'YE$ is $\gamma$. So $\angle Y'YA=180^\circ-\gamma$. Observe now that $\angle Y'XA=\angle EFA=\gamma$, so $XY'YA$ is inscriptible. So the five points $A,A', Y, Y'=X^*$ and $X$ are on a circle. Its center is on $(s)$, the axis of symmetry of $XX^*$. Reflecting the first four points w.r.t. $(s)$ we get all eight points on the circle!

$\bullet$ (6) From $NE^2=NY\cdot NY'$ the circle $(E,Y,Y')$ is tangent to $NE$ in $E$. (Reciprocal to the power of a point, here $N$, w.r.t. a circle, here $(E,Y,Y')$.) Same argument for the other two circles.

$\bullet$ (7) From $2z = \overset\frown{X'X} = \overset\frown{Y'Y} = \overset\frown{AT} $ we get $AX\|TX'$ and $AY\|TY'$. From $2(y+z) = \overset\frown{A'Y} = \overset\frown{WA} $ we get $AA'\|WY$. This was easy. Let us show the colinearity $(TAZ)$. This follows from $$ \begin{aligned} \angle AZF &= \angle FEA -\angle EAZ\\ &=\beta-\gamma\\ &=\angle X'YA - \angle AXY'\\ &=\angle A^*X'Y - \angle ATY'\\ &=\frac12 \Big(\ \operatorname{arclength}(YATT^*A^*) - \operatorname{arclength}(YAT)\ )\\ &=\frac12 \operatorname{arclength}(TT^*A^*)\\ &=\angle TAA^*\ . \end{aligned} $$ We conclude because of $AA^*\|ZF$, so $(TAZ)$ are colinear.

$\bullet$ (8) The inversion $\mathcal I$ with center $N$ maps $A'$, $Y'$, $E=E'$ respectively to $A$, $Y$, $E$, which are on a line. This line transforms back into a circle through $N$.

$\bullet$ (9) By construction of $W$, $WYA'A$ is isosceles, so $KA'A$ is isosceles. Le us show that $WA'$ is the tangent in $A'$ to $(O_E)=(NA'Y'E)$. For this, it is enough to show that the angle between the to-be-tangent $A'W$ and the chord $A'Y'$ in $(O_E)$ is "the degenerated case" of an angle against $\overset\frown{A'Y'}$ and has the same length as $\angle A'NY'=\angle A'EY'$. For this we compute: $$ \begin{aligned} \angle A'NY' &= \angle ANY \\ &= \frac 12\Big(\ \overset\frown{AY}-\overset\frown{A'Y'}\ \Big) \\ &= \frac 12\Big(\ \overset\frown{A'W}-\overset\frown{A'Y'}\ \Big) \\ &= \frac 12\overset\frown{Y'W} \\ &= \angle WA'Y\ . \end{aligned} $$

$\bullet$ (10-14) By construction, $(A'Z'ZA)$ is inscriptible, since the outer angle in $A'$ of this quadrilateral is $\beta-\gamma$, equalt to the interior angle in $Z$. (So $NA'\cdot NA =NZ'\cdot NZ$, the power of $N$ w.r.t. $(A'Z'ZA)$.)

We know that $N,A',A$ are colinear, and that $\angle T^*A'A=\angle T^*A^*A=\beta-\gamma$, so $\angle T^*A'A$ is the opposite angle for $\angle Z'A'N$, so $T^*,A',Z'$ are colinear.

For the same reason, one can use the colinearity of $N, T', T$, and the opposite angles $$ \angle NT'Z'= \angle NA'Z'= \angle AA'T^* \angle TT'A^* $$ to conclude that $A^',T',Z'$ are colinear.

Note that the angle between the circles $(NA'Y'E)$ and $(NZ'A'T')$ is $\gamma$, since it is fo the transformed objects. But here, we also know the tangents explicitly.

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I have to stop here, jax is to slow, typing is practically impossible.

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