Nine points circle theorem

by Nikpour, Apr 25, 2013, 2:06 PM

In this post, I will explore many results regarding the altitudes, the medians, the orthocenter and the centroid of a triangle $\triangle ABC$.

I will denote by $H_A, H_B, H_C$ the feet of the altitudes with respect to $A, B, C$ and $M_A, M_B, M_C$ the midpoints of sides $a, b, c$. Also let $H$ and $G$ denote the orthocenter and centroid of $\triangle ABC$.

Let's start with a well-known result:

1:Nine-point circle: The points $H_A, H_B, H_C, M_A, M_B, M_C$ are all concyclic.

Proof: Since $\triangle ABH_A$ is right and $M_C$ is the midpoint of $AB$, we have that $\triangle M_CBH_A$ is isosceles, thus $\angle BH_AM_C = \angle B and \angle M_CH_AM_A = 180 - \angle B = 180 - \angle M_CM_BM_A$. Therefore, $H_A, M_A, M_B, M_C$ are concyclic, and similarly all six points lie on a circle.

Remark. The midpoints of $AH, BH, CH$ are also in this circle, thus the name nine-point circle. The proof is analogous angle-chasing.

2:Prove that the ray $H_AG$ intersects the circumcircle of $\triangle ABC$ at a point $T'_A$ such that $AT'_A$ is parallel to $BC$.


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Proof: Let $\mathcal{H}$ denote the homothety centered at $G$ which sends $\triangle ABC$ to $\triangle M_AM_BM_C$. Then the image of $T'_A$ is the intersection of the nine-point circle of $\triangle ABC$ and $GH_A$, which is just $H_A$. But then $H_A M_A \parallel AT'_A$, so $BC \parallel AT'_A$.

3:Let $\mathcal{C}$ denote the circle passing through $M_B$, $M_C$ tangent to the circumcircle $\omega$ of $ABC$ at a point $T_A \neq A$. Then $T_A, H_A, G$ are collinear.

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Proof: Let $\mathcal{C}_2$ denote the circumcircle of $AM_CM_B$, which is also tangent to $\omega$. Let $P_A$ be the radical center of $\mathcal{C}, \mathcal{C}_2, \omega$, and let $T_1, T_2$ be the intersections of $M_BM_C$ and $\omega$, as above. Then since $P_AA$ and $P_AT_A$ are tangent to $\omega, AT_1T_AT_2$ is a harmonic quadrilateral. Let $T_A^*$ denote the second intersection of $T'_AG$ and $\omega$, and $H'_A$ denote the image of $H_A$ under the homothety $\mathcal{H}$ which sends $\triangle ABC$ to $\triangle M_AM_BM_C$. Then $M_AH'_A \perp T_1T_2$, so $OH'_A \perp T_1T_2$ and $H'_A$ is the midpoint of $T_1T_2$. We have shown in (2) that $AT'_A \parallel T_1T_2$, so $T'_A(T_2, T_1; H'_A, A)$ is a harmonic pencil, which implies that $T'_A(T_2, T_1; T_A^*, A)$ is a harmonic pencil, and $AT_1T_A^*T_2$ is a harmonic quadrilateral $\implies T_A^* \equiv T_A$, as desired.

Now, a simple but very useful lemma about the orthocenter:

4: Let $A'$ be the point diametrically opposite to $A$ w.r.t. $\omega$. Prove that the ray $HA' $ intersects $BC$ at $M_A$, and that $M_A$ is also the midpoint of $HA'$.


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Proof: I claim that $BHCA'$ is a parallelogram. Indeed, $BH \parallel A'C$ since it is immediate that $\angle HBC = \angle A'CB = 90 - \angle C$. Analogously, $HC \parallel BA'$ since we have $\angle BCH = \angle CBA' = 90- \angle B$. Therefore, $BHCA'$ is a parallelogram, thus $HA'$ passes through $M_A$, and $M_A$ is the midpoint of $HA'$.

5: (USA TSTST 2011 #4 restatement) Let rays $M_B H$ and $M_C H$ intersect the circumcircle $\omega$ of $\triangle ABC$ at $B_1$ and $C_1$ respectively. Then $P_A, B_1, C_1$ are collinear, where $P_A$ is defined as in (3).

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Proof: Let $B'$ and $C'$ denote the antipodes of $B$ and $C$ w.r.t. $\omega$. From (4) we have that $M_B$ and $M_C$ are the midpoints of $HB'$ and $HC'$, thus $B'C' \parallel M_BM_C$, so $\angle C_1M_CM_B = \angle C_1C'B' = \angle C_1B_1M_B$, hence $M_BM_CB_1C_1$ is cyclic with circumcircle $\mathcal{C}_1$. Let $\mathcal{C}_2$ be the circumcircle of $AM_BM_C$. Then $P_A$ is the radical center of $\omega, \mathcal{C}_1, \mathcal{C}_2$, therefore $P_A, B_1, C_1$ are collinear, as desired.

6: (APMO 2012 #4 restatement) Let $\triangle ABC$ be an acute triangle with circumcircle $\omega$. Denote by $A_1$ the intersection of the ray $M_AH$ with $\omega$ and by $A_2$ the second intersection of $A_1H_A$ and $\omega$. Prove that the quadrilateral $ABA_2C$ is harmonic.

First, I will prove a lemma about harmonic quadrilaterals:

$\bullet$ $\blacktriangleright$ Let $ABCD$ be a harmonic quadrilateral with circumcircle $\omega$ and let $P$ be any point in the plane different from $A,B,C,D$. Denote by $A', B', C', D'$ the second intersections of $PA, PB, PC, PD$ and $\omega$. Then $A'B'C'D'$ is also a harmonic quadrilateral.

Proof of lemma: Using similar triangles, it's not hard to find that:
$A'B'\cdot C'D' = A'D'\cdot B'C' = \frac{AB \cdot CD \cdot \mathcal{P}(P)^2}{EA \cdot EB \cdot EC \cdot ED}$,
where $\mathcal{P}(P)$ denotes the power of $P$ w.r.t. $\omega$. Thus $A'B'C'D'$ is harmonic.

Main Proof: Let $A'$ denote the point diametrically opposite to $A$ w.r.t. $\omega$, and $H'$ the reflection of $H$ w.r.t. $BC$. It is easy to prove that $H' \in \omega$, and since $M_A$ is the midpoint of $HA'$, we have $A'H' \parallel BC \implies A'(B,C;M_A,H')$ is harmonic, so $CH'BA_1$ is a harmonic quadrilateral. Applying the above lemma to $CH'BA_1$ and taking $P \equiv H_A$ gives the desired result.

7: (OBM (Brazil MO) 2011 #5 restatement) Let $\triangle ABC$ be an acute triangle with circumcircle $\omega$. Let the circumcircle of $AH_BH_C$ intersect $\omega$ at $F$. Prove that the angle bisectors of $\angle BFC$ and $\angle BHC$ intersect at a point on side $BC$.

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Proof: From the angle bisector theorem, the problem statement is equivalent to proving that $\frac{BH}{CH} = \frac{BF}{CF}$. This looks a lot like a harmonic quadrilateral, but of course $H \notin \omega$. Fortunately, we can take the reflection $H'$ of $H$ w.r.t. $BC$, so that $\frac{BH'}{CH'} = \frac{BH}{CH}$ and $H' \in \omega$. Now, it remains to prove that $CH'BF$ is a harmonic quadrilateral. But in (6) we have shown that $CH'BA_1$ is a harmonic quadrilateral, where $A_1$ is the intersection of the ray $M_AH$ and $\omega$.
Thus, it suffices to prove $F \equiv A_1 \iff A_1, H_C, H_B, A$ are concyclic. Notice that $H$ is clearly in the circumcircle of $AH_BH_C$. But then $\angle AA_1H = \angle AH_BH = 90$, hence $A_1$ is also in the circumcircle of $AH_BH_C$, so $F \equiv A_1$, as desired.

8: Let $A, B, C,D$ be four concyclic points. Let $s_a$ denote the Simson line of $A$ w.r.t. $BCD$, and define $s_b, s_c, s_d$ similarly. Let $h_a$ denote the line connecting $A$ to the orthocenter of $BCD$, and define $h_b, h_c, h_d$ similarly. Prove that the eight lines $h_a, h_b, h_c, h_d, s_a, s_b, s_c, s_d$ are all concurrent.


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draw(circle((3.45,1.27), 2.92), blue);
draw((1.7,3.6)--(6.26,0.48), ccwwqq);
draw((4.97,3.76)--(0.64,0.5), ccwwqq);
draw((6.02,5.3)--(6.26,0.48), wwwwww);
draw((0.4,5.32)--(0.64,0.5), wwwwww);
draw((0.4,5.32)--(2.8,2.13), wwwwww);
draw((0.64,0.5)--(2.42,3.1), wwwwww);
draw((4.21,3.19)--(6.26,0.48), wwwwww);
draw((0.4,5.32)--(4.97,3.76), wwwwww);
draw((5.5,2.41)--(0.64,0.5), wwwwww);
draw((6.02,5.3)--(1.7,3.6), wwwwww);
draw((6.02,5.3)--(3.85,2.13), wwwwww);
draw((4.97,3.76)--(4.96,0.48), wwwwww);
draw((6.26,0.48)--(1.22,2.2), wwwwww);
draw((0.49,3.53)--(6.09,3.82), ccwwqq);
draw((0.64,0.5)--(6.26,0.48), ccwwqq);
draw((6.26,0.48)--(4.58,4.73), ccwwqq);
draw((0.4,5.32)--(6.26,0.48), linewidth(1.6) + ffcctt);
draw((6.02,5.3)--(0.64,0.5), linewidth(1.6) + ffcctt);
draw((1.7,3.6)--(4.96,2.2), linewidth(1.6) + ffcctt);
draw((4.97,3.76)--(1.69,2.04), linewidth(1.6) + ffcctt);
draw((4.58,4.73)--(1.68,0.5), linewidth(1.6) + qqwwqq);
draw((0.49,3.53)--(5.5,2.41), linewidth(1.6) + qqwwqq);
draw((1.22,2.2)--(6.09,3.82), linewidth(1.6) + qqwwqq);
draw((1.7,3.6)--(1.68,0.5), wwwwww);
draw((2.09,4.74)--(0.64,0.5), ccwwqq);
draw((2.09,4.74)--(4.96,0.48), linewidth(1.6) + qqwwqq);
draw(circle((1.17,2.05), 1.64), red);
draw(circle((3.33,3.68), 1.64), red);
draw(circle((5.61,2.12), 1.76), red);
draw(shift((3.45,0.49))*xscale(2.81)*yscale(2.81)*arc((0,0),1,-0.2,179.8), red);
/* dots and labels */
dot((1.7,3.6),dotstyle);
label("$A$", (1.74,3.66), NE * labelscalefactor);
dot((0.64,0.5),dotstyle);
label("$B$", (0.68,0.56), NE * labelscalefactor);
dot((6.26,0.48),dotstyle);
label("$C$", (6.31,0.54), NE * labelscalefactor);
dot((4.97,3.76),dotstyle);
label("$D$", (5.01,3.82), NE * labelscalefactor);
dot((3.85,2.13),dotstyle);
label("$H_{DB}$", (3.9,2.19), NE * labelscalefactor);
dot((4.96,0.48),dotstyle);
label("$H_{DA}$", (5,0.55), NE * labelscalefactor);
dot((1.69,2.04),dotstyle);
label("$H_D$", (1.73,2.1), NE * labelscalefactor);
dot((1.68,0.5),dotstyle);
label("$H_{AD}$", (1.73,0.56), NE * labelscalefactor);
dot((2.42,3.1),dotstyle);
label("$H_{BD}$", (2.46,3.17), NE * labelscalefactor);
dot((1.22,2.2),dotstyle);
label("$H_{CD}$", (1.26,2.26), NE * labelscalefactor);
dot((2.8,2.13),dotstyle);
label("$H_{AC}$", (2.84,2.2), NE * labelscalefactor);
dot((4.58,4.73),dotstyle);
label("$H_{AB}$", (4.63,4.79), NE * labelscalefactor);
dot((4.21,3.19),dotstyle);
label("$H_{CA}$", (4.25,3.25), NE * labelscalefactor);
dot((5.5,2.41),dotstyle);
label("$H_{BA}$", (5.54,2.48), NE * labelscalefactor);
dot((4.96,2.2),dotstyle);
label("$H_{A}$", (5.01,2.27), NE * labelscalefactor);
dot((0.49,3.53),dotstyle);
label("$H_{BC}$", (0.53,3.6), NE * labelscalefactor);
dot((6.09,3.82),dotstyle);
label("$H_{CB}$", (6.13,3.88), NE * labelscalefactor);
dot((6.02,5.3),dotstyle);
label("$H_{B}$", (6.06,5.36), NE * labelscalefactor);
dot((0.4,5.32),dotstyle);
label("$H_{C}$", (0.44,5.38), NE * labelscalefactor);
dot((3.33,2.9),dotstyle);
dot((2.09,4.74),dotstyle);
label("$H_{DC}$", (2.13,4.81), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]



Proof: Denote by $H_{AB}$ the foot of the perpendicular from $A$ to the segment $CD$ (the one which does not contain $B$), and similarly define the $12$ other points of this type, as above. Denote by $H_A$ the orthocenter of $BCD$, and define $H_B, H_C, H_D$ similarly. Observe that the red circles illustrated above with diameters $AB, BC, CD, DA$ contain each four other points.

From Pascal's Theorem on $AH_{AC}H_{AD}BH_{BC}H_{BD}$, we get that $s_a, s_b, h_c$ are concurrent. Analogously, Pascal's Theorem on $AH_{AD}H_{AC}BH_{BD}H_{BC}$ gives that $s_a, s_b, h_d$ are concurrent. Similarly, all eight lines are concurrent, so done.
This post has been edited 2 times. Last edited by Nikpour, Jul 9, 2013, 5:14 PM

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