Circles through the Orthocenter

The applet below illustrates Problem 1 from the 2008 International Mathematics Olympiad:

An acute-angled triangle ABC has orthocentre H. The circle passing through H with centre the midpoint of BC intersects the line BC at A₁ and A₂. Similarly, the circle passing through H with centre the midpoint of CA intersects the line CA at B₁ and B₂, and the circle passing through H with centre the midpoint of AB intersects the line AB at C₁ and C₂. Show that A₁, A₂, B₁, B₂, C₁, C₂ lie on a circle.

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Solution

Solution

One solution stems from the observation that the orthocenter serves as the radical center of the three circles whilst the altitudes are the radical axes of the circles taken in pairs. The situation is reminscent of the circles constructed on the sides of a triangle as diameters.

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Consider the circles centered at the midpoints M_a and M_b of sides BC and AC, respectively. Since CH⊥AB we also have M_aM_b⊥CH. But the line joining the centers of two intersecting circles is perpendicular to their common chord. In particular, this is what CH is - the line containing the common chord of the two circles. Which exactly means that CH is their radical axis.

The same holds for the other two pairs of circles.

Let r_a, r_b, r_c be the radii of the circles centered at M_a, M_b, M_c, which we denote C(M_a, r_a), C(M_b, r_b), C(M_c, r_c), respectively. Let also set a = BC, b = AC, c = AB.

Since C lies on the radical axis of C(M_a, r_a) and C(M_b, r_b), we may apply the Pythagorean theorem to the triangles formed by tangents of common length t from C to the two circles, half-sides of BC and AC and the radii of the circles:

(1)	(a/2)² - r_a² = t² = (b/2)² - r_b².

On the other hand, if the six points A₁, A₂, B₁, B₂, C₁, C₂ lie on a circle, the latter is bound to be centered at the circumcenter O of ΔABC. So lets show that O is equidistant from the six points. To this end, denote s_a = OM_a, s_b = OM_b, s_c = OM_c. Since, O is the circumcenter of ΔABC,

(2)	(a/2)² + s_a² = R² = (b/2)² + s_b²,

R being the circumradius of ΔABC. Subtracting (1) from (2) we see that

(3)	r_a² + s_a² = r_b² + s_b²,

implying that O is equidistant from A₁, A₂, B₁, B₂. The four points A₁, A₂, B₁, B₂ are thus concyclic. Similarly, A₁, A₂, C₁, C₂ are equidistant from O which puts C₁ and C₂ on the same circle.

Observe now that the requirement for ΔABC to be acute is spurious. The six points A₁, A₂, B₁, B₂, C₁, C₂ are concyclic for any ΔABC. A further generalization replaces the altitudes with arbitrary concurrent cevians. Which shows that both requirements are nothig but red herrings.

Red Herrings: a Sample List

Radical Axis and Radical Center

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