If sin² A+sin² B+sin² C=1.Show that circumscribed circle of triangle ABC cuts its nine-point circle orthogonally.

To show that the circumcircle of triangle ABC intersects its nine-point circle orthogonally, given the condition \( \sin^2 A + \sin^2 B + \sin^2 C = 1 \), we need to delve into some properties of triangles and circles. Let's break this down step by step.

Understanding the Key Concepts

First, let's clarify what the circumcircle and the nine-point circle are:

• Circumcircle: This is the circle that passes through all three vertices of triangle ABC.
• Nine-point Circle: This circle passes through nine significant points of the triangle, including the midpoints of each side, the feet of the altitudes, and the midpoints of the segments from each vertex to the orthocenter.

Using the Given Condition

The equation \( \sin^2 A + \sin^2 B + \sin^2 C = 1 \) is a crucial piece of information. This condition holds true for a triangle where the angles A, B, and C are related to a right triangle. Specifically, if we consider a triangle where one angle is \( 90^\circ \), the sine of that angle is 1, and the sines of the other two angles will satisfy the equation.

Orthogonality of Circles

Two circles are said to intersect orthogonally if at the points of intersection, the tangents to the circles are perpendicular to each other. Mathematically, this can be expressed using the radii and the distance between the centers of the circles.

Proving the Orthogonality

Let’s denote the circumradius of triangle ABC as \( R \) and the radius of the nine-point circle as \( r \). The center of the circumcircle is the circumcenter \( O \), and the center of the nine-point circle is the nine-point center \( N \).

From the properties of triangles, we know that the radius of the nine-point circle is half the circumradius, so:

r = R/2

Now, we need to find the distance \( ON \) between the circumcenter \( O \) and the nine-point center \( N \). For any triangle, this distance can be expressed as:

ON = R \cdot \cos A \cdot \cos B \cdot \cos C

Given our condition \( \sin^2 A + \sin^2 B + \sin^2 C = 1 \), we can derive that:

\( \cos^2 A + \cos^2 B + \cos^2 C = 1 - \sin^2 A - \sin^2 B - \sin^2 C = 0 \)

This implies that at least one of the angles A, B, or C must be \( 90^\circ \). Without loss of generality, let’s assume angle A is \( 90^\circ \). In this case, the circumcenter \( O \) lies at the midpoint of the hypotenuse, and the nine-point center \( N \) also lies on the line segment connecting the midpoint of the hypotenuse to the right angle vertex.

Final Steps to Show Orthogonality

To establish that the circles intersect orthogonally, we can use the relationship:

R^2 = ON^2 + r^2

Substituting \( r = R/2 \) into the equation gives:

R^2 = ON^2 + (R/2)^2

Which simplifies to:

R^2 = ON^2 + R^2/4

Rearranging this leads to:

ON^2 = R^2 - R^2/4 = 3R^2/4

Since \( ON^2 \) is positive, this confirms that the circles intersect orthogonally. Thus, we have shown that the circumcircle of triangle ABC intersects its nine-point circle orthogonally under the given condition.

Conclusion

In summary, the relationship between the angles of triangle ABC and the properties of the circumcircle and nine-point circle leads us to the conclusion that they intersect orthogonally when \( \sin^2 A + \sin^2 B + \sin^2 C = 1 \). This elegant connection highlights the beauty of geometry in triangles.