Euler Line & Nine‑Point Circle
Geometry – IMO Study Notes
(Prepared for advanced high‑school students preparing for the International Mathematical Olympiad)
1. Simple Explanation
-
Euler Line – In any non‑degenerate triangle (ABC), the orthocenter (H), the circumcenter (O), and the centroid (G) are collinear.
That straight line is called the Euler line. -
Nine‑Point Circle – In the same triangle (ABC), the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments joining each vertex to the orthocenter all lie on a single circle.
This circle is called the Nine‑Point Circle (NPC).
Why these objects are important?
- They reveal hidden symmetries of the triangle.
- Many Olympiad problems ask you to prove collinearity or concyclicity, or to compute distances involving these points.
2. Key Points
| Concept | Definition | Key Properties |
|---|---|---|
| Centroid (G) | Intersection of the three medians. | (G) divides each median in the ratio (2:1) (vertex : midpoint). |
| Circumcenter (O) | Intersection of perpendicular bisectors of sides. | Equidistant from (A,B,C). Lies on the Euler line. |
| Orthocenter (H) | Intersection of the three altitudes. | For acute triangles, inside; obtuse, outside. Lies on the Euler line. |
| Nine‑Point Center (N) | Center of the Nine‑Point Circle. | Midpoint of (OH). Lies on the Euler line. |
| Nine‑Point Circle (NPC) | Circle passing through 9 special points: midpoints of sides, feet of altitudes, midpoints of (AH, BH, CH). | Radius (= \frac{R}{2}) where (R) is circumradius. |
| Euler Line Relation | (OG : GH = 1 : 2). | So (G) divides (OH) in the ratio (1:2). |
| Coordinates | In barycentric or vector form, formulas for (O, H, G, N) are linear combinations of vertices. | Useful for calculations. |
| Orthocenter Reflection | Reflect (H) across each side → points lie on circumcircle. | Useful for proving properties of (N). |
3. Worked Example
Problem
Let (ABC) be a triangle with (AB = 13), (BC = 14), (CA = 15).
Find the coordinates (in Cartesian system with (A(0,0)), (B(13,0)), (C(0,14))) of:
- Circumcenter (O).
- Orthocenter (H).
- Centroid (G).
- Show that (O,G,H) are collinear and compute the ratio (\frac{OG}{GH}).
- Find the radius of the Nine‑Point Circle.
Step 1 – Set up coordinates
Given vertices:
- (A(0,0))
- (B(13,0))
- (C(0,14))
Step 2 – Find (O) (circumcenter)
The perpendicular bisector of (AB):
- Midpoint (M_{AB} = (6.5, 0)).
- Slope of (AB) = 0 → perpendicular slope = ∞ → vertical line (x = 6.5).
The perpendicular bisector of (AC):
- Midpoint (M_{AC} = (0,7)).
- Slope of (AC) = (14/0 = \infty) → perpendicular slope = 0 → horizontal line (y = 7).
Intersection gives (O = (6.5, 7)).
Step 3 – Find (H) (orthocenter)
Equation of altitude from (A): line through (A) perpendicular to (BC).
- Slope of (BC): ((14-0)/(0-13) = -14/13).
- Perpendicular slope = (13/14).
- Altitude from (A): (y = \frac{13}{14}x).
Equation of altitude from (B): line through (B) perpendicular to (AC).
- Slope of (AC): (14/0 = \infty) → perpendicular slope = 0 → horizontal line (y = 0).
Intersection: solve (y = 0 = \frac{13}{14}x) → (x=0).
Thus (H = (0,0) = A).
(Indeed, in a right triangle the orthocenter is the right‑angle vertex; here (C) is the right angle? Wait: (AB^2 + AC^2 = 13^2 + 14^2 = 169 + 196 = 365). (BC^2 = 14^2 + 13^2? Actually (BC=14), so (BC^2 = 196). Not a right triangle. Let's recompute altitudes properly. Mistake: altitude from (B) perpendicular to (AC) has slope 0? (AC) is vertical line (x=0) so altitude from (B) is horizontal (y=0). Ok. Altitude from (A) is (y = \frac{13}{14}x). Intersection gives (H=(0,0)). So (A) is orthocenter. That means triangle is right at (A). Check: (AB^2 + AC^2 = 13^2 + 0^2? Actually AC=14, AB=13, so (AB^2 + AC^2 = 169+196=365). (BC^2 = 14^2? Wait BC=?? We set B(13,0), C(0,14). So BC^2 = (13-0)^2 + (0-14)^2 = 169 + 196 = 365