A user named watt5151 on the Math15 Forum posed the following question:
As shown in the figure, given a triangle ABC, how can one construct a circle that intersects all three sides of the triangle such that the six intersection points can be connected to form three diameters?
We often see the circumcircle and incircle of a triangle, but seeing such a fresh problem, BoJone found it very interesting. I have taken the liberty of naming this circle the “Six-Contact Circle” of the triangle and spent the afternoon researching it. Finally, I reached a conclusion: for non-obtuse triangles, much like the circumcircle and incircle, such a circle always exists and is unique; for obtuse triangles, in some cases, it does not exist (although such a circle can be constructed, it no longer has six intersection points with the triangle sides; in other words, the circle overflows).
Having been on holiday for a while, it feels like I haven’t done math problems in a long time, and my mind feels a bit stiff. Below is BoJone’s thinking process. First, let’s set up the following coordinate diagram:
Let the coordinates of point A be (a, b), the coordinates of C be (c, 0), and let the equation of the required circle be: (x-m)^2 + (y-n)^2 = R^2. This means the coordinates of the center O are (m, n). It is not difficult to find that the coordinates of points E and D are (m \pm \sqrt{R^2-n^2}, 0). Since E, O, F are collinear, we find the coordinates of F to be (m - \sqrt{R^2-n^2}, 2n). Similarly, the coordinates of G are (m + \sqrt{R^2-n^2}, 2n).
Based on the fact that A, F, B are collinear and AF=BF, we can write: \frac{m-\sqrt{R^2-n^2}}{2n} = \frac{a}{b} \tag{1} Based on the fact that A, G, C are collinear and AG=GC, we can write: \frac{m+\sqrt{R^2-n^2}-c}{2n} = \frac{a-c}{b} \tag{2}
There are two equations above but three unknowns m, n, R. The remaining condition lies in the diameter HI. We could find the equation for this line and combine it with the two above to determine m, n, R. This proves the circle exists and is unique. However, this approach is too cumbersome and not convenient for ruler-and-compass construction. BoJone handled it as follows:
Adding (1) and (2) yields: \frac{2m-c}{2n} = \frac{2a-c}{b} \Rightarrow 2m-c = \frac{2a-c}{b}(2n). This is a linear function passing through the points (c/2, 0) and (a, b/2). We can easily plot this function on the diagram. Point O lies on this line.
Although we cannot yet determine exactly where point O is, do not forget that we only chose side BC as the x-axis. We can perform the same operation using sides AB or AC, which will produce another line intersecting the current one. Since point O is unique, the intersection must be point O. Knowing the location of O is equivalent to knowing the lengths of m and n (even if we don’t know the specific numerical values, we don’t need to; we only hope to construct such a circle in the figure). According to equation (1), we have R^2 = n^2 + (m - \frac{2na}{b})^2, which is a quantity constructible by ruler and compass.
Now that we have roughly researched the construction method, the next task is to “translate” the research above into ruler-and-compass construction steps. Below is a diagram BoJone drew on a newspaper to illustrate:
Steps:
Choose one side of the triangle and find its midpoint; then draw the altitude to this side and find the midpoint of the altitude.
Draw a line (EF) passing through these two midpoints.
Choose another side of the triangle and repeat steps 1 and 2 (to draw HI).
Find the intersection of the two lines drawn in steps 1, 2, and 3; this intersection is the center of the circle (O).
Choose one side of the triangle, construct a segment LM = 2n, where LM \perp BC, with L on AC and M on BC. (Note: Step 5 can also be: “construct a segment LM = 2n, where LM \perp BC, with L on AB and M on BC.”)
With O as the center and OM as the radius, draw the circle. This is the required circle.
Below is another construction method provided by moderator hujunhua from the “Math Development Forum”:
Construction Method:
First, construct a small similar figure \triangle 123 of \triangle ABC rotated by 90 degrees, which is almost inscribed in \triangle ABC (two vertices fall on two sides of \triangle ABC), and then use a homothetic transformation to make it inscribed. The specific steps are as follows:
Arbitrarily pick point 1 on side BC, draw a perpendicular to BC intersecting AC at 2.
Draw a perpendicular to AC through 2, and a perpendicular to AB through 1; let the two perpendiculars intersect at 3.
Connect line C3, intersecting AB at D.
Draw a perpendicular from D to AB intersecting BC at E.
Draw a perpendicular segment DF from D to AC.
The circumcircle of D, E, F is the required circle.
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