Medians and the Circle

Triangle △ABC is any arbitrary triangle. The medians of the triangle have been drawn (in green). The medians intersect the circumscribed circle at points PA, PB, and PC. The medians intersect the opposite sides of the triangle at points A', B', and C'.

The altitudes, BA2 and CA1, of △CAPA and △BAPA have been drawn (in orange). Note that the areas of △BA'PA and △CA'PA are equal since PAA' is a median of △BPAC. Therefore, areas of △BAPA and △CAPA are equal. This means that the altitudes that have been drawn are equal in length.

Let APA, BPB, and CPC equal NA, NB, and NC, respectively. We will find the length of these in terms of a, b, and c.

Knowing that the areas of △BAPA and △CAPA are equal, we can find an interesting relationship between u and v.

First, the areas of △BAPA and △CAPA can be given by \(\frac{1}{2}cu\cdot \sin(\angle \mathrm{ABP_{A}})\) and \(\frac{1}{2}bv\cdot \sin(\angle \mathrm{ACP_{A}})\), respectively. Angles ABPA and ACPA are opposite angles of a cyclic quadrilateral and are complementary. Therefore, the sine of these angles is equal. Therefore, we have the following relationship (and similar relationships from the other sides):

(1) \(\frac{b}{c} = \frac{u}{v}\), \(\frac{a}{b}= \frac{y}{z}\), and \(\frac{c}{a}= \frac{w}{x}\).

We can multiply all the relationships in (i) and obtain a curious relationship:

(2) \(\frac{u}{v} \cdot \frac{w}{x} \cdot \frac{y}{z} = 1\)

Now, from the cyclic quadrilateral ABPAC, we can state the following: \(ub+vc = a\cdot N_{A}\). Since \(u=\frac{b}{c}\cdot v\), this becomes \(\frac{b^{2}}{c}\cdot v+vc = a\cdot N_{A}\) or

(i) \(v(b^{2}+c^{2})= ac\cdot N_{A}\)

Similarly,

(ii) \(u(b^{2}+c^{2})= ab\cdot N_{A}\)

The equalities in (i) and (ii) are important later on.

Now, using the law of cosines on △ABPA and △ACPA, the following two relationships are true:

(iii)\(N_{A}^{2}=c^{2}+u^{2}-2uc\cdot \cos (\angle \mathrm{ABP_{A}})\)

(iv) \(N_{A}^{2}=b^{2}+v^{2}-2vb\cdot \cos (\angle \mathrm{ACP_{A}})\) = \(b^{2}+v^{2}+2vb \cdot \cos(\angle \mathrm{ABP_{A}})\)

Substituting the value of cos(∠ABPA) from one into the other, we have:

(v) \(N_{A}^{2} = c^{2}+u^{2} - \frac{2uc}{2vb}\left [ N_{A}^{2} - (b^{2}+v^{2}) \right ]\)

Since uc = vb, we have

(vi) \(2N_{A}^{2}=b^2+c^2+u^2+v^2\)

Making the substitution from (i) and (ii) into (vi), we have:

(vii) \(2N_{A}^{2}=b^2+c^2+\frac{a^2b^2N_{A}^{2}}{(b^2+c^2)^2}+\frac{a^2c^2N_{A}^{2}}{(b^2+c^2)^2}\)

(viii) \(2N_{A}^{2}=b^2+c^2+\frac{a^2N_{A}^{2}}{b^2+c^2}\)

We can solve for the desired length NA in (viii):

(3) \(N_{A}=\dfrac{b^2+c^2}{\sqrt{2(b^2+c^2)-a^2}}\).

We also note that the length of the median (this is the segment AA' and its length is denoted mA) is given by:

The length of the median, mA, from vertex A to side BC is given by the formula \(m_{A}=\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}\).

Therefore, the following is also true:

(ix) \(m_{A}\cdot N_{A}=\frac{1}{2}(b^2+c^2)\)

We also have the following relationships with u and v derived from (i), (ii), and (ix):

(x) \(u=\frac{ab}{2m_{A}}\)

(xi) \(v=\frac{ac}{2m_{A}}\)

[When we divide u by v from these equalities, we obtain the original ratio of b/c.]

The areas of the triangles △BPAC, △CPBA, and △APCB are given by the formulas: \(\frac{a^2K}{4m_{A}^2}\), \(\frac{b^2K}{4m_{B}^2}\), and \(\frac{c^2K}{4m_{C}^2}\), respectively, where K is the area of △ABC.

The relationship in (iv) gives us an interesting inequality. This relationship can be stated as:

(xii) \(\frac{2m_{A}^{2}}{b^2+c^2}=\frac{m_{A}}{N_{A}}\)

(xiii) \(\frac{2(b^2+c^2)-a^2}{4(b^2+c^2)} \) = \(\frac{1}{2}-\frac{a^2}{4(b^2+c^2)}=\frac{m_{A}}{N_{A}}\)

Now, we know that \(0 \lt \frac{m_{A}}{N_{A}} \lt 1\). This can be deduced from the fact that there is not a triangle where MA > NA or MA = NA. Also, the point of concurrency of the medians always lies inside the triangle.

Therefore, we can rewrite (e) as: \(0 \lt \frac{1}{2}-\frac{a^2}{4(b^2+c^2)} \lt 1\) or \(0 \lt 2-\frac{a^2}{b^2+c^2} \lt 4\) or

(vi) \(\dfrac{a^2}{b^2+c^2} \lt 2\).