A median of a triangle is the geometric mean of the adjacent sides; find the cosine of one angle in terms of the others

Question

$AD$ is a median of $\triangle ABC$. $|AD|$ is the geometric mean of $|AB|$ and $|AC|$.

Find $1+\cos A$ in terms of $\cos B$ and $\cos C$.

Edit

This is the second part of the question

Also prove that 1+cosA=√2|cosB-cosC|.

Answer 1

$$\frac{1}{2}\sqrt{2b^2+2c^2-a^2}=\sqrt{bc},$$ which gives $$a^2=2b^2+2c^2-4bc.$$ Thus, $b\neq c$ and $$\cos\beta\cos\gamma=\frac{a^2+c^2-b^2}{2ac}\cdot\frac{a^2+b^2-c^2}{2ab}=$$ $$=\frac{(2b^2+2c^2-4bc+c^2-b^2)(2b^2+2c^2-4bc+b^2-c^2)}{4(2b^2+2c^2-4bc)bc}=$$ $$=\frac{(3c^2-4bc+b^2)(3b^2-4bc+c^2)}{8(b-c)^2bc}=\frac{(b-c)^2(3b-c)(b-3c)}{8bc(b-c)^2}=$$ $$=\frac{3(b^2+c^2)-10bc}{8bc}=\frac{3}{8}\left(\frac{b}{c}+\frac{c}{b}\right)-\frac{5}{4}.$$ Thus, $$\frac{b}{c}+\frac{c}{b}=\frac{10+8\cos\beta\cos\gamma}{3}.$$ Id est, $$1+\cos\alpha=1+\frac{b^2+c^2-a^2}{2bc}=1+\frac{b^2+c^2-2b^2-2c^2+4bc}{2bc}=$$ $$=3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{b}\right)=3-\frac{5+4\cos\beta\cos\gamma}{3}=\frac{4}{3}(1-\cos\beta\cos\gamma).$$

About your second problem. $$1+\cos\alpha=1+\frac{b^2+c^2-a^2}{2bc}=$$ $$=1+\frac{b^2+c^2-2b^2-2c^2+4bc}{2bc}=3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{a}\right).$$ Also, we see that $3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{a}\right)\geq0$.

In another hand, $$\sqrt2|\cos\beta-\cos\gamma|=\sqrt2\left|\frac{a^2+c^2-b^2}{2ac}-\frac{a^2+b^2-c^2}{2ab}\right|=$$ $$=\sqrt2\left|\frac{2b^2+2c^2-4bc+c^2-b^2}{2ac}-\frac{2b^2+2c^2-4bc+b^2-c^2}{2ab}\right|=$$ $$=\sqrt2\left|\frac{b^2-4bc+3c^2}{2ac}-\frac{3b^2-4bc+c^2}{2ab}\right|=$$ $$=\sqrt2|b-c|\cdot\left|\frac{b-3c}{2ac}-\frac{3b-c}{2ab}\right|=$$ $$=\frac{\sqrt2|b-c|}{2abc}\cdot\left|(b-3c)b-(3b-c)c\right|=$$ $$=\frac{\sqrt2|b-c|}{2\sqrt2|b-c|bc}\cdot\left|b^2+c^2-6bc\right|=$$ $$=3-\frac{1}{2}\left(\frac{b}{c}+\frac{c}{a}\right)$$ and we are done!

Answer 2

From these easily-proven relations (for any $\triangle ABC$) ...

$$\begin{align} b &= a \cos C + c \cos A \\ c &= a \cos B + b \cos A \end{align}\tag{1}$$

... we deduce ... $$(b - c)(1+\cos A) = a(\cos C - \cos B) \tag{2}$$

For a triangle whose median from $A$ has a length, $d$, equal to the geometric mean of $b$ and $c$ (that is, $d^2 = b c$), Stewart's Theorem (for instance) yields ... $$b^2\cdot \frac{a}{2} + c^2\cdot\frac{a}{2} = a\left(d^2 + \frac{a}{2}\cdot\frac{a}{2}\right)\quad\to\quad a^2 = 2 (b-c)^2 \tag{3}$$

Therefore, incorporating $(2)$, we can write (since $b \neq c$, and $1+\cos A > 0$ (why?)) ... $$(1 + \cos A)^2 = 2\;(\cos C - \cos B)^2 \quad\to\quad 1 + \cos A = \sqrt{2}\;|\cos B - \cos C| \tag{4}$$ ... as desired. $\square$

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Note. I'm posting this answer because I think it's worth sharing. I do not condone OP's ploy of accepting @Michael's answer in order to get more effort from him, and then subsequently un-accepting it in order to get more effort from the community.

Answer 3

Taking into account the new, more relaxed context, here is a proof by analytic geometry.

It connects the issue with conic curves, simplifying the problem by using a single variable as we are going to see it (see figure at the bottom).

WLOG, one can assume that $B(-1,0), C(1,0)$ are fixed (thus their midpoint is $D=O$), and $A$ is variable with coordinates $(x,y)$.

The geometric mean condition, written under the form:

$$(AB.AC)^2=AO^4$$ is equivalent to

$$[(x+1)^2+y^2][(x-1)^2+y^2]=(x^2+y^2)^2$$

itself equivalent to

$$\tag{1}x^2-y^2=\tfrac12,$$

meaning that point $A$ must belong to a certain hyperbola $(H)$.

Let us assume, again WLOG, that point $A$ is on the right branch of the hyperbola (i.e., with $x>\tfrac12$).

Then using relationship $a^2=b^2+c^2-2bc \cos(A)$ and the two others by cyclic permutation, replacing $y^2$ by $x^2-\dfrac12$ (see (1)), one obtains

$$\cos(A)=\dfrac{4x^2-3}{4x^2-1}, \ \ \cos(B)=\sqrt{2}\dfrac{1+x}{2x+1}, \ \ \cos(C)=\sqrt{2}\dfrac{1-x}{2x-1}.$$

yielding the final relationship:

$\sqrt{2}(\cos(B)-\cos(C)) = (\sqrt{2})^2\dfrac{2-4x^2}{1-4x^2}=\dfrac{4-8x^2}{1-4x^2}=1+\dfrac{3-4x^2}{1-4x^2}=1+\cos(A)$

(no need for absolute values because of the choice $x>0$ for the abscissa of $A$).

Remark :

It can be easily proven that hyperbola $(H)$ has $B$ and $C$ as its foci! Moreover $(H)$ can be described as the locus of points $A$ such that

$$|AB-AC|=constant=(1+\dfrac{\sqrt{2}}{2})(1-\dfrac{\sqrt{2}}{2})=\dfrac12.$$