# Geometric Inequality

I discovered a nice inequality while playing on Geogebra again. I found that it can be proved quite quickly by making a simple construction.
The triangle ABC, with sides of length a, b and c opposite vertices A, B and C respectively, satisfies a√2 ≥ b+ c. M is the midpoint of AB and P is on BC such that ∠MPA = ∠BCA. Then BC ≥ 2AP.

Proof
The condition that a√2 ≥ b+ c is merely necessary to guarantee the existence of at least one such P. A derivation of this inequality will be included at the end of this post.
Let Q be the point on AP extended such that AP = PQ.
Triangle ABQ is an enlargement of triangle AMP, since AB = 2AM and AQ = 2AP, so MP is parallel to BQ and ∠MPA = ∠BQA.
So BQCA is a cyclic quadilateral by the converse of the same segment theorem.

Applying intersecting chords theorem, AP · PQ = BP · PC; that is, AP² = BP · PC
Let BP = x and let PC = y.
$AP = \sqrt{xy}$
But from the AM-GM inequality,
$\frac{x+y}{2}\geq \sqrt{xy}$
*This is also apparent from rearranging
$(\sqrt{x}-\sqrt{y})^{2}\geq 0$
So
$\frac{BP+PC}{2}\geq AP$
And as BP + PC = BC, this means BC ≥ 2AP as desired.
Equality occurs when
$x=y$
which is when P is the midpoint of BC. In this case, it is easy to show that triangle ABC is right angled at A.

Derivation of condition for existence of at least one such P
Let N be the midpoint of AC. Then triangle ABC is an enlargement of triangle AMN, as AB = 2AM and AC = 2AN, so ∠MNA = ∠BCA.

The set of points P such that ∠MPA = ∠BCA = ∠MNA is the set of points located on the circumcircle of triangle AMN, on the same side of AM as N; this is due to the same segment theorem.
Hence, for there to be such a P located on BC, the circumcircle of AMN must touch BC at least once.
Clearly this does not always happen:

Sometimes there are two such P:

And sometimes, in the case we will now examine, there is exactly one such P.

In this case we can apply secant-tangent theorem:
$BP^{2}=BM\cdot BA=\frac{c^{2}}{2}$
$CP^{2}=CN\cdot CA=\frac{b^{2}}{2}$
$BP^{2}\cdot CP^{2}=\frac{b^{2}\cdot c^{2}}{4}$
$BP\cdot CP=\frac{bc}{2}$
Also
$BP+PC=a$
$BP^{2}+PC^{2}+2BP\cdot PC=a^{2}$
$\frac{b^{2}+c^{2}+2bc}{2}=a^{2}$
$(b+c)^{2}=2a^{2}$
$b+c=a\sqrt{2}$
Fix the side BC and consider for which points A there exists exactly one P. a√2 is constant so the locus of A is an ellipse, since the sum of the distances to each of two foci is constant for every point on the locus. Here, the foci are B and C, and the constant sum is a√2.

For all points A on the ellipse, the circumcircle of triangle AMN is tangent to BC.
We can now consider the cases where the circle AMN is not tangent to BC.
If we move A within the ellipse, the sum of the distances from A to the foci decreases to become less than a√2, and the circumcircle of triangle AMN moves towards BC so that it cuts it twice. Hence there are two such P when a√2 > b+ c.

If we move A out of the ellipse, the sum of the distances from A to the foci increases to become more than a√2, and the circumcircle of triangle AMN moves away from BC so that it no longer touches it. Hence there is no such P when a√2 < b+ c.

Therefore, for there to exist at least one such P, we must have a√2 ≥ b+ c.
I am grateful to my friend Jeremy Ho for helping to check this result.