Track / Stage 6 / 227 of 400 #1707 of 2444

Problem 1707

National Olympiad, first round

Geometry Difficulty 6.4 Prove it Philippine Mathematical Olympiad · Philippines

Let $PMO$ be a triangle with $PM = 2$ and $\angle PMO = 120^\circ$ . Let $B$ be a point on $PO$ such that $PM$ is perpendicular to $MB$ , and suppose that $PM = BO$ . The product of the lengths of the sides of the triangle can be expressed in the form $a + b \sqrt[3]{c}$ , where $a, b, c$ are positive integers, and $c$ is minimized. Find $a + b + c$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:
Extend $PM$ to a point $C$ such that $PC \perp OC$ . Since $\angle PMO = 120^\circ$ , $\angle CMO = 60^\circ$ and $\angle COM = 30^\circ$ . Let $PB = x$ and $MC = a$ . Then $CO = a \sqrt{3}$ and $OM = 2a$ . Moreover, $\triangle PMB$ and $\triangle PCO$ are similar triangles. Thus, we have
$\frac{2}{2+a} = \frac{x}{x+2}$
so $x = \frac{4}{a}$ .

Furthermore, by the Cosine Law on side $PO$ of $\triangle PMO$ , we have
$(x+2)^2 = 4 + 4a^2 + 2a$
Plugging in $x = \frac{4}{a}$ and expanding, we have
$\frac{16}{a^2} + \frac{16}{a} + 4 = 4 + 4a^2 + 4a$
and so $4 + 4a = a^4 + a^3$ . Hence $a^3 = 4$ and $a = \sqrt[3]{4}$ . Thus, $x = \frac{4}{\sqrt[3]{4}}$ .

It follows that the product of the lengths of the sides of the triangle is
$(2a)(2)(x+2) = (2\sqrt[3]{4})(2)(2\sqrt[3]{2} + 2) = 16 + 8\sqrt[3]{4}$
so $a + b + c = 16 + 8 + 4 = 28$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.