Perpendicular medians

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Source: Finnish Mathematics Competition 2006, Final Round, Problem 4

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#1 h Nov 14, 2011, 10:04 PM • 2 Y

Y by Adventure10, Mango247

Two medians of a triangle are perpendicular. Prove that the medians of the triangle are the sides of a right-angled triangle.

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#2 h Nov 15, 2011, 12:05 AM • 2 Y

Y by Adventure10, Mango247

I am tempted to move this from Geometry Unsolved to Pre-olympiad, but am refraining from doing so as it is from an official olympiad. If a different mod (probably Luis) sees fit, go ahead and move it.

Proof. Let $\triangle ABC$ be an arbitrary triangle with sides $a,b,c$ and respectively corresponding medians $m_a,m_b,m_c$ which concur at the centroid $G$ .
Without loss of generality, suppose that $m_a\perp m_b$ . Then $AB^2=AG^2+BG^2$

$\implies c^2=\left(\frac{2}{3}\cdot m_a\right)^2+\left(\frac{2}{3}\cdot m_b\right)^2$

$=\frac{4}{9}\cdot\left(\frac{2b^2+2c^2-a^2}{4}+\frac{2c^2+2a^2-b^2}{4}\right)=\frac{a^2+b^2+4c^2}{9}$

$\implies a^2 +b^2=5c^2\implies a^2+b^2+4c^2=2a^2+2b^2-c^2$

Finally, note that $m_a^2+m_b^2=\frac{2b^2+2c^2-a^2}{4}+\frac{2c^2+2a^2-b^2}{4}$

$=\frac{a^2+b^2+4c^2}{4}=\frac{2a^2+2b^2-c^2}{4}=m_c^2$ . $\square$

Thus, $m_a\perp m_b\implies m_a^2+m_b^2=m_c^2$ , and clearly the cyclic versions hold.

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#3 h Dec 22, 2013, 7:10 PM • 2 Y

Y by Adventure10 and 1 other user

Remark: Let $\triangle$ be the area of the original triangle, then the right triangle in question has area $\tfrac{3}{4}\triangle$ .

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#4 h May 4, 2019, 8:02 PM • 1 Y

Y by Adventure10

socrates wrote:

Two medians of a triangle are perpendicular. Prove that the medians of the triangle are the sides of a right-angled triangle.

The vectors represented by the medians are $\frac{B+C}{2} - A, \frac{C+A}{2}-B,$ and $\frac{A+B}{2} - C$ , and they add up to $0$ . This means we can translate the medians to form a triangle. Since translation preserves angles between lines, we are done.

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