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Geometry Level 2

An equilateral triangle has a vertex at $\left( 5,10 \right)$ . If this triangle has an orthocenter at $\left( 3,4 \right)$ , find the area of the triangle.

If your answer is of the form $a\sqrt{b}$ , where $b$ is square free, find $a+b$ .

The answer is 33.

1 solution

Akeel Howell
Apr 5, 2018

For any equilateral triangle, the orthocenter is in the same position as the centriod, whose distance from any of the vertices is twice as far as it is from the side opposite that vertex. Hence, the altitudes of this triangle have length $\dfrac{3}{2}\sqrt{\left( 10 - 4 \right) + \left( 5 - 3 \right)^2} = 3\sqrt{10}$ .

Let the length of each side be $L$ . By Pythagoras' theorem, $3\sqrt{10} = \sqrt{L^2 - \dfrac{L^2}{4}} = \dfrac{L}{2}\sqrt{3} \ \implies \ L = 6\sqrt{\dfrac{10}{3}}$ .

Therefore, the area is $\dfrac{1}{2} \times 6\sqrt{\dfrac{10}{3}} \times 3\sqrt{10} = 30\sqrt{3}$ and hence, $a+b = \boxed{33}$ .

Nice question and solution. As the Euler "line" of an equilateral triangle degenerates to a single point, (i.e., is not actually a line), you probably don't need to mention the Euler line in your question, (it is not a necessary piece of information anyway).

Brian Charlesworth - 3 years, 2 months ago

Point taken (pun intended). I’ve updated the question.

Akeel Howell - 3 years, 2 months ago

Is this enough info?

The answer is 33.

1 solution

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