Go back to 1950 for the answer

Geometry Level 5

$\large (1950\cos \alpha, 1950\sin \alpha) , \quad (1950\cos \beta, 1950\sin \beta) , \quad (1950\cos \gamma, 1950\sin \gamma)$

A triangle has vertices on the coordinates as described above, where $\alpha,\beta$ and $\gamma$ satisfy the following system of equations,

$\large \cos \alpha + \cos \beta + \cos \gamma = \dfrac{411}{325} , \qquad \sin \alpha + \sin \beta + \sin \gamma = \dfrac{872}{325}.$

If the orthocenter of this triangle has a coordinate of $(a,b)$ , compute $a+b$ .

Give your answer to 2 decimal places.

The answer is 7698.00.

1 solution

Shourya Pandey
May 15, 2016

Let $A = (1950cos\alpha,1950sin\alpha), B= (1950cos\beta, 1950sin\beta), C=(1950cos\gamma,1950sin\gamma)$ .

Clearly, the given points lie on the circle $x^2+y^2 = 1950^2$ , so that the circumcentre $O$ of $\Delta ABC$ is the origin $(0,0)$ . Also, if $G, H$ respectively denote the centroid and the orthocentre of the triangle, then using vertices as vectors, we get

$H- O = 3 (G-O)$ , or $H= 3G -2O = 3 (\frac {A+B+C}{3})- 2(O)$ = $A+B+C-O$ .

$(a,b)=H = (1950cos\alpha +1950cos\beta+1950cos\gamma, 1950sin\alpha+1950sin\beta+1950sin\gamma) = (2466,5232)$ , so that $a+b=2466+5232 =7698$ .

Exactly the intended solution! +1

Nihar Mahajan - 5 years, 1 month ago

Bro i did the same just forgot to multiply 1950 at last😭😭😭😭

Varun Rustagi - 4 years, 9 months ago

Nice solution! can also be further solved(after establishing circumcentre is origin) by finding coordinates of centroid using the same ratio relation that has been used in determing position vector.

Deepak Kumar - 5 years ago

Go back to 1950 for the answer

The answer is 7698.00.

1 solution

0 pending reports