Not enough info...

Geometry Level 5

An acute isoceles triangle ABC has integer side lengths of a,b,c. The perpendicular feet are drawn from each angle and intersect at H. Then, each side is perpendicularly bisected to meet at O. Point N is drawn such that N is on segment $\overline{OH}$ . Point L is drawn at the midpoint of line $\overline{AH}$ such that $\overline{LN}=\frac{1}{2}\overline{OA}$ . Triangle $\triangle ABC$ has an area of 48. If the sum of the two smallest possible lengths of $\overline{NH}$ can be expressed as $\frac{x}{y}$ for positive co-prime $integers$ x,y, find x+y.

Also, this is probably the worst problem that I've made cuz this is just stupid hard and uses practically unknown formulas.

HINT: try to see if you can do it without this information below

One side length is 10 of both triangles.

The answer is 71.

1 solution

Trevor Arashiro
Oct 14, 2014

Google 9-point center and read the wolfram mathworld entry it helps to under stand this solution.

This is the lazy way to do this problem, there is a way that is much more mathematical that requires much less knowledge than this one, see if you can figure it out.

The 9-pointcenter is halfway between the ortho center and circum center. The distance between the orthocenter (H) and circumcenter (O) is $\overline{OH}^2=9\left(\dfrac{abc}{4\triangle}\right)^2-(a^2+b^2+c^2)$ where a,b,c are the sides and $\triangle$ is the area of the triangle (the first half of the equation in parentheses=the circum radius). The first few heron triangles that have an area of 48 are (10,10,12) and (10,10,16). Plugging them in and dividing the answers by two we get

$11/8+13/2=63/8$ .

$63+8=71$

Not enough info...

The answer is 71.

1 solution

0 pending reports