Need help with this SEA Math Competition question! Geometry

Question

Two of the altitudes of a scalene triangle with integer-length sides are 6 and 14 units. Given that the third altitude is also an integer, what is the maximum it can be?

This was a multiple choice question, and the options were; A.7

B.9

C.11

D.12

E.13

The strategy I used was; Derive an equation or an inequality to represent the third altitude and maximise the variables inside the equation, I tried doing this by naming all lengths, including the ones that were split up by the altitude. Then creating inequalities based of off the triangle inequality. I tried reducing them down into a few inequalities so that I could maximise the length of the other altitude, but it didn't work :(. I realised there were too many variables to deal with and gave up.

Answer 1

Inequalities, yes. But which ones?

We must know two basic facts about triangles to solve this problem:

$\color{blue}{FIRST.-}$ THE PRODUCT OF THE LENGTHS OF A SIDE AND THE ALTITUDE TO THAT SIDE EQUALS TWICE THE AREA.

So if denote twice the area by K and let a, b, and c be the sides corresponding to the altitudes of length 4, 12, and h, we get the formulas,

$a = K/4$, $b = K/12$, $c = K/h$

$\color{blue}{SECOND.-}$ A TRIANGLE EXISTS WITH SIDE LENGTHS a, b, AND c IF AND ONLY IF THEY SATISFY THE THREE TRIANGLE INEQUALITIES:

$a < b + c$, $b < c + a$, and $c < a + b$.

At this point, with some algebra knowledge, it is possible to deduce the largest possible value of h immediately.

If you are not confident with algebra, you can test the integer values of h starting at h = 1 (and skipping h = 4 and h = 12 to avoid isosceles triangles) to discover the largest value of h for which a, b, and c satisfy the triangle inequalities.

Answer 2

Revision: The triangle of interest is $ABC$ with given elevations of $14$ and $6$. My solution is based on conjecture but I do use two extreme examples to determine the nearest integer from the answer options.

Because $BD = 6$ is less than half the elevation $14$, $DC$ will always be shorter than $AD$. As angle $\alpha$ approaches $90$ degrees and angle $\theta$ becomes smaller, the ratio of $DC$ to $AD$ reaches a maximum.

For an extreme case where $\theta = 0.1$ then $\alpha = 89.76667$. $AC = 6016.04$ and $AD = 3437.74$ The elevation $EC \approx 6\cdot \frac{6016.04}{3437.74} \approx 10.5$ max. An extreme in the other direction where $\theta = 25$ then $\alpha = 9.561$ with a resulting elevation $EC \approx 7.18$ Between these extremes answer $B = 9$ is therefore the max integer possible.