Jon P. answered 07/04/15
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Knowledgeable Math, Science, SAT, ACT tutor - Harvard honors grad
The algebra to do this problem is messy, but it works. I'm not going to work out every step for you, but I can show you how to approach the problem.
1. Draw the points on a coordinate axis. Notice that A is in quadrant II and B is in quadrant I. That means that AB crosses the y-axis. Call the point where it crosses the axis M.
2. Notice that the y-axis divides Δ AOB into two triangles, AMO and BMO. You can find the area of Δ AOB by finding the areas of AMO and BMO separately and adding them.
3. MO is shared by both triangles AMO and BMO. In fact it is a base of both triangles. The height of AMO is 1, because A is 1 unit from MO (the y-axis). And the height of BMO is 1/λ. To find the area of both triangles, you need to find the length of MO.
4. Consider AB and a line with an equation, y = mx + b. You know two points on the line, A and B, so you can find the slope. And using the slope and one of the points, you can find the y-intercept, b. But b is just the same as the length of MO! So by finding the equation of the line for AB, you've found the length of MO.
5. Once you have MO, you can find the area of Δ's AMO and BMO and add them to find the area of Δ AOB. You will be able to show that it is equal to (1+λ) / (2λ2).
6. Notice that the area of Δ AOB is also equal to 1/2 bh, where AB is the base, and the "height from O to the line joining A and B" is the height.
7. Calculate b, the length of AB, using the distance formula.
8. Divide the area you got in step 5 by 1/2 b and that will give you h, which is the second thing you were asked for. You will be able to show that it is equal to 1/(√((2λ2) - 2λ + 1))
9. If you have studied calculus, you will be able to maximize the height using derivatives. If not, then...
10. Notice that to find the value of λ that gives the LARGEST value of h is the same as finding the value of λ that gives the SMALLEST value of the denominator of 1/(√((2λ2) - 2λ + 1)). In fact, it's the same as finding the value of λ that gives the smallest value of the expression inside the square root sign -- (2λ2) - 2λ + 1. That's equal to 4λ2 - 2λ + 1.
11. Change λ to x and think of this as a quadratic function that you want to graph on the x-y plane. It will form a parabola that opens upward, and the vertex will be the point where the function reaches its minimum. The x coordinate of the vertex will be the value of λ that gives you the maximum value of the height from O to AB.
Complicated? Yes. But do each step one at a time, and the answer will be there.
If you can find a simpler way to do this, then by all means try it!!

Jon P.
tutor
Well, first of all, you have to do steps 1-3 yourself, because I can't draw anything here. Just follow what I said and make a picture. Pick a couple of values for λ, like 1 and 1/2, so you can see how the picture looks in a couple of cases. Then I'll start with step 4.
AB is a line with an equation y = mx + b. We know the coordinates for A and B, so we can find the slope, m:
m = (1/λ2 - 1) / (1/λ - (-1)) =
(1/λ2 - 1) / (1/λ + 1) =
[(1 - λ2) / λ2] / [(1 + λ) / λ] =
[(1 - λ2) / λ2] * [λ / (1 + λ)] =
[(1 - λ)(1 + λ) / λ2] * [λ / (1 + λ)] =
(1 - λ) / λ
So m = (1 - λ) / λ. What's b? Since we know that (-1, 1) is on the line, we can set up an equation to find b.
(See next comment for that.)
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07/05/15

Jon P.
tutor
Here's the equation for b:
y = mx + b
y = x (1 - λ) / λ + b
1 = (-1)(1 - λ) / λ + b
1 = (λ - 1) / λ + b
1 - (λ - 1) / λ = b
[λ - (λ - 1)] / λ = b
1 / λ = b
So now we know b is 1 / λ. Remember, b is the same as the length of MO. That takes care of Step 4.
Do you see how to use 1 / λ as the length of MO to find the area of Δ's AMO and BMO?
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07/05/15
David K.
07/05/15