Question

Hello,

I propose this exercise if anyone can answer me.
Let ABC be a right triangle, of hypotenuse BC = a, where
`A\widehat{B}C=x`.
Let A' be the symmetric of A with respect to BC.
Determine x so that:
-the area of ​​the triangle AA'C is maximum
Show and justify all the steps (take the picture).

Thank you.

Answer 1

• for fixed AC, x = 45° maximizes the area
• for fixed AB, x → 90° maximizes the area

Explanation:

Call the point of intersection of AA' and BC point X. Then ...

CX = AC·cos(90°-x) = AC·sin(x)

and the area of AA'C is ...

area = AC·CX·sin(90°-x) = AC²·sin(x)cos(x) = (1/2)AC²·sin(2x)

Obviously, area is maximized for 2x = π/2, or x = π/4 when AC is fixed.

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On the other hand, ...

AC = AB·tan(x)

so the area of the triangle is ...

area = (1/2)AC²·sin(2x) = (1/2)(AB·tan(x))²·sin(2x) = AB²·sin(x)³/cos(x)

For fixed AB, area approaches infinity as x approaches 90°.

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Comment on the attachments

The attached diagrams show AC=1 and B free to move. Values of x around 45° are shown. The number in the middle of the figure is the approximate area of ΔAA'C.