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

Accepted Solution

Answer:for fixed AC, x = 45° maximizes the areafor fixed AB, x → 90° maximizes the areaStep-by-step 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.___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°._____Comment on the attachmentsThe 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.