You're not saying that because it looks that way, right? Where is thy evidence :) Let the one who survives, survive with an evidence! Let the one who dies, die with an evidence!

I can immediately reject 1/√2 . imagine it this way: if it was a quarter circle at the same position, intuitively sin(Lamedh=45)= 1/√2 is the answer due to symmetry. having a semi-circle must change the natural of the problem. most likely the angle be lower than 45 degrees. that is because areas A and B are no longer symmetric. i agree with newzad, 1/2 looks like the answer. in that case angle Lamedh=30 degrees.

Here'e the proof: Let the triangle sides along the x-axis and y-axis be x & y resply and let the hypotenuse be z. If R be the radius of the semi-circle then we have: z^2=x^2+y^2 and y/z=R/(x-R) by figure geometry. Now the area of the triangle is xy/2 and A+B = xy/2-π*R^2/2. We have,therefore, to minmize a=xy=Rx√(x/(x-2R)). After doing the usual da/dx=0, we obtain x=3R from which z=2R√3 and y=R√3 while sin(ξ)=y/z= 1/2.

What you've actually proved is that given is the radius of the incircle the area/perimeter of the circumscribed triangle is minimal iff the triangle is equilateral. (Just reflect the diagram through the x-axis) Somewhat expected... A standard corollary as it stands of the fact that the cotangent function is convex on the interval [0 Pi].

So this is a nice extremum problem, indeed, and you can even watch it from beyond standard calculus.

ל

ReplyDeleteis provided for those, seeking the character. It must be Hebrew.

D

ReplyDeleteYou're not saying that because it looks that way, right? Where is thy evidence :)

ReplyDeleteLet the one who survives, survive with an evidence!

Let the one who dies, die with an evidence!

I have the solution, but it is long, i don't like long solutions:) if i found a short one i will write it

ReplyDeleteMy take is that (A+B) is minimum when the triangle is isosceles and minimum itself which gives us sin(ξ)= 1/√2 or Option E

ReplyDeleteI can immediately reject 1/√2 . imagine it this way: if it was a quarter circle at the same position, intuitively sin(Lamedh=45)= 1/√2 is the answer due to symmetry. having a semi-circle must change the natural of the problem. most likely the angle be lower than 45 degrees. that is because areas A and B are no longer symmetric.

ReplyDeletei agree with newzad, 1/2 looks like the answer. in that case angle Lamedh=30 degrees.

-binary

I checked numerically. sin(ξ)= 1/2 or Option D as suggested by Binary seems right

ReplyDeleteHere'e the proof: Let the triangle sides along the x-axis and y-axis be x & y resply and let the hypotenuse be z. If R be the radius of the semi-circle then we have: z^2=x^2+y^2 and y/z=R/(x-R) by figure geometry. Now the area of the triangle is xy/2 and A+B = xy/2-π*R^2/2. We have,therefore, to minmize a=xy=Rx√(x/(x-2R)). After doing the usual da/dx=0, we obtain x=3R from which z=2R√3 and y=R√3 while sin(ξ)=y/z= 1/2.

ReplyDeletethank you Joe 4 this nice work!

ReplyDelete-binary

What you've actually proved is that given is the radius of the incircle the area/perimeter of the circumscribed triangle is minimal iff the triangle is equilateral. (Just reflect the diagram through the x-axis) Somewhat expected... A standard corollary as it stands of the fact that the cotangent function is convex on the interval [0 Pi].

ReplyDeleteSo this is a nice extremum problem, indeed, and you can even watch it from beyond standard calculus.

Sorry, sorry, the cotangent function is convex on (0, Pi/2] but that's what is needed.

ReplyDelete