@@stavkomarnitsky8466 Area = base x perpendicular height, base is AB so doesn’t change, So we only care about minimising perpendicular height of C (from AB). Suppose gradient of tangent at C is not equal to gradient of AB. Then we can shift C a little bit to the left or to the right to decrease perpendicular height and thus area. Therefore the minimum is when the gradient of tangent is same as gradient of AB
Rotate everything so that line AB is horizontal. Then, moving C left or right doesnt change triangle area, however moving it higher (away from AB) increases area since it increases height of the triangle. So you need to minimize distance between C and closest point on AB(→minimize height). and if you understand how derivatives work with minimization you know that slope of x² at point C must be equal to slope of line AB. So 2x = 4/6→x=1/3, y=1/9@@stavkomarnitsky8466
My method: the area of the triangle is the base |AB| times the height, and the heigh is just the distance from C(t,t²) to the line AB. AB is given by -2x+3y+4=0, so the distance from C to AB is just (-2t+3t²+4)/sqrt(13). This is always positive, and minimal when t = 1/3 (by taking the derivative or completing the square)
Rotate everything so that AB becomes a horizontal line. Then we can see that we must find the minimum of the rotated parabola to achieve the minimum height of a triangle ABC. This will only happen when the slope of the rotated parabola at the minimum point is 0 (same as rotated AB). And if we rotate back to original position, we see that the slope at the point is the same as AB, aka find the point in parabola such that the slope at that point is the same as slope AB.
Calling the point on the parabola C, with co-ordinates (t, t^2), the angle CA makes with the x-axis has sine t^2/CA and cosine (6-t)/CA. Meanwhile the angle BA makes with the axis has sine 4/BA and cosine 6/BA. From this, sine of combined angle CAB is (6t^2-4t+24)/(CA*BA). Plugging this into the general area of a triangle the area is 3t^2-2t+12, which has a minimum of 35/3 at t=1/3. You don't even need calculus, can just complete the square.
Omg, i solved this in maybe the hardest possible way 😭😭 I spent the last hour writing a function of the distance from the line to the parabola, taking its derivative, setting it equal to zero and solving for x. My largest simplified coefficient in my derivative was 5 digits long. It was a pain but somehow once I started I kinda felt like I was at war with it. Omg.
@@graf_paper interesting! Definitely some tricky ways to solve this! In an interview they'd definitely steer you into trying to solve this in a slightly more easy way!
Easier solution: we know gradient of tangent at C must be gradient of AB, so easy to see that C = (1/3,1/9)
@@amansparekh bingo! Nice!!
Could you explain how we know that?
@@stavkomarnitsky8466 Area = base x perpendicular height, base is AB so doesn’t change, So we only care about minimising perpendicular height of C (from AB). Suppose gradient of tangent at C is not equal to gradient of AB. Then we can shift C a little bit to the left or to the right to decrease perpendicular height and thus area. Therefore the minimum is when the gradient of tangent is same as gradient of AB
Rotate everything so that line AB is horizontal. Then, moving C left or right doesnt change triangle area, however moving it higher (away from AB) increases area since it increases height of the triangle. So you need to minimize distance between C and closest point on AB(→minimize height). and if you understand how derivatives work with minimization you know that slope of x² at point C must be equal to slope of line AB. So 2x = 4/6→x=1/3, y=1/9@@stavkomarnitsky8466
Ohhh that's smart. Thanks@alphazero339
Nice problem, love these videos!
My method: the area of the triangle is the base |AB| times the height, and the heigh is just the distance from C(t,t²) to the line AB. AB is given by -2x+3y+4=0, so the distance from C to AB is just (-2t+3t²+4)/sqrt(13). This is always positive, and minimal when t = 1/3 (by taking the derivative or completing the square)
@@skylardeslypere9909 nice solve!
Rotate everything so that AB becomes a horizontal line. Then we can see that we must find the minimum of the rotated parabola to achieve the minimum height of a triangle ABC. This will only happen when the slope of the rotated parabola at the minimum point is 0 (same as rotated AB). And if we rotate back to original position, we see that the slope at the point is the same as AB, aka find the point in parabola such that the slope at that point is the same as slope AB.
@@warmpianist ah nice solve! I like it!
Calling the point on the parabola C, with co-ordinates (t, t^2), the angle CA makes with the x-axis has sine t^2/CA and cosine (6-t)/CA. Meanwhile the angle BA makes with the axis has sine 4/BA and cosine 6/BA. From this, sine of combined angle CAB is (6t^2-4t+24)/(CA*BA). Plugging this into the general area of a triangle the area is 3t^2-2t+12, which has a minimum of 35/3 at t=1/3. You don't even need calculus, can just complete the square.
@@hugh081 yep, that works nicely!
y=2/3x-4 C:y=-3/2(x-x0)+x0² x=24/13+9/13x0+6/13x0² h=√13|11 2/3+3(x0-1/3)²|/13≥35√13/39 S=35/3
Omg, i solved this in maybe the hardest possible way 😭😭
I spent the last hour writing a function of the distance from the line to the parabola, taking its derivative, setting it equal to zero and solving for x.
My largest simplified coefficient in my derivative was 5 digits long. It was a pain but somehow once I started I kinda felt like I was at war with it.
Omg.
@@graf_paper interesting! Definitely some tricky ways to solve this! In an interview they'd definitely steer you into trying to solve this in a slightly more easy way!
2 vids, 1 day. nice
Thank you!!
@@JPiMaths upload more so i can watch as many as i can in 1 day before my interview lol