Geometry problem



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If that what you mean by similar yes.
And these angles are always equal because the lines are parallel, there is no other option.

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Thanks guys for all the replies!
I’m studying all of them, i’ll get back asap!

Pure gold! Thanks!
I did not understand the formulas, you worked only with areas and no trigonometry at all! Well done!

I’ve added a switch on the quadratic formula and calculated the slope:

Re describing the situation (i’ve renamed the parameters (why didn’t i use such at start?)):
this is a slider left-right animation where many content is seen, at each user input it slides left or right with a smooth animation.
If the user press the buttons during the animation, i wanted the animation to keep current velocity and adapt the acceleration to reach the target distance in the wanted time.

2021-06-17 12_37_02-Window
Given initial velocity V, distance to travel S and time T, find the acceleration.
The animation will speed up if current speed is too low, or bounce back if it was too fast.
Now it will also work if initial speed is negative, which can happens.

find top point

find top (22.8 KB)

Thanks again guys!

I’ll convert this to javascript and i’ll try to link here a working page.

A = (H+(T*b-H)/4)*(T-H/b)+H^2/(2*b)

This is an equation for the area using trigonometry. b is my simplification for tanβ which is the inside angle at the right end of the shape. Instead of b you could also use half of the the angle at the tip which is always 90° - b. I’ve used this website to solve the equation for b.

b = (((4*A^2-4*A*H*T+2*H^2*T^2)/T^4)^0.5)+(2*A-H*T)/T^2 (36.4 KB)

Below is a simplified version: (19.3 KB)


No need for this because the second solution always < 0 which leave only the first one

Another way to find the same equation: 2*T*M²+4*(T*H-A)*M+(T*H-2*A)*H = 0
And always: (T/N)-(H/M) = 2 , [N and M > 0]

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Well, modern days need modern tools…

The upper point would be (take the first point as 0,0):
image (7.0 KB)

By the way the equation of the upper point, while H/T is fixed & A is changing is:


so the asymptote is x=T/2

while T is changing is:


while H is changing is:



Damn guys! Y’all are going all in here! :smiley:

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If you noticed with this distance is fixed

The area always the same when we move the point along the curve, maybe there is a geometrical solution.

image (19.2 KB)

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This is another solution using Trigonometry to find z (23.3 KB)

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I don’t have anything to add rather than to say that I love this forum :rofl: :heart_eyes:


Another way to get at the solution while avoiding trig functions (done on a post-it note, the way math was meant to be done :grin:): complete the left triangle to get two mirror-image triangles each of height y and width x. The total area of both triangles is xy and the newly added triangle piece has area \frac{1}{2}H(2x-T). Putting the areas together and using rise/run for similar triangles, we get two equations in our two unknowns x and y:

  • xy = A + \frac{1}{2}H(2x-T)
  • \frac{y}{x} = \frac{H}{2x-T}

This can be simplified down to a quadratic equation for x, with y given in terms of x:

  • Hx^2 - 2(HT-A)x + T(\frac{1}{2}HT-A) = 0
  • y = \frac{1}{x}[A + H(x-\frac{1}{2}T)]

So apply the quadratic formula to get the width x, and then plug it into the second equation to get the height y.