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This problem appeared on the network, and although it looks simple I am not sure of the result. enter image description here

The polygon $ABCD$ is a square with side $4$ cm and the curve inscribed inside the square is a parabola, determine the length of the segment $AP$, if $QC$ is $3$ cm.

problem

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  • $\begingroup$ Welcome to MathSE. Can you show your attempts for to solve the problem? Maybe a construction in geoebra? $\endgroup$
    – user798113
    Commented Jan 3, 2021 at 23:48
  • $\begingroup$ Where did you see this problem "n the network"? Can you link to it? $\endgroup$
    – Calvin Lin
    Commented Jan 4, 2021 at 0:45
  • $\begingroup$ @ Calvin Lin on the page that I don't remember said admission problem To a university in South Korea (Csat) $\endgroup$
    – BeTDa
    Commented Jan 4, 2021 at 2:57
  • $\begingroup$ @ Александр Пальма , is a problem of admission to a university, I am trying to solve it by hand $\endgroup$
    – BeTDa
    Commented Jan 4, 2021 at 2:59

2 Answers 2

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The quadratic

  • Passes through the coordinates $(0,0), (4, 3)$.
  • Has a maximum at $X < 4$
  • Has the form $ y = - A (x-X)^2+4$.

Can you take it from here? If not, explain what you've tried and where you're stuck.

$y = -\frac{9}{16}(x-\frac{8}{3})^2 + 4$.
Hence $AP = \frac{8}{3}$.

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  • $\begingroup$ @ Calvin Lin 1) 0=-A(0+B/2A) $\endgroup$
    – BeTDa
    Commented Jan 4, 2021 at 2:12
  • $\begingroup$ @BeTDa What are you trying to say? What is B? What is A? There are 2 (linear) equations with 2 unknowns, so you should be able to solve for $A$ and $X$. $\endgroup$
    – Calvin Lin
    Commented Jan 4, 2021 at 2:30
  • $\begingroup$ @ Calvin Lin this I wanted to put 1) 0=-A(0-B/2A)^2 +4 -> B^2=4A , 2) 3=-A(4-B/2A)^2 ->B=3 and A =9/4, maximum atx= -B/2A ......The value of x should be positive, I get lost there $\endgroup$
    – BeTDa
    Commented Jan 4, 2021 at 2:41
  • $\begingroup$ Ah, I see what your B/2A is. How did you get $ B^2 = 4A$ from the first equation? Shouldn't it be $ B^2 = 16A $? $\endgroup$
    – Calvin Lin
    Commented Jan 4, 2021 at 3:27
  • $\begingroup$ The second equation should be $ 3 = -A ( 4 - B/2A) ^2 + 4 $ right? So that's an error in that too. Solving these 2 linear equations, you should get $A = 9/16, B/2A = 8/3$ (reject the $B/2A > 4$ case. $\endgroup$
    – Calvin Lin
    Commented Jan 4, 2021 at 4:13
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Hope you know differentiation to find maximum/minimum of standard parabola curve of form $y = ax^2+ bx + c $.

Taking B as origin since it passes through (0,0), we get c=0

$$ y =a x^2+bx \tag1 $$

Differentiate to get max tangent point

$$ 2 ax +b=0\to x_{t} =- b/(2a) \tag 2$$

plug into (1) and simplify

$$ y_t= -b^2/(4a) = 4 \text{ (given)} \tag3$$

Point $(4,3)$ is given lying on the parabola

$$ 3= 16 a +4 b \tag4 $$

Eliminate $a$ between (3) and (4) and simplify resulting in the quadratic equation

$$ b^2-4b+3=0,\; (b-1)(b-3)=0 ; b=(1,3);\;\tag5 $$

From(3) corresponding $a$ values are

$$ \left(\dfrac{-1}{16},\dfrac{-9}{16} \right) \tag 6$$

Plug $(a,b) $ values into (2) we get two values of tangent point

$$ x_t=(8,8/3) \tag7$$

The first value lies outside interval $0<x_t<4$ so is discarded, second value is taken. The parabola has equation

$$y=3 x\left(1-\dfrac{3x}{16} \right) \tag 8$$

whose graph is the one given in your question.

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  • $\begingroup$ @Narasimham-hi, Thanks for your time, as you would do the problem without using derivatives, the concepts of maximum and minimum are known to me $\endgroup$
    – BeTDa
    Commented Jan 4, 2021 at 17:06

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