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I know you can prove that the product of the roots of the monic quadratic $x^2+a_1x+a_0$ equals the $y$-intercept $a_0$ by comparing its coefficients to the coefficients of $(x-m)(x-c)$ where $m$ and $c$ are the roots. So $a_0 = mc$. This is how Vieta's formulas are derived.

However, I was wondering if there was a geometric proof of why this is true.

I drew the diagram below: enter image description here

In the diagram the roots are $(m, 0)$ and $(c, 0)$ while the y-intercept is $(0, b)$. I also drew the point directly above the vertex of the parabola (the midpoint of the roots) and created a few triangles. I tried using Stewart's Theorem on some of the triangles but couldn't seem to get the desired result that $b = mc$.

Can anyone provide some insight on how to prove this fact geometrically? Would I need to also draw the focus and directrix and do some geometry using those?

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    $\begingroup$ This answer of mine about the geometry of the quadratic formula may give some insights. Or not. The answer is a couple of years old now, and I've forgotten the details. :) $\endgroup$
    – Blue
    Aug 13 '20 at 17:46
  • $\begingroup$ @mihirb, what I noticed (don't know if correct or not) that: let the roots be point $A$ and $B$ and $y$-intercept be $C$. We've to prove $OC=OA.OB$. Now, the RHS is, like, $m^2$ and LHS $m$. How can they be equal? $\endgroup$
    – SarGe
    Aug 13 '20 at 18:23
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    $\begingroup$ @mihirb: See the proof of Property 1 from my other answer to that question. In the context of Property 1a, Property 1 tells us that the square erected on a semi-chord has area $4fp$, where $4f$ is the latus rectum, and $p$ is the distance from vertex to chord. So, the difference between two such areas is the product of $4f$ and the difference of two such distances ... but the latter is simply the distance (say, $g$) between the chords. Thus, the difference of squares is $4f\cdot g$, which we may interpret as the area the rectangle I described. $\endgroup$
    – Blue
    Aug 13 '20 at 23:00
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    $\begingroup$ @mihirb: It's always satisfying to learn that that an old answer still comes in handy. :) If you get from "I think [it proves my question]" to "I know", then I recommend expanding your comment into a complete answer. Cheers! $\endgroup$
    – Blue
    Aug 14 '20 at 1:19
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    $\begingroup$ @Blue Yeah it proves it. I added a complete answer $\endgroup$
    – mihirb
    Aug 14 '20 at 17:21
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Thanks to @Blue for showing me his old answer here which essentially answers my question. The property used from his answer (which is proved in his answer) is as follows:

Property 1. If $P$ is a point on a vertical opening parabola, then the point's horizontal displacement from the vertex is the geometric mean of the parabola's latus rectum and the point's vertical distance from the vertex.

Now here is a diagram from one of @Blue 's answers:

enter image description here

By Property 1. $|KV|^2 = |AK||KC|$ and $|VS|^2 = |AK||KO|$.

So $|KV|^2-|VS|^2 = |AK|(|KC|-|KO|) = |AK||OC|$.

Thus, $(|KV|-|VS|)(|KV|+|VS|) = |AK||OC|$.

But $|KV|-|VS| = OR_{-}$ and $|KV|+|VS| = OR_{+}$.

Which means that $|OR_{-}||OR_{+}| = |AK||OC|$.

For a monic quadratic, $|AK| = 1$ So we get that $|OR_{-}||OR_{+}| = |OC| = c$ as desired.

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I will use $y = (x - 3)(x - 7)$ to illustrate the idea.

The lines $y = 3$ and $x = 1$ cut the large $7 \times 21$ rectangle into $4$ sub-rectangles.

enter image description here

The diagonal will further cut the rectangle into a $(21 - 3) \times (7 - 6)$ rectangle and a $(7 - 1) \times (3 - 0)$ rectangle. The two will be equal in area.

Adding the $1 \times 3$ rectangle to both will give the required result.

Added: The sequence of construction is (1) draw the largest rectangle; (2) Let the circle (O, radius = 3) cut the y-axis at (0, 3); (3) draw the line y = 3; (4) Let the ine x = 1 cut y = 3 at (1, 3); (5) join O(0, 0) with P(1, 3) and join P(1, 3) with Q(7, 21). OPQ will then be the diagonal of the largest rectangle.

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  • $\begingroup$ Where does this use a property of the parabola? $\endgroup$
    – aschepler
    Aug 13 '20 at 19:21
  • $\begingroup$ @aschepler It doesn't seem to use any parabola properties anywhere yet it seems to be correct and give the right answer $\endgroup$
    – mihirb
    Aug 13 '20 at 19:23
  • $\begingroup$ I mean, does it just prove that $3 \cdot 7 = 3 \cdot 7$ and $rs = rs$? Is it assuming or proving that the $y$-intercept actually equals the product of the roots? $\endgroup$
    – aschepler
    Aug 13 '20 at 19:30
  • $\begingroup$ It seems to show it without assuming the y-intercept is the product of roots. You do have to able to construct a unit length segment though. But it seems this argument could work for any shape not just a parabola or polynomial. Not sure what's up with that? $\endgroup$
    – mihirb
    Aug 13 '20 at 19:56
  • $\begingroup$ @aschepler I think the issue is we don't know if the diagonal of the red subrectangle in the bottom left and diagonal of the white subrectangle on the top right are collinear. So the splitting in half with the diagonal of the whole rectangle doesn't work necessarily. Unless there is a parabola property that guarantees it. The only property I can see right now that would guarantee it is the product of roots equalling y-intercept (so slopes are equal) so it is a circular argument. $\endgroup$
    – mihirb
    Aug 13 '20 at 20:17

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