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Help me solve this problem mathbros

Anonymous Wed 09 Mar 21:55:18 2022 No.14291715 View Reply Original Report

Quoted By: >>14291905 >>14291940 >>14291987

I heard that there's a lot of mathematicians and geniuses here so maybe you wouldn't mind helping a polish HS student with this problem? description is as follows:

You are given parabola with an equation of y = x^2 and points A = (0, 2) and B = (1,3), whose node C is lying on that parabola. Let "m" be first coordinate of C node.

a) Determine the area of a triangle ABC as a function of m variable.
b) Determine all m values for which triangle ABC is acute-angled.

Anonymous

Anonymous Wed 09 Mar 2022 22:20:52 No.14291755 Report

Quoted By: >>14291765

>If you have all three sides, you'll use Heron's formula, and the formula for the area of a triangle.

Anonymous

Anonymous Wed 09 Mar 2022 22:26:00 No.14291765 Report

Quoted By:

I'll 100% get back to you but
>>14291755
has the right idea.

Anonymous

Anonymous Wed 09 Mar 2022 23:13:41 No.14291864 Report

Quoted By:

$Area = \sqrt{\frac{3}{8}m^4 - \frac{22}{9}m^2 + \frac{17}{2}}$

For acute angles, $-\sqrt{17}-1 \leq 2m \leq \sqrt{17} - 1$ , or for integer solutions

$m = \{-2,-1,0,1\}$

t. physicist

Anonymous

Anonymous Wed 09 Mar 2022 23:27:10 No.14291905 Report

Quoted By: >>14291908

>>14291715
Alternatively, consider this approach, which does not require the use of Heron's formula.

Let P be a point that varies across the line AB. Therefore, the line perpendicular to AB through P may be given by the equation y = c - x, for some varying value of c.
This line intersects the parabola at x^2 = c - x, or x^2 + x - c = 0. The area of ABC may be written 1/2 * sqrt(2) * d, where d is the distance from P to the intersection point. The quadratic equation may be solved to give x = -1/2 +/- sqrt(1 + 4c) which is defined in the reals for c >= -0.25. You should be able to find the c value in terms of the m value, for in this case (m + 1/2)^2 = 1 + 4c.

Anonymous

Anonymous Wed 09 Mar 2022 23:29:31 No.14291908 Report

Quoted By:

>>14291905
Meant sqrt(1+4c)/2, but you get the method of approach

Anonymous

Anonymous Wed 09 Mar 2022 23:46:40 No.14291940 Report

Quoted By: >>14291959

>>14291715
Let P be given by the intersection of y = c-x with a varying c, such that c-x intersects 1+x at x=(c-1)/2, therefore P has the coordinates ((c-1)/2, (c+1)/2)

Now let y = c-x intersect the parabola such that m^2+m-c = 0, such that c=m^2+m. The distance d, from P to (m m^2), is given by sqrt(((c-1)/2 - m)^2 + ((c+1)/2 - m^2)^2) = 1/2sqrt((m^2-m-1)^2 + (-m^2 + m + 1)^2) = sqrt(2)/2 sqrt((m^2-m-1)^2).

You can find the area by A = 1/2 * b*h but this looks more like the desired answer as it simplifies nicely.

Anonymous

Anonymous Wed 09 Mar 2022 23:53:28 No.14291959 Report

Quoted By: >>14291963

>>14291940
P actually has the coordinates ((c-2)/2, (c+2)/2), but c=m^2+m still and you get something more like d^2 = (m^2-m-2)^2 + (-m^2+m+2) so I believe the final answer is along the lines A = 1/2 (m^2-m-2)

Anonymous

Anonymous Wed 09 Mar 2022 23:55:05 No.14291963 Report

Quoted By: >>14291971

>>14291959
Of course, the absolute must be taken, but one may verify it works for m=-1 (0) and m=1 (1) these trivial cases are always good to check to validate. But yes, the answer seems to be A = |1/2(m^2-m-2)|

Anonymous

Anonymous Wed 09 Mar 2022 23:57:43 No.14291971 Report

Quoted By: >>14291973

>>14291963
In addition, i dont know where the phys guy messed up, but his area formula does not work for m=-1 (one can see the area here is clearly 0). I'm sure Heron's makes things simpler but I'm not sure if it is taught at Polish HS (it should be, however), the above approach was probably intended

Anonymous

Anonymous Thu 10 Mar 2022 00:00:01 No.14291973 Report

Quoted By:

>>14291971
Furthermore, they seemed to have messed up the angles too. This part is not too hard, just erect perpendiculars from A and B respectively, for the triangle to be acute the vertex C must lie strictly between these erected perpendiculars

Anonymous

Anonymous Thu 10 Mar 2022 00:09:08 No.14291986 Report

Quoted By: >>14291996 >>14292000

The easiest approach would be to use vector algebra. Define two vectors $u = C - B$ and [/math] v = A - B [/math]. Then we have

$\begin{align}<br />\vec{u} &= (m-1,m^2-3,0) \\<br />\vec{v} &= (-1,-1,0)<br />\end{align}<br />$

We turn them into 3D vectors so that we can take the cross product.

$\text{Area} = \frac{1}{2} |\vec{u}\times \vec{v}| = \frac{|m^2 - m - 2|}{2}.$

Next, for the acute angle, take the dot product. We demand the dot product be positive for acute angles.

$\vec{u} \cdot \vec{v} = 3 - m - m^2 > 0$

Matura to Chuj

Matura to Chuj Thu 10 Mar 2022 00:09:48 No.14291987 Report

Quoted By: >>14291991

>>14291715
Nie chce mi si?tego teraz rozwi?zywa?, jak nie jeste?kompletna pizda to zrobisz to samemu.
do a) potrzebujesz pierwsze okre?li? d?ugo?? podstawy, która wynosi pierwiastek z 2, punkt C okre?lony wed?ug m to b?dzie (m, m^2), chyba nie musz?wyja?nia? dlaczego. potem mo?esz z którego? wzoru w karcie obliczy?d?ugo?ci pozosta?ych dwóch boków ze wspó?rz?dnych, albo przynajmniej wyrazi? to przy u?yciu m. https://www.matemaks.pl/wzory-na-pole-trojkata.html wzór trzeci, masz wszystkie boki i obwód, wi?c mo?na opisa? pole za pomoc?zmiennej m.

do b) najlepiej narysowa? sobie dwie proste prostopad?e do prostej przechodz?cej przez A i B, rysujesz je tak zeby jedna przechodzi?a przez A a druga przez B, wtedy widzisz na jakim wycinku paraboli musi znajdowa? si?punkt m ?eby wszystkie k?ty by?y <90 stopni. Rozwi?zujesz uk?ad równa? ?eby znale?? punkty przeci?cia paraboli z prostymi, i masz granice przedzia?ów, w których trójk?t b?dzie ostrok?tny.

Matura to Chuj

Matura to Chuj Thu 10 Mar 2022 00:10:53 No.14291991 Report

Quoted By:

>>14291987
S? na to lepsze sposoby, ale tego od ciebie oczekuj?z?amasy z centralej komiski egzaminacyjnej

Anonymous

Anonymous Thu 10 Mar 2022 00:13:27 No.14291996 Report

Quoted By:

>>14291986
Again, a simple substitution of m=1 shows your angle test fails. A right angled triangle is present.

Anonymous

Anonymous Thu 10 Mar 2022 00:14:33 No.14292000 Report

Quoted By: >>14292008

>>14291986
You've made the assumption that one angle being acute implies the whole triangle is acute, which is of course absurd

Anonymous

Anonymous Thu 10 Mar 2022 00:22:22 No.14292008 Report

Quoted By: >>14292027

>>14292000
What? All I did was assume angle ABC is acute, lol.

Anonymous

Anonymous Thu 10 Mar 2022 00:29:34 No.14292027 Report

Quoted By: >>14292034

>>14292008
If you read the question, it asks for ABC to be acute, which is not the case under your criterion.

An acute triangle is a triangle for which ALL of its angles are acute. Unless you're referring to a non-euclidean geometry, any given triangle will have atleast two acute angles.

Anonymous

Anonymous Thu 10 Mar 2022 00:34:08 No.14292034 Report

Quoted By:

>>14292027
I read the question, I just didn't remember that faggy high school term, and assumed that it just meant angle B was acute. I have a PhD in math and never once had to care about that stupid term.

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