TITLE: Ten natural numbers such that the sums of each choice of nine of them are $82, 83, 84, 85, 87, 89, 90, 91, 92$
QUESTION [1 upvotes]: In a Whats App group the following puzzle was asked

Ten (not necessarily different) natural number are such that if all but one of them is added, the possible results (depending on which one is omitted) are 82, 83, 84, 85, 87, 89, 90, 91, 92.

Non-uniqueness of the answer was also  mentioned there. Naively, I set up 9 linear equations on 10 variables. By fixing, one of them I could get
the set of 10 numbers as 1,2,3,4,6,8,9,10,11,39. If we leave out first 9 numbers one by one to get the sum of remaining 9  as required. But leaving out 39  does not do so.
What could be the proper approach and the answer?

REPLY [0 votes]: Let ten natural numbers be $a,b,c,d,e,f,g,h,k,z$, If $S$ denotes  their}sum then we have $$S-a=82, S-b=83, S-c=84, S-d=85, S-e=87, S-f=89, S-g=90, S-h=91, S-k=92.$$
Adding these 9 equations we $$9S-(S-z)=783\implies S=\frac{783-z}{8}$$
Using this in the 9 equations, we get 9 variables in terms of 10th variab;w $z$ in descending order
$$\frac{127-z}{8},\frac{119-z}{8},\frac{111-z}{8},\frac{103-z}{8},\frac{87-z}{8}, \frac{71-z}{8}, \frac{63-z}{8}, \frac{55-z}{8}, \frac{47-z}{8}.
$$
Let us choose  a natural number $z$ such that  the lowest one attains the  largest value, with all being natural numbers. For $z=7,$
we get ten natural numbers $15,14,13,12,10,8,7,6,5$ and $z=7$ as the required answer. Interestingly, the double entry of 7, occurs by itself.


TITLE: Parabola features
QUESTION [0 upvotes]: Define curve type given by equation. Find focus, directrix, vertex, center
The equation is $16y^2-100x-16y+17=0$.
I defined curve type by using invariant matrix. That is parabola.
Is it possible to find curve features without converting to canonical equation?

REPLY [1 votes]: For this particular parabola, I think the easiest approach is to convert the equation into standard form, but there are other ways to extract the parameters from the general equation.
To find the axis direction, take two points $P_1$ and $P_2$ on the parabola and compute the tangents at those points. Let $Q_1$ be the intersection of those tangents and $Q_2$ be the fourth vertex of the parallelogram with sides $P_1Q_1$ and $P_2Q_1$. The line $\overline{Q_1Q_2}$ is parallel to the parabola’s axis.
Another way to find this axis direction is to compute the pole of the line at infinity. This will be a point at infinity that lies on the parabola’s axis, and thus corresponds to its (inhomogeneous) direction vector. If $C$ is the matrix that represents the general equation of a conic and $C^{\tiny\triangle}$ its adjugate, then the pole of a line $\mathbf l$ is $C^{\tiny\triangle}\mathbf l$. For this parabola, $$C = \begin{bmatrix}0&0&-50\\0&16&-8\\-50&-8&17\end{bmatrix}$$ and the line at infinity is $\mathbf l_\infty=[0:0:1]$, so its pole is the last column of $$C^{\tiny\triangle}=\begin{bmatrix}208&400&800\\400&-2500&0\\800&0&0\end{bmatrix},$$ namely $[1:0:0]$—the parabola’s axis is parallel to the $x$-axis. We already knew that, of course, because there are no second-degree terms that involve $x$ in this parabola’s equation, but the method described here is generally applicable.
Equivalently, compute the intersection $\mathbf p$ of the parabola with the line at infinity. Continuing with homogeneous matrix methods, this involves finding an $\alpha$ such that $\mathcal M_{\mathbf l_\infty}^T C \mathcal M_{\mathbf l_\infty} + \alpha \mathcal M_{\mathbf l_\infty}$, where $$\mathcal M_{\mathbf l_\infty} = \begin{bmatrix}0&-1&0\\1&0&0\\0&0&0\end{bmatrix},$$ the “cross-product matrix” of $\mathbf l_\infty$, has rank one. The resulting matrix will be a scalar multiple of $\mathbf p\mathbf p^T$, so you can read this point directly from the matrix. For this parabola, $\mathcal M_{\mathbf l_\infty}^T C \mathcal M_{\mathbf l_\infty} = \operatorname{diag}(16,0,0)$ is already a rank-one matrix, so there’s nothing more to do and we again have the intersection point $[1:0:0]$. In general, you’ll end up with a quadratic equation in $\alpha$ to solve. (This method of finding the intersection of a line and conic is generally applicable—you end up with a multiple of $\mathbf p\mathbf q^T$ or $\mathbf q\mathbf p^T$, where $\mathbf p$ and $\mathbf q$ are the intersection points.) This approach can be less work than the previous one as it was in this case, but an advantage of the first method is that it produces the answer with a direct calculation, which can make it more suitable for automation.
Once you know the axis direction, you can find the vertex. It’s the point at which the normal to the parabola is parallel to its axis. Finding this point requires simultaneously solving a linear equation and the equation of the parabola, which you can then reduce to solving a quadratic equation by back-substitution. Continuing from above, the gradient at $(x,y)$ is $(-100,32y-16)$. For this to be parallel to the $x$-axis we must have $32y-16=0$, or $y=\frac12$. Plugging this into the parabola’s equation and solving for $x$ produces $x={13\over100}$.
With the vertex and axis direction in hand, you can compute the focal distance via the property described here in the Wikipedia article on parabolas: Compute the intersection of the parabola with a line perpendicular to its axis. Let $c$ be the chord length and $d$ the distance of the vertex to this line. The focal distance is then $f={c^2\over16d}$. For this parabola, we can try $x=1$ as the line, which gives us $y=\frac14\left(2\pm\sqrt{87}\right)$, so $c=\frac12\sqrt{87}$, $d={87\over100}$ and $f={25\over16}$. You now have the vertex, axis direction and focal distance, and in the process of computing the latter you will have determined the direction in which the parabola opens. The focus and directrix are easily computed from here.