Below is a puzzle of counting triangles.How to solve such puzzle ?
source:
http://gpuzzles.com/mind-teasers/how-many-triangles-challenge/?source=stackmath
[Math] how to solve triangles count puzzle
geometrypuzzlerecreational-mathematics
Related Solutions
Say that instead of four triangles along each edge we have $n$. First count the triangles that point up. This is easy to do if you count them by top vertex. Each vertex in the picture is the top of one triangle for every horizontal grid line below it. Thus, the topmost vertex, which has $n$ horizontal gridlines below it, is the top vertex of $n$ triangles; each of the two vertices in the next row down is the top vertex of $n-1$ triangles; and so on. This gives us a total of
$$\begin{align*} \sum_{k=1}^nk(n+1-k)&=\frac12n(n+1)^2-\sum_{k=1}^nk^2\\ &=\frac12n(n+1)^2-\frac16n(n+1)(2n+1)\\ &=\frac16n(n+1)\Big(3(n+1)-(2n+1)\Big)\\ &=\frac16n(n+1)(n+2)\\ &=\binom{n+2}3 \end{align*}$$
upward-pointing triangles.
The downward-pointing triangles can be counted by their by their bottom vertices, but it’s a bit messier. First, each vertex not on the left or right edge of the figure is the bottom vertex of a triangle of height $1$, and there are $$\sum_{k=1}^{n-1}=\binom{n}2$$ of them. Each vertex that is not on the left or right edge or on the slant grid lines adjacent to those edges is the bottom vertex of a triangle of height $2$, and there are
$$\sum_{k=1}^{n-3}k=\binom{n-2}2$$ of them. In general each vertex that is not on the left or right edge or on one of the $h-1$ slant grid lines nearest each of those edges is the bottom vertex of a triangle of height $h$, and there are
$$\sum_{k=1}^{n+1-2h}k=\binom{n+2-2h}2$$ of them.
Algebra beyond this point corrected.
The total number of downward-pointing triangles is therefore
$$\begin{align*} \sum_{h\ge 1}\binom{n+2-2h}2&=\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-2k}2\\ &=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}(n-2k)(n-2k-1)\\ &=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}\left(n^2-4kn+4k^2-n+2k\right)\\ &=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+2\sum_{k=0}^{\lfloor n/2\rfloor-1}k^2-(2n-1)\sum_{k=0}^{\lfloor n/2\rfloor-1}k\\ &=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+\frac13\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\left(2\left\lfloor\frac{n}2\right\rfloor-1\right)\\ &\qquad\qquad-\frac12(2n-1)\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\;. \end{align*}$$
Set $\displaystyle m=\left\lfloor\frac{n}2\right\rfloor$, and this becomes
$$\begin{align*} &m\binom{n}2+\frac13m(m-1)(2m-1)-\frac12(2n-1)m(m-1)\\ &\qquad\qquad=m\binom{n}2+m(m-1)\left(\frac{2m-1}3-n+\frac12\right)\;. \end{align*}$$
This simplifies to $$\frac1{24}n(n+2)(2n-1)$$ for even $n$ and to
$$\frac1{24}\left(n^2-1\right)(2n+3)$$ for odd $n$.
The final figure, then is
$$\binom{n+2}3+\begin{cases} \frac1{24}n(n+2)(2n-1),&\text{if }n\text{ is even}\\\\ \frac1{24}\left(n^2-1\right)(2n+3),&\text{if }n\text{ is odd}\;. \end{cases}$$
They can be counted quite easily by systematic brute force.
All of the triangles are isosceles right triangles; I’ll call the vertex opposite the hypotenuse the peak of the triangle. There are two kinds of triangles:
- triangles whose hypotenuse lies along one side of the square;
- triangles whose legs both lie along sides of the square and whose peaks are at the corners of the square.
The triangles of the second type are easy to count: each corner is the peak of $4$ triangles, so there are $4\cdot4=16$ such triangles.
The triangles of the first type are almost as easy to count. I’ll count those whose hypotenuses lie along the bottom edge of the square and then multiply that by $4$. Such a triangle must have a hypotenuse of length $1,2,3$, or $4$. There $4$ with hypotenuse of length $1$, $3$ with hypotenuse of length $2$, $2$ with hypotenuse of length $3$, and one with hypotenuse of length $4$, for a total of $10$ triangles whose hyponenuses lie along the base of the square. Multiply by $4$ to account for all $4$ sides, and you get $40$ triangles of the second type and $40+16=56$ triangles altogether.
Added: This approach generalizes quite nicely to larger squares: the corresponding diagram with a square of side $n$ will have $4n$ triangles of the first type and $$4\sum_{k=1}^nk=4\cdot\frac12n(n+1)=2n(n+1)$$
of the second type, for a total of $2n^2+6n=2n(n+3)$ triangles.
Best Answer
Be systematic. Label all your points. Choose a unique label for each triangle, e.g. by listing point labels in alphabetic order. Enumerate triangles in some order so you can check for them one at a time. I'd suggest lexicographic order.
Exploit symmetry. Many triangles will occur four times, or even eight times, throughout the figure in rotated and posibly reflected versions of one another. So you can keep the work down if you find only one representative of each such group, as long as you make sure to get the associated count right.
Combining these ideas, I'd label the figure like this:
Then you can enumerate triangles like this:
$\qquad 8ACF, 0ADD, 0ADE, 8ADF, 0AEE, 0AEF, 4AFF$
$\qquad 8BCF,0BDD,0BDE,8BDF,0BEE,8BEF,4BFF$
So you get a total of
$$(4+8+8+8+8+4)+(4+4+8+8+8+4)+(8+4)+4=92$$
unless I (still) made a mistake. But since this solution now agrees with the $92$ stated on the original problem statement question, I trust it might (finally) be correct now. Thanks to TonyK for spotting the one I had missed! My first attempt was way lower, so you really have to be very systematic to get this even close to correct.
Originally I had a higher count, namely $104$ (would be $108$ by now with my other fixes), but that's because I'd assumed $BCF$ to be collinear so I had $8ABC$ and $8BBC$. Jyrki pointed out in a comment that it doesn't look like that in your original post, even though it's close.
The same approach can be applied to other, similar tasks, as demonstrated in this post.