I understand I have the right to do this difficulty by just counting every the triangles but how carry out I execute this faster?

$\begingroup$ yes, yet I don't know just how to usage it because that this question. I recognize I have to pick 3 segment to type a triangle yet I don't recognize where to select them indigenous $\endgroup$

You are watching: If all of the diagonals are drawn from a vertex of a pentagon, how many triangles are formed?

because that a continuous pentagon over there is more than likely nothing faster than just counting the triangles. However counting i do not care increasingly complicated as the number of sides that a regular polygon increases. The problem is facility by the fact that, for constant polygons of also $n>4$ sides, one or more concurrencies of 3 or an ext diagonals occur.

But the consistent pentagon and all constant polygons v an odd number of sides are more law-abiding, because their diagonals do not create three-line concurrencies. If we begin with consistent pentagon $ABCDE$, and draw the diagonals in alphabetical order, we have actually $AC$, $AD$, $BD$, $BE$, $CE$. The variety of triangles developed by $AC$, $AD$ is $1+2=3$. $BD$, $BE$ kind $6+11=17$ triangles. $CE$ develops $15$ triangles, making$$3+17+15=35$$ in all.

II. In a polygon, illustration in alphabet order all diagonals native $A$, then all those indigenous $B$, and so on, we find that diagonals native a suggest to that alphabetically nearest neighbor, i.e. $AC$, $BD$, $CE$, $DF$, $EG$, ... Generate $1$, $6$, $15$, $28$, $45$,... Triangles, respectively. For the $rth$ such diagonal, the number of triangles produced is therefore $2r^2-r$. But because $EG$, the 5th such diagonal line here, very first appears in the regular heptagon, substituting $n-2$ for $r$ we get $$2n^2-9n+10$$ with $n\ge3$. The amount $\Sigma_1$ of this triangles is therefore $$2\left(\fracn^33+\fracn^22+\fracn6\right)-9\left(\fracn^2+n2\right)+10n=\frac4n^3-21n^2+35n-186$$(It is crucial to deduct $\frac186=3$, because $n\ge3$ and also $2n^2-9n+10=3$ because that $n=1$, and also $0$ for $n=2$. A comparable adjustment is do in determining every of the sums below.)

Diagonals attracted to vertices which are two vertices away from $A$, $B$, $C$, ... , i.e. $AD$, $BE$, $CF$, $DG$, $EH$, ... Produce $1$, $11$, $29$, $55$, $89$, ... Triangles, respectively. The legislation of this succession is $4n^2-18n+19$, and also the amount $\Sigma_2$, over denominator $6$ and also again through $n\ge3$, is$$\frac8n^3-42n^2+64n-246$$Since in the pentagon no diagonal join vertices much more than two vertices apart, the preceding two sums suffice for calculating how countless triangles the diagonals produce. Because that $CE$, the critical diagonal join in the pentagon, and also the greatest term in the an initial sequence, $n=r+2=5$, and also $$\frac4n^3-21n^2+35n-186=22$$ for $BE$, the best term in the 2nd sequence, $n=4$ and $$\frac8n^3-42n^2+64n-246=12$$ $AD$ is the last diagonal attracted from $A$, but note that every diagonal drawn from $A$ generates one triangle in a polygon, other than the last, i beg your pardon generates two. Hence again, the variety of triangles developed by diagonals in the consistent pentagon is$$22+12+1=35$$

III. Similarly, methodical counting of triangles in odd constant polygons created by diagonals authorized vertices separated by $3$, $4$, $5$, $6$ vertices, i.e.$$AE, BF, CG,...; AF, BG, CH,...; AG, BH, CI...; AH, BI, CJ...$$yields sequences$$1, 18, 48, 91, 147,...$$$$1, 27, 72, 136, 219,...$$$$1, 38, 101, 190, 305,...$$$$1, 51, 135, 253, 405,...$$whose laws, with common denominator $2$, are$$\frac13n^2-57n+562$$$$\frac19n^2-81n+742$$$$\frac26n^2-108n+922$$$$\frac34n^2-138n+1102$$and who sums $\Sigma_3-6$, with typical denominator $6$, are$$\frac13n^3-66n^2+89n-186$$$$\frac19n^3-93n^2+110n6$$$$\frac26n^3-123n^2+127n+306$$$$\frac34n^3-156n^2+140n+726$$As us calculated the variety of triangles in the continual pentagon making use of the first two of the coming before six sums, for this reason we can calculate the triangle in the heptagon using the very first four, and also generally the number of triangles in one odd $n$-agon using $n-3$ together sums. Thus letting $n=7$, $6$, $5$, $4$, respectively, in the very first four sums provided above, we calculate the variety of triangles in the constant heptagon as$$95+96+67+28+1=287$$Again, letting $n=9$, $8$, $7$, $6$, $5$, $4$ in the 6 sums above, we discover that the continuous nonagon contains$$252+316+305+236+140+52+1=1302$$triangles.

IV. Inspecting the molecule of the expression above, because that the variety of triangles created by diagonals joining point out separated by $1$, $2$, $3$, ... Vertices, we deserve to discern a general rule. Over usual denominator $2$ us have$$4n^2-18n+20$$$$8n^2-36n+38$$$$13n^2-57n+56$$$$19n^2-81n+74$$$$26n^2-108n+92$$$$34n^2-138n+110$$Thus it is straightforward to extrapolate to matching sequences for polygons of ever before larger weird $n$.

The summations that these succeeding sequences, over usual denominator $6$, are likewise predictable, if slightly much more complex:$$4n^3-21n^2+35n-18$$$$8n^3-42n^2+64n-24$$$$13n^3-66n^2+89n-18$$$$19n^3-93n^2+110n$$$$26n^3-123n^2+127n+30$$$$34n^3-156n^2+140n+72$$

If this extrapolations are warranted, climate diagonals in the $11$, $13$, and also $15$-agon create $4257$, $11297$, and also $25935$, triangles, respectively. Lastly, the diagonals that the commemorated $17$-agon that Carl Friedrich Gauss create $53516$ triangles!

Can us extract a solitary general ascendancy for the variety of triangles, as a function of odd $n$?


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Well, the sequence obtained so far, because that the number of triangles created by the diagonals of consistent polygons of $3$, $5$, $7$, $9$, $11$, $13$, $15$, $17$ sides, namely$$1, 35, 287, 1302, 4257, 11297, 25935, 53516, ...$$is given by the expression$$\fracn^6+15n^5-95n^4+225n^3-266n^2+120n720$$This seems to it is provided what OP requests: a fast way to calculate the number of triangles, not simply for the continuous pentagon but for any regular polygon with an odd variety of sides. For consistent polygons with even $n>4$ sides, the expression is almost accurate, yet over-counts by a little due come one or an ext concurrencies of 3 or an ext diagonals. Because that example, it over-counts the constant hexagon, which has one 3-line concurrency, by $111-110=1$ triangle, and also the octagon, which has eight 3-line concurrencies and one 4-line concurrency, by $644-626=18$ triangles.