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arunspanda
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Let T_n be the number of all possible triangles formed by Updated on: 26 Mar 2014, 07:19
Originally posted by arunspanda on 26 Mar 2014, 06:17.
Last edited by Bunuel on 26 Mar 2014, 07:19, edited 1 time in total.
Renamed the topic and edited the question.
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Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10
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A
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Bunuel
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Let T_n be the number of all possible triangles formed by 26 Mar 2014, 07:21
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arunspanda
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10
Show more

In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).

We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) --> \(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.

Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rule 3. Thank you.
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Let T_n be the number of all possible triangles formed by 26 Feb 2019, 07:44
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Bunuel
arunspanda
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10

In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).

We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) -->\(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\)--> \(n=5\).

Answer: A.

Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.
Show more

Bunuel can you please expand upon the highlighted part? I get that we are cross multiplying for LCD, but I am getting lost.

Also, is there any way this is related to n-2 (number of triangles from a regular polygon)? I tried to use the idea that n(n-2) would give us the number of vertices*number of triangles formed from each.

Tn = n(n-2)
Tn+1 = n(n-2) + 10
... ?
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Let T_n be the number of all possible triangles formed by 26 Feb 2019, 08:07
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Bunuel
arunspanda
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If \(T_{n+1} - T_n = 10\), then the value of n is

A. 5
B. 6
C. 7
D. 8
E. 10

In a plane if there are n points of which no three are collinear, then the number of triangles that can be formed by joining them is \(C^3_n\).

We are given that \(T_{n+1} - T_n =C^3_{n+1}-C^3_n= 10\) --> \(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\) -->\(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\) --> \((n-1)n=20\)--> \(n=5\).

Answer: A.

Or: one additional point gives 10 more triangles, so when we add one point there are 10 different pairs of points which make triangles with that additional point --> \(C^2_n=10\) --> \((n-1)n=20\) --> \(n=5\).

Answer: A.

Bunuel can you please expand upon the highlighted part? I get that we are cross multiplying for LCD, but I am getting lost.

Also, is there any way this is related to n-2 (number of triangles from a regular polygon)? I tried to use the idea that n(n-2) would give us the number of vertices*number of triangles formed from each.

Tn = n(n-2)
Tn+1 = n(n-2) + 10
... ?
Show more


\(C^3_{n+1}-C^3_n= \frac{(n+1)!}{3!(n-2)!}-\frac{n!}{3!(n-3)!}=10\);

Express (n+1)! as (n-2)!*(n-1)n(n+1) and n! as (n-3)!*(n-2)(n-1)n:

\(\frac{(n-2)!*(n-1)n(n+1)}{3!(n-2)!}-\frac{(n-3)!*(n-2)(n-1)n}{3!(n-3)!}=10\)

\(\frac{(n-1)n(n+1)}{6}-\frac{(n-2)(n-1)n}{6}=10\)

\(\frac{(n-1)n(n+1)-(n-2)(n-1)n}{6}=10\)

\(\frac{(n-1)n(n+1-n+2)}{6}=10\)

\(\frac{(n-1)n*3}{6}=10\)

\((n-1)n=20\)

\(n=5\).
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Let T_n be the number of all possible triangles formed by 26 Feb 2019, 11:08
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We could also solve this by directly inputting the numbers. 6c3-5c3=10.Hence value of n is 5

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Let T_n be the number of all possible triangles formed by 06 Oct 2023, 08:05
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Given: Let \(T_n\) be the number of all possible triangles formed by joining vertices of an n-sided regular polygon.
Asked: If \(T_{n+1} - T_n = 10\), then the value of n is

T_n = nC3 = n(n-1)(n-2)/6

\(T_{n+1} - T_n = (n+1)n(n-1)/6 - n(n-1)(n-2)/6 = 10\)
\(T_{n+1} - T_n = n(n-1)/6 {(n+1)-(n-2)} = n(n-1)/2 = 10\)
n(n-1) = 20 = 5*4
n = 5

IMO A
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