Question

Triangle ∆ABC is given. At segment AC point M is chosen so that
|AM| : |MC| = 1 : m and at segment BC point N is chosen so that
|BN| : |NC| = 1 : n. Find all natural numbers m and n if the ratio of
areas AΔABC : ΑΔΜΝΟ = 2 : 1 is known. (Solve this task without using
Trigonometry.)​

Answer 1

Given :- Triangle ∆ABC is given. At segment AC point M is chosen so that |AM| : |MC| = 1 : m and at segment BC point N is chosen so that |BN| : |NC| = 1 : n. Find all natural numbers m and n if the ratio of areas AΔABC : ΑΔΜΝC = 2 : 1 is known. ?

Answer :-

we have,

→ AM/MC = 1/m

so,

→ MC/AC = m/(m + 1)

and,

→ BN/NC = 1/n

so,

→ NC/BC = n/(n + 1)

then,

→ Area ∆MNC/ Area ∆ABC = (1/2)

→ (m/m + 1) * (n/n + 1) = (1/2)

→ (mn)/(m + 1)(n + 1) = (1/2)

→ 2mn = (m + 1)(n + 1)

→ 2mn = mn + m + n + 1

→ 2mn - mn = m + n + 1

→ mn = m + n + 1

→ mn - m - n = 1

adding 1 both sides ,

→ mn - m - n + 1 = 2

→ mn - n - m + 1 = 2

→ n(m - 1) - (m - 1) = 2

→ (m - 1)(n - 1) = 2

now, since m and n both are natural numbers , so,

• when m = 2 , n = 3
• when m = 3 , n = 2

therefore, we can conclude that, only 2 pair of values of m and n are possible and they are (2,3) or (3,2) .

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