Let A B C be a triangle with A B=5, A C=4, B C=6 . The internal angle bisector of C intersects the side A B at D . Points M and N are taken on sides B C and A C , respectively, such that D M \| A C and D N \| B C . If

Correct answer: 02 (A D)/(B D)=(A C)/(B C)=(2)/(3)=(C M)/(B M)=(A N)/(C N) So, C M=(2)/(5) × 6=(12)/(5) and, C M=(3)/(5) × 4=(12)/(5) aligned & C=(B C^2+A C^2-A B^2)/(2 B C · A C)=(C M^2+C M^2-M N^2)/(2 C N · C M) & (6^2+4^2-5^2)/(2 · 4 · 6)=(((12)/(5))^2+((12)/(5))^2-M N^2)/(2 ·((12)/(5))((12)/(5))) & M N^2=(126)/(25)=(p)/(q) aligned Sum of digits of |p-q|=2

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

Let A B C be a triangle with A B=5, A C=4, B C=6 . The internal...

Let

A B C

be a triangle with

A B=5, A C=4, B C=6

. The internal angle bisector of

C

intersects the side

A B

D

. Points

M

and

N

are taken on sides

B C

and

A C

, respectively, such that

D M \| A C

and

D N \| B C

. If

(M N)^2=\frac{p}{q}

where

p

and

q

are relatively prime positive integers then what is the sum of the digits of

|p-q| ?

Let A B C be a triangle with A B=5, A C=4, B C=6 . The internal angle bisector of C intersects the side A B at D .

Correct answer

Step-by-step solution

\therefore \quad \frac{A D}{B D}=\frac{A C}{B C}=\frac{2}{3}=\frac{C M}{B M}=\frac{A N}{C N}

So,

C M=\frac{2}{5} \times 6=\frac{12}{5}

and,

C M=\frac{3}{5} \times 4=\frac{12}{5}

\begin{aligned} & \cos C=\frac{B C^2+A C^2-A B^2}{2 B C \cdot A C}=\frac{C M^2+C M^2-M N^2}{2 C N \cdot C M} \\ & \Rightarrow \quad \frac{6^2+4^2-5^2}{2 \cdot 4 \cdot 6}=\frac{\left(\frac{12}{5}\right)^2+\left(\frac{12}{5}\right)^2-M N^2}{2 \cdot\left(\frac{12}{5}\right)\left(\frac{12}{5}\right)} \\ & \Rightarrow \quad M N^2=\frac{126}{25}=\frac{p}{q} \end{aligned}

Sum of digits of

|p-q|=2

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