How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $a,b,c$ be the sides of a triangle and $A,B,C$ be the angles opposite to those sides respectively. If $\sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$, then prove that the triangle is isosceles.

I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 6 from 2016

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Let $a,b,c$ be the sides of a triangle and $A,B,C$ be the angles opposite to those sides respectively. Given $\sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B$

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

$\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})$ $\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B$

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

$\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B$ $\Rightarrow b\sin A\cos B=a\cos A \sin B$ $\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A}$

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

$\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A}$ $\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}$  [since $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$] $\Rightarrow \frac{\cos B}{\cos A}=1$ $\cos B=\cos A$ $A=B$.

$\Rightarrow \Delta ABC$ is isosceles  (Proved).

# Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

# Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="9" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]