Cool Geometry

Geometry Level 3

In △ A B C \triangle ABC , ∠ A = 2 ∠ B \angle A=2\angle B . What is the value of B C 2 BC^{2} ?

3 A C − A B 3AC-AB None of these A C ( A C + A B ) AC(AC+AB) A B 2 AB^{2} A B + A C AB+AC

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Aditya Lalbondre by Aditya Lalbondre , 22, India

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2 solutions

Extend C A CA to D D Such that ∠ C B D = ∠ C A B \angle CBD=\angle CAB

△ C B A \triangle CBA is similar to △ C D B \triangle CDB

C B C D \frac{CB}{CD} = C A C B \frac{CA}{CB}

So,

B C 2 = C D ⋅ C A BC^{2}=CD \cdot CA

Also,

A B = A D AB=AD ..........................Isoceles triangle theorem in △ B A D \triangle BAD

So,

B C 2 = A C ( A C + A B ) BC^{2}=AC(AC+AB)

16 Helpful 2 Interesting 2 Brilliant 2 Confused

It's an rmo problem

Aarush Priyankaj - 2 years, 9 months ago

How is AB = AD?

Shubhrajit Sadhukhan - 7 months ago
Timoy Cheng
Apr 15, 2015

You can use a combination of the law of cosines and law of sines.

Let the length of AB = c, the length of BC = a, the length of AC = b, the measure of angle B be x, and the measure of angle A be 2x.

By the Law of Sines,

sin(A)/sin(B) = sin(2x)/sinx = 2 cosx = a/b

We get:

2cosx = a/b (equation 1)

However, the Law of Cosines states that

cos(B)

= cos x

= (a^2 + c^2 - b^2)/2ac

Substituting this into equation 1, we get the equation

(a^2 + c^2 - b^2)/ac = a/b

(a^2)b + (c^2)b - b^3 = (a^2)c

(a^2)b - (a^2)c = b^3 - (c^2)b

(a^2)(b-c) = b(b^2 - c^2) = b (b+c)(b-c)

(a^2 - b(b+c))(b-c) = 0

So it's either a^2 = b(b+c) (or BC^2 = AC(AC+AB)) or b - c = 0

If b - c = 0, or b = c, then we get that the triangle would be a 45-45-90 triangle, with AC and BC as the legs (using the Law of Sines). Using the Pythagorean theorem, we arrive at BC^2 = AC^2 + AB^2 = AC(AC+AB) (since AB = AC).

Thus, BC^2 = AC(AC+AB)

2 Helpful 1 Interesting 1 Brilliant 0 Confused

In latex:

From Cosine rule, b 2 = a 2 + c 2 − 2 a c cos ⁡ x b^2 = a^2 + c^2 - 2ac \cos x cos ⁡ x = a 2 + c 2 − b 2 2 a c \textcolor{#624F41}{\cos x} = \dfrac{a^2 + c^2 - b^2}{2ac}

Using Sine rule, a sin ⁡ 2 x = b sin ⁡ x \dfrac{a}{\sin 2x} = \dfrac{b}{\sin x} a b = sin ⁡ 2 x sin ⁡ x \dfrac{a}{b} = \dfrac{\sin 2x}{\sin x} a b = 2 cos ⁡ x \dfrac{a}{b} = 2 \textcolor{#624F41}{\cos x} a b = 2 ( a 2 + c 2 − b 2 2 a c ) \dfrac{a}{b} = 2 \left(\dfrac{a^2 + c^2 - b^2}{2ac}\right) a 2 c = a 2 b + c 2 b − b 3 a^2c = a^2b + c^2b - b^3 a 2 c − a 2 b = c 2 b − b 3 a^2c - a^2b = c^2b - b^3 a 2 ( c − b ) = b ( c 2 − b 2 ) a^2(c - b) = b(c^2 - b^2) a 2 = b ( c + b ) \boxed{a^2 = b(c + b)} Or, c − b = 0 c - b = 0 that makes the triangle isosceles and right angled, satisfying the required conditions.

Shubhrajit Sadhukhan - 7 months ago

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