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Geometry Level 4

If in a triangle A B C ABC an angle bisector is drawn from A A to B C BC such that it cuts B C BC at D D . If ∠ B A D = 1 5 ∘ \angle BAD = 15^{\circ} and ∠ A B C = 4 5 o \angle ABC =45^{o} . If A D = 5 6 AD=5\sqrt{6} and the area of the triangle can be expressed as m n + p \large{ \frac{m}{\sqrt{n}+p} } , for some positive integers ( m , n , p ) (m, n, p) . Find m + n + p m+n+p .


The answer is 239.

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Department 8 by Department 8 , 27, Israel

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2 solutions

Aditya Chauhan
Dec 27, 2015

Angles of the triangle: A=30 , B= 45,C= 105.

By sine rule a sin ⁡ A = b sin ⁡ B = c sin ⁡ C \dfrac{a}{\sin{A}}=\dfrac{b}{\sin{B}}=\dfrac{c}{\sin{C}}

2 a = 2 b = 2 2 c 3 + 1 2a=\sqrt{2}b=\dfrac{2\sqrt{2}c}{\sqrt{3}+1}

c = 3 + 1 2 b c=\dfrac{\sqrt{3}+1}{2}b

Length of angle bisector

AD= 2 b c b + c cos ⁡ ( A 2 ) = 5 6 \dfrac{2bc}{b+c}\cos(\dfrac{A}{2}) = 5\sqrt{6}

= 2 × b × b ( 3 + 1 2 ) b + b 3 + 1 2 \dfrac{2 \times b \times b(\dfrac{\sqrt{3}+1}{2})}{b+b\dfrac{\sqrt{3}+1}{2}}

b = 30 3 + 1 \dfrac{30}{\sqrt{3}+1}

Area = b 2 × sin ⁡ A × sin ⁡ C 2 × sin ⁡ B \dfrac{b^{2} \times \sin{A} \times \sin{C}}{2 \times \sin{B}}

= 225 2 3 + 2 = 225 12 + 2 \dfrac{225}{2\sqrt{3}+2}=\dfrac{225}{\sqrt{12}+2}

Therefore , m + n + p = 225 + 12 + 2 = 239 m+n+p = 225 + 12 + 2 =\boxed{239}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Your way is very nice, but I first determined Base BC, and dropped Perpendicular from A, which created an isosceles right-triangle.

Department 8 - 5 years, 5 months ago

same way buddy

Kaustubh Miglani - 5 years, 5 months ago

Typo in last line. 225+12+ 2 =239.

Niranjan Khanderia - 5 years, 3 months ago

∠ s . . . A D C = 60 , C A D = 15 , D C A = 105 S i n 105 = 6 + 2 4 A p p l y i n g S i n R u l e b = S i n B ∗ a S i n A : − Δ A D C . . . A C = S i n 60 ∗ 5 6 S i n 105 , Δ A B C . . . A C = S i n 45 ∗ B C S i n 30 , ∴ B C = S i n 60 ∗ 5 6 S i n 105 ∗ S i n 30 S i n 45 h , t h e ⊥ f r o m A o n B C , = 5 6 ∗ S i n 60. A r e a Δ A B C = 1 2 ∗ B C ∗ h = 1 2 ∗ S i n 60 ∗ 5 6 S i n 105 ∗ S i n 30 S i n 45 ∗ 5 6 ∗ S i n 60 A r e a Δ A B C = 225 2 ( 3 + 1 ) = 225 12 + 2 . m + n + p = 225 + 12 + 2 = 239 \angle s ~~...ADC=60, ~~~~CAD=15, ~~~~DCA=105~ ~~Sin105=\dfrac{\sqrt6 +\sqrt2} 4~~\\ Applying~ Sin ~Rule~~b=SinB*\dfrac a {SinA}:-\\ \Delta ~ADC~...AC=Sin60*\dfrac{5\sqrt6}{Sin105}, ~~~~\Delta ~ABC~... AC=Sin45*\dfrac {BC}{Sin30}, \\ \therefore~BC= Sin60*\dfrac{5\sqrt6}{Sin105}*\dfrac{Sin30}{Sin45}\\ h, ~the ~ \bot ~from ~ A ~ on ~BC, =5\sqrt6*Sin60.\\ Area ~\Delta ~ABC=\frac 1 2 *BC*h\\ =\frac 1 2* Sin60* \dfrac{5\sqrt6}{Sin105}*\dfrac{Sin30}{Sin45}*5\sqrt6*Sin60\\ Area ~\Delta ~ABC= \dfrac{225}{2(\sqrt3+1)}= \dfrac{225}{\sqrt{12}+2}.~~\\ m+n+p=225+12+2=\Large~~~\color{#D61F06}{239}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Your way is very nice, but I first determined Base BC, and dropped Perpendicular from A, which created an isosceles right-triangle.

Department 8 - 5 years, 3 months ago

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