angle chaser

Geometry Level 3

The figure shows an isosceles right triangle ABC at B . AC = 2 , an other two lines AE & AD are drawn such that one of them is a median line ( ie bisect the opposite line CB) , and the other line is the angle bisector of angle A . the angle between the two lines AE and AD is α , if tan α = √p -√q , where p and q are positive integers , then find the value of p – q .


The answer is 1.

Aziz Alasha by Aziz Alasha , 59, Sudan

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1 solution

David Vreken
Sep 11, 2018

As an isosceles right triangle, ∠ C A B = 180 ° − 90 ° 2 = 45 ° \angle CAB = \frac{180° - 90°}{2} = 45° , so the angle bisector A E AE makes ∠ E A B = 45 ° 2 \angle EAB = \frac{45°}{2} , and the median A D AD makes ∠ D A B = tan ⁡ − 1 1 2 \angle DAB = \tan^{-1} \frac{1}{2} .

Therefore, α = tan ⁡ − 1 1 2 − 45 ° 2 \alpha = \tan^{-1} \frac{1}{2} - \frac{45°}{2} and tan ⁡ α = tan ⁡ ( tan ⁡ − 1 1 2 − 45 ° 2 ) \tan \alpha = \tan(\tan^{-1} \frac{1}{2} - \frac{45°}{2}) .

Using the tangent addition formula, tan ⁡ α = tan ⁡ ( tan ⁡ − 1 1 2 ) − tan ⁡ ( 45 ° 2 ) 1 + tan ⁡ ( tan ⁡ − 1 1 2 ) tan ⁡ ( 45 ° 2 ) \tan \alpha = \frac{\tan(\tan^{-1} \frac{1}{2}) - \tan(\frac{45°}{2})}{1 + \tan(\tan^{-1} \frac{1}{2})\tan(\frac{45°}{2})} = = 1 2 − tan ⁡ ( 45 ° 2 ) 1 + 1 2 tan ⁡ ( 45 ° 2 ) \frac{\frac{1}{2} - \tan(\frac{45°}{2})}{1 + \frac{1}{2}\tan(\frac{45°}{2})} .

Using the tangent half-angle formula, tan ⁡ ( 45 ° 2 ) = 1 − cos ⁡ 45 ° sin ⁡ 45 ° = 1 − 2 2 2 2 = 2 − 1 \tan(\frac{45°}{2}) = \frac{1 - \cos 45°}{\sin 45°} = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \sqrt{2} - 1 .

Substituting tan ⁡ ( 45 ° 2 ) = 2 − 1 \tan(\frac{45°}{2}) = \sqrt{2} - 1 into tan ⁡ α = 1 2 − tan ⁡ ( 45 ° 2 ) 1 + 1 2 tan ⁡ ( 45 ° 2 ) \tan \alpha = \frac{\frac{1}{2} - \tan(\frac{45°}{2})}{1 + \frac{1}{2}\tan(\frac{45°}{2})} gives tan ⁡ α = 1 2 − ( 2 − 1 ) 1 + 1 2 ( 2 − 1 ) ) \tan \alpha = \frac{\frac{1}{2} - (\sqrt{2} - 1)}{1 + \frac{1}{2}(\sqrt{2} - 1))} which simplifies to tan ⁡ α = 5 2 − 7 = 50 − 49 \tan \alpha = 5 \sqrt{2} - 7 = \sqrt{50} - \sqrt{49} .

Therefore, p = 50 p = 50 and q = 49 q = 49 and p − q = 1 p - q = \boxed{1} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Why does the median make angle DAB as tan inverse 1/2 ?

Anurag Bisht - 2 years, 8 months ago

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As an isosceles triangle, A B = B C AB = BC , and as a median, D B = 1 2 B C = 1 2 A B DB = \frac{1}{2}BC = \frac{1}{2}AB . So tan ⁡ ∠ D A B = D B A B = 1 2 A B A B = 1 2 \tan \angle DAB = \frac{DB}{AB} = \frac{\frac{1}{2}AB}{AB} = \frac{1}{2} .

David Vreken - 2 years, 8 months ago

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Got It! Thanks!

Anurag Bisht - 2 years, 8 months ago

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