3D and Limits

Geometry Level 4

The incenter I I of the triangle P Q R PQR is the foot of the normal from the point M = ( 1 , 2 , 6 ) M = (1,2,6) to the x y xy -plane, where P , Q , R P,Q,R are the feet of altitudes of an isosceles triangle A B C ABC whose vertex is A A and base B C BC of 6 unit length.

Let lim A π 2 + e v e k 1 sin A = e k L \displaystyle \lim_{A\to {\frac \pi 2}^+} \dfrac{e^v - e^k}{\sqrt{1- \sin A}} = \dfrac{e^k}L for integer k k , where v v is the volume of the tetrahedron M I B C MIBC .

Find the value of 1 k 2 L 2 \dfrac1{k^2 L^2 } .


The answer is 2.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Avi Solanki
Apr 24, 2017

This is a solution by my friend

@Devansh Jain .

Δ P Q R \Delta PQR is the pedal triangle of Δ A B C \Delta ABC . Incentre of the pedal triangle is the orthocentre of triangle Δ A B C \Delta ABC . I don't see why ' I I is the same for Δ P Q R \Delta PQR and Δ A B C \Delta ABC . Also,I think u need to mention that Δ A B C \Delta ABC is in the xy plane .

Sumanth R Hegde - 4 years, 1 month ago

Though the answer is correct . But I think there is a major flaw in the solution . Area of IBC is not equal to 1/3 area ABC as this triangle is not equilateral.

Ankit Kumar Jain - 2 years, 8 months ago

@Md Zuhair @Thomas Jacob @Aaron Jerry Ninan What do you guys think? Tell me if I am wrong.

Ankit Kumar Jain - 2 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...