 ## (De high school Toan 2023) Let the pyramid S.ABC whose base ABC is a right triangle at A. Side is AC = a, (BC = asqrt 5 ).

• Question:

Let ABC be a pyramid whose base ABC is a right triangle at A. Side AC = a, $$BC = a\sqrt 5$$. The plane (SAB) is perpendicular to the base plane and the equilateral triangle SAB. Call K points on the side SC such that SC = 3SK. Calculate the distance $$d$$ between the two lines AC and BK according to a.

Reference explanation:

Let H be the midpoint of AB $$\Rightarrow SH \bot AB$$ (due to equilateral triangle SAB)

Do $$(SAB) \bot (ABC) \Rightarrow SH \bot (ABC)$$

Since triangle ABC is right-angled at A, AB=2a$$\Rightarrow SH = a\sqrt 3 .$$

$${S_{ABC}} = \frac{1}{2}AB.AC = \frac{1}{2}.2a.a = {a^2}$$

Draw KM parallel to AC cutting SA at M. Then AC//KM deduces AC//(BKM)

So d(AC,BK)=d(AC,(BKM))

We have $$AC \bot AB;AC \bot SH$$ so $$AC \bot (SAB)$$

The guy $$AI \bot BM,$$ is done by KM//AC so $$AI \bot KM$$ infers $$AI \bot \left( {BKM} \right)$$ Infer d(AC,BK)=d(AC,(BKM))=d(A,(BKM))=AI

We have: $$\frac{{MA}}{{SA}} = \frac{{KC}}{{SC}} = \frac{2}{3} \Rightarrow {S_{AMB}} = \frac {2}{3}{S_{SAB}} = \frac{2}{3}{(2a)^2}\frac{{\sqrt 3 }}{4} = \frac{2}{3}{ a^2}\sqrt 3 .$$

We have again $$BM = \sqrt {A{B^2} + A{M^2} – AB.AM.\cos {{60}^0}} = \frac{{2a\sqrt 7 }}{ 3}$$

So $$AI = \frac{{2{S_{ABM}}}}{{BM}} = \frac{{2\sqrt {21} a}}{7}.$$ So $$d(AC) ,BK) = \frac{{2\sqrt {21} a}}{7}.$$