The Minimum Possible Surface Area of A Tetrahedron with A Fixed Face and Volume
Let be a triangle such that , , and . Furthermore, let be a point in space such that the volume of tetrahedron is . The minimum possible surface area of tetrahedron can be expressed in the form , where and are positive integers and is a squarefree integer. Find .
The answer is 122.
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1 solution
Minimum area is obtained when D is just above the incenter I of ABC.
ABC is made up of two right angled triangles.
M is a point on AC such that AM=9 and MC=5. Right triangle CBM has BM=12, with angle BMC=90,
so also triangle ABM has BM=12 with angle BMA =90.
So BM=12 is the altitude and triangle area =1/2 * 14 * 12=84.
inradius of ABC, r=2 * area/perimeter= 2 * 84/42=4.
Volume=1/3 * area * height=1/3 * 84 * DI=28. So height h, DI=1.
On the sides, the slant height, perpendicular to the side will be at the point of tangentcy of the incircle.
Slant height, inradius of ABC and ID form a right triangle with slant height as hypotenuse.
This slant height is of same length for all three sides. S l a n t h e i g h t = h 2 + r 2 = 1 7 . S o t h e s i d e s u r f a c e a r e a = 2 1 ∗ p e r i m e t e r ∗ s l a n t h e i g h t = 2 1 ∗ 4 2 ∗ 1 7 . T o t a l s u r f a c e a r e a = 2 1 ∗ 1 7 + 8 4 = p q + r . p + q + r = 2 1 + 1 7 + 8 4 = 1 2 2 .