Difference between revisions of "2022 SSMO Speed Round Problems/Problem 4"
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
| + | Let <math>r</math> be the inradius of <math>ABCD</math>. Note that <cmath>\frac{1}{2}r(AB+BC+CD+DA)=[ABCD]\implies r = \frac{2\cdot120}{24+24} = 5.</cmath> We have <cmath>h^2+5^2 = 13^2\implies h = 12,</cmath> meaning the volume of <math>PABCD</math> is <cmath>\frac{1}{3}h[ABCD] = \frac{1}{3}(12)(120) = \boxed{480}.</cmath> | ||
| + | |||
| + | |||
| + | ~pinkpig | ||
Latest revision as of 17:37, 13 September 2025
Problem
Consider a quadrilateral 
with area 
and satisfying 
. There exists a point 
in 3D space such that the distances from to 
, 
, 
, and 
are all equal to 
. Find the volume of 
.
Solution
Let 
be the inradius of . Note that ![\[\frac{1}{2}r(AB+BC+CD+DA)=[ABCD]\implies r = \frac{2\cdot120}{24+24} = 5.\]](http://latex.artofproblemsolving.com/6/3/0/6306ec7774e7090ac64376b231f6f36a368f844c.png)
We have ![\[h^2+5^2 = 13^2\implies h = 12,\]](http://latex.artofproblemsolving.com/5/9/3/593c0af89eceb7863ac0cb05e3165a17fa55a2f7.png)
meaning the volume of is ![\[\frac{1}{3}h[ABCD] = \frac{1}{3}(12)(120) = \boxed{480}.\]](http://latex.artofproblemsolving.com/f/0/6/f066135769627f54faa5bf2b8ae2cd6701dfbdb7.png)
~pinkpig
