In right angled ΔABC, ∠ABC=90 and ∠BCA=60. The incircle of ΔABC touches sides BC and AB at points D and E respectively. Let F be the midpoint of DE. P, Q, R are the feet of perpendiculars from F on BC, CA, AB respectively.
Find (AB+BC+CA)/(PQ+QR+RP).
AC=l AB=
√3/2*l BC=l/2
Then AB+AC+BC= (3+√3)/2 * l
(r is the radio of the incircle and l= (1+√3)*r (after applying A(ABC)=s*r (s=semiperimeter)
So: AB+AC+BC= (6 + 4√3)*r
FR=FP= r/2
For FQ: If M is the interseccion of RF (extended) with AC, then FQ=√3/2 *FM
and FM= (3+5√3)/6)*r (after some calculation)
Then apply cosine rule to calculate PQ, QR, RP
I got: PQ+QR+RP=4.8187*r
Then if I get all things all right:
(AB+AC+BC)/(PQ+QR+RP)= 2.6828
This is not a difficult problem, as we are in a right triangle...
Edited on April 17, 2019, 4:28 am
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Posted by armando
on 2019-04-17 04:21:33 |
Solution
