Let D, E, F be points on the sides BC, CA, AB of a triangle ABC, respectively. Suppose AD, BE, CF are concurrent at P. If PF/PC = 2/3, PE/PB = 2/7 and PD/PA = m/n, where m, n are positive integers with gcd(m, n) = 1, find m + n.
Plz answer
Plz answer
Answers & Comments
Answer:
∠AFE. Prove that ΔABC is equilateral.
Medium
Solution
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Let BD=CE=AF=x;∠BDF=∠CED=∠AFE=θ.
Note that ∠AFD=B+θ,
and hence ∠DFE=B. Similarly, ∠EDF=Cand∠FED=A. ∴EFD is similar to ABC
We may take FD=ka,DE=kbandEF=kc, for some positive real constant k.
Applying sine rule to triangle BDF, we obtain
sinθ
c−x
=
sinB
ka
=
b
2Rka
,
where R is the circum-radius of ABC.
∴2Rksinθ=b(c−x)/a.
Similarly, we obtain 2Rksinθ=c(a−x)/band2Rksinθ=a(b−x)/c.
We therefore get
a
b(c−x)
=
b
c(a−x)
=
c
a(b−x)
(1)
If some two sides are equal, say, a=b, then a(c−x)=c(a−x) giving a=c; we get a=b=c and ABC is equilateral.
Suppose no two sides of ABC are equal.
We may assume a is the least.
Since (1) is cyclic in a,b,c, we have to consider two cases : a<b<canda<c<b.
Case 1.
a<b<c
.
In this case a<c and hence b(c−x)<a(b−x), form (1). Since b>aandc−x>b−x, we get b(c−x)>a(b−x), which is a contradiction.
Case 2.
a<c<b
.
We may write (1) in the form
a/b
(c−x)
=
b/c
(a−x)
=
c/a
(b−x)
(2).
Now a<c gives a−x<c−x so that
c
b
<
b
a
. This gives b
2
<ac. But b>aandb>c, so that b
2
>ac, which again leads to a contradiction
Thus Case 1 and Case 2 cannot occur.
∴a=b=c.