362. Fermat's problem and other metrical problems.
by Virgil Nicula, Nov 8, 2012, 9:36 AM
Fermat's problem. Let 
be a rectangle and let 
be the circle with the diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
so that 
. Consider
such that 
doesn't separate 
, 
. Let 
, 
. Prove that 
.
Proof (synthetic).. Suppose w.l.o.g. that
and 
. Denote the projection 
of on 
and 
, 
. Observe that 
and 
.
Thus,
, where 
. Therefore, 
An easy extension of the Fermat's problem. Let a rectangle and the circle with the diameter . Let for which ![$N\in [AB]\ ,\ MN\perp AB$](//latex.artofproblemsolving.com/2/9/d/29d3aaf2f66e12c1c01dca9b148d34c563cc8afc.png)
and doesn't
separate and . Denote , and 
. Prove that 
.
Proof (metric).
, 
and 
, 
and 
,
where
. Hence 
.
PP1. Let
be a circle with the diameter and let be a rectangle with 
. The tangent
to from cut in 
. Denote 
, 
, 
. Prove that :
is the midpoint of the segment ![$[EF]$](//latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
and 
.
and 
(Fermat's problem).
Proof.
PP2. Solve the system
, where 
and show that 
.
Proof (algebraic). Prove easily that
. Our system 
and 
. But 
, Therefore,
and 
and 
. 
, 
, 
and from the relation
obtaim that 
. Add the equations of the initial system : 
.
Therefore,
. In conclusion, 
and 
.
Proof 2 (geometric). Take a triangle
of sides 
where 
. Let the point 
inside it so that 
, 
, 
and
(Fermat point). See that is right-angled at 
and consequently has area 
. But its area 
. Equalizing the two area relations we get the desired 
.
PP3. Let
be an isosceles -right triangle. For a given point 
denote 
so that 
and 
. Prove that 
.
Proof.
.
Remark. An interesting particular case is when
. Therefore, 
the point 
is the midpoint of .
PP4. Let be an -right triangle and let be a point of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Denote 
. Prove that 
.
Proof 1 (metric). Denote
. Apply the Stewart's relation to the cevian 
in 
.
Thus,
, what is the relation 
. Observe that
.
Proof 2 (trigonometric). Denote
and apply Sinus' theorem in 
, where 
,
i.e.
. Therefore, 
, what is truly.
Remark. I"ll apply to the degenerate
an well-known property: if has 
, then there is the relation 
, where
. Obtain that 
. Otherwise. 
.
PP5. is a trapezoid with 
and 
. Prove that 
belongs to the radical axis of the circles 
, 
with the diameters ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, respectively.
Proof 1. Denote the midpoints
, 
of and respectively and 
, 
. Define the power 
of 
w.r.t. . Thus,
.
.
Since
and 
get 
belongs to the radical axis of and .
Proof 2. Denote the projections
. Thus,
![$\left\{\begin{array}{c}
[AOD]=[BOC]\implies OD\cdot AM=OC\cdot BN\\\\
\frac {MA}{MO}=\tan\widehat{AOD}=\tan\widehat{BOC}=\frac {NB}{NO}\end{array}\right\|\implies$](//latex.artofproblemsolving.com/d/f/a/dfa4b8272f62e0b3fcdee7ce6f7af61c59be3d5a.png)
.
PP6. Let a square and 
for which 
and the incircles 
, 
of 
, 
respectively. Find 
.
Proof 1 (trigonometric). Let
, 
and 
. Thus, 
, 
and 
, i.e.
, 
. Therefore, 
.
Proof 2 (synthetic). Let: the circles , 
; the projections , 
of 
, 
on ; 
. Thus, 
, 
, 
and 
.
Prove easily that
![$[c+(a+b)]^2=2(c+a)(c+b)\iff$](//latex.artofproblemsolving.com/1/9/e/19e4e07da3053eee2f939d55aff8263c32c6dce5.png)
.
Remark (equivalent). Let an isosceles
-right 
. For 
let the feet , of the -bisectors in 
, 
respectively. Prove that 
.
Extension.
is convex and cyclic so that 
. Let 
and
. Prove that , 
is cyclic and 
.
Proof.
and 
, 
, where
and 
, i.e. is cyclic and 
. Prove easily that 
, i.e. 
.
be a rectangle and let 
be the circle with the diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
so that 
. Consider
such that 
doesn't separate 
, 
. Let 
, 
. Prove that 
.Proof (synthetic).. Suppose w.l.o.g. that
and 
. Denote the projection 
of on 
and 
, 
. Observe that 
and 
.Thus,
, where 
. Therefore, 
An easy extension of the Fermat's problem. Let a rectangle
and the circle with the diameter . Let for which ![$N\in [AB]\ ,\ MN\perp AB$](http://latex.artofproblemsolving.com/2/9/d/29d3aaf2f66e12c1c01dca9b148d34c563cc8afc.png)
and doesn'tseparate
and . Denote , and 
. Prove that 
.Proof (metric).
, 
and 
, 
and 
,where
. Hence 
.PP1. Let
be a circle with the diameter and let be a rectangle with 
. The tangentto
from cut in 
. Denote 
, 
, 
. Prove that :
is the midpoint of the segment ![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
and 
.
and 
(Fermat's problem).Proof.
PP2. Solve the system
, where 
and show that 
.Proof (algebraic). Prove easily that
. Our system 
and 
. But 
, Therefore,
and 
and 
. 
, 
, 
and from the relation
obtaim that 
. Add the equations of the initial system : 
.Therefore,
. In conclusion, 
and 
.Proof 2 (geometric). Take a triangle
of sides 
where 
. Let the point 
inside it so that 
, 
, 
and
(Fermat point). See that is right-angled at 
and consequently has area 
. But its area 
. Equalizing the two area relations we get the desired 
.PP3. Let
be an isosceles -right triangle. For a given point 
denote 
so that 
and 
. Prove that 
.Proof.
.Remark. An interesting particular case is when
. Therefore, 
the point 
is the midpoint of .PP4. Let
be an -right triangle and let be a point of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Denote 
. Prove that 
.Proof 1 (metric). Denote
. Apply the Stewart's relation to the cevian 
in 
.Thus,
, what is the relation 
. Observe that
.Proof 2 (trigonometric). Denote
and apply Sinus' theorem in 
, where 
,i.e.
. Therefore, 
, what is truly.Remark. I"ll apply to the degenerate
an well-known property: if has 
, then there is the relation 
, where
. Obtain that 
. Otherwise. 
.PP5.
is a trapezoid with 
and 
. Prove that 
belongs to the radical axis of the circles 
, 
with the diameters ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, respectively.Proof 1. Denote the midpoints
, 
of and respectively and 
, 
. Define the power 
of 
w.r.t. . Thus,
.
.Since
and 
get 
belongs to the radical axis of and .Proof 2. Denote the projections
. Thus,![$\left\{\begin{array}{c}
[AOD]=[BOC]\implies OD\cdot AM=OC\cdot BN\\\\
\frac {MA}{MO}=\tan\widehat{AOD}=\tan\widehat{BOC}=\frac {NB}{NO}\end{array}\right\|\implies$](http://latex.artofproblemsolving.com/d/f/a/dfa4b8272f62e0b3fcdee7ce6f7af61c59be3d5a.png)
.PP6. Let a square
and 
for which 
and the incircles 
, 
of 
, 
respectively. Find 
.Proof 1 (trigonometric). Let
, 
and 
. Thus, 
, 
and 
, i.e.
, 
. Therefore, 
.Proof 2 (synthetic). Let: the circles
, 
; the projections , 
of 
, 
on ; 
. Thus, 
, 
, 
and 
.Prove easily that
![$[c+(a+b)]^2=2(c+a)(c+b)\iff$](http://latex.artofproblemsolving.com/1/9/e/19e4e07da3053eee2f939d55aff8263c32c6dce5.png)
.Remark (equivalent). Let an isosceles
-right 
. For 
let the feet , of the -bisectors in 
, 
respectively. Prove that 
.Extension.
is convex and cyclic so that 
. Let 
and
. Prove that , 
is cyclic and 
.Proof.
and 
, 
, where
and 
, i.e. is cyclic and 
. Prove easily that 
, i.e. 
.This post has been edited 74 times. Last edited by Virgil Nicula, Nov 16, 2015, 5:22 PM
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