332. Some metrical problems from the contests I.
by Virgil Nicula, Jan 14, 2012, 9:32 AM
PP1 (2012 Japan Mathematical Olympiad). Given a triangle 
with the circumcircle 
. Let the points 
and 
such
that
, 
and 
is the midpoint of ![$[DE]$](//latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Denote the diameter ![$[AS]$](//latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
of 
and 
. Determine 
and the ratio 
.
Remark. I denoted the tangent line
to the circle in the point 
.
Proof. Denote the power
of the point 
w.r.t. the circle , i.e. 
. Observe that 
. Apply theorem of median to ![$[AO]$](//latex.artofproblemsolving.com/f/2/e/f2e2fb80c3e5e642dfc8b634152b9c973d507fa1.png)
in 
. Apply the power of the point 
w.r.t. the circle 
. Apply an well-known property (or can prove easily) 
and the
Menelaus theorem to
.
PP2. A line drawn from the vertex
of an equilateral triangle meets the side 
at 
and circumcircle at 
. Show that 
.
Proof 1 (with areas).
. Thus, ![$[BPC]=[DPB]+[DPC]\iff$](//latex.artofproblemsolving.com/a/7/3/a734b1ef44e15de45735c55cc84a148106fbd679.png)
.
Proof 2.
(Virgil Nicula - nice !).
Proof 3. From the Pompeiu's relation obtain that
. Observe that 
. From the relations 
and 
obtain that 
(Sunken Rock - very nice !).
An easy extension. A line from the vertex of an -isosceles triangle cut the side at and circumcircle at . Show that 
.
Proof. From the Pompeiu's relation obtain that
. Observe that
. From the relations and obtain that 
.
PP3 (Shortlist IMO - 1984) Let be a triangle with the incicle 
. Consider the circle 
which is exterior
tangent to and to the sidelines 
. Define analogously the circles 
and 
. Prove that 
.
Proof. Prove easily that
. Observe that
. I"ll apply the conditioned identity 
. In conclusion,
. Remark that 
.
PP4. Denote the area
of the triangle with the sides of lengths 
, 
, 
. Ascertain
, 
and 
so that 
.
Proof. Denote
. Thus, ![$\left\|\begin{array}{ccc}
BC'=2m_a & ; & [ABC']=S=S(7,b,c)\\\\
CA'=2m_b & ; & [BCA']=S=S(a,8,c)\\\\
AB'=2m_c & ; & [CAB']=S=S(a,b,10)\end{array}\right\|$](//latex.artofproblemsolving.com/5/4/f/54f8e97d2cec0091faf429ab4c06b77d0c53238a.png)
.
Therefore,
PP5. Let be a triangle with the incircle . Denote 
. Prove that 
.
Proof 1 (trigonometric). From an well-known property
obtain that 
. Apply the Sinus' theorem in 
.
Since
obtain that 
, i.e. 
Proof 2 (synthetic). From an well-known property obtain that the quadrilateral 
is inscribed in the circle 
with the diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
.
Prove easily that
.
PP6. Let be an -isosceles triangle. Prove that for any 
exists the relation 
.
Denote the midpoint
of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and the second intersection of the line 
with the circumcircle 
of the triangle .
Proof 1. Observe that
is an antiparallel to the side ![$[BE]$](//latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
in 
, i.e. 
. Using the power 
of w.r.t. obtain that 
. Therefore,
.In conclusion, .
Proof 2..
. The power 
. In conclusion, .
Proof 3. Suppose w.l.o.g.
. Thus, 
.
Proof 4. Suppose w.l.o.g. that
. Thus, 
.
Proof 5. Suppose w.l.o.g. that . Apply the generalized Pythagoras' teorem :
.
Proof 6. Denote
, 
, 
, 
and 
, 
, 
.
Thus,
. In conclusion, 
, what is truly.
Proof 7. Stewart relation
.
An easy extension. Let be an -isosceles triangle. Prove that for any 
exists the relation 
.
Proof. Apply Stewart's relation
because 
.
Proposed problem 1. Let
for which exists so that 
. Prove that is - isosceles or 
is the bisector of 
.
Proof. Exists so that 
. Apply Stewart's relation 
, i.e. 
is the bisector of .
Proposed problem 2. Prove that for exists so that 
and the point is uniquelly.
Proof. Apply Stewart's relation
. Since find so that obtain that
, i.e. is unique and it is the foot of bisector in .
PP7. Let
be an trapezoid for which 
, 
and 
. Suppose that 
. Find the value of 
.
Proof. I"ll construct the trapezoid with the mentioned property. Since can suppose w.l.o.g. 
and 
. Thus, 
and since 
, then 
so that
, where 
. Denote the midpoint of ![$[D_1D_2]$](//latex.artofproblemsolving.com/b/b/4/bb48cd6ac5a67967d55a8c150da0081fb97bc8c1.png)
, i.e. 
. Since 
and 
obtain that 
. Therefore,
.
PP8. Acute is given. A line 
passing through vertex 
is drawn. Let the bisectors of 
and 
intersect and 
at and 
and line 
at points and 
respectively. Prove 
.
Proof. Define the following relation between any two real numbers
, 
(same sign) .
Denote the length
of the -bisector a.s.o. Prove easily that 
and 
,
i.e.
. In conclusion, 
==================================================================================================
Indeed, 
a.s.o. and 
![$a(s-b)(b+c)^2=(b-a)\left[s(c^2-ab)+ab(a+b+2c)\right]\equiv (b-a)\cdot E$](//latex.artofproblemsolving.com/d/2/1/d21220e76417e41a567929cb5a89525b359359a0.png)
. Observe that 
because
.
and 
.
Lemma (Gergonne). Let and 
, 
, 
so that 
. Prove that 
.
Proof. From Aubel's relation get
a.s.o. Thus, 
a.s.o. So, 
.
PP9. Inside the triangle a point is marked. Three straight lines, which are parallel to the sides of triangle are drawn from that
point. Sections of the lines, which are inside the triangle are the same length . Find if triangle side lengths are 
.
Proof. Let be that point. It is well-known that if 
, , being obtained in a similar way, then (Gergonne). This implies 
.
PP10. Let be a triangle . For 
, 
and 
denote 
. Prove that 
both in magnitude and sign.
Proof. Denote
. Therefore, 
. Particular cases. 
See here. Remark.
means that the point belongs to the line 
and ![$[XY]$](//latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
is the notation of a segment. In upper proof means 
.
PP11. Let
so that 
and 
. Let equilateral so that separates and 
and
. Then 
.Particular case. 
.
Proof. Denote
. Apply property in 
![$\tan\phi \tan\theta\left[1+\sqrt 3\cdot\tan (\phi -\theta )\right]=$](//latex.artofproblemsolving.com/a/2/4/a24d50e0005bc15af26148f2da4ae3f304030d59.png)
. Apply generalized
Pytagoras' theorem in
. From
obtain 
.
with the circumcircle 
. Let the points 
and 
suchthat
, 
and 
is the midpoint of ![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Denote the diameter ![$[AS]$](http://latex.artofproblemsolving.com/8/1/f/81f842d87b22553e4b674bb2bf46a080da785009.png)
of 
and 
. Determine 
and the ratio 
.Remark. I denoted the tangent line
to the circle in the point 
.Proof. Denote the power
of the point 
w.r.t. the circle , i.e. 
. Observe that 
. Apply theorem of median to ![$[AO]$](http://latex.artofproblemsolving.com/f/2/e/f2e2fb80c3e5e642dfc8b634152b9c973d507fa1.png)
in 
. Apply the power of the point 
w.r.t. the circle 
. Apply an well-known property (or can prove easily) 
and theMenelaus theorem to
. PP2. A line drawn from the vertex
of an equilateral triangle meets the side 
at 
and circumcircle at 
. Show that 
.Proof 1 (with areas).
. Thus, ![$[BPC]=[DPB]+[DPC]\iff$](http://latex.artofproblemsolving.com/a/7/3/a734b1ef44e15de45735c55cc84a148106fbd679.png)
.Proof 2.
(Virgil Nicula - nice !).Proof 3. From the Pompeiu's relation obtain that
. Observe that 
. From the relations 
and 
obtain that 
(Sunken Rock - very nice !).An easy extension. A line from the vertex
of an -isosceles triangle cut the side at and circumcircle at . Show that 
.Proof. From the Pompeiu's relation obtain that
. Observe that . From the relations and obtain that 
.PP3 (Shortlist IMO - 1984) Let
be a triangle with the incicle 
. Consider the circle 
which is exteriortangent to
and to the sidelines 
. Define analogously the circles 
and 
. Prove that 
.Proof. Prove easily that
. Observe that
. I"ll apply the conditioned identity 
. In conclusion,
. Remark that 
.PP4. Denote the area
of the triangle with the sides of lengths 
, 
, 
. Ascertain
, 
and 
so that 
.Proof. Denote
. Thus, ![$\left\|\begin{array}{ccc}
BC'=2m_a & ; & [ABC']=S=S(7,b,c)\\\\
CA'=2m_b & ; & [BCA']=S=S(a,8,c)\\\\
AB'=2m_c & ; & [CAB']=S=S(a,b,10)\end{array}\right\|$](http://latex.artofproblemsolving.com/5/4/f/54f8e97d2cec0091faf429ab4c06b77d0c53238a.png)
.Therefore,
PP5. Let
be a triangle with the incircle . Denote 
. Prove that 
.Proof 1 (trigonometric). From an well-known property
obtain that 
. Apply the Sinus' theorem in 
.Since
obtain that 
, i.e. 
Proof 2 (synthetic). From an well-known property
obtain that the quadrilateral 
is inscribed in the circle 
with the diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
.Prove easily that
.PP6. Let
be an -isosceles triangle. Prove that for any 
exists the relation 
.Denote the midpoint
of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and the second intersection of the line 
with the circumcircle 
of the triangle .Proof 1. Observe that
is an antiparallel to the side ![$[BE]$](http://latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
in 
, i.e. 
. Using the power 
of w.r.t. obtain that 
. Therefore,
.In conclusion, .Proof 2..
. The power 
. In conclusion, .Proof 3. Suppose w.l.o.g.
. Thus, 
.Proof 4. Suppose w.l.o.g. that
. Thus, 
.Proof 5. Suppose w.l.o.g. that
. Apply the generalized Pythagoras' teorem :
.Proof 6. Denote
, 
, 
, 
and 
, 
, 
.Thus,
. In conclusion, 
, what is truly.Proof 7. Stewart relation
.An easy extension. Let
be an -isosceles triangle. Prove that for any 
exists the relation 
.Proof. Apply Stewart's relation
because 
. Proposed problem 1. Let
for which exists so that 
. Prove that is - isosceles or 
is the bisector of 
.Proof. Exists
so that 
. Apply Stewart's relation 
, i.e. 
is the bisector of .Proposed problem 2. Prove that for
exists so that 
and the point is uniquelly.Proof. Apply Stewart's relation
. Since find so that obtain that
, i.e. is unique and it is the foot of bisector in . PP7. Let
be an trapezoid for which 
, 
and 
. Suppose that 
. Find the value of 
.Proof. I"ll construct the trapezoid with the mentioned property. Since
can suppose w.l.o.g. 
and 
. Thus, 
and since 
, then 
so that
, where 
. Denote the midpoint of ![$[D_1D_2]$](http://latex.artofproblemsolving.com/b/b/4/bb48cd6ac5a67967d55a8c150da0081fb97bc8c1.png)
, i.e. 
. Since 
and 
obtain that 
. Therefore,
.PP8. Acute
is given. A line 
passing through vertex 
is drawn. Let the bisectors of 
and 
intersect
and 
at and 
and line 
at points and 
respectively. Prove 
.Proof. Define the following relation between any two real numbers
, 
(same sign) .Denote the length
of the -bisector a.s.o. Prove easily that 
and 
,i.e.
. In conclusion, 
==================================================================================================
Indeed, 
a.s.o. and 
![$a(s-b)(b+c)^2=(b-a)\left[s(c^2-ab)+ab(a+b+2c)\right]\equiv (b-a)\cdot E$](http://latex.artofproblemsolving.com/d/2/1/d21220e76417e41a567929cb5a89525b359359a0.png)
. Observe that 
because
.
and 
.Lemma (Gergonne). Let
and 
, 
, 
so that 
. Prove that 
.Proof. From Aubel's relation get
a.s.o. Thus, 
a.s.o. So, 
.PP9. Inside the triangle a point is marked. Three straight lines, which are parallel to the sides of triangle are drawn from that
point. Sections of the lines, which are inside the triangle are the same length
. Find if triangle side lengths are 
.Proof. Let
be that point. It is well-known that if 
, , being obtained in a similar way, then (Gergonne). This implies 
.PP10. Let
be a triangle . For 
, 
and 
denote 
. Prove that 
both in magnitude and sign.Proof. Denote
. Therefore, 
. Particular cases. 
See here. Remark.
means that the point belongs to the line 
and ![$[XY]$](http://latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
is the notation of a segment. In upper proof means 
.PP11. Let
so that 
and 
. Let equilateral so that separates and 
and
. Then 
.Particular case. 
.Proof. Denote
. Apply property in 
![$\tan\phi \tan\theta\left[1+\sqrt 3\cdot\tan (\phi -\theta )\right]=$](http://latex.artofproblemsolving.com/a/2/4/a24d50e0005bc15af26148f2da4ae3f304030d59.png)
. Apply generalizedPytagoras' theorem in
. From
obtain 
.This post has been edited 220 times. Last edited by Virgil Nicula, Nov 19, 2015, 1:51 PM
August 2018
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