424. Some nice problems of E. E. Q. Rodriguez and others.
by Virgil Nicula, May 4, 2015, 2:40 PM
PP1. Let 
be a triangle with the incenter 
. Prove that 
.
Proof 1 (synthetic - Sunken Rock). Denote
so that 
and the intersection 
. Prove easily that 
is
-isosceles, i.e. 
is 
-isosceles, i.e. 
is a cyclical quadrilateral 
is 
-isosceles, i.e. 
with the incenter
. In conclusion, 
and 
, i.e. 
.
Proof 2 (metric).
.
Observe that
. In conclusion, using the relations 
and 
obtain that
![$\frac {c(2a-c)\left[a+\frac {c(a-c)}a\right]}{a+\frac {a^2}c}=$](//latex.artofproblemsolving.com/6/2/4/62477a3b4ed95b3941a49d41cc2f33fcf07a2b38.png)
, i.e. .
Proof 3 (trigonometric). Apply the theorem of Sines in
.
Therefore,
. Observe that 
verify the relation 
.
PP2. Let be a triangle with the -excenter 
. Prove that 
.
Proof 2 (metric).
.
Observe that
. In conclusion, using the relations and obtain that 
![$\frac {a(2c+a)\left[c-\frac {a(a+c)}c\right]}{\frac {c^2}a-c}=$](//latex.artofproblemsolving.com/1/8/2/18218d69a8f6c98e344c2dd813a7af6bd7c0e412.png)
![$\frac {a^2\left[2\left(3a^2c+a^3\right)-ac^2-3a^2c-a^3\right]}{\left(3a^2c+a^3\right)-ac^2}=$](//latex.artofproblemsolving.com/5/0/4/50463492cac271dc71567cbf9f815381fde7f703.png)
, i.e. 
.
Proof 3 (trigonometric). Apply the theorem of Sines in
. Therefore, 
. Observe that 
verify the relation .
Lemma.
.
Proof. Prove easily that
, i.e. 
. Thus, 
. Apply the theorem of Sines
in
and 
. Now can prove easily the problems PP3 and PP4.
PP3. Let be a triangle with the -excenter and 
. Ascertain the value of the angle 
.
Proof 3 (trigonometric). Apply the theorem of Sines in
. Therefore, 
.
PP4. Let be a triangle with the -excenter and 
. Ascertain the value of the angle .
Proof 1 (synthetic).
and 
, i.e. 
.
Proof 2 (metric). Denote
, where 
. Thus, 
. Is well-known that 
and 
.
Therefore,
.
On other hand,
because 
. In conclusion, 
.
Proof 3 (trigonometric). Apply the theorem of Sines in
. Thus, 
.
PP5. Let
and 
for which construct 
. Denote 
and 
. Prove that 
.
Proof. Denote
. Thus, 
. Apply the Ceva' theorem to the point 
and 
. Apply the Menelaus' theorem to the transversals 
and 
. Is well-known or prove easily that the division 
is a harmonical division,
i.e.
. Thus, 
. In conclusion, 
.
Apply an well-known relation for the estimate of the ratio
.
Remark.![$\frac {[BAC]}{[EAF]}=\frac {bc}{xy}=\frac {PB\cdot PC}{PE\cdot PF}=\frac {[BPC]}{[EPF]} \implies$](//latex.artofproblemsolving.com/f/e/8/fe8ba4833b375cf339ab902194a0e89d36df4f26.png)
![$[BAC]\cdot [EPF]=[EAF]\cdot [BPC]$](//latex.artofproblemsolving.com/c/8/f/c8fd3d4cba08eba423d693d23cd0010ee3e71461.png)
. For any the ratio 
, i.e. it is constant.
PP6 (Edson Curahua). Let with the -bisector 
, where 
and 
. Prove that 
.
Proof 1 (trigonometric). Apply the theorem of Sines in
and 
![$\left[\cos\left(30^{\circ}-A\right)+\cos\left(30^{\circ}+A\right)\right]\sin A=\sin 60^{\circ}\sin \left(60^{\circ}-A\right)\iff$](//latex.artofproblemsolving.com/8/1/2/812adcade321fa6ccc446452cf7bf43d9b9f7d57.png)
.
PP7 (Ruben Dario). Let an equilateral with 
. Prove that 
.
Proof (metric). Denote
, 
and 
. Apply the generalized Pythagoras' theorem to 
, where 
. Apply the theorem of median in the mentioned triangles 
In conclusion, 
.
Remark.
. Prove
easily that
. Indeed, 
.
PP8. Let and the interior point so that 
. Prove that 
.
Proof 1 (Sunken Rock). Let
and the reflection of in 
. Triangle 
is equilateral and, since 
we infer that 
is cyclic.
Also
is angle bisector of 
, wherefrom due to angles equality we get 
, hence is the circumcenter of , consequently 
, giving 
Proof 2. Denote
and apply the trigonometric form of the Ceva's theorem 
.
PP9. Let an-rightangled with 
. Let the incircles 
, 
and
of the triangles , 
and 
respectively and 
. Prove that 
.
Proof. Let
. Thus, 
. In the trapezoid 
there is the relation 
.
Therefore,
and 
. Since 
obtain that
, i.e. .
An easy extension. Let with 
. Let the incircles , and
of the triangles , and respectively and . Prove that 
.
Proof. Let . Thus, in the trapezoid there is the relation . Thus, and
. Denote 
, 
and 
, where 
. Observe that 
. Therefore,
and 
. Observe that 
and 
. So 
.
In conclusion, using the relations , and 
obtain that 
. Are well-known the relations 
and
. Thus, the relation 
becomes 
![$\frac 1{AE}=\frac 1{2S}\left[(b+c)+a\left(1-\cot\frac A2\right)\right]$](//latex.artofproblemsolving.com/2/4/3/2433a7c4bfc2635018f757db8b8c50ac0b9ed118.png)
.
Particular case.
.
PP10 (JBMO - team selection test 2015). Let an- rightangled with the -bisector 
, where . The sideline 
cut again the circumcircle of
at the point 
. Consider the point 
for which 
. Prove that 
.
Proof (Gabi Cuc). Apply the theorem of the-bisector 
. Therefore,
is cyclic 
. Thus, 
. In conclusion, 
and 
, i.e. .
PP11 (Ruben Dario). Let with 
and 
. Prove that 
Proof 1.
.Thus, 
. Otherwise. The relation 
where 
- length of the circumradius for 
.
Proof 2.
Get similarly that 
Therefore, 
be a triangle with the incenter 
. Prove that 
.Proof 1 (synthetic - Sunken Rock). Denote
so that 
and the intersection 
. Prove easily that 
is
-isosceles, i.e. 
is 
-isosceles, i.e. 
is a cyclical quadrilateral 
is 
-isosceles, i.e. 
with the incenter
. In conclusion, 
and 
, i.e. 
.Proof 2 (metric).
.Observe that
. In conclusion, using the relations 
and 
obtain that
![$\frac {c(2a-c)\left[a+\frac {c(a-c)}a\right]}{a+\frac {a^2}c}=$](http://latex.artofproblemsolving.com/6/2/4/62477a3b4ed95b3941a49d41cc2f33fcf07a2b38.png)
, i.e. .Proof 3 (trigonometric). Apply the theorem of Sines in
.Therefore,
. Observe that 
verify the relation 
.PP2. Let
be a triangle with the -excenter 
. Prove that 
.Proof 2 (metric).
.Observe that
. In conclusion, using the relations and obtain that 
![$\frac {a(2c+a)\left[c-\frac {a(a+c)}c\right]}{\frac {c^2}a-c}=$](http://latex.artofproblemsolving.com/1/8/2/18218d69a8f6c98e344c2dd813a7af6bd7c0e412.png)
![$\frac {a^2\left[2\left(3a^2c+a^3\right)-ac^2-3a^2c-a^3\right]}{\left(3a^2c+a^3\right)-ac^2}=$](http://latex.artofproblemsolving.com/5/0/4/50463492cac271dc71567cbf9f815381fde7f703.png)
, i.e. 
.Proof 3 (trigonometric). Apply the theorem of Sines in
. Therefore, 
. Observe that 
verify the relation .Lemma.
. Proof. Prove easily that
, i.e. 
. Thus, 
. Apply the theorem of Sinesin
and 
. Now can prove easily the problems PP3 and PP4. PP3. Let
be a triangle with the -excenter and 
. Ascertain the value of the angle 
.Proof 3 (trigonometric). Apply the theorem of Sines in
. Therefore, 
.PP4. Let
be a triangle with the -excenter and 
. Ascertain the value of the angle .Proof 1 (synthetic).
and 
, i.e. 
.Proof 2 (metric). Denote
, where 
. Thus, 
. Is well-known that 
and 
.Therefore,
.On other hand,
because 
. In conclusion, 
.Proof 3 (trigonometric). Apply the theorem of Sines in
. Thus, 
.PP5. Let
and 
for which construct 
. Denote 
and 
. Prove that 
. Proof. Denote
. Thus, 
. Apply the Ceva' theorem to the point 
and 
. Apply the Menelaus' theorem to the transversals 
and 
. Is well-known or prove easily that the division 
is a harmonical division,i.e.
. Thus, 
. In conclusion, 
.Apply an well-known relation for the estimate of the ratio
.Remark.
![$\frac {[BAC]}{[EAF]}=\frac {bc}{xy}=\frac {PB\cdot PC}{PE\cdot PF}=\frac {[BPC]}{[EPF]} \implies$](http://latex.artofproblemsolving.com/f/e/8/fe8ba4833b375cf339ab902194a0e89d36df4f26.png)
![$[BAC]\cdot [EPF]=[EAF]\cdot [BPC]$](http://latex.artofproblemsolving.com/c/8/f/c8fd3d4cba08eba423d693d23cd0010ee3e71461.png)
. For any the ratio 
, i.e. it is constant.PP6 (Edson Curahua). Let
with the -bisector 
, where 
and 
. Prove that 
.Proof 1 (trigonometric). Apply the theorem of Sines in
and 
![$\left[\cos\left(30^{\circ}-A\right)+\cos\left(30^{\circ}+A\right)\right]\sin A=\sin 60^{\circ}\sin \left(60^{\circ}-A\right)\iff$](http://latex.artofproblemsolving.com/8/1/2/812adcade321fa6ccc446452cf7bf43d9b9f7d57.png)
.PP7 (Ruben Dario). Let an equilateral
with 
. Prove that 
. Proof (metric). Denote
, 
and 
. Apply the generalized Pythagoras' theorem to 
, where 
. Apply the theorem of median in the mentioned triangles 
In conclusion, 
.Remark.
. Proveeasily that
. Indeed, 
.PP8. Let
and the interior point so that 
. Prove that 
.Proof 1 (Sunken Rock). Let
and the reflection of in 
. Triangle 
is equilateral and, since 
we infer that 
is cyclic.Also
is angle bisector of 
, wherefrom due to angles equality we get 
, hence is the circumcenter of , consequently 
, giving 
Proof 2. Denote
and apply the trigonometric form of the Ceva's theorem 
. Quote:
Balkan Mathematical Olympiad - Athens, Hellas, 2015. Let ABC be a scalene triangle with incentre I and circumcircle (ω). The lines AI,BI,CI intersect
again (ω) at D,E,F respectively. The lines through I parallel to BC,CA,AB intersect EF,FD,DE at K,L,M respectively. Prove that K,L,M are collinear.
Proof. First we will prove that KA is tangent to (ω). Indeed, it is a well-known fact that FA=FB=FI and EA=EC=EI, so EF is the perpendicular bisector of AI. It follows that KA=KI and ∠KAF=∠KIF=∠FCB=∠FEB=∠FEA, so KA is tangent to (ω). Similarly we can prove that LB,MC are tangent to (ω) as well. Let A',B',C' the intersections of AI,BI,CI with BC,CA,AB respectively. From Pascal’s theorem on the cyclic hexagon AABEFC we get K,C',B' collinear. Similarly L,C',A' collinear and M,B',A' collinear. Then from Desargues’ theorem for DEF and A'B'C' which are perspective from I, we get that K,L,M of the intersection of their corresponding sides are collinear as wanted.
Remark. After proving that KA,LB,MC are tangent to (ω), we can argue as follows: it readily follows that KAF~KEA, i.e. KA/KE=KF/KA=AF/EA, thus KF/KE=(AF/AE)^2. In a similar way we can find that ME/MD=(CE/CD)^2 and LD/LF=(BD/BF)^2. Multiplying we obtain (KF/KE).(ME/MD).(LD/LF)=1, so by the converse of Menelaus theorem applied in DEF we get that K,L,M are collinear.
again (ω) at D,E,F respectively. The lines through I parallel to BC,CA,AB intersect EF,FD,DE at K,L,M respectively. Prove that K,L,M are collinear.
Proof. First we will prove that KA is tangent to (ω). Indeed, it is a well-known fact that FA=FB=FI and EA=EC=EI, so EF is the perpendicular bisector of AI. It follows that KA=KI and ∠KAF=∠KIF=∠FCB=∠FEB=∠FEA, so KA is tangent to (ω). Similarly we can prove that LB,MC are tangent to (ω) as well. Let A',B',C' the intersections of AI,BI,CI with BC,CA,AB respectively. From Pascal’s theorem on the cyclic hexagon AABEFC we get K,C',B' collinear. Similarly L,C',A' collinear and M,B',A' collinear. Then from Desargues’ theorem for DEF and A'B'C' which are perspective from I, we get that K,L,M of the intersection of their corresponding sides are collinear as wanted.
Remark. After proving that KA,LB,MC are tangent to (ω), we can argue as follows: it readily follows that KAF~KEA, i.e. KA/KE=KF/KA=AF/EA, thus KF/KE=(AF/AE)^2. In a similar way we can find that ME/MD=(CE/CD)^2 and LD/LF=(BD/BF)^2. Multiplying we obtain (KF/KE).(ME/MD).(LD/LF)=1, so by the converse of Menelaus theorem applied in DEF we get that K,L,M are collinear.
PP9. Let an
-rightangled with 
. Let the incircles 
, 
and
of the triangles , 
and 
respectively and 
. Prove that 
. Proof. Let
. Thus, 
. In the trapezoid 
there is the relation 
.Therefore,
and 
. Since 
obtain that
, i.e. .An easy extension. Let
with 
. Let the incircles , and of the triangles
, and respectively and . Prove that 
.Proof. Let
. Thus, in the trapezoid there is the relation . Thus, and 
. Denote 
, 
and 
, where 
. Observe that 
. Therefore,
and 
. Observe that 
and 
. So 
.In conclusion, using the relations
, and 
obtain that 
. Are well-known the relations 
and
. Thus, the relation 
becomes 
![$\frac 1{AE}=\frac 1{2S}\left[(b+c)+a\left(1-\cot\frac A2\right)\right]$](http://latex.artofproblemsolving.com/2/4/3/2433a7c4bfc2635018f757db8b8c50ac0b9ed118.png)
.Particular case.
.PP10 (JBMO - team selection test 2015). Let an
- rightangled with the -bisector 
, where . The sideline 
cut again the circumcircle of
at the point 
. Consider the point 
for which 
. Prove that 
.Proof (Gabi Cuc). Apply the theorem of the
-bisector 
. Therefore,
is cyclic 
. Thus, 
. In conclusion, 
and 
, i.e. .PP11 (Ruben Dario). Let
with 
and 
. Prove that 
Proof 1.
.Thus, 
. Otherwise. The relation 
where 
- length of the circumradius for 
.Proof 2.
Get similarly that 
Therefore, 
This post has been edited 252 times. Last edited by Virgil Nicula, Sep 28, 2015, 5:35 PM
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