389. Some geometric properties in a triangle I.
by Virgil Nicula, Nov 9, 2013, 9:19 AM
PP1. Let a 
-rightangled 
with 
and 
. For a mobile ![$M\in [AC]$](//latex.artofproblemsolving.com/7/b/1/7b1a07ab551be77c55343ddd04e6923c5592b7ab.png)
define ![$P\in [BC]$](//latex.artofproblemsolving.com/0/8/7/087dcfc6e817021079bdd2436e980d096613bc7a.png)
so that 
. Finf the range of 
.
Proof. Denote
and 
. Prove easily that 
. Apply the theorem of Sines in 
. Thus, 
![$\frac 32\left[\cos \left(2x+30^{\circ}\right)+\cos 30^{\circ}\right]+$](//latex.artofproblemsolving.com/1/9/3/193be3e99339b6f7da8884eabba102efed202e8a.png)
![$2\left[\sin\left (2x+30^{\circ} \right)-\sin 30^{\circ}\right]=$](//latex.artofproblemsolving.com/4/3/3/433b826352ed99a1f6cf8b9aefe592d1aaea4c49.png)
![$\left(\frac {3\sqrt 3}{4}-1\right)+\frac 12\left[3\cos \left(2x+30^{\circ}\right)+4\sin \left(2x+30^{\circ}\right)\right]\le$](//latex.artofproblemsolving.com/7/0/6/7066aaebc6a236a15cc837bc1b4ca6492951d400.png)
. Thus, 
with equality iff 
. In conclusion, 
.
Remark. I used only the inequality
with equality iff 
.
PP2. Let an acute with the orthic 
, where 
, 
, 
. Let the midpoint 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the projection 
of
the orthocenter
on 
and the 
-bisector 
of , where 
. Prove that 
and 
.
Proof.
.
.
Thus,
, i.e. 
is the -bisector of 
.
PP3.
and 
exists relation 
. Find 
where 
Proof.
![$R\left[2\sin x\sin\frac x2+\cos\frac x2\left(\sum\cos A-1\right)+\sin\frac x2\cdot\sum\sin A\right]=$](//latex.artofproblemsolving.com/c/e/c/cec3cc739d86543d041287581f9a7fc3223a0071.png)
![$R\left[2\sin x\sin\frac x2-\cos \frac x2+\cos \left(A-\frac x2\right)+\cos\left(B-\frac x2\right)+\cos\left(C-\frac x2\right)\right]=$](//latex.artofproblemsolving.com/5/2/2/5222a2f2d310986a15e6e49a2807982d09d2924f.png)
.
Remark.
. For 
obtain 
.
, i.e. 
. Example. 
PP4. Solve the equation
in . See here
PP5. Let , 
with 
, circle 
which is interior tangent to 
in 
and 
. Prove that 
.
Proof. From an well-known property obtain that the rays
and 
are -bisectors of 
and 
respectively. Thus, 
and 
.
Let
, i.e. 
. Therefore, 
,
Remark. Observe that
and 
are cyclically. Thus, 
, i.e. the ray 
is the bisector of 
(Julio Orihuela's proof).
PP6. Let with incenter 
which touches the at , , . Let the midpoint of . Prove that 
.
Proof.
and
. In conclusion, 
.
PP7. Let and for 
, 
denote 
. Prove that 
is tangential 
Proof. Let the incircle of and 
. Thus, 
and 
.
PP8. Let a convex
for which 
, 
and exists circle what is tangent to 
, 
, 
, 
. Prove that 
.
Proof. Let
. Thus, 
and 
.
PP9. Let with 
. Prove that ![$[BDM]=[CEN]\iff AB=AC$](//latex.artofproblemsolving.com/c/8/1/c81f809e83e9491292d1bf7367a2d84e4cedacd0.png)
.
Proof. Is evidently that![$AB=AC\implies [BDM]=[CEN]$](//latex.artofproblemsolving.com/e/9/3/e93791c511cd5653cb05a32e98cd151f278ccfe6.png)
. Suppose ![$[BDM]=[CEN]=x$](//latex.artofproblemsolving.com/1/b/9/1b9fba31a63d423a8a1322ed1eb5866b0ee1e885.png)
and let 
.
From![$\frac {x+u}{x+v}=\frac {[ABM]}{[ACN]}=\frac {BM}{CN}=\frac mn$](//latex.artofproblemsolving.com/f/f/7/ff7f40714065ba535c18c7e8439a9f3545e84c27.png)
obtain 
and 
![$(m-n)[2mn+a(m+n)]=0\implies$](//latex.artofproblemsolving.com/3/7/a/37a0a87f12c4e9834501fa732fd72c1e16423888.png)
.
PP10. Let with 
. Find 
and ascertain 
(standard notations).
Proof 1. I"ll use some remarkable properties of
, where 
is the Nagel's point w.r.t. 
. Is well-known that 
with 
, i.e. 
is a trapezoid and 
and 
. Let 
, where 
and apply
the generalized Pythagoras' theorem to the mentioned corresponding sides/triangles
. Remark that 
and 
.
Proof 2 (Israel Diaz, Peru). Denote the Euler's circle
w.r.t. and 
. From the Euler's relation 
obtain that
. Thus, 
and 
. From the Feuerbach's theorem (the incircle is interior tangent to the Euler's circle) obtain that 
, i.e.
. Hence the relations 
. Apply the Stewart's relation to :
Using the relation
obtain that 
.
PP11. The incircle of touches it at 
, 
, 
. Suppose that circumcenter of belongs to circumcircle of . Prove 
Proof. Denote the circumcenter
of and suppose w.l.o.g. 
, separates , and 
. Is well-known that 
.
Prove easily that
(s.s.a. - obtuse) , i.e. 
and 
. Apply Stewart's relation
to
.
Apply the Stewart's relation to
in 
![$ad^2+a(s-b)(s-c)=l^2[(s-b)+(s-c)]\implies$](//latex.artofproblemsolving.com/d/1/a/d1a2cf55671afec48755af8cd7c4bc685b131189.png)
. From 
get 
, i.e. 
.
PP12 (Julio Orihuela, Peru). Let with the circumcircle 
. Consider and 
so that 
and 
. Let 
so that 
and doesn't separate , 
and 
. Prove that 
.
Proof.
and 
. Hence from 
obtain that 
is -bisector of 
and .
PP13. Let with the orthocenter . Denote 
and 
. Prove that 
.
Proof.
Denote the orthic triangle 
, where , and . Observe that 
and 
. Apply the Menelaus' theorem to the transversal 
and
, i.e. the division 
is harmonically and 
.
Denote 
. Apply a well-known property (or you can prove easily it) of a harmonical division 
. Therefore,
is a cyclical quadrilateral 
. See PP1 from here.
PP14. Prove that
(Carlos Olivera & Ruben Auqui, PERU).
Proof. Denote
so that 
the incircle of 
. Observe that 
and
. Therefore, 
.
Hence
. On other hand 
. Using the relations 
and 
obtain easily that 
, what is truly.
PP15 (Stan Fulger's lemma). Let the circle and 
so that 
. For a point 
so that doesn't
separate and denote its projections 
on , , 
respectively. Prove that 
.
Proof. Thus,
.
Application 1.-excircle 
touches at 
, where 
. Let the distance 
. Define analogously the distancies 
and 
. Prove that 
.
Proof 1. Appy the upper lemma and obtain that
. In conclusion, 
.
Proof 2. Suppose w.l.o.g. that
and 
. Prove easily that 
and 
. Thus
and 
and 
a.s.o. Obtain quickly that using the well-known identities 
and 
.
Application 2. Let be a triangle with the incircle 
and the -excircle 
. The circle touches at 
and the -excircle touches at 
. Denote the areas ![$S=[DEF]$](//latex.artofproblemsolving.com/b/c/d/bcd97a7380624fb73fbef9040305e12cbe3a28db.png)
and ![$S'=[D'E'F']$](//latex.artofproblemsolving.com/0/2/1/021ccb17f6887ec779b36cc974968355b620c16d.png)
. Prove that 
.
Proof 1. Prove easily that
and 
. Therefore, using the identities from 
obtain that
.
Remark. Can use the Stan Fulger's lemma. Denote the projections
, 
and 
of 
on 
, and respectively. Thus, 
and
![$S'=[E'D'F']=s\sin\frac A2\cdot\sqrt{(s-b)(s-c)\sin B\sin C}=$](//latex.artofproblemsolving.com/a/9/3/a934e07cec37411fdb6c72d834ed2ad2e9b796c2.png)
.
Otherwise. Denote the diameter![$[DS]$](//latex.artofproblemsolving.com/d/3/e/d3edda0cd1b9b44cc8c2bfe6b4cc240f712aaf9f.png)
of . Is well-known 
and prove easily that 
, 
, 
, i.e.
the triangles
and 
are homothetically with the vertex and the ratio 
. Therefore, ![$S'=\left(\frac s{s-a}\right)^2\cdot [SEF]$](//latex.artofproblemsolving.com/0/f/9/0f968269df5ee6664ac09865191a1e131a53f50e.png)
, where
![$[SEF]=2r^2\sin\frac C2\sin\frac B2\cos\frac A2=$](//latex.artofproblemsolving.com/d/f/4/df413249991c515a0bcfcd215ac95537cc5d943e.png)
a.s.o.
Proof 2. I"ll use the relation
, where is the area of . Thus, 
and 
. Hence:
In conclusion,
.
PP16 (O.M. IberoAm., 2014). Let an acute with the orthocenter and the orthic , where 
.
Denote
. Prove that 
is cyclically.
Proof. Apply the Menelaus theorem to the mentioned transversals/triangles
and 
are isosceles trapezoids because 
and 
respectively. Since
is cyclically.
PP17. Let and and the bisectors 
of 
respectively,
where
. Prove that 
.
Proof.
. Apply Stewart's to 
.
Therefore,
.
PP18. Let parallelogram with 
and 
. Prove that 
.
Proof. Let
. Apply the generalized Pytagoras' theorem to 
the relation .
PP19. Let the circle , 
and the midpoint of the small arc ![$\overarc[]{AB}$](//latex.artofproblemsolving.com/7/d/7/7d78b2263c122b211ec9a6ebd0a804354ec087b6.png)
. Prove that ![$(\forall )\ P\in \overarc[]{AB}$](//latex.artofproblemsolving.com/b/7/7/b779ec50af49132e538700f866d8c048d4997405.png)
have 
.
Proof. Suppose w.l.o.g.![$P\in\overarc[]{MB}$](//latex.artofproblemsolving.com/3/0/3/30328124fc3d98b4ec5b37375e21dfc275f94424.png)
and let ![$R\in \overarc[]{MA}$](//latex.artofproblemsolving.com/1/6/4/1643de826691d509d786eb1cf16223384ecb5687.png)
so that 
. Apply the Ptolemy's theorem to the isosceles trapezoid 
. Since 
obtain that .
PP20. Aratati ca in exista echivalenta 
Demonstratie 1.
are solutiile 
Asadar, 
Intr-adevar, 
si 
si 
Se arata usor ca 
Discriminantul in raport cu 
este 
si singura radacina pozitiva este
Asadar 
Demonstratie 2. Si acum o solutie foarte simpla pentru problema propusa la nivelul clasei a VII - a !
Se observa ca
![$A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right]\ \Longrightarrow\ b^2+c^2\ \ge\ a^2+4S\sqrt 3\ .$](//latex.artofproblemsolving.com/c/f/4/cf436b07b6eb2902089f0f5a4b7fa06b3f2abfe5.png)
Iata o problema similara 
Gasiti
o extindere pastrand expresia
Observatie. 
unde am notat 
Deci,
Extindere.
Demonstratie. Se observa ca
-rightangled 
with 
and 
. For a mobile ![$M\in [AC]$](http://latex.artofproblemsolving.com/7/b/1/7b1a07ab551be77c55343ddd04e6923c5592b7ab.png)
define ![$P\in [BC]$](http://latex.artofproblemsolving.com/0/8/7/087dcfc6e817021079bdd2436e980d096613bc7a.png)
so that 
. Finf the range of 
.Proof. Denote
and 
. Prove easily that 
. Apply the theorem of Sines in 
. Thus, 
![$\frac 32\left[\cos \left(2x+30^{\circ}\right)+\cos 30^{\circ}\right]+$](http://latex.artofproblemsolving.com/1/9/3/193be3e99339b6f7da8884eabba102efed202e8a.png)
![$2\left[\sin\left (2x+30^{\circ} \right)-\sin 30^{\circ}\right]=$](http://latex.artofproblemsolving.com/4/3/3/433b826352ed99a1f6cf8b9aefe592d1aaea4c49.png)
![$\left(\frac {3\sqrt 3}{4}-1\right)+\frac 12\left[3\cos \left(2x+30^{\circ}\right)+4\sin \left(2x+30^{\circ}\right)\right]\le$](http://latex.artofproblemsolving.com/7/0/6/7066aaebc6a236a15cc837bc1b4ca6492951d400.png)
. Thus, 
with equality iff 
. In conclusion, 
.Remark. I used only the inequality
with equality iff 
.PP2. Let an acute
with the orthic 
, where 
, 
, 
. Let the midpoint 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the projection 
ofthe orthocenter
on 
and the 
-bisector 
of , where 
. Prove that 
and 
.Proof.
.
.Thus,
, i.e. 
is the -bisector of 
. PP3.
and 
exists relation 
. Find 
where 
Proof.
![$R\left[2\sin x\sin\frac x2+\cos\frac x2\left(\sum\cos A-1\right)+\sin\frac x2\cdot\sum\sin A\right]=$](http://latex.artofproblemsolving.com/c/e/c/cec3cc739d86543d041287581f9a7fc3223a0071.png)
![$R\left[2\sin x\sin\frac x2-\cos \frac x2+\cos \left(A-\frac x2\right)+\cos\left(B-\frac x2\right)+\cos\left(C-\frac x2\right)\right]=$](http://latex.artofproblemsolving.com/5/2/2/5222a2f2d310986a15e6e49a2807982d09d2924f.png)
.Remark.
. For 
obtain 
.
, i.e. 
. Example. 
PP4. Solve the equation
in . See herePP5. Let
, 
with 
, circle 
which is interior tangent to 
in 
and 
. Prove that 
.Proof. From an well-known property obtain that the rays
and 
are -bisectors of 
and 
respectively. Thus, 
and 
.Let
, i.e. 
. Therefore, 
,Remark. Observe that
and 
are cyclically. Thus, 
, i.e. the ray 
is the bisector of 
(Julio Orihuela's proof).PP6. Let
with incenter 
which touches the at , , . Let the midpoint of . Prove that 
.Proof.
and
. In conclusion, 
.PP7. Let
and for 
, 
denote 
. Prove that 
is tangential 
Proof. Let the incircle
of and 
. Thus, 
and 
.PP8. Let a convex
for which 
, 
and exists circle what is tangent to 
, 
, 
, 
. Prove that 
. Proof. Let
. Thus, 
and 
.PP9. Let
with 
. Prove that ![$[BDM]=[CEN]\iff AB=AC$](http://latex.artofproblemsolving.com/c/8/1/c81f809e83e9491292d1bf7367a2d84e4cedacd0.png)
.Proof. Is evidently that
![$AB=AC\implies [BDM]=[CEN]$](http://latex.artofproblemsolving.com/e/9/3/e93791c511cd5653cb05a32e98cd151f278ccfe6.png)
. Suppose ![$[BDM]=[CEN]=x$](http://latex.artofproblemsolving.com/1/b/9/1b9fba31a63d423a8a1322ed1eb5866b0ee1e885.png)
and let 
.From
![$\frac {x+u}{x+v}=\frac {[ABM]}{[ACN]}=\frac {BM}{CN}=\frac mn$](http://latex.artofproblemsolving.com/f/f/7/ff7f40714065ba535c18c7e8439a9f3545e84c27.png)
obtain 
and 
![$(m-n)[2mn+a(m+n)]=0\implies$](http://latex.artofproblemsolving.com/3/7/a/37a0a87f12c4e9834501fa732fd72c1e16423888.png)
.PP10. Let
with 
. Find 
and ascertain 
(standard notations).Proof 1. I"ll use some remarkable properties of
, where 
is the Nagel's point w.r.t. 
. Is well-known that 
with 
, i.e. 
is a trapezoid and 
and 
. Let 
, where 
and applythe generalized Pythagoras' theorem to the mentioned corresponding sides/triangles
. Remark that 
and 
.Proof 2 (Israel Diaz, Peru). Denote the Euler's circle
w.r.t. and 
. From the Euler's relation 
obtain that
. Thus, 
and 
. From the Feuerbach's theorem (the incircle is interior tangent to the Euler's circle) obtain that 
, i.e.
. Hence the relations 
. Apply the Stewart's relation to :
Using the relation
obtain that 
.PP11. The incircle
of touches it at 
, 
, 
. Suppose that circumcenter of belongs to circumcircle of . Prove 
Proof. Denote the circumcenter
of and suppose w.l.o.g. 
, separates , and 
. Is well-known that 
.Prove easily that
(s.s.a. - obtuse) , i.e. 
and 
. Apply Stewart's relationto
.Apply the Stewart's relation to
in 
![$ad^2+a(s-b)(s-c)=l^2[(s-b)+(s-c)]\implies$](http://latex.artofproblemsolving.com/d/1/a/d1a2cf55671afec48755af8cd7c4bc685b131189.png)
. From 
get 
, i.e. 
.PP12 (Julio Orihuela, Peru). Let
with the circumcircle 
. Consider and 
so that 
and 
. Let 
so that 
and doesn't separate , 
and 
. Prove that 
.Proof.
and 
. Hence from 
obtain that 
is -bisector of 
and .PP13. Let
with the orthocenter . Denote 
and 
. Prove that 
.Proof.
Denote the orthic triangle 
, where , and . Observe that 
and 
. Apply the Menelaus' theorem to the transversal 
and
, i.e. the division 
is harmonically and 
. Denote 
. Apply a well-known property (or you can prove easily it) of a harmonical division 
. Therefore,
is a cyclical quadrilateral 
. See PP1 from here.PP14. Prove that
(Carlos Olivera & Ruben Auqui, PERU).
Proof. Denote
so that 
the incircle of 
. Observe that 
and
. Therefore, 
.Hence
. On other hand 
. Using the relations 
and 
obtain easily that 
, what is truly.PP15 (Stan Fulger's lemma). Let the circle
and 
so that 
. For a point 
so that doesn'tseparate
and denote its projections 
on , , 
respectively. Prove that 
.Proof. Thus,
.Application 1.
-excircle 
touches at 
, where 
. Let the distance 
. Define analogously the distancies 
and 
. Prove that 
.Proof 1. Appy the upper lemma and obtain that
. In conclusion, 
.Proof 2. Suppose w.l.o.g. that
and 
. Prove easily that 
and 
. Thus
and 
and 
a.s.o. Obtain quickly that using the well-known identities 
and 
.Application 2. Let
be a triangle with the incircle 
and the -excircle 
. The circle touches at 
and the -excircle touches at 
. Denote the areas ![$S=[DEF]$](http://latex.artofproblemsolving.com/b/c/d/bcd97a7380624fb73fbef9040305e12cbe3a28db.png)
and ![$S'=[D'E'F']$](http://latex.artofproblemsolving.com/0/2/1/021ccb17f6887ec779b36cc974968355b620c16d.png)
. Prove that 
.Proof 1. Prove easily that
and 
. Therefore, using the identities from 
obtain that
.Remark. Can use the Stan Fulger's lemma. Denote the projections
, 
and 
of 
on 
, and respectively. Thus, 
and
![$S'=[E'D'F']=s\sin\frac A2\cdot\sqrt{(s-b)(s-c)\sin B\sin C}=$](http://latex.artofproblemsolving.com/a/9/3/a934e07cec37411fdb6c72d834ed2ad2e9b796c2.png)
.Otherwise. Denote the diameter
![$[DS]$](http://latex.artofproblemsolving.com/d/3/e/d3edda0cd1b9b44cc8c2bfe6b4cc240f712aaf9f.png)
of . Is well-known 
and prove easily that 
, 
, 
, i.e.the triangles
and 
are homothetically with the vertex and the ratio 
. Therefore, ![$S'=\left(\frac s{s-a}\right)^2\cdot [SEF]$](http://latex.artofproblemsolving.com/0/f/9/0f968269df5ee6664ac09865191a1e131a53f50e.png)
, where![$[SEF]=2r^2\sin\frac C2\sin\frac B2\cos\frac A2=$](http://latex.artofproblemsolving.com/d/f/4/df413249991c515a0bcfcd215ac95537cc5d943e.png)
a.s.o.Proof 2. I"ll use the relation
, where is the area of . Thus, 
and 
. Hence:
In conclusion,
.PP16 (O.M. IberoAm., 2014). Let an acute
with the orthocenter and the orthic , where 
.Denote
. Prove that 
is cyclically.Proof. Apply the Menelaus theorem to the mentioned transversals/triangles
and 
are isosceles trapezoids because 
and 
respectively. Since
is cyclically.PP17. Let
and and the bisectors 
of 
respectively,where
. Prove that 
. Proof.
. Apply Stewart's to 
.Therefore,
.PP18. Let parallelogram
with 
and 
. Prove that 
.Proof. Let
. Apply the generalized Pytagoras' theorem to 
the relation .PP19. Let the circle
, 
and the midpoint of the small arc ![$\overarc[]{AB}$](http://latex.artofproblemsolving.com/7/d/7/7d78b2263c122b211ec9a6ebd0a804354ec087b6.png)
. Prove that ![$(\forall )\ P\in \overarc[]{AB}$](http://latex.artofproblemsolving.com/b/7/7/b779ec50af49132e538700f866d8c048d4997405.png)
have 
. Proof. Suppose w.l.o.g.
![$P\in\overarc[]{MB}$](http://latex.artofproblemsolving.com/3/0/3/30328124fc3d98b4ec5b37375e21dfc275f94424.png)
and let ![$R\in \overarc[]{MA}$](http://latex.artofproblemsolving.com/1/6/4/1643de826691d509d786eb1cf16223384ecb5687.png)
so that 
. Apply the Ptolemy's theorem to the isosceles trapezoid 
. Since 
obtain that .PP20. Aratati ca in
exista echivalenta 
Demonstratie 1.
are solutiile 
Asadar, 
Intr-adevar, 
si 
si 
Se arata usor ca 
Discriminantul in raport cu 
este 
si singura radacina pozitiva este
Asadar 
Demonstratie 2. Si acum o solutie foarte simpla pentru problema propusa la nivelul clasei a VII - a !
Se observa ca![$A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right]\ \Longrightarrow\ b^2+c^2\ \ge\ a^2+4S\sqrt 3\ .$](http://latex.artofproblemsolving.com/c/f/4/cf436b07b6eb2902089f0f5a4b7fa06b3f2abfe5.png)
Iata o problema similara 
Gasitio extindere pastrand expresia
Observatie. 
unde am notat 
Deci,
Extindere.
Demonstratie. Se observa ca
This post has been edited 349 times. Last edited by Virgil Nicula, May 30, 2016, 10:28 PM
August 2018
January 2018