436. Two characterizations of A-right triangle ABC.
by Virgil Nicula, Dec 9, 2015, 8:16 AM
Lemma 1. Let a convex 
and 
so that 
. Denote 
and 
. Then 
.
Proof. Suppose w.l.o.g.
and let 
so that 
. Thus, 
and
.
Remark. Prove easily that![$\boxed{[APB]=\frac {m\cdot [ANB]+n\cdot [AMB]}{m+n}}\ (2)$](//latex.artofproblemsolving.com/d/9/5/d957130448e7e27481e1b76e63ac134a650a6bf1.png)
. Indeed, ![$\frac {[AMB]}a=\frac {[ANB]}b=\frac {[APB]}c=\frac 12\cdot\sin\phi$](//latex.artofproblemsolving.com/e/1/6/e164f8f2c758e460a22c7fd4bc8fdc85ebd2b8c2.png)
, where 
.
Lemma 2.
P1 . Let
with the incenter 
and 
. Prove that 
.
Proof. For the first equivalence see PP8 from here. Denote
I"ll use the identities (here is the trick) from the lemma 2. Thus,
Analogously prove that 
and 
P2 (Edson Curahua). Let with incenter 
and ![$\left\{\begin{array}{cccc}
m & = & [BIF]\\\\
n & = & [CIE]\\\\
p & = & [EIF]\end{array}\right\|$](//latex.artofproblemsolving.com/b/2/3/b2394ae94d55a2af9c3659afd0dd0e001d27f9b2.png)
. Prove that 
.
Proof 1 (Franzua Ynfanzon). Denote
and ![$[MIN]=q$](//latex.artofproblemsolving.com/5/6/0/560d3fa70025857e8c71334d292d8ff75dc7ec7e.png)
. Observe that ![$\left\{\begin{array}{ccc}
BIM\sim BIF & \implies & [BIM]=m\\\\
CIN\sim CIE & \implies & [CIN]=n\end{array}\right\|$](//latex.artofproblemsolving.com/8/7/8/878ef44b994f86596310bf09fddfd4b192f4d94c.png)
and
![$\frac pq=\frac {[EIF]}{[MIN]}=$](//latex.artofproblemsolving.com/0/d/f/0dfa736bbedc75753c6a098105ca10181a75ca87.png)
![$p[3-2(1+\cos A)]=$](//latex.artofproblemsolving.com/0/5/9/059801dff82bf2705398133dbe5bb98193ab9d1c.png)
. In conclusion, ![$[BIC]=m+n+q\ \stackrel{(1)}{\implies}$](//latex.artofproblemsolving.com/8/8/1/881320afe8a6d754a590a9a8843f5e24a17e5684.png)
![$\boxed{[BIC]=m+n+p(1-2\cos A)}\ (2)$](//latex.artofproblemsolving.com/3/9/8/39827c8df28fa64d42ac2922bd118509759912ca.png)
and ![$[EIF]\cdot [BIC]=[BIF]\cdot [CIE]\ \stackrel{(2)}{\iff}\ p[m+n+p(1-2\cos A)]=mn\iff$](//latex.artofproblemsolving.com/1/8/8/1888ee850c5ce683084fd8bd696c268634b8fcf1.png)
Proof 2. Van Aubel's theorem
. I"ll use identity from lemma 2 
,
i.e.
. Particular cases: 
.
Examples
P3 (S.S. Palacios Paulino). Let with the 
-excircle 
what touches the sidelines 
,
, 
at 
, 
, 
respectively. Denote 
so that 
. Prove that 
.
Proof 1. Apply in
an well-known or prove easily the identity 
(standard notations). Hence 
.
Proof 2. Denote
so that 
. Observe that 
, 
and 
. Apply lemma 1
to the trapezoid
. In conclusion, , where 
is the length of the -altitude in the triangle 
(Franzua Ynfanzon).
P4 (S.S. Palacios Paulino). Let with the circumcircle 
and the midpoints 
, 
of the sides ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
,
![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
respectively. Suppose that 
so that 
. Prove that 
is cyclically .
Proof 1.
.
Therefore,
Therefore, 
is cyclically .
Proof 2 (Stan Fulger). Let the symmetric
of w.r.t. 
is cyclic 
isosceles
and 
are similarly 
belongs
to the circumcircle of the triangle
, i.e. the quadrilateral is cyclically. 
P5. Let
and midpoint 
of ![$ [BC]$](//latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
. 
, 
so that 
and 
. Prove that 
or 
Lemma (trigonometric form of the Ptolemy's theorem). Let
be a triangle. Then for any 
exists the identity 
Particular cases :
Proof. Denote
. Thus, 
and 
. Apply the Sinus' theorem in 
and 
.
From relation
obtain that ![$ \frac {c}{\sin (B + x)}\cdot[a\sin x - c\sin (B + x)] + \frac {b}{\sin (C + y)}\cdot[a\sin y - b\sin (C + y)] = 0$](//latex.artofproblemsolving.com/d/4/4/d447ea425356674f98a09f4fe3841fdcffa52379.png)
. Using upper lemma obtain 
or
. Prove easily that 
and in this case ![$ [MN]$](//latex.artofproblemsolving.com/b/6/d/b6d9ce210c89713ee87840b48f477522ee16651e.png)
is the 
- middleline of , i.e. 
, 
.
P6. Let a convex
with 
and 
. Denote the projections , on 
of , 
respectively. Prove that 
.
Proof 1. Suppose w.l.o.g.
, 
, 
, 
, 
and 
Therefore,
. In conclusion, from the relations 
, 
and 
obtain that 
Proof 2. Suppose w.l.o.g. and denote 
Thus, 
. Therefore, 
. In conclusion, 
P7. Let and the points 
so that the ray 
is the common bisector of the angles 
and 
. Prove that 
.
Proof. Let
. Thus, 
.
P8. Let with 
and 
so that 
and 
. Prove that 
and 
is cyclically.
Proof 1. Construct the equilateral
so that 
separates 
, 
. Thus, is the circumcenter of 
and is the circumcenter of 
isosceles 
is cyclically 
(Stan Fulger).
Proof 2.
. Apply Stewart's theorem in 
, i.e. or 
, i.e. 
, what means 
, i.e. in 
we have 
a.s.o.
Proof 3.
and 
. Apply the theorem of Sines to 
. So 
. Prove easily that if , then 
and if , then 
. (İxtiyar İsmayilov).
Remark. If
, then prove easily that 
, where 
, i.e. 
and is cyclically .
P9. Let with the points 
and 
, where 
. Prove that 
.
Proof. Denote
. Apply the Menelaus' theorem to and the transversals 
. Apply the Desargues' theorem to the triangles and 
, i.e. .
P10 Let a square with 
and the circles 
, 
where 
. For 
where 
let 
be the circle
what is exterior tangent to
. Denote 
such that 
such that 
. Prove that 
and 
.
Proof. Denote
. Observe that 
and 
From
and obtain that 
Particular case.
, i.e. 
and 
(Cristian Tello).
P11 Let be a square with . Construct the circles 
, where ![$r\in\left(\frac 12,1\right]$](//latex.artofproblemsolving.com/7/5/f/75f66d788689f7d4aa6caab8d2a5c1607ffff010.png)
and the following circles
what is tangent to 
and exterior tangent to , 
what is tangent to 
, interior tangent to and exterior tangent to 
what is tangent to and interior tangent to 
Prove that ![$(\forall )\ r\in\left(\frac 12,1\right]$](//latex.artofproblemsolving.com/8/c/9/8c970ebcc590b8c296bb1cd27eafa50906cc7788.png)
there is the relation 
.
Proof.
I"ll eliminate the variable
between these three upper relations, i.e. 
P12. Let an-right with the incircle 
and the centroid 
Ascertain the semiperimeter 
where 
Proof 1.
and 
Observe that 
Denote the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
Thus, 
Apply the
Stewart's relation to the cevian
in 
where 
![$9\cdot IG^2+2m_a^2=3\cdot \left[IA^2+2\left(ID^2+DM^2\right)\right]\iff$](//latex.artofproblemsolving.com/e/b/3/eb3175da63be939229a79776eb0394722494765e.png)
![$9r^2+\frac 12=3\cdot \left[2r^2+2r^2+\frac {(b-c)^2}2\right]\iff$](//latex.artofproblemsolving.com/b/5/4/b541215c61123185145bb6bba86a064d23cacb63.png)
i.e. 
Remark. Prove similarly that generally
Since 
a.s.o., then in the particular case
for
obtain that 
Our concrete problem becomes 
If 
then 
and 
i.e. 
Proof 2. Suppose w.l.o.g.
Observe that 
and 
Apply the Pythagoras' theorem to the side ![$[IG]$](//latex.artofproblemsolving.com/9/1/7/917e276dc5e3493e398d9adbbbbfc6517777c0f0.png)
of
when 
i.e. 
P13. Let an-right with the incircle 
Prove that there is the relations 
(standard notations).
Proof.
Observe that
i.e. is the circumcenter of 
and 
Therefore,
Observe that
Remark. Prove easily that
there are the relations 
a.s.o. Observe that
P14 (Miguel Ochoa Sanchez). Let an-right with the -internal angle-bisector 
where 
and 
Ascertain the length of the cathetus ![$[AB]\ .$](//latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Proof (Barış Altay). Denote
Hence 
and 
I"ll use the remarkable identity 
Hence:
![$y-1=1+\frac 2y\implies y^2-2y-2=0\implies y=1+\sqrt 3\implies c^2=y-1=\left(1+\sqrt 3\right)-1=\sqrt 3\implies
c^2=\sqrt 3\implies\boxed{c=\sqrt[4]{3}}\ .$](//latex.artofproblemsolving.com/b/9/c/b9c2e6859eb682e38188c651791752cf131c6866.png)
Extension. Let with the internal angle-bisector where and Ascertain the length 
of the cathetus depending on the value of the angle 
Proof. Denote Hence and I"ll use the remarkable identity or prove easily that 
Hence:
where
Thus, 
Hence 
and 
(standard notations).
.
and 
so that 
. Denote 
and 
. Then 
.Proof. Suppose w.l.o.g.
and let 
so that 
. Thus, 
and
.Remark. Prove easily that
![$\boxed{[APB]=\frac {m\cdot [ANB]+n\cdot [AMB]}{m+n}}\ (2)$](http://latex.artofproblemsolving.com/d/9/5/d957130448e7e27481e1b76e63ac134a650a6bf1.png)
. Indeed, ![$\frac {[AMB]}a=\frac {[ANB]}b=\frac {[APB]}c=\frac 12\cdot\sin\phi$](http://latex.artofproblemsolving.com/e/1/6/e164f8f2c758e460a22c7fd4bc8fdc85ebd2b8c2.png)
, where 
.Lemma 2.
P1 . Let
with the incenter 
and 
. Prove that 
.Proof. For the first equivalence see PP8 from here. Denote
I"ll use the identities (here is the trick) from the lemma 2. Thus,
Analogously prove that 
and 
P2 (Edson Curahua). Let
with incenter 
and ![$\left\{\begin{array}{cccc}
m & = & [BIF]\\\\
n & = & [CIE]\\\\
p & = & [EIF]\end{array}\right\|$](http://latex.artofproblemsolving.com/b/2/3/b2394ae94d55a2af9c3659afd0dd0e001d27f9b2.png)
. Prove that 
.Proof 1 (Franzua Ynfanzon). Denote
and ![$[MIN]=q$](http://latex.artofproblemsolving.com/5/6/0/560d3fa70025857e8c71334d292d8ff75dc7ec7e.png)
. Observe that ![$\left\{\begin{array}{ccc}
BIM\sim BIF & \implies & [BIM]=m\\\\
CIN\sim CIE & \implies & [CIN]=n\end{array}\right\|$](http://latex.artofproblemsolving.com/8/7/8/878ef44b994f86596310bf09fddfd4b192f4d94c.png)
and
![$\frac pq=\frac {[EIF]}{[MIN]}=$](http://latex.artofproblemsolving.com/0/d/f/0dfa736bbedc75753c6a098105ca10181a75ca87.png)
![$p[3-2(1+\cos A)]=$](http://latex.artofproblemsolving.com/0/5/9/059801dff82bf2705398133dbe5bb98193ab9d1c.png)
. In conclusion, ![$[BIC]=m+n+q\ \stackrel{(1)}{\implies}$](http://latex.artofproblemsolving.com/8/8/1/881320afe8a6d754a590a9a8843f5e24a17e5684.png)
![$\boxed{[BIC]=m+n+p(1-2\cos A)}\ (2)$](http://latex.artofproblemsolving.com/3/9/8/39827c8df28fa64d42ac2922bd118509759912ca.png)
and ![$[EIF]\cdot [BIC]=[BIF]\cdot [CIE]\ \stackrel{(2)}{\iff}\ p[m+n+p(1-2\cos A)]=mn\iff$](http://latex.artofproblemsolving.com/1/8/8/1888ee850c5ce683084fd8bd696c268634b8fcf1.png)
Proof 2. Van Aubel's theorem
. I"ll use identity from lemma 2 
,i.e.
. Particular cases: 
.Examples
P3 (S.S. Palacios Paulino). Let
with the 
-excircle 
what touches the sidelines 
,
, 
at 
, 
, 
respectively. Denote 
so that 
. Prove that 
.Proof 1. Apply in
an well-known or prove easily the identity 
(standard notations). Hence 
.Proof 2. Denote
so that 
. Observe that 
, 
and 
. Apply lemma 1to the trapezoid
. In conclusion, , where 
is the length of the -altitude in the triangle 
(Franzua Ynfanzon).P4 (S.S. Palacios Paulino). Let
with the circumcircle 
and the midpoints 
, 
of the sides ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
,![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
respectively. Suppose that 
so that 
. Prove that 
is cyclically . Proof 1.
.Therefore,
Therefore, 
is cyclically .Proof 2 (Stan Fulger). Let the symmetric
of w.r.t. 
is cyclic 
isosceles
and 
are similarly 
belongsto the circumcircle of the triangle
, i.e. the quadrilateral is cyclically. 
P5. Let
and midpoint 
of ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
. 
, 
so that 
and 
. Prove that 
or 
Lemma (trigonometric form of the Ptolemy's theorem). Let
be a triangle. Then for any 
exists the identity 
Particular cases :
Proof. Denote
. Thus, 
and 
. Apply the Sinus' theorem in 
and 
.From relation
obtain that ![$ \frac {c}{\sin (B + x)}\cdot[a\sin x - c\sin (B + x)] + \frac {b}{\sin (C + y)}\cdot[a\sin y - b\sin (C + y)] = 0$](http://latex.artofproblemsolving.com/d/4/4/d447ea425356674f98a09f4fe3841fdcffa52379.png)
. Using upper lemma obtain 
or
. Prove easily that 
and in this case ![$ [MN]$](http://latex.artofproblemsolving.com/b/6/d/b6d9ce210c89713ee87840b48f477522ee16651e.png)
is the 
- middleline of , i.e. 
, 
. P6. Let a convex
with 
and 
. Denote the projections , on 
of , 
respectively. Prove that 
.Proof 1. Suppose w.l.o.g.
, 
, 
, 
, 
and 
Therefore,
. In conclusion, from the relations 
, 
and 
obtain that 
Proof 2. Suppose w.l.o.g.
and denote 
Thus, 
. Therefore, 
. In conclusion, 
P7. Let
and the points 
so that the ray 
is the common bisector of the angles 
and 
. Prove that 
.Proof. Let
. Thus, 
.P8. Let
with 
and 
so that 
and 
. Prove that 
and 
is cyclically.Proof 1. Construct the equilateral
so that 
separates 
, 
. Thus, is the circumcenter of 
and is the circumcenter of 
isosceles 
is cyclically 
(Stan Fulger).Proof 2.
. Apply Stewart's theorem in 
, i.e. or 
, i.e. 
, what means 
, i.e. in 
we have 
a.s.o.Proof 3.
and 
. Apply the theorem of Sines to 
. So 
. Prove easily that if , then 
and if , then 
. (İxtiyar İsmayilov).Remark. If
, then prove easily that 
, where 
, i.e. 
and is cyclically .P9. Let
with the points 
and 
, where 
. Prove that 
. Proof. Denote
. Apply the Menelaus' theorem to and the transversals 
. Apply the Desargues' theorem to the triangles and 
, i.e. .P10 Let a square
with 
and the circles 
, 
where 
. For 
where 
let 
be the circlewhat is exterior tangent to
. Denote 
such that 
such that 
. Prove that 
and 
. Proof. Denote
. Observe that 
and 
From and obtain that 
Particular case.
, i.e. 
and 
(Cristian Tello).P11 Let
be a square with . Construct the circles 
, where ![$r\in\left(\frac 12,1\right]$](http://latex.artofproblemsolving.com/7/5/f/75f66d788689f7d4aa6caab8d2a5c1607ffff010.png)
and the following circles 
what is tangent to 
and exterior tangent to , 
what is tangent to 
, interior tangent to and exterior tangent to 
what is tangent to and interior tangent to 
Prove that ![$(\forall )\ r\in\left(\frac 12,1\right]$](http://latex.artofproblemsolving.com/8/c/9/8c970ebcc590b8c296bb1cd27eafa50906cc7788.png)
there is the relation 
.Proof.
I"ll eliminate the variable
between these three upper relations, i.e. 
P12. Let an
-right with the incircle 
and the centroid 
Ascertain the semiperimeter 
where 
Proof 1.
and 
Observe that 
Denote the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
Thus, 
Apply theStewart's relation to the cevian
in 
where 
![$9\cdot IG^2+2m_a^2=3\cdot \left[IA^2+2\left(ID^2+DM^2\right)\right]\iff$](http://latex.artofproblemsolving.com/e/b/3/eb3175da63be939229a79776eb0394722494765e.png)
![$9r^2+\frac 12=3\cdot \left[2r^2+2r^2+\frac {(b-c)^2}2\right]\iff$](http://latex.artofproblemsolving.com/b/5/4/b541215c61123185145bb6bba86a064d23cacb63.png)
i.e. 
Remark. Prove similarly that generally
Since 
a.s.o., then in the particular casefor
obtain that 
Our concrete problem becomes 
If 
then 
and 
i.e. 
Proof 2. Suppose w.l.o.g.
Observe that 
and 
Apply the Pythagoras' theorem to the side ![$[IG]$](http://latex.artofproblemsolving.com/9/1/7/917e276dc5e3493e398d9adbbbbfc6517777c0f0.png)
of
when 
i.e. 
P13. Let an
-right with the incircle 
Prove that there is the relations 
(standard notations). Proof.
Observe that
i.e. is the circumcenter of 
and 
Therefore,
Observe that
Remark. Prove easily that
there are the relations 
a.s.o. Observe that
P14 (Miguel Ochoa Sanchez). Let an
-right with the -internal angle-bisector 
where 
and 
Ascertain the length of the cathetus ![$[AB]\ .$](http://latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Proof (Barış Altay). Denote
Hence 
and 
I"ll use the remarkable identity 
Hence:
![$y-1=1+\frac 2y\implies y^2-2y-2=0\implies y=1+\sqrt 3\implies c^2=y-1=\left(1+\sqrt 3\right)-1=\sqrt 3\implies
c^2=\sqrt 3\implies\boxed{c=\sqrt[4]{3}}\ .$](http://latex.artofproblemsolving.com/b/9/c/b9c2e6859eb682e38188c651791752cf131c6866.png)
Extension. Let
with the internal angle-bisector where and Ascertain the length 
of the cathetus depending on the value of the angle 
Proof. Denote
Hence and I"ll use the remarkable identity or prove easily that 
Hence:
where
Thus, 
Hence 
and 
(standard notations)..
This post has been edited 435 times. Last edited by Virgil Nicula, Jun 28, 2017, 6:33 PM
August 2018
January 2018