409. Geometry 2.
by Virgil Nicula, Mar 20, 2015, 1:40 PM
P1. Let 
with ![$A_k\in [BC]\ ,\ k\in \overline {0,n}$](//latex.artofproblemsolving.com/9/2/f/92f7b7733585ddad29308aa4cae81b66842837c6.png)
, where 
and 
so that 
for any 
.
For any
, where 
denote the length 
of its inradius and the length 
of its 
-exradius. Prove that 
.
Proof. In any there is the relation 
. Indeed, let the tangent points 
and 
of 
and -excircle 
with 
.
Then
and 
. Let 
,
where
and 
. Hence 
.
P2 (Miguel Ochoa Sanchez). For denote the lengths 
of the inradii for 
respectively. Prove that 
.
Proof.
PP3 (metric). Let with 
so that 
. Let 
and 
. Prove that 
.
Proof (own). Denote
and 
. Thus, 
. Observe that 
. Apply the Menelaus' theorem to the transversal 
. With 
apply the generalized Pythagoras' theorem in triangles 
.
Therefore,
, where
, i.e. 
. Can prove easily that 
verifies the equation .
Indeed,
. In conclusion, 
.
P4 (M.O.Sanchez). Let an-isosceles 
with the circumcircle 
. For 
denote its symmetrical
w.r.t. and 
. Prove that 
, i.e. the division 
is harmonically.
Proof.
and 
is cyclic 
is harmonic.
P5 (BMO, 2013). The-exincircle 
of is tangent to 
, 
in 
, 
respectively. The 
-exincircle 
is tangent to
, in 
, 
respectively. Let 
. Prove that 
is cyclic.
Proof. Denote
. Observe that 
, 
and 
. Apply the theorem of Sines in 
. Apply the theorem of Cosines in the triangles 
and 
. Observe that 
and 
. Thus, 
.
Denote
. Since 
and 
are cyclically obtain that 
is cyclically.
P6. Let-right-angled with incircle 
which touches , 
, at , 
, respectively. Let 
. Prove that 
.
Proof. Denote
, 
and 
. Thus, 
and 
. From the well-known property 
obtain that 
is
tangent to and 
. Since 
is an harmonical division and 
obtain that the ray 
is the bisector of the angle 
, i.e. 
. Observe that 
. In conclusion, 
.
P7 (Ruben Auqui). Let an-rightangled . Denote the points 
and 
. Prove that 
.
Proof 1 (metric). Denote
and 
. Thus, 
and 
. Apply the Menelaus' theorem to the transversals 
.
Proof 2 (trigonometric). Denote
, where and 
. Apply the theorem of Sines in the triangles
.
Proof 3 (synthetic).
Remark.
. Prove easily that the following interesting relation 
.
P8. Let with 
and 
so that 
. Denote the intersection between the -bisector of 
and the
-bisector of . Prove that 
and ascertain the maximum value of the angle 
.
Proof. Observe that
. Denote 
and 
. Thus, 
. In this case 
.
Two particular cases
.
P9. Let the semicircle
with the diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and the quadrant 
with the arc ![$\overarc[]{OC}$](//latex.artofproblemsolving.com/c/a/9/ca9a55dbb5bedba66550bdbfb6b8021e3c226251.png)
,
where
so that doesn't separate and . Prove that ![$\left[B\overarc[]{OE}B\right]=2\left[A\overarc[]{CE}\overarc[]{EA}\right]$](//latex.artofproblemsolving.com/6/6/c/66c88be71b869ae3627d2fcd162c04f987dc0a2f.png)
.
Proof. Denote![$\left\{\begin{array}{ccc}
\left[A\overarc[]{CE}\overarc[]{EA}\right] & = & S_1\\\\
\left[B\overarc[]{OE}B\right] & = & S_2\end{array}\right\|\ ,\ y=[AOE]$](//latex.artofproblemsolving.com/3/3/1/331fb68374af36daba974723e0ed6f1a1592848c.png)
and ![$\left[\overarc[]{AE}A\right]=x=\left[\overarc[]{OE}O\right]$](//latex.artofproblemsolving.com/d/a/6/da6c7f1fabcde6635116f474f65822799c084bf2.png)
![$\left\{\begin{array}{ccccc}
[OEA]=[OEB] & \implies & y=x+S_2\\\\
\left[A\overarc[]{EO}A\right]=2\left[A\overarc[]{CE}A\right] & \implies & y+x=2\left(S_1+x\right)\end{array}\right\|$](//latex.artofproblemsolving.com/2/9/7/297992671a9e0191bee575eff6265b7bac8a5b76.png)
.
P10 (Miguel Ochoa Sanchez). Let
where denote: the exterior -bisector of 
cut at 
the exterior -bisector of cut
at
the projections , and of a mobile point 
on the sidelines , and respectively. Prove that 
.
Proof 1. Denote: the projections
, 
of on , respectively; the projections 
, 
of on , respectively 
.
Observe that
. Prove easily that in the trapezoid 
, where 
,
and 
there is the well-known relation 
(J. D. Valdivia Fuentes' proof).
Proof 2 (own). Denote the area
of and 
, Observe that ![$[BPC]=[ABC]+[PAB]+[PAC]\iff$](//latex.artofproblemsolving.com/5/b/f/5bfa6230979322c8982f79175c0fe099599b321c.png)
. From 
![$\frac {[EAF]}{[ABC]}=\frac {AE}{AC}\cdot\frac{AF}{AB}\implies$](//latex.artofproblemsolving.com/9/9/c/99cc56f8a833e3150b29dec6f1decdbc938bc6ac.png)
![$\boxed{[EAF]=\frac {bcS}{(a-b)(a-c)}}\ (1)$](//latex.artofproblemsolving.com/3/9/0/390c66a2bae4b1ebd4c111657e3f629cd82538b1.png)
.
Since![$[EAF]=[PAE]+[PAF]=y\cdot AE+z\cdot AF$](//latex.artofproblemsolving.com/9/7/0/9700f2e90c697545b71dd4770d96c1047ef0f99d.png)
obtain that 
.
From the sum of the relations
and 
obtain that 
.
Lemma 1. Let
be a trapezoid where 
and 
, 
so that 
. Denote 
. Then 
.
Lemma 2. Let a quadrilateral and a point . Then there is the relation ![$MA\cdot [BCD]+MB\cdot [ACD]=AB\cdot [MCD]$](//latex.artofproblemsolving.com/c/6/c/c6c7e44c8076778cc9ab429a24045ddf82a9b291.png)
, where ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of 
.
P11 (Sunken Rock). Let where denote: the interior -bisector of cut at the interior -bisector of cut at
the projections , and of a mobile point on the sidelines , and respectively. Prove that .
Proof (own). Denote the area of and , Observe that ![$[BPC]=[ABC]-[PAB]-[PAC]\iff$](//latex.artofproblemsolving.com/1/4/9/149dd1333f548a1b04d0b3541c73bd2cb7d884e5.png)
. From 
![$\boxed{[EAF]=\frac {bcS}{(a+b)(a+c)}}\ (1)$](//latex.artofproblemsolving.com/9/e/d/9ed46b00fc5c9e7c100fa7d8a7136d58fafc652b.png)
.
Since obtain that 
.
From the sum of the relations and obtain that 
.
An easy extension. Let and 
. For a point 
denote
. Prove that ![$[PBC]=\frac 1n\cdot [PAC]+\frac 1m\cdot [PAB]$](//latex.artofproblemsolving.com/e/d/2/ed28c6bbc8d98f3c47b44d626e1d86c2ffe1bdfd.png)
, i.e. 
.
P12. The figure is an equilateral triangle surrounded by
congruent rectangles of width 
all inscribed in a circle of radius . Find the length of the side of the equilateral triangle.
Proof. Let
and 
, where 
equilateral which is inscribed in with 
so that 
is a rectangle with 
the midpoint of ![$[DE]\ .$](//latex.artofproblemsolving.com/1/d/a/1da2706ed712f1a6100a01df004875c2de6a8679.png)
Prove easily that 
. Apply Pythagora's theorem in the -right 
. For 
obtain that 
i.e. 
.
P13. Let a semicircle
with diameter and the circles 
what are tangent to and interior tangent to the semicircle 
so
that
i.e. 
and the circle 
is exterior tangent to the circles 
and 
Prove that there is the relation 
Proof.Denote the tangent points
of the circles 
with the line respectively. Apply the Pythagoras' theorem to the triangles
and to the triangle 
![$\left[(y-z)R-2yz\right]^2=4yz\left(R^2-2yR\right)$](//latex.artofproblemsolving.com/7/0/c/70ce95bb6867683ea2144c1cd63cd5751bb0f6e1.png)
Observe that 
![$\left(R^2-2xR\right)\left(R^2-2yR\right)=\left[R(x+y)+2xy-R^2\right]^2\iff$](//latex.artofproblemsolving.com/8/9/b/89b2c089f91fef3f4ae8eb705b50c3d20d88ed94.png)
Let the matrix 
The elimination of 
between the relations
and 
becomes the following relation with the determinants 
P14 (GMB, Batinetu-Giurgiu si Neculai Stanciu). Let with the area 
Prove that 
Proof 1. Denote the projection of on and the midpoint of the side ![$[AB]\ .$](//latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Thus, 
Apply the Pythagoras' theorem in -right-angled 
![$\boxed{16S^2+\left|a^2-b^2\right|^2=c^2\left[2\left(a^2+b^2\right)-c^2\right]}\ (2)\ .$](//latex.artofproblemsolving.com/f/c/8/fc8cfc8601585d37ebd6d0bf2837a0b1aed80cd7.png)
Thus, 
![$ab(a+b)^2=c^2\left[2\left(a^2+b^2\right)-c^2\right]-$](//latex.artofproblemsolving.com/e/b/d/ebdfcdcd640d7a5e9efa017d97d7aba4284484d1.png)
![$\left[\left(a^2+b^2\right)-c^2\right]^2+ab\left[(a+b)^2-4ab\right]=0\iff$](//latex.artofproblemsolving.com/f/5/2/f52451443718c26e1d68ee4e176df771aad59278.png)
i.e. 
and 
Proof 2. I"ll use
Thus, 
From the previous relation 
and the inequality
obtain that 
and from the relation obtain finally that 
, i.e. 
Proof 3. I"ll use the identity
Therefore, 
![$\left[\left(a^2+b^2\right)-c^2\right]^2-$](//latex.artofproblemsolving.com/d/f/8/df81a0e295155b68240c34eed79d520e5551420e.png)
and 
i.e. 
and 
Proof 4.
![$4a^2b^2\left(\cancel 1-\sin^2C\right)+ab\left[(a+b)^2-\cancel{4ab}\right]=$](//latex.artofproblemsolving.com/0/7/b/07b48df1fab5cdc012fef11533151e2586b347ed.png)
P15 (Miguel Ochoa Sanchez). Let
be a triangle with the incircle what touches at 
and
For the midpoint point of 
denote 
and 
Prove that 
Proof 1. Let
where 
Prove easily that 
is the bisector of 
i.e. 
Apply the Menelaus' theorem to the transversals
Therefore,
i.e. 
and 
with ![$A_k\in [BC]\ ,\ k\in \overline {0,n}$](http://latex.artofproblemsolving.com/9/2/f/92f7b7733585ddad29308aa4cae81b66842837c6.png)
, where 
and 
so that 
for any 
.For any
, where 
denote the length 
of its inradius and the length 
of its 
-exradius. Prove that 
.Proof. In any
there is the relation 
. Indeed, let the tangent points 
and 
of 
and -excircle 
with 
.Then
and 
. Let 
,where
and 
. Hence 
.P2 (Miguel Ochoa Sanchez). For
denote the lengths 
of the inradii for 
respectively. Prove that 
. Proof.
PP3 (metric). Let
with 
so that 
. Let 
and 
. Prove that 
.Proof (own). Denote
and 
. Thus, 
. Observe that 
. Apply the Menelaus' theorem to the transversal 
. With 
apply the generalized Pythagoras' theorem in triangles 
.Therefore,
, where
, i.e. 
. Can prove easily that 
verifies the equation .Indeed,
. In conclusion, 
.P4 (M.O.Sanchez). Let an
-isosceles 
with the circumcircle 
. For 
denote its symmetrical
w.r.t. and 
. Prove that 
, i.e. the division 
is harmonically.Proof.
and 
is cyclic 
is harmonic.P5 (BMO, 2013). The
-exincircle 
of is tangent to 
, 
in 
, 
respectively. The 
-exincircle 
is tangent to
, in 
, 
respectively. Let 
. Prove that 
is cyclic.Proof. Denote
. Observe that 
, 
and 
. Apply the theorem of Sines in 
. Apply the theorem of Cosines in the triangles 
and 
. Observe that 
and 
. Thus, 
.Denote
. Since 
and 
are cyclically obtain that 
is cyclically.P6. Let
-right-angled with incircle 
which touches , 
, at , 
, respectively. Let 
. Prove that 
.Proof. Denote
, 
and 
. Thus, 
and 
. From the well-known property 
obtain that 
istangent to
and 
. Since 
is an harmonical division and 
obtain that the ray 
is the bisector of the angle 
, i.e. 
. Observe that 
. In conclusion, 
.P7 (Ruben Auqui). Let an
-rightangled . Denote the points 
and 
. Prove that 
. Proof 1 (metric). Denote
and 
. Thus, 
and 
. Apply the Menelaus' theorem to the transversals 
.Proof 2 (trigonometric). Denote
, where and 
. Apply the theorem of Sines in the triangles 
.Proof 3 (synthetic).
Remark.
. Prove easily that the following interesting relation 
. P8. Let
with 
and 
so that 
. Denote the intersection between the -bisector of 
and the
-bisector of . Prove that 
and ascertain the maximum value of the angle 
.Proof. Observe that
. Denote 
and 
. Thus, 
. In this case 
.Two particular cases
.P9. Let the semicircle
with the diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and the quadrant 
with the arc ![$\overarc[]{OC}$](http://latex.artofproblemsolving.com/c/a/9/ca9a55dbb5bedba66550bdbfb6b8021e3c226251.png)
,where
so that doesn't separate and . Prove that ![$\left[B\overarc[]{OE}B\right]=2\left[A\overarc[]{CE}\overarc[]{EA}\right]$](http://latex.artofproblemsolving.com/6/6/c/66c88be71b869ae3627d2fcd162c04f987dc0a2f.png)
.Proof. Denote
![$\left\{\begin{array}{ccc}
\left[A\overarc[]{CE}\overarc[]{EA}\right] & = & S_1\\\\
\left[B\overarc[]{OE}B\right] & = & S_2\end{array}\right\|\ ,\ y=[AOE]$](http://latex.artofproblemsolving.com/3/3/1/331fb68374af36daba974723e0ed6f1a1592848c.png)
and ![$\left[\overarc[]{AE}A\right]=x=\left[\overarc[]{OE}O\right]$](http://latex.artofproblemsolving.com/d/a/6/da6c7f1fabcde6635116f474f65822799c084bf2.png)
![$\left\{\begin{array}{ccccc}
[OEA]=[OEB] & \implies & y=x+S_2\\\\
\left[A\overarc[]{EO}A\right]=2\left[A\overarc[]{CE}A\right] & \implies & y+x=2\left(S_1+x\right)\end{array}\right\|$](http://latex.artofproblemsolving.com/2/9/7/297992671a9e0191bee575eff6265b7bac8a5b76.png)
.P10 (Miguel Ochoa Sanchez). Let
where denote: the exterior -bisector of 
cut at 
the exterior -bisector of cut at
the projections , and of a mobile point 
on the sidelines , and respectively. Prove that 
.Proof 1. Denote: the projections
, 
of on , respectively; the projections 
, 
of on , respectively 
.Observe that
. Prove easily that in the trapezoid 
, where 
,
and 
there is the well-known relation 
(J. D. Valdivia Fuentes' proof).Proof 2 (own). Denote the area
of and 
, Observe that ![$[BPC]=[ABC]+[PAB]+[PAC]\iff$](http://latex.artofproblemsolving.com/5/b/f/5bfa6230979322c8982f79175c0fe099599b321c.png)
. From 
![$\frac {[EAF]}{[ABC]}=\frac {AE}{AC}\cdot\frac{AF}{AB}\implies$](http://latex.artofproblemsolving.com/9/9/c/99cc56f8a833e3150b29dec6f1decdbc938bc6ac.png)
![$\boxed{[EAF]=\frac {bcS}{(a-b)(a-c)}}\ (1)$](http://latex.artofproblemsolving.com/3/9/0/390c66a2bae4b1ebd4c111657e3f629cd82538b1.png)
.Since
![$[EAF]=[PAE]+[PAF]=y\cdot AE+z\cdot AF$](http://latex.artofproblemsolving.com/9/7/0/9700f2e90c697545b71dd4770d96c1047ef0f99d.png)
obtain that 
.From the sum of the relations
and 
obtain that 
. Lemma 1. Let
be a trapezoid where 
and 
, 
so that 
. Denote 
. Then 
.Lemma 2. Let a quadrilateral
and a point . Then there is the relation ![$MA\cdot [BCD]+MB\cdot [ACD]=AB\cdot [MCD]$](http://latex.artofproblemsolving.com/c/6/c/c6c7e44c8076778cc9ab429a24045ddf82a9b291.png)
, where ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of 
. P11 (Sunken Rock). Let
where denote: the interior -bisector of cut at the interior -bisector of cut at the projections
, and of a mobile point on the sidelines , and respectively. Prove that .Proof (own). Denote the area
of and , Observe that ![$[BPC]=[ABC]-[PAB]-[PAC]\iff$](http://latex.artofproblemsolving.com/1/4/9/149dd1333f548a1b04d0b3541c73bd2cb7d884e5.png)
. From 
![$\boxed{[EAF]=\frac {bcS}{(a+b)(a+c)}}\ (1)$](http://latex.artofproblemsolving.com/9/e/d/9ed46b00fc5c9e7c100fa7d8a7136d58fafc652b.png)
.Since
obtain that 
.From the sum of the relations
and obtain that 
. An easy extension. Let
and 
. For a point 
denote 
. Prove that ![$[PBC]=\frac 1n\cdot [PAC]+\frac 1m\cdot [PAB]$](http://latex.artofproblemsolving.com/e/d/2/ed28c6bbc8d98f3c47b44d626e1d86c2ffe1bdfd.png)
, i.e. 
.P12. The figure is an equilateral triangle surrounded by
congruent rectangles of width 
all inscribed in a circle of radius . Find the length of the side of the equilateral triangle.Proof. Let
and 
, where 
equilateral which is inscribed in with 
so that 
is a rectangle with 
the midpoint
of ![$[DE]\ .$](http://latex.artofproblemsolving.com/1/d/a/1da2706ed712f1a6100a01df004875c2de6a8679.png)
Prove easily that 
. Apply Pythagora's theorem in the -right 
. For 
obtain that 
i.e. 
.P13. Let a semicircle
with diameter and the circles 
what are tangent to and interior tangent to the semicircle 
sothat
i.e. 
and the circle 
is exterior tangent to the circles 
and 
Prove that there is the relation 
Proof.Denote the tangent points
of the circles 
with the line respectively. Apply the Pythagoras' theorem to the triangles 
and to the triangle 
![$\left[(y-z)R-2yz\right]^2=4yz\left(R^2-2yR\right)$](http://latex.artofproblemsolving.com/7/0/c/70ce95bb6867683ea2144c1cd63cd5751bb0f6e1.png)
Observe that 
![$\left(R^2-2xR\right)\left(R^2-2yR\right)=\left[R(x+y)+2xy-R^2\right]^2\iff$](http://latex.artofproblemsolving.com/8/9/b/89b2c089f91fef3f4ae8eb705b50c3d20d88ed94.png)
Let the matrix 
The elimination of 
between the relations
and 
becomes the following relation with the determinants 
P14 (GMB, Batinetu-Giurgiu si Neculai Stanciu). Let
with the area 
Prove that 
Proof 1. Denote the projection
of on and the midpoint of the side ![$[AB]\ .$](http://latex.artofproblemsolving.com/f/4/3/f43654eb147de5ab7da28700a60e256a697a2e10.png)
Thus, 
Apply the Pythagoras' theorem in -right-angled 
![$\boxed{16S^2+\left|a^2-b^2\right|^2=c^2\left[2\left(a^2+b^2\right)-c^2\right]}\ (2)\ .$](http://latex.artofproblemsolving.com/f/c/8/fc8cfc8601585d37ebd6d0bf2837a0b1aed80cd7.png)
Thus, 
![$ab(a+b)^2=c^2\left[2\left(a^2+b^2\right)-c^2\right]-$](http://latex.artofproblemsolving.com/e/b/d/ebdfcdcd640d7a5e9efa017d97d7aba4284484d1.png)
![$\left[\left(a^2+b^2\right)-c^2\right]^2+ab\left[(a+b)^2-4ab\right]=0\iff$](http://latex.artofproblemsolving.com/f/5/2/f52451443718c26e1d68ee4e176df771aad59278.png)
i.e. 
and 
Proof 2. I"ll use
Thus, 
From the previous relation 
and the inequality
obtain that 
and from the relation obtain finally that 
, i.e. 
Proof 3. I"ll use the identity
Therefore, 
![$\left[\left(a^2+b^2\right)-c^2\right]^2-$](http://latex.artofproblemsolving.com/d/f/8/df81a0e295155b68240c34eed79d520e5551420e.png)
and 
i.e. 
and 
Proof 4.
![$4a^2b^2\left(\cancel 1-\sin^2C\right)+ab\left[(a+b)^2-\cancel{4ab}\right]=$](http://latex.artofproblemsolving.com/0/7/b/07b48df1fab5cdc012fef11533151e2586b347ed.png)
P15 (Miguel Ochoa Sanchez). Let
be a triangle with the incircle what touches at 
and
For the midpoint point of 
denote 
and 
Prove that 
Proof 1. Let
where 
Prove easily that 
is the bisector of 
i.e. 
Apply the Menelaus' theorem to the transversals 
Therefore,
i.e. 
and 
This post has been edited 247 times. Last edited by Virgil Nicula, Jul 15, 2017, 5:33 PM
August 2018
January 2018