419. Some geometry problems for hight school I
by Virgil Nicula, Mar 28, 2015, 4:11 PM
PP1. In the 
-right 
denote 
and 
for which denote 
. Prove that 
.
Proof 1. Observe that
is -isosceles triangle, I,e. 
. Let 
so that 
. Thus, 
is a rhombus, i.e. 
and 
.
In conclusion,
.
Proof 2.
. Observe that
is -isosceles, i.e. . Apply the theorem of Cosines to 
.
PP2. Let. Consider 
so that 
. Prove that 
Remark. Some interesting particular cases
and 
. Vezi aici / aici / aici cateva aplicatii.
Proof. Let the circle
with the diameter ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
. Observe that 
and 
. Let 
and 
. Thus,
and 
. From the power of the point 
w,r,t, the circle obtain that 
. Since 
obtain that 
. Therefore, 
![$\frac {b\left(a^2+c^2-b^2\right)}{2a^2\left[b^2\left(a^2+c^2-b^2\right)+2a^2c^2\right]}\implies$](//latex.artofproblemsolving.com/7/d/f/7df008eca0ddd6db403d129e89e0f00ac46b1852.png)
Extension. Let and 
. Prove that 
Proof. Let the circumcircle of 
and 
. Apply theorem of Sines in 
and
. From
the power of the point w,r,t, the circle obtain that 
. Since
obtain that 
. Therefore, 
![$\frac {b\left(a^2+c^2-b^2-\Phi\right)}{2a^2\left[2a^2c^2+b^2\left(a^2+c^2-b^2-\Phi\right)\right]}\implies$](//latex.artofproblemsolving.com/a/0/8/a080b8cfbb1ab1cf6121869182d085a970675adf.png)
.
PP3. Let and 
for which 
is the bisector of 
. For 
denote 
and 
. Prove that
. Remark. If the point 
is the incenter of the triangle 
then obtain an well-known classical problem ==> here
PP4. Let
be a triangle and let 
be a point in its interior. Lines 
, 
, 
intersect sides 
, 
, 
at 
, 
, 
respectively. Prove that![$[PAF]+[PBD]+[PCE]=\frac S2\iff P$](//latex.artofproblemsolving.com/0/f/f/0ffb1377009ffadfc3ab36b85f51f5ea3e150c5b.png)
lies on at least one of the medians of triangle .
Proof. Denote the area![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
of the triangle 
. I will suppose w.l.o.g. that ![$[ABC]=2$](//latex.artofproblemsolving.com/c/1/4/c14299dcf775aef4150e578061733fd76e2d8772.png)
and I note:
- the midpoints of the sides ![$[BC],[CA],[AB]$](//latex.artofproblemsolving.com/9/0/7/907e51f409d3d7ab73812cf20b2d5bfd529245bb.png)
respectively;
![$\bullet\ x_1=[PAF]\ ,\ x_2=[PAE]\ ;\ y_1=[PBD]\ ,\ y_2=$](//latex.artofproblemsolving.com/a/6/9/a690bbdaa1f3aa23e67cf4f08b204ea6e0d28342.png)
![$[PBF]\ ;\ z_1=[PCE]\ ,\ z_2=[PCD]\ ;$](//latex.artofproblemsolving.com/d/5/5/d55e8ffec581e0daf4e3f13c200b268f0d92887a.png)
Thus, 
From the Ceva's theorem results the relation
, i.e. 
![$P\in [AM]\cup [BN]\cup [CP]\ .$](//latex.artofproblemsolving.com/6/a/5/6a58a82d97ecc99388e0e3205217148e8e33bf43.png)
PP5. Let with the incentre 
and the midpoint 
of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. The implication 
is evidently. Prove that its reverse implication isn't truly.
Proof 1. Suppose that
isn't obtuse 
. So 
, where is midpoint of ![$[BC]\ ,\ OM=R\cos A\iff$](//latex.artofproblemsolving.com/f/f/8/ff8c9e324da337c8c247034fc711d32d63870762.png)
.
Proof 2. 
![$2\left[\frac {ac(s-b)}{s}+\frac {ab(s-c)}{s}\right]-a^2=$](//latex.artofproblemsolving.com/5/b/5/5b503c5567172914bea11f46c67dad2dd8257272.png)
![$2[a(b+c)-8Rr]-a^2\iff$](//latex.artofproblemsolving.com/2/5/4/2543c463e46722b9631b81da2ea0f3e38f72d278.png)
. In conclusion, 
. Examples 
.
PP6. Let with the centroid 
, the circumcircle 
, the symmedian point 
(Lemoine's centre) and 
. Denote the midpoints 
of
the sides![$[BC]\ ,\ [CA]\ ,\ [AB]$](//latex.artofproblemsolving.com/e/9/9/e994825632d7880f34a3e58a3d95a6e2d9f9830f.png)
. Prove that 
is cyclically.
Proof.
Denote 
. Are well-known 
i.e.
. Thus, 
![$b^2\left[2\left(a^2+c^2\right)-b^2\right]=$](//latex.artofproblemsolving.com/e/b/d/ebdb1c02f204bed8cd2e73ee9b5f796b57aa95df.png)
![$c^2\left[2\left(b^2+a^2\right)-c^2\right]\iff$](//latex.artofproblemsolving.com/b/b/d/bbd7d5d6a8c94261cb2c25650b3fd83e586d948e.png)
.
is harmonically, i.e. 
and 
. Therefore, 
.
is cyclically 
.
![$b^2\left[2\left(a^2+c^2\right)-b^2\right]=c^2\left[2\left(b^2+a^2\right)-c^2\right]\iff$](//latex.artofproblemsolving.com/0/a/c/0ac6a76cf41fd005f05999e5996de110a235402c.png)
Remark.
(for 
) 
(for 
) 
. Thus 
![$\frac {a^2b^2c^2\left[\left(b^2+c^2\right)^2-4b^2c^2\right]}{\left(b^4-c^4\right)^2}=\frac {b^2c^2\left(b^2+c^2-a^2\right)\left[2\left(b^2+c^2\right)-a^2\right]}{\left(a^2+b^2+c^2\right)\left(b^2+c^2\right)^2}\iff$](//latex.artofproblemsolving.com/b/b/b/bbb5ff78a5c6fbab1d55029413a47c9b8cd0637e.png)
![$\frac {a^2\left[\left(b^2+c^2\right)^2-4b^2c^2\right]}{\left(b^2-c^2\right)^2}=$](//latex.artofproblemsolving.com/e/b/1/eb13e514a1530a829d3d0eea94937fe2227a1455.png)
![$\frac {\left(b^2+c^2-a^2\right)\left[2\left(b^2+c^2\right)-a^2\right]}{a^2+b^2+c^2}\iff$](//latex.artofproblemsolving.com/5/8/6/586445543985ad0a9f748d95ff8b0d5a4ad45101.png)
![$a^2\left(a^2+b^2+c^2\right)=\left(b^2+c^2-a^2\right)\left[2\left(b^2+c^2\right)-a^2\right]\iff$](//latex.artofproblemsolving.com/c/2/0/c20a96d147903b65c785674567e987a873f968c9.png)
.
PP7. Let
with the circumcircle 
the midpoint of ![$[BC]\ ;\ \left\{\begin{array}{ccc}
\{A,N\} & = & AM\cap w\\\\\
P & \in & BN\cap AC\\\\
Q & \in & CN\cap AB\end{array}\right\|$](//latex.artofproblemsolving.com/8/9/b/89b3264efd88a92da7345d0c34d35a45c6e9cb2b.png)
. Prove that is the centroid of 
.
Proof. From the power of w.r.t. obtain that 
.
Therefore, is the centroid of 
![$3\left(b^2+c^2\right)=2\left[2\left(b^2+c^2\right)-a^2\right]\iff$](//latex.artofproblemsolving.com/f/7/4/f74cb45f81dc83c427c28c1ea8ea63c2e37aec33.png)
.
PP8. Let with the circumcircle 
and the incircle 
. Denote 
. Prove that ![$\left\{\begin{array}{cccc}
(1)\ : & IY\cdot IZ & = & R\cdot IA\\\\
(2)\ : & IB\cdot IC & = & 2r\cdot IX\\\\
(3)\ : & 2r\cdot [XYZ] & = & R\cdot [ABC]\end{array}\right\|$](//latex.artofproblemsolving.com/7/e/8/7e80744ecdc39096b5fe000133e11a22ac68e485.png)
.
Proof. Prove easily that
and 
a.s.o. Thus, 
. Therefore 
![$\frac {[XYZ]}{[ABC]}=\frac {XY\cdot YZ\cdot ZX}{AB\cdot BC\cdot CA}=$](//latex.artofproblemsolving.com/8/e/9/8e9078b7cb29e78f1321a2139d8d050969cebaa0.png)
![$\boxed{2r\cdot [XYZ]=R\cdot [ABC]}$](//latex.artofproblemsolving.com/6/f/d/6fd040fb2fae9ed4ac8e8c39c0902409dd79efee.png)
.
Remark. Can use the trigonometrical identities
and 
.
PP9. Let with the circumcircle and the - bisector 
, where 
. Denote the bisector
of ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
and 
. Prove that the sideline is tangent to the circumcircle of 
.
Proof. Denote
, where 
is the tangent line to at 
. Observe that 
is 
-isosceles, i.e. 
. In conclusion, 
the sideline is tangent to the circumcircle of .
PP11 (Ruben Dario Augui). Let the circle
and two points 
so that 
, where 
- the tangent to the circle in the point 
.
For the point which belongs to the small arc 
denote the intersections 
. Prove that is the midpoint of .
Proof.
and 
are cyclically with diameters 
. Apply well-known relation to 
in 
Apply another well-known relation to
and the points 
.
PP12 (Ruben Dario). Let with the incircle 
and 
and the midpoint of ![$[EF]$](//latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
. Prove that 
.
Proof 1 (metric). I"ll use the well-known identities
and the equivalence 
. Therefore,
, i.e. 
. Observe that 
and apply the theorem of median in the triangles 
, i.e. 
. Therefore, 
. In conclusion, .
Proof 2 (metric shorter).
. Therefore, 
![$\left[a-b+\frac {c^2}{a+b}\right]-\left[a-c+\frac {b^2}{a+c}\right]=2(c-b)\iff$](//latex.artofproblemsolving.com/9/5/1/9513b2dba8156be1101b76324fc61970a64d0c8d.png)
. In conclusion, .
Proof 3 (metric shortest). Suppose w.l.o.g. and let 
Apply the Menelaus' theorem to 
In conclusion, 
An easy extension. Let and for an interior point denote 
. Prove that 
Proof. Denote
and apply the Menelaus' theorem to 
In conclusion, 
Particular case. If is the midpoint of ![$[EF]\ ,$](//latex.artofproblemsolving.com/0/b/c/0bc54c04b2e9be880bb9e24cab1ca0a2858859d1.png)
i.e. 
, then 
Therefore
Prove easily that 
(Nagel's point)
PP13 (Cristian Tello). Let be the Fermat's point and be the centroid of . Prove that 
, where 
.
Proof (metric). Construct outside of the equilateral triangles 
, 
and 
Are well-known or prove easily that 
, 
and 
. Apply the theorem of Cosines in 
. From the relations 
and 
obtain that 
. Apply the well-known Leibniz's identity
PP14 (Ruben Dario). Let-right with 
so that 
. For a mobile point 
define
. Prove that the difference 
is constant.
Proof. Denote
and observe that 
. Apply an well-known relation 
. Now I"ll use the Ceva's theorem 
(constant).
PP15 (Edson Curahua Ortega). Let a square 
two points 
so that 
, 
and 
the projection 
of on 
. Prove that 
.
Proof 1. Suppose w.l.o.g.
and denote 
and 
. Observe that 
and 
. Therefore,
.
Proof 2 (Ruben Dario).
. Thus, 
.
Proof 3.
. From the relations and obtain that 
.
PP16. In denote 
. Prove that 
is the -bisector of .
Proof. Denote
so that 
. Thus, 
. Observe that 
, i.e. 
, From
obtain that 
and 
, i.e. 
. Therefore, 
.
In conclusion,
, i.e. 
and 
. Nice !
PP17 (Edson Curahua). Let the circle
and 
so that 
. For 
construct the rhombus 
Denote
. Prove that 
, i.e. 
Proof 1. Prove easily that
. Apply the theorem of Sines in 
![$\frac {2r\sin (45-x)}{\sin [45-(x+y)]}=\frac {r}{\sin y}\iff$](//latex.artofproblemsolving.com/9/7/e/97e5e468ddf6233f9b7a24cabfb79a8b53d2cc61.png)
![$2\sin y\sin(45-x)=\sin [45-(x+y)]\iff$](//latex.artofproblemsolving.com/8/2/2/822a4e2d0ad01cc7f43d89d2fa7ca6b2c06f86b0.png)
Proof 2. Suppose w.l.o.g. that
. Prove easily that 
. Apply the theorem of Sines in 
![$\frac {\sqrt 2}{\sin [45+(x-y)]}=\frac 1{\sin (x+y)}\iff$](//latex.artofproblemsolving.com/a/f/f/aff2df95f6e75a33454798b4608b9149d15ec8ee.png)
Remark. In the particular case
have 
and 
PP18 (Carlos Olivera). Let an-right with -excircle 
and 
-excircle 
. Denote 
Prove that 
.
Proof 1. Is well-known that
. Apply Menelaus' theorem to transversal 
.
Remark. Is well-known or prove easily the remarkable identity
and the particular case 
Proof 2. Let
so that 
, i.e. 
. Prove easily that 
and 
. Thus, 
. Hence, 
is isosceles 
what is truly.
Proof 3. Let
. Apply theorem of Sines to 
, Prove easily that 
, i.e. 
, what is truly. Thus, 
, i.e. 
.
PP19 (Ruben Dario).-isosceles and 
so that 
Proof 1 (Nicolás Vilches). Let
, where 
and 
. Construct the point 
so that 
the line separates 
, .
Prove easily that
, where 
Apply the theorem of Sines in 
In conclusion, The relation 
is true 
I"ll can use the well
known identity
. Indeed, 
![$\sin v[\sin v+2\sin u\cos (u+v)]\iff$](//latex.artofproblemsolving.com/9/0/7/907310444069b49365ddb2f88f155151be8f1f79.png)
![$\sin (2u+v)\sin v=\sin v[\sin v+2\sin u\cos (u+v)]\iff$](//latex.artofproblemsolving.com/7/1/7/717e2c68ebabf6cfbbc354de2a30a024b8f6c1e7.png)
, what is true.
Proof 2. Let the midpoint of and suppose w.l.o.g. 
. Let 
, where and . Thus, 
and 
. Observe that 
, 
and
. Therefore, the relation is equivalently with the relation 
![$2mn=(p-m)(p-n)\cdot\left[1+\frac {1-p^2}{1+p^2}\right]\iff$](//latex.artofproblemsolving.com/9/5/f/95fcb66ef6347cb949231f0f2eb3f296b4255843.png)
the true relation .
Proof 3. Let
, where and 
. Apply the theorem of Sines to 
Apply the theorem of Sines to 
The relation is true 
![$\sin^22z=\sin^22x+\sin^22y+\cos 2z\left[\cos 2(x-y)-\cos 2z\right]\iff$](//latex.artofproblemsolving.com/5/4/7/547ce8109f33db7dab00fedc54fce1ae3ffd3d32.png)
what is true.
PP20. Let a rectangle
with 
and a circle with the diameter ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
. For a mobile point so that the doesn't separate , let 
and the intersections , of with 
, 
respectively. Prove that 
. For 
obtain the Fermat's problem.
Proof. See here.
-right 
denote 
and 
for which denote 
. Prove that 
.Proof 1. Observe that
is -isosceles triangle, I,e. 
. Let 
so that 
. Thus, 
is a rhombus, i.e. 
and 
.In conclusion,
.Proof 2.
. Observe that 
is -isosceles, i.e. . Apply the theorem of Cosines to 
.PP2. Let
. Consider 
so that 
. Prove that 
Remark. Some interesting particular cases
and 
. Vezi aici / aici / aici cateva aplicatii. Proof. Let the circle
with the diameter ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
. Observe that 
and 
. Let 
and 
. Thus,
and 
. From the power of the point 
w,r,t, the circle obtain that 
. Since 
obtain that 
. Therefore, 
![$\frac {b\left(a^2+c^2-b^2\right)}{2a^2\left[b^2\left(a^2+c^2-b^2\right)+2a^2c^2\right]}\implies$](http://latex.artofproblemsolving.com/7/d/f/7df008eca0ddd6db403d129e89e0f00ac46b1852.png)
Extension. Let
and 
. Prove that 
Proof. Let the circumcircle
of 
and 
. Apply theorem of Sines in 
and 
. Fromthe power of the point
w,r,t, the circle obtain that 
. Since obtain that
. Therefore, 
![$\frac {b\left(a^2+c^2-b^2-\Phi\right)}{2a^2\left[2a^2c^2+b^2\left(a^2+c^2-b^2-\Phi\right)\right]}\implies$](http://latex.artofproblemsolving.com/a/0/8/a080b8cfbb1ab1cf6121869182d085a970675adf.png)
.PP3. Let
and 
for which 
is the bisector of 
. For 
denote 
and 
. Prove that
. Remark. If the point 
is the incenter of the triangle 
then obtain an well-known classical problem ==> herePP4. Let
be a triangle and let 
be a point in its interior. Lines 
, 
, 
intersect sides 
, 
, 
at 
, 
, 
respectively. Prove that
![$[PAF]+[PBD]+[PCE]=\frac S2\iff P$](http://latex.artofproblemsolving.com/0/f/f/0ffb1377009ffadfc3ab36b85f51f5ea3e150c5b.png)
lies on at least one of the medians of triangle .Proof. Denote the area
![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
of the triangle 
. I will suppose w.l.o.g. that ![$[ABC]=2$](http://latex.artofproblemsolving.com/c/1/4/c14299dcf775aef4150e578061733fd76e2d8772.png)
and I note:
- the midpoints of the sides ![$[BC],[CA],[AB]$](http://latex.artofproblemsolving.com/9/0/7/907e51f409d3d7ab73812cf20b2d5bfd529245bb.png)
respectively;![$\bullet\ x_1=[PAF]\ ,\ x_2=[PAE]\ ;\ y_1=[PBD]\ ,\ y_2=$](http://latex.artofproblemsolving.com/a/6/9/a690bbdaa1f3aa23e67cf4f08b204ea6e0d28342.png)
![$[PBF]\ ;\ z_1=[PCE]\ ,\ z_2=[PCD]\ ;$](http://latex.artofproblemsolving.com/d/5/5/d55e8ffec581e0daf4e3f13c200b268f0d92887a.png)
Thus, 
From the Ceva's theorem results the relation
, i.e. 
![$P\in [AM]\cup [BN]\cup [CP]\ .$](http://latex.artofproblemsolving.com/6/a/5/6a58a82d97ecc99388e0e3205217148e8e33bf43.png)
PP5. Let
with the incentre 
and the midpoint 
of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. The implication 
is evidently. Prove that its reverse implication isn't truly.Proof 1. Suppose that
isn't obtuse 
. So 
, where is midpoint of ![$[BC]\ ,\ OM=R\cos A\iff$](http://latex.artofproblemsolving.com/f/f/8/ff8c9e324da337c8c247034fc711d32d63870762.png)
.Proof 2.
![$2\left[\frac {ac(s-b)}{s}+\frac {ab(s-c)}{s}\right]-a^2=$](http://latex.artofproblemsolving.com/5/b/5/5b503c5567172914bea11f46c67dad2dd8257272.png)
![$2[a(b+c)-8Rr]-a^2\iff$](http://latex.artofproblemsolving.com/2/5/4/2543c463e46722b9631b81da2ea0f3e38f72d278.png)
. In conclusion, 
. Examples 
.PP6. Let
with the centroid 
, the circumcircle 
, the symmedian point 
(Lemoine's centre) and 
. Denote the midpoints 
ofthe sides
![$[BC]\ ,\ [CA]\ ,\ [AB]$](http://latex.artofproblemsolving.com/e/9/9/e994825632d7880f34a3e58a3d95a6e2d9f9830f.png)
. Prove that 
is cyclically.Proof.
Denote 
. Are well-known 
i.e.
. Thus, 
![$b^2\left[2\left(a^2+c^2\right)-b^2\right]=$](http://latex.artofproblemsolving.com/e/b/d/ebdb1c02f204bed8cd2e73ee9b5f796b57aa95df.png)
![$c^2\left[2\left(b^2+a^2\right)-c^2\right]\iff$](http://latex.artofproblemsolving.com/b/b/d/bbd7d5d6a8c94261cb2c25650b3fd83e586d948e.png)
.
is harmonically, i.e. 
and 
. Therefore, 
.
is cyclically 
.
![$b^2\left[2\left(a^2+c^2\right)-b^2\right]=c^2\left[2\left(b^2+a^2\right)-c^2\right]\iff$](http://latex.artofproblemsolving.com/0/a/c/0ac6a76cf41fd005f05999e5996de110a235402c.png)
Remark.
(for 
) 
(for 
) 
. Thus 
![$\frac {a^2b^2c^2\left[\left(b^2+c^2\right)^2-4b^2c^2\right]}{\left(b^4-c^4\right)^2}=\frac {b^2c^2\left(b^2+c^2-a^2\right)\left[2\left(b^2+c^2\right)-a^2\right]}{\left(a^2+b^2+c^2\right)\left(b^2+c^2\right)^2}\iff$](http://latex.artofproblemsolving.com/b/b/b/bbb5ff78a5c6fbab1d55029413a47c9b8cd0637e.png)
![$\frac {a^2\left[\left(b^2+c^2\right)^2-4b^2c^2\right]}{\left(b^2-c^2\right)^2}=$](http://latex.artofproblemsolving.com/e/b/1/eb13e514a1530a829d3d0eea94937fe2227a1455.png)
![$\frac {\left(b^2+c^2-a^2\right)\left[2\left(b^2+c^2\right)-a^2\right]}{a^2+b^2+c^2}\iff$](http://latex.artofproblemsolving.com/5/8/6/586445543985ad0a9f748d95ff8b0d5a4ad45101.png)
![$a^2\left(a^2+b^2+c^2\right)=\left(b^2+c^2-a^2\right)\left[2\left(b^2+c^2\right)-a^2\right]\iff$](http://latex.artofproblemsolving.com/c/2/0/c20a96d147903b65c785674567e987a873f968c9.png)
.PP7. Let
with the circumcircle 
the midpoint of ![$[BC]\ ;\ \left\{\begin{array}{ccc}
\{A,N\} & = & AM\cap w\\\\\
P & \in & BN\cap AC\\\\
Q & \in & CN\cap AB\end{array}\right\|$](http://latex.artofproblemsolving.com/8/9/b/89b3264efd88a92da7345d0c34d35a45c6e9cb2b.png)
. Prove that is the centroid of 
.Proof. From the power of
w.r.t. obtain that 
.Therefore,
is the centroid of 
![$3\left(b^2+c^2\right)=2\left[2\left(b^2+c^2\right)-a^2\right]\iff$](http://latex.artofproblemsolving.com/f/7/4/f74cb45f81dc83c427c28c1ea8ea63c2e37aec33.png)
.PP8. Let
with the circumcircle 
and the incircle 
. Denote 
. Prove that ![$\left\{\begin{array}{cccc}
(1)\ : & IY\cdot IZ & = & R\cdot IA\\\\
(2)\ : & IB\cdot IC & = & 2r\cdot IX\\\\
(3)\ : & 2r\cdot [XYZ] & = & R\cdot [ABC]\end{array}\right\|$](http://latex.artofproblemsolving.com/7/e/8/7e80744ecdc39096b5fe000133e11a22ac68e485.png)
.Proof. Prove easily that
and 
a.s.o. Thus, 
. Therefore 
![$\frac {[XYZ]}{[ABC]}=\frac {XY\cdot YZ\cdot ZX}{AB\cdot BC\cdot CA}=$](http://latex.artofproblemsolving.com/8/e/9/8e9078b7cb29e78f1321a2139d8d050969cebaa0.png)
![$\boxed{2r\cdot [XYZ]=R\cdot [ABC]}$](http://latex.artofproblemsolving.com/6/f/d/6fd040fb2fae9ed4ac8e8c39c0902409dd79efee.png)
.Remark. Can use the trigonometrical identities
and 
.PP9. Let
with the circumcircle and the - bisector 
, where 
. Denote the bisector
of ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
and 
. Prove that the sideline is tangent to the circumcircle of 
.Proof. Denote
, where 
is the tangent line to at 
. Observe that 
is 
-isosceles, i.e. 
. In conclusion, 
the sideline is tangent to the circumcircle of .PP11 (Ruben Dario Augui). Let the circle
and two points 
so that 
, where 
- the tangent to the circle in the point 
.For the point
which belongs to the small arc 
denote the intersections 
. Prove that is the midpoint of . Proof.
and 
are cyclically with diameters 
. Apply well-known relation to 
in 
Apply another well-known relation to
and the points 
.PP12 (Ruben Dario). Let
with the incircle 
and 
and the midpoint of ![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
. Prove that 
. Proof 1 (metric). I"ll use the well-known identities
and the equivalence 
. Therefore,
, i.e. 
. Observe that 
and apply the theorem of median in the triangles 
, i.e. 
. Therefore, 
. In conclusion, .Proof 2 (metric shorter).
. Therefore, 
![$\left[a-b+\frac {c^2}{a+b}\right]-\left[a-c+\frac {b^2}{a+c}\right]=2(c-b)\iff$](http://latex.artofproblemsolving.com/9/5/1/9513b2dba8156be1101b76324fc61970a64d0c8d.png)
. In conclusion, .Proof 3 (metric shortest). Suppose w.l.o.g.
and let 
Apply the Menelaus' theorem to 
In conclusion, 
An easy extension. Let
and for an interior point denote 
. Prove that 
Proof. Denote
and apply the Menelaus' theorem to 
In conclusion,
Particular case. If
is the midpoint of ![$[EF]\ ,$](http://latex.artofproblemsolving.com/0/b/c/0bc54c04b2e9be880bb9e24cab1ca0a2858859d1.png)
i.e. 
, then 
Therefore 
Prove easily that 
(Nagel's point)
PP13 (Cristian Tello). Let
be the Fermat's point and be the centroid of . Prove that 
, where 
.Proof (metric). Construct outside of
the equilateral triangles 
, 
and 
Are well-known or prove easily that 
, 
and 
. Apply the theorem of Cosines in 
. From the relations 
and 
obtain that 
. Apply the well-known Leibniz's identity 
PP14 (Ruben Dario). Let
-right with 
so that 
. For a mobile point 
define
. Prove that the difference 
is constant. Proof. Denote
and observe that 
. Apply an well-known relation 
. Now I"ll use the Ceva's theorem 
(constant).PP15 (Edson Curahua Ortega). Let
a square 
two points 
so that 
, 
and 
the projection 
of on 
. Prove that 
. Proof 1. Suppose w.l.o.g.
and denote 
and 
. Observe that 
and 
. Therefore,
.Proof 2 (Ruben Dario).
. Thus, 
.Proof 3.
. From the relations and obtain that 
.PP16. In
denote 
. Prove that 
is the -bisector of .Proof. Denote
so that 
. Thus, 
. Observe that 
, i.e. 
, From obtain that 
and 
, i.e. 
. Therefore, 
.In conclusion,
, i.e. 
and 
. Nice ! PP17 (Edson Curahua). Let the circle
and 
so that 
. For 
construct the rhombus 
Denote
. Prove that 
, i.e. 
Proof 1. Prove easily that
. Apply the theorem of Sines in 
![$\frac {2r\sin (45-x)}{\sin [45-(x+y)]}=\frac {r}{\sin y}\iff$](http://latex.artofproblemsolving.com/9/7/e/97e5e468ddf6233f9b7a24cabfb79a8b53d2cc61.png)
![$2\sin y\sin(45-x)=\sin [45-(x+y)]\iff$](http://latex.artofproblemsolving.com/8/2/2/822a4e2d0ad01cc7f43d89d2fa7ca6b2c06f86b0.png)
Proof 2. Suppose w.l.o.g. that
. Prove easily that 
. Apply the theorem of Sines in 
![$\frac {\sqrt 2}{\sin [45+(x-y)]}=\frac 1{\sin (x+y)}\iff$](http://latex.artofproblemsolving.com/a/f/f/aff2df95f6e75a33454798b4608b9149d15ec8ee.png)
Remark. In the particular case
have 
and 
PP18 (Carlos Olivera). Let an
-right with -excircle 
and 
-excircle 
. Denote 
Prove that 
.Proof 1. Is well-known that
. Apply Menelaus' theorem to transversal 
.Remark. Is well-known or prove easily the remarkable identity
and the particular case 
Proof 2. Let
so that 
, i.e. 
. Prove easily that 
and 
. Thus, 
. Hence, 
is isosceles 
what is truly.Proof 3. Let
. Apply theorem of Sines to 
, Prove easily that 
, i.e. 
, what is truly. Thus, 
, i.e. 
.PP19 (Ruben Dario).
-isosceles and 
so that 
Proof 1 (Nicolás Vilches). Let
, where 
and 
. Construct the point 
so that 
the line separates 
, .Prove easily that
, where 
Apply the theorem of Sines in 
In conclusion, The relation 
is true 
I"ll can use the wellknown identity
. Indeed, 
![$\sin v[\sin v+2\sin u\cos (u+v)]\iff$](http://latex.artofproblemsolving.com/9/0/7/907310444069b49365ddb2f88f155151be8f1f79.png)
![$\sin (2u+v)\sin v=\sin v[\sin v+2\sin u\cos (u+v)]\iff$](http://latex.artofproblemsolving.com/7/1/7/717e2c68ebabf6cfbbc354de2a30a024b8f6c1e7.png)
, what is true.Proof 2. Let the midpoint
of and suppose w.l.o.g. 
. Let 
, where and . Thus, 
and 
. Observe that 
, 
and
. Therefore, the relation is equivalently with the relation 
![$2mn=(p-m)(p-n)\cdot\left[1+\frac {1-p^2}{1+p^2}\right]\iff$](http://latex.artofproblemsolving.com/9/5/f/95fcb66ef6347cb949231f0f2eb3f296b4255843.png)
the true relation .Proof 3. Let
, where and 
. Apply the theorem of Sines to 
Apply the theorem of Sines to 
The relation
is true 
![$\sin^22z=\sin^22x+\sin^22y+\cos 2z\left[\cos 2(x-y)-\cos 2z\right]\iff$](http://latex.artofproblemsolving.com/5/4/7/547ce8109f33db7dab00fedc54fce1ae3ffd3d32.png)
what is true. PP20. Let a rectangle
with 
and a circle with the diameter ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
. For a mobile point so that the doesn't separate , let 
and the intersections
, of with 
, 
respectively. Prove that 
. For 
obtain the Fermat's problem.Proof. See here.
This post has been edited 350 times. Last edited by Virgil Nicula, Oct 26, 2015, 12:16 PM
August 2018
January 2018