408. Geometry 1.
by Virgil Nicula, Mar 4, 2015, 1:13 PM
PP1. Let an 
-isosceles 
with the point 
so that 
and the incircle 
of 
. Prove that 
.
Proof. Denote
, 
and the midpoint 
of 
. Observe that 
. Thus, 
. Otherwise. Apply the Pythagoras' theorem
to
![$(r+x)^2+r^2(m+2)^2=[r(m+1)+x]^2\iff$](//latex.artofproblemsolving.com/0/8/8/088afaf95257b00c840a0e04260643eefc489593.png)
a.s.o.
PP2. Let an equilateral with 
so that 
and 
. Denote 
and
with 
. Prove that 
, where 
is the area of the triangle 
.
Proof. Denote
. Apply the Stewart's relation to 
![$2\left[c^2+(a+b)^2+2c(a+b)\right]-2ab-c(a+b)=$](//latex.artofproblemsolving.com/7/4/8/748e956712e3b17d1f0f1384041de4186a42dcd5.png)
. Thus, 
. Apply the well-known identity
![$4[MAN]=\left(AM^2+AN^2-MN^2\right)\tan\widehat{MAN}$](//latex.artofproblemsolving.com/8/f/5/8f539a9e4ee299d065294e086401faff48d1f832.png)
, i.e. 
![$2\left[c^2-\left(a^2+b^2+ab\right)\right]=$](//latex.artofproblemsolving.com/0/b/e/0be15c0fd23ec565dfde4472214c3c1e1f3116db.png)
. In the particular case 
obtain 
.
PP3. Prove that in a there is the Ionescu - Weitzenbock's inequality 
, where is the area of .
Proof. I"ll use the well-known identity
. Denote 
. The identity 
becomes
. Hence 
.
PP4. Let an-right with the incircle 
and the excircles 
, 
. Denote 
and 
. Prove that 
.
Proof. Is well-known that
and 
. Denote the projections 
, 
of 
on the sidelines 
, 
respectively.
Apply the Menelaus' theorem to
.
, i.e. .
Remark 1.
.
Remark 2. with midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and -symmedian 
, where 
, where 
is the distance of to the line 
.
An easy extension. Let with the incircle and the excircles , . Denote 
. Prove that 
.
Remark. In the particular case
, i.e. 
obtain that 
![$\frac {c\left[c^2+c(b-a)\right]}{b\left[b^2+b(c-a)\right]}=\frac {c^2(b+c-a)}{b^2(b+c-a)}\implies$](//latex.artofproblemsolving.com/0/8/9/089a78ed81bc43894e9f657b71924eb6c53c41c0.png)
, i.e. is the -symmedian. Since obtain that is the -altitude, i.e. 
.
Proof. Is well-known that and . Apply the Menelaus' theorem to the transversal 
. Apply an well-known identity 
. Remark that ![$\frac {SB}{SC}=\frac {c\left[\left(a^2-b^2\right)+c(b-a)\right]}{b\left[\left(a^2-c^2\right)+b(c-a)\right]}=$](//latex.artofproblemsolving.com/5/1/6/51656467e1ff4e180e50cdd9824b8ed5a95ca1b8.png)
![$\frac {c\left[c^2-2bc\cos A+c(b-a)\right]}{b\left[b^2-2bc\cos A+b(c-a)\right]}\implies$](//latex.artofproblemsolving.com/e/6/7/e67d943a06bb9e70cdf06b17455fc4316e3c138c.png)
. In the particular case , i.e. obtain that 
is the -altitude, i.e. .
Lemma. Let
be a trapezoid so that 
and 
, 
so that 
. Then there is the relation 
.
Proof. Suppose w.l.o.g.
and let 
, 
so that 
and 
. Thus, 
, 
and 
.
Therefore,
.
PP5. Let and the points 
. Denote 
. Prove that 
.
Proof 1. Let
so that 
. Thus, 
.
Proof 2. Let
and 
so that 
. Thus, 
.
Proof 3. Apply the Menelaus' theorem to the transversals
.
Proof of the second relation. Apply an well-known relation to the cevians
.
Proof of the third relation. The third relation (an implication) is also simply. Denote the intersection
and apply the Menelaus' theorem to the transversals 
. The required relation becomes 
what is truly because 
are collinearly and this relation is the Menelaus' condtion for it.
Remark. If
, then 
and our relation becomes 
. From Aubel's relation 
,
where
, obtain that 
, what means that the division 
is harmonically.
PP6. Let a square
and the circle 
, where 
. Consider the points 
and 
so that the line
is tangent to in the point 
. Prove that 
is cyclically, where is the midpoint of ![$[EF]$](//latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
.
Proof 1 (metric). Denote and 
, where 
. Observe that 
. Prove easily that 
. Therefore, 
is a cyclical quadrilateral 
. Analogously obtain that
is a cyclical quadrilateral 
.
In conclusion, is cyclically and its circumcircle is the Euler's circle of the triangle 
.
Proof 2 (synthetic). Prove easily that
are cyclical deltoids 
(with common value 
(with common value 
where 
with common value 
Therefore,
a.s.o.
Remark. The length
of the circumradius for 
.
The minimum area of
is touched when 
and in this case 
, i.e. ![$\boxed{\ \sqrt 2-1\ \le\ \frac {[EAF]}{a^2}\ < \frac 12\ }$](//latex.artofproblemsolving.com/6/a/d/6ad9f025c4adf7162c0061bf1c40302aa87d19cc.png)
.
PP7. Construct
and 
outside of where 
. Let 
so that 
. Prove that 
.
Proof 1 (synthetic). Denote the midpoints
of , ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
respectively. Prove easily that 
. Thus, 
and 
, i.e. 
is
-isosceles right triangle. Denote the midpoint 
of ![$[MN]$](//latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
. So 
, i.e. ![$[DE]$](//latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
is the midline of the trapezoid 
.
Proof 2 (trigonometric). Denote
where 
and apply the theorem of Sines in
and 
what is truly.
PP8. Let a
-isosceles , the midpoint of and the projection 
of on 
. Prove that 
.
Proof. Denote:
, i.e. 
; the midpoint of , i.e. 
; the intersection 
, i.e. 
is the centroid of and
the incircle of
; the projection of on , i.e. 
. Prove easily that 
and 
, i.e. 
and
. Thus, and belong to the circle with the diameter , i.e. 
. Therefore, 
. Thus, 
and 
. Since 
obtain that 
.
PP9. Let
be a triangle with 
and the incircle which touches at , 
and 
. Prove that 
.
Proof 1.
.
Remark 1. Can prove easily the relation . Indeed, let 
so that 
and 
. Observe that 
is an equilateral triangle, i.e.
and 
.
Remark 2. Denote
. Is well-known that the division 
is harmonically and 
.
Proof 2 (Mihai Miculita). Denote
and 
so that 
. Prove easily that 
is the 
-excenter of
and is the - excenter of 
. In conclusion, the ray 
is the bisector of 
and the ray 
is the bisector of 
, i.e. 
. Very nice !
PP10. An acute with 
so that 
, 
which is tangent to at and 
.
Proof 1 (Sunken Rock). Let , 
be the midpoints of , ![$[FG]$](//latex.artofproblemsolving.com/0/f/9/0f9b357c5e6613e1cba5fb221ffcccd888d76b34.png)
respectively. Since 
, 
are cyclically obtain that 
.
In conclusion,
.
Proof 2. Denote
. Thus, 
. In conclusion, we"ll get the following implication chains
.
An easy extension. Let an acute and a circle which is tangent to at 
.
Denote
. Prove that 
.
Proof 1. Let , be the midpoints of the segments , respectively. Apply the Menelaus' theorem to the transversal 
and the triangle
. Since 
, 
are cyclically obtain that
.
Proof 2. Prove easily that
.
PP11. Let a cyclical quadrilateral where 
. Denote the midpoints 
of the sides ![$([AB],[BC],[CD],[DA])$](//latex.artofproblemsolving.com/8/6/6/866a1d9bd29a04196d9f1cb516181e689cccfb67.png)
respectively. Construct the circumcircles of the triangles
, 
and denote - their radical axis. Prove that 
(O.N.M. Serbia, 2015).
Nice proof (Mihai Miculita) ! Denote
. Therefore, 
.
Denote the circumcircle of 
the circumcircle 
of 
and the circumcircle 
of 
.
Therefore, is cyclic from the power 
of 
w.r.t. obtain that 
. Thus,
is cyclic 
.
is cyclic 
.
From
obtain that 
. Since 
, 
are cyclical quadrilaterals obtain 
and 
.
PP12. Let two exterior tangent circles 
with 
their an exterior common tangent 
where 
the circle
which is tangent in to and which is exterior tangent to the circles 
in 
respectively. Prove that 
and 
the quadrilaterals
, 
and 
are cyclically 
is the bisector of the angle 
, i.e. 
.
Proof.
Let 
and 
. Therefore
.
the quadrilateral is cyclic.
the quadrilateral is cyclic.
the quadrilateral is cyclic.
Let 
. Thus, 
and 
.
Denote 
and apply Menelaus' theorem to 
and 
.
Application. Construct at least a circle so that a given point belongs to and is tangent to given circles 
and 
.
Proof. Let
, 
so that is a common exterior tangent of the circles 
, 
. Denote and 
, 
. Apply the Menelaus' theorem to 
and the points 
. Suppose w.l.o.g. that 
. Denote
. Thus, 
and 
the quadrilateral 
is cyclically. Denote
. Hence from the powers of 
w.r.t. the circumcircles of and obtain that 
, i.e. belongs to the circumcircle of 
. In conclusion, 
- the required circle and is exterior tangent to a.s.o. 
PP13 (Miguel Ochoa Sanchez). Let with 
and the incircle 
which touches
at , and . Prove that 
Proof. Let
. Apply the theorem of Sines in and 
![$\cos\frac {x+15^{\circ}}2\left[\sin 15^{\circ}+\sin (2x-15^{\circ})\right]=\sin (2x-15^{\circ})\cos\frac {x-45^{\circ}}2\iff$](//latex.artofproblemsolving.com/7/b/5/7b5c9b37ce43cc999d04346d7d95afaa8c85d7e1.png)
![$\sin x\left[\cos\left(\frac {3x-15^{\circ}}2+\cos\frac {x-45^{\circ}}2\right)\right]=\sin (2x-15^{\circ})\cos\frac {x-45^{\circ} }2\iff$](//latex.artofproblemsolving.com/1/9/0/19046874d4c66b7ddf042e6a84405c00b7f618a2.png)
![$\sin x\cos\frac {3x-15^{\circ}}2=\cos\frac {x-45^{\circ}}2\left[\sin (2x-15^{\circ})-\sin x\right]\iff$](//latex.artofproblemsolving.com/a/1/1/a1177682cded42ffc9f8c22b6011c7b25bd400ef.png)
.
Extension. Is well-known that
. Thus, 
. For get 
and
.
PP14 (M.O. Sanchez). Let and 
so that 
and 
, where 
. Prove that 
(degrees) .
Proof 1. Observe that
is -isosceles 
and 
. Apply the theorem of Sines in 
.
PP15 (M.O. Sanchez). Let with incenter 
for which let 
Prove that 
Proof.
Apply the Menelaus' theorem 
and 
-isosceles 
with the point 
so that 
and the incircle 
of 
. Prove that 
.Proof. Denote
, 
and the midpoint 
of 
. Observe that 
. Thus, 
. Otherwise. Apply the Pythagoras' theoremto
![$(r+x)^2+r^2(m+2)^2=[r(m+1)+x]^2\iff$](http://latex.artofproblemsolving.com/0/8/8/088afaf95257b00c840a0e04260643eefc489593.png)
a.s.o.PP2. Let an equilateral
with 
so that 
and 
. Denote 
and
with 
. Prove that 
, where 
is the area of the triangle 
.Proof. Denote
. Apply the Stewart's relation to 
![$2\left[c^2+(a+b)^2+2c(a+b)\right]-2ab-c(a+b)=$](http://latex.artofproblemsolving.com/7/4/8/748e956712e3b17d1f0f1384041de4186a42dcd5.png)
. Thus, 
. Apply the well-known identity![$4[MAN]=\left(AM^2+AN^2-MN^2\right)\tan\widehat{MAN}$](http://latex.artofproblemsolving.com/8/f/5/8f539a9e4ee299d065294e086401faff48d1f832.png)
, i.e. 
![$2\left[c^2-\left(a^2+b^2+ab\right)\right]=$](http://latex.artofproblemsolving.com/0/b/e/0be15c0fd23ec565dfde4472214c3c1e1f3116db.png)
. In the particular case 
obtain 
.PP3. Prove that in a
there is the Ionescu - Weitzenbock's inequality 
, where is the area of .Proof. I"ll use the well-known identity
. Denote 
. The identity 
becomes
. Hence 
.PP4. Let an
-right with the incircle 
and the excircles 
, 
. Denote 
and 
. Prove that 
.Proof. Is well-known that
and 
. Denote the projections 
, 
of 
on the sidelines 
, 
respectively.Apply the Menelaus' theorem to
.
, i.e. .Remark 1.
.Remark 2.
with midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and -symmedian 
, where 
, where 
is the distance of to the line 
. An easy extension. Let
with the incircle and the excircles , . Denote 
. Prove that 
.Remark. In the particular case
, i.e. 
obtain that 
![$\frac {c\left[c^2+c(b-a)\right]}{b\left[b^2+b(c-a)\right]}=\frac {c^2(b+c-a)}{b^2(b+c-a)}\implies$](http://latex.artofproblemsolving.com/0/8/9/089a78ed81bc43894e9f657b71924eb6c53c41c0.png)
, i.e. is the -symmedian. Since obtain that is the -altitude, i.e. 
.Proof. Is well-known that
and . Apply the Menelaus' theorem to the transversal 
. Apply an well-known identity 
. Remark that ![$\frac {SB}{SC}=\frac {c\left[\left(a^2-b^2\right)+c(b-a)\right]}{b\left[\left(a^2-c^2\right)+b(c-a)\right]}=$](http://latex.artofproblemsolving.com/5/1/6/51656467e1ff4e180e50cdd9824b8ed5a95ca1b8.png)
![$\frac {c\left[c^2-2bc\cos A+c(b-a)\right]}{b\left[b^2-2bc\cos A+b(c-a)\right]}\implies$](http://latex.artofproblemsolving.com/e/6/7/e67d943a06bb9e70cdf06b17455fc4316e3c138c.png)
. In the particular case , i.e. obtain that 
is the -altitude, i.e. .Lemma. Let
be a trapezoid so that 
and 
, 
so that 
. Then there is the relation 
.Proof. Suppose w.l.o.g.
and let 
, 
so that 
and 
. Thus, 
, 
and 
.Therefore,
.PP5. Let
and the points 
. Denote 
. Prove that 
.Proof 1. Let
so that 
. Thus, 
.Proof 2. Let
and 
so that 
. Thus, 
.Proof 3. Apply the Menelaus' theorem to the transversals
.Proof of the second relation. Apply an well-known relation to the cevians
.Proof of the third relation. The third relation (an implication) is also simply. Denote the intersection
and apply the Menelaus' theorem to the transversals 
. The required relation becomes 
what is truly because 
are collinearly and this relation is the Menelaus' condtion for it.Remark. If
, then 
and our relation becomes 
. From Aubel's relation 
,where
, obtain that 
, what means that the division 
is harmonically.PP6. Let a square
and the circle 
, where 
. Consider the points 
and 
so that the line
is tangent to in the point 
. Prove that 
is cyclically, where is the midpoint of ![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
. Proof 1 (metric). Denote
and 
, where 
. Observe that 
. Prove easily that 
. Therefore, 
is a cyclical quadrilateral 
. Analogously obtain that
is a cyclical quadrilateral 
.In conclusion,
is cyclically and its circumcircle is the Euler's circle of the triangle 
.Proof 2 (synthetic). Prove easily that
are cyclical deltoids 
(with common value 
(with common value 
where 
with common value 
Therefore,
a.s.o.Remark. The length
of the circumradius for 
.The minimum area of
is touched when 
and in this case 
, i.e. ![$\boxed{\ \sqrt 2-1\ \le\ \frac {[EAF]}{a^2}\ < \frac 12\ }$](http://latex.artofproblemsolving.com/6/a/d/6ad9f025c4adf7162c0061bf1c40302aa87d19cc.png)
.PP7. Construct
and 
outside of where 
. Let 
so that 
. Prove that 
.Proof 1 (synthetic). Denote the midpoints
of , ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
respectively. Prove easily that 
. Thus, 
and 
, i.e. 
is
-isosceles right triangle. Denote the midpoint 
of ![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
. So 
, i.e. ![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
is the midline of the trapezoid 
.Proof 2 (trigonometric). Denote
where 
and apply the theorem of Sines in 
and 
what is truly.PP8. Let a
-isosceles , the midpoint of and the projection 
of on 
. Prove that 
. Proof. Denote:
, i.e. 
; the midpoint of , i.e. 
; the intersection 
, i.e. 
is the centroid of andthe incircle of
; the projection of on , i.e. 
. Prove easily that 
and 
, i.e. 
and
. Thus, and belong to the circle with the diameter , i.e. 
. Therefore, 
. Thus, 
and 
. Since 
obtain that 
.PP9. Let
be a triangle with 
and the incircle which touches at , 
and 
. Prove that 
.Proof 1.
.Remark 1. Can prove easily the relation
. Indeed, let 
so that 
and 
. Observe that 
is an equilateral triangle, i.e.
and 
.Remark 2. Denote
. Is well-known that the division 
is harmonically and 
.Proof 2 (Mihai Miculita). Denote
and 
so that 
. Prove easily that 
is the 
-excenter of and
is the - excenter of 
. In conclusion, the ray 
is the bisector of 
and the ray 
is the bisector of 
, i.e. 
. Very nice ! PP10. An acute
with 
so that 
, 
which is tangent to at and 
.Proof 1 (Sunken Rock). Let
, 
be the midpoints of , ![$[FG]$](http://latex.artofproblemsolving.com/0/f/9/0f9b357c5e6613e1cba5fb221ffcccd888d76b34.png)
respectively. Since 
, 
are cyclically obtain that 
.In conclusion,
.Proof 2. Denote
. Thus, 
. In conclusion, we"ll get the following implication chains 
.An easy extension. Let an acute
and a circle which is tangent to at 
.Denote
. Prove that 
.Proof 1. Let
, be the midpoints of the segments , respectively. Apply the Menelaus' theorem to the transversal 
and the triangle
. Since 
, 
are cyclically obtain that 
.Proof 2. Prove easily that
.PP11. Let a cyclical quadrilateral
where 
. Denote the midpoints 
of the sides ![$([AB],[BC],[CD],[DA])$](http://latex.artofproblemsolving.com/8/6/6/866a1d9bd29a04196d9f1cb516181e689cccfb67.png)
respectively. Construct the circumcircles of the triangles
, 
and denote - their radical axis. Prove that 
(O.N.M. Serbia, 2015).Nice proof (Mihai Miculita) ! Denote
. Therefore, 
.Denote the circumcircle
of 
the circumcircle 
of 
and the circumcircle 
of 
.Therefore,
is cyclic from the power 
of 
w.r.t. obtain that 
. Thus,
is cyclic 
.
is cyclic 
.From
obtain that 
. Since 
, 
are cyclical quadrilaterals obtain 
and 
.PP12. Let
two exterior tangent circles 
with 
their an exterior common tangent 
where 
the circle
which is tangent in to and which is exterior tangent to the circles 
in 
respectively. Prove that 
and 
the quadrilaterals
, 
and 
are cyclically 
is the bisector of the angle 
, i.e. 
.Proof.
Let 
and 
. Therefore 
.
the quadrilateral is cyclic.
the quadrilateral is cyclic.
the quadrilateral is cyclic.
Let 
. Thus, 
and 
.
Denote 
and apply Menelaus' theorem to 
and 
.Application. Construct at least a circle
so that a given point belongs to and is tangent to given circles 
and 
. Proof. Let
, 
so that is a common exterior tangent of the circles 
, 
. Denote and 
, 
. Apply the Menelaus' theorem to 
and the points 
. Suppose w.l.o.g. that 
. Denote 
. Thus, 
and 
the quadrilateral 
is cyclically. Denote
. Hence from the powers of 
w.r.t. the circumcircles of and obtain that 
, i.e. belongs to the circumcircle of 
. In conclusion, 
- the required circle and is exterior tangent to a.s.o. 
PP13 (Miguel Ochoa Sanchez). Let
with 
and the incircle 
which touches at
, and . Prove that 
Proof. Let
. Apply the theorem of Sines in and 
![$\cos\frac {x+15^{\circ}}2\left[\sin 15^{\circ}+\sin (2x-15^{\circ})\right]=\sin (2x-15^{\circ})\cos\frac {x-45^{\circ}}2\iff$](http://latex.artofproblemsolving.com/7/b/5/7b5c9b37ce43cc999d04346d7d95afaa8c85d7e1.png)
![$\sin x\left[\cos\left(\frac {3x-15^{\circ}}2+\cos\frac {x-45^{\circ}}2\right)\right]=\sin (2x-15^{\circ})\cos\frac {x-45^{\circ} }2\iff$](http://latex.artofproblemsolving.com/1/9/0/19046874d4c66b7ddf042e6a84405c00b7f618a2.png)
![$\sin x\cos\frac {3x-15^{\circ}}2=\cos\frac {x-45^{\circ}}2\left[\sin (2x-15^{\circ})-\sin x\right]\iff$](http://latex.artofproblemsolving.com/a/1/1/a1177682cded42ffc9f8c22b6011c7b25bd400ef.png)
.Extension. Is well-known that
. Thus, 
. For get 
and
.PP14 (M.O. Sanchez). Let
and 
so that 
and 
, where 
. Prove that 
(degrees) . Proof 1. Observe that
is -isosceles 
and 
. Apply the theorem of Sines in 
.PP15 (M.O. Sanchez). Let
with incenter 
for which let 
Prove that 
Proof.
Apply the Menelaus' theorem 
and 
This post has been edited 333 times. Last edited by Virgil Nicula, Jul 30, 2016, 2:30 PM
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