370. Some metrical relations in a triangle.
by Virgil Nicula, Feb 14, 2013, 11:19 AM
PP1 (Hanoi Open Mathematical Olympiad 2010). Let 
with the incenter 
. Denote 
. Prove that 
.
Proof 1.
.
Proof 2. Suppose
and let 
. Apply Menelaus' theorem to 
and 
. Prove
that
, i.e. 
is tangent to the circumcircle of 
. In conclusion,
Proof 3. Let
. From 
, where 
and 
obtain 
. Thus, 
.
Proof 4. Apply theorem of Sinus in
.
Remark. Prove that
. Thus, 
![$\frac {(s-a)\left[s(s-a)+(s-b)(s-c)\right]}{s}\ \stackrel{(1)}{\implies}$](//latex.artofproblemsolving.com/b/c/8/bc83a23d781aa456ef82684664dff48cada07465.png)
. Otherwise, 
.
PP2 (Hanoi OMO 2009). Let acute with area 
. Let 
be located as follows: 
is the point where altitude from 
on 
meets the outwards
facing semicirle drawn on![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
as diameter. 
and 
are located similarly. Evaluate the sum ![$T=[BCA_1]^2+[CAB_1]^2+[ABC_1]^2$](//latex.artofproblemsolving.com/c/b/f/cbf63b73acd4959c04538c0d447e051516a3fd56.png)
, where denoted ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
- the area of .
Proof 1.![$4\cdot \sum\sin A\cos B\cos C=2\cdot \sum\sin A\left[\cos (B+C)+\cos (B-C)\right]=$](//latex.artofproblemsolving.com/f/4/6/f46411773081f30cabc9df6ebfff1f5c5964f90c.png)
![$2\cdot \sum\sin A\left[-\cos A+\cos (B-C)\right]=$](//latex.artofproblemsolving.com/1/2/0/120d35382ab946fbceb8202916b3acb82ecb0e77.png)
![$\sum \left[-\sin 2A+2\sin (B+C)\cos (B-C)\right]=$](//latex.artofproblemsolving.com/e/3/0/e30f439cadb9cc95cbebff8352fb5dbd2355a834.png)
. Let 
. So 
![$[BCA_1]^2=$](//latex.artofproblemsolving.com/1/a/b/1ab556e4ed63071e057c81a2f447117536c314be.png)
![$\sum [BCA_1]^2=\frac {abc}{4}\cdot\sum a\cdot\cos B\cos C=$](//latex.artofproblemsolving.com/7/c/d/7cdb7c672f76e1fbbd15a386d968b96c3117321c.png)
![$\boxed{\sum [BCA_1]^2=S^2}$](//latex.artofproblemsolving.com/d/2/4/d24dec9a546c3eff71271a876c9d2761693e1634.png)
.
Proof 2. Let
and orth. 
of 
![$\frac{[A_1BC]^2}{[ABC]^2}=\frac{A_1D^2}{AD^2}=$](//latex.artofproblemsolving.com/9/b/d/9bd0441ff974dc2fa0486dd0cc943737857f6600.png)
![$\frac{DH}{AD}=\frac{[BCH]}{[ABC]}$](//latex.artofproblemsolving.com/b/e/e/beefba73d485e53cf7b216593ad726d754b510a3.png)
. Thus, ![$\sum [BCH]=S\implies$](//latex.artofproblemsolving.com/8/f/f/8ffc9d48a907f14e7c8e6ef7b3f934714600eae6.png)
![$\sum [BCA_1]^2=S^2$](//latex.artofproblemsolving.com/c/3/c/c3ce38628039ecfcf1101bd123d5204c9b615fc9.png)
Remark.
.
PP3. Let and suppose that its incircle 
touch , 
, 
at 
and 
. Prove that 
.
Proof. From a well-known property (or prove easily)
, 
. Thus, 
, 
, 
are cyclicaly. Thus, 
is the radical axis of the circumcircles
for , 
, i.e. 
and 
is the bisector of 
. Thus, 
, i.e. . Remark that 
.
PP4. The inscribed circle of
touches , 
, in 
, in . ![$[GF]$](//latex.artofproblemsolving.com/a/4/b/a4b6b1ae1f5f71c9f547186e5bf566eaef35c6c0.png)
is a diameter of 
and 
. Prove that 
.
Proof 1.
.
Proof 2. Is well-known that
, where 
. Indeed, apply Pascal's theorem to 
. , 
belong to the circle with
the diameter![$[PR]$](//latex.artofproblemsolving.com/a/d/2/ad2ecef541359529c80aaf94c8c550c9e3bb1e7c.png)
because 
, 
and 
is its center because 
. So, , , 
, 
belong to the circle 
, where .
Remark. After having shown
collinear, 
is the orthocenter of 
. Hence 
and 
. As 
is isosceles, the conclusion follows.
PP5. Let
be a triangle with the incircle and the -exincircle 
. Let 
and 
. Prove that 
is the bisector of 
.
Proof 1. Suppose w.l.o.g.
. Let 
, the diameter ![$[DL]$](//latex.artofproblemsolving.com/b/4/a/b4ad5e4c7fd7c6a45982d8341cc6b0fdf4fa3488.png)
of 
and 
. Thus, 
the division 
is harmonically. Since 
obtain that the ray is the bisector of the angle .
Proof 2. Suppose w.l.o.g. . Let , the diameter of and the midpoints 
, 
of , ![$[ID]$](//latex.artofproblemsolving.com/3/9/c/39c0124bdd40abf098bae12cd10f3b575581439f.png)
respectively. Prove easily that 
,
. Thus, 
is cyclically. Since 
obtain that 
the ray 
, i.e. the ray is the bisector of .
An easy extension. Let
be a circle what is tangent to 
in the point 
. Consider 
in this order so that 
.
Denote the point for which 
and 
. Define the intersection 
. Prove that the ray 
is the bisector of the angle 
.
Proof. Denote the diameter![$[AL]$](//latex.artofproblemsolving.com/7/3/7/737a9b49658cbe239832952bb575c8bbdf7dbb35.png)
of and the midpoint , of the segments ![$[AE]$](//latex.artofproblemsolving.com/0/e/e/0ee81ab9421e97824f7785dc2c44c0f5b04a03b1.png)
, respectively. Thus,
is an cyclical quadrilateral. Since 
is the bisector of the segment obtain that 
is the bisector of the angle .
Lemma (Freeman). Let acute with circumcircle 
. Denote the midpoints , of ![$[OA]$](//latex.artofproblemsolving.com/c/8/8/c88c6759afc11d97267e95d4023b4419386ea20f.png)
and respectively. Then 
.
Proof. Let
PP6. Let an acute with circumcircle . Let midpoints , of and and 
. Suppose that 
and 
. Find 
.
Proof 1. Since from upper lemma obtain and 
.
Proof 2. Let , 
, 
. So 
, where 
. Therefore, 
, i.e. 
.
PP7. Let with orthocentre and circumcircle 
and the parallelogram 
, where 
and 
. Prove that 
.
Proof.
.
An easy extension. Let with the circumcenter 
and let its interior point so that 
. Define 
. Prove that .
Proof.
.
PP8. Let an equilateral and which belongs to the small 
. Denote 
. Prove that 
.
Proof 1.
. Ptolemy's theorem
. Thus, 
.
Proof 2. From the second theorem of the-bisector in 
obtain tat 
. Using the
Ptolemy's theoremin
obtain that . Thus, .
Proof 3. Apply in relation for length of the -bisector: 
.
PP9. Let
be three circles so that 
is interior tangent to the circles in the points 
,
where
is equilateral. Ascertain the radius 
of the circle 
which is inside of and is exterior tangent to the circles 
.
Proof. Apply the generalized Pythagoras' theorem:
Denote
. Apply theorem of Cosine in 
![$\cos \phi_3=\frac {\left(r+r_1\right)^2+\left(r+r_2\right)^2-\left[3R^2-3R\left(r_1+r_2\right) +r_1^2+r_2^2+r_1r_2\right]}{2\left(r+r_1\right)\left(r+r_2\right)}=$](//latex.artofproblemsolving.com/c/a/9/ca92da3e1695ab252382777324bcf03752663698.png)
. Prove easily that
.
PP10. Let with 
and 
. Denote 
. Prove that 
.
Proof 1 (metric). Apply the Stewart's relation to the cevian
. Thus, 
, what is the relation
. Observe that 
.
Proof 2. Denote
and apply Sinus' theorem in 
, where 
.
Therefore,
, what is truly.
Remark. I"ll apply to the degenerate quadrilateral
one well-known property: if 
is a quadrilateral with 
, then there is the relation
, where 
. Obtain that 
. Otherwise. 
.
PP11. In
denote the points of tangenty 
, 
, 
of the incircle 
.
The bisector of
meet 
at 
and denote 
. Prove that 
bisects 
.
Proof.
bisects .
An easy extension. Let
be a triangle with incenter 
. For an interior point 
of the angle 
denote its projections , ,
to the sides of . The bisector of 
meet at and denote . Prove that bisects 
.
Proof.
. Therefore, bisects 
. In conclusion, bisects . Remark that for 
obtain another interesting particular cases (Stergiu's problems) .
PP12. Let a
-rightangled with 
and 
. For a mobile point ![$M\in [AC]$](//latex.artofproblemsolving.com/7/b/1/7b1a07ab551be77c55343ddd04e6923c5592b7ab.png)
define the point ![$P\in [BC]$](//latex.artofproblemsolving.com/0/8/7/087dcfc6e817021079bdd2436e980d096613bc7a.png)
so that 
. Find 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)
. Therefore, 
with equality iff 
. In conclusion, 
.
Remark. I used only inequality
with equality iff 
. A shorter way to obtain relation 
is application of known relation for 
.
Extension. Let a-right with 
and 
. For a mobile let so that 
, so that 
. Find the range of
Proof. Let
and 
. Apply an well-known relation for 
![$\frac {2ab\sin\phi}{b\left[\cos\phi -\cos (2x+\phi )\right]+a\left[\sin (2x+\phi )+\sin\phi\right]}\implies$](//latex.artofproblemsolving.com/2/4/2/24221b5e48c94948bcacc3759fcbe1fe8bdc7b32.png)
![$\boxed{BP=\frac {2ab\sin\phi}{(b\cos\phi +a\sin\phi )+\left[a\sin (2x+\phi )-b\cos (2x+\phi )\right]}}$](//latex.artofproblemsolving.com/f/5/6/f56fd19b47ccc590a046773483f8c1a9e029f81a.png)
. Thus, 
because 
.
An interesting case. If
, then 
. If and 
, then 
.
Remark.
.
PP13. Let a square and so that 
and which doesn't separate , . Define 
and 
. Prove that 
.
Proof 1 (metric).
.
.
Thus, the relations
mean that is the orthocenter of 
.
Proof 2 (analytic). Let
, 
, 
and 
and 
, 
so that 
, i.e. 
.
Thus,
. Thus, 
. Thus, 
.
PP14. Let an-isosceles with the incenter such that 
and 
. Find the length 
of ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
.
Proof 1. Let
, i.e. 
. Apply the theorem of Sines in 
. Since 
obtain that 
. Let 
, where 
, i.e.
.
Prove easily
and 
.
Proof 2 (metric).
. Thus,
.
PP15. In the ray 
is the bisector of the angle 
, where 
. For 
denote
and 
. Prove that 
.
Proof. Let
. Apply Ceva's theorem to and 
.
Thus,
.
PP16. Let be a rectangle with 
. For a point 
denote 
,
the incircle of 
and the -excircle 
of 
. Prove that 
.
Proof.
.
On other hands
.
PP17 (IMO shortlist 1998). Let be the incenter of . Let 
and be the points of tangency of the incircle of with 
and
respectively. The line 
passes through and is parallel to 
. The lines 
and 
intersect at and respectively. Prove that 
is acute.
Lemma.
Proof.
Therefore, 
.
Proof of the proposed problem.
If 
, then 
is harmonic and belongs to the exterior of the circle with the diameter 
PP18. Prove that in any triangle there are the identities 
.
Proof.
because 
.
In conclusion,
. Remark that 
. I"ll use the identity 
.
Thus,
![$2R^3\cdot\left[3\sum_{\mathrm{cyc}}\sin A\cos (B-C)-\sum_{3\mathrm{cyc}}\sin 3A\cos (B-C)\right]\ \stackrel{1\wedge 2}{=}$](//latex.artofproblemsolving.com/a/9/3/a93c4e237c35ea68d199022e57d86b1bb4084fa4.png)
.
PP19. Let be an -right triangle with the incenter . Prove that 
.
Proof 1. Let
so that 
. Thus, 
and prove easily that 
. In conclusion, 
.
Proof 2. I"ll use a known relation a.s.o 
.
Proof 3. Is well-known
. Thus,
, what is truly.
Proof 4.
and ![$ar=2\cdot [BIC]=IB\cdot IC\cdot\sin\widehat{BIC}\implies$](//latex.artofproblemsolving.com/9/f/e/9fe81ebe652a5334657b5a6b14b9a56c84d5f2d4.png)
.
Proof 5. Apply theorem of Sines in
.
Remark. Denote the tangent points of the incircle with 
respectively. Thus, 
is a square and 
. Apply an well-known property:
.
.
Remark.
there is the chain 
.
PP20. Let an isosceles with 
and the midpoint of ![$[BC]\ ,\ AM=11\ .$](//latex.artofproblemsolving.com/f/7/6/f765b2eb7030f765e494a744c8097d717635dba0.png)
Suppose that
there is
with 
and 
Find the perimeter of 
Proof (without trigonometry). Denote
and 
. Thus, 
, 
and
. Thus, 
. So the perimeter of is 
, i.e. 
.
Remark. Prove easily that 
is true the equiuvalence : 
. See PP2 from here
with the incenter 
. Denote 
. Prove that 
.Proof 1.
.Proof 2. Suppose
and let 
. Apply Menelaus' theorem to 
and 
. Provethat
, i.e. 
is tangent to the circumcircle of 
. In conclusion, Proof 3. Let
. From 
, where 
and 
obtain 
. Thus, 
.Proof 4. Apply theorem of Sinus in
.Remark. Prove that
. Thus, 
![$\frac {(s-a)\left[s(s-a)+(s-b)(s-c)\right]}{s}\ \stackrel{(1)}{\implies}$](http://latex.artofproblemsolving.com/b/c/8/bc83a23d781aa456ef82684664dff48cada07465.png)
. Otherwise, 
.PP2 (Hanoi OMO 2009). Let acute
with area 
. Let 
be located as follows: 
is the point where altitude from 
on 
meets the outwardsfacing semicirle drawn on
![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
as diameter. 
and 
are located similarly. Evaluate the sum ![$T=[BCA_1]^2+[CAB_1]^2+[ABC_1]^2$](http://latex.artofproblemsolving.com/c/b/f/cbf63b73acd4959c04538c0d447e051516a3fd56.png)
, where denoted ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
- the area of .Proof 1.
![$4\cdot \sum\sin A\cos B\cos C=2\cdot \sum\sin A\left[\cos (B+C)+\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/f/4/6/f46411773081f30cabc9df6ebfff1f5c5964f90c.png)
![$2\cdot \sum\sin A\left[-\cos A+\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/1/2/0/120d35382ab946fbceb8202916b3acb82ecb0e77.png)
![$\sum \left[-\sin 2A+2\sin (B+C)\cos (B-C)\right]=$](http://latex.artofproblemsolving.com/e/3/0/e30f439cadb9cc95cbebff8352fb5dbd2355a834.png)
. Let 
. So 
![$[BCA_1]^2=$](http://latex.artofproblemsolving.com/1/a/b/1ab556e4ed63071e057c81a2f447117536c314be.png)
![$\sum [BCA_1]^2=\frac {abc}{4}\cdot\sum a\cdot\cos B\cos C=$](http://latex.artofproblemsolving.com/7/c/d/7cdb7c672f76e1fbbd15a386d968b96c3117321c.png)
![$\boxed{\sum [BCA_1]^2=S^2}$](http://latex.artofproblemsolving.com/d/2/4/d24dec9a546c3eff71271a876c9d2761693e1634.png)
.Proof 2. Let
and orth. 
of 
![$\frac{[A_1BC]^2}{[ABC]^2}=\frac{A_1D^2}{AD^2}=$](http://latex.artofproblemsolving.com/9/b/d/9bd0441ff974dc2fa0486dd0cc943737857f6600.png)
![$\frac{DH}{AD}=\frac{[BCH]}{[ABC]}$](http://latex.artofproblemsolving.com/b/e/e/beefba73d485e53cf7b216593ad726d754b510a3.png)
. Thus, ![$\sum [BCH]=S\implies$](http://latex.artofproblemsolving.com/8/f/f/8ffc9d48a907f14e7c8e6ef7b3f934714600eae6.png)
![$\sum [BCA_1]^2=S^2$](http://latex.artofproblemsolving.com/c/3/c/c3ce38628039ecfcf1101bd123d5204c9b615fc9.png)
Remark.
.PP3. Let
and suppose that its incircle 
touch , 
, 
at 
and 
. Prove that 
.Proof. From a well-known property (or prove easily)
, 
. Thus, 
, 
, 
are cyclicaly. Thus, 
is the radical axis of the circumcirclesfor
, 
, i.e. 
and 
is the bisector of 
. Thus, 
, i.e. . Remark that 
.PP4. The inscribed circle of
touches , 
, in 
, in . ![$[GF]$](http://latex.artofproblemsolving.com/a/4/b/a4b6b1ae1f5f71c9f547186e5bf566eaef35c6c0.png)
is a diameter of 
and 
. Prove that 
.Proof 1.
.Proof 2. Is well-known that
, where 
. Indeed, apply Pascal's theorem to 
. , 
belong to the circle withthe diameter
![$[PR]$](http://latex.artofproblemsolving.com/a/d/2/ad2ecef541359529c80aaf94c8c550c9e3bb1e7c.png)
because 
, 
and 
is its center because 
. So, , , 
, 
belong to the circle 
, where .Remark. After having shown
collinear, 
is the orthocenter of 
. Hence 
and 
. As 
is isosceles, the conclusion follows.PP5. Let
be a triangle with the incircle and the -exincircle 
. Let 
and 
. Prove that 
is the bisector of 
.Proof 1. Suppose w.l.o.g.
. Let 
, the diameter ![$[DL]$](http://latex.artofproblemsolving.com/b/4/a/b4ad5e4c7fd7c6a45982d8341cc6b0fdf4fa3488.png)
of 
and 
. Thus, 
the division 
is harmonically. Since 
obtain that the ray is the bisector of the angle .Proof 2. Suppose w.l.o.g.
. Let , the diameter of and the midpoints 
, 
of , ![$[ID]$](http://latex.artofproblemsolving.com/3/9/c/39c0124bdd40abf098bae12cd10f3b575581439f.png)
respectively. Prove easily that 
,
. Thus, 
is cyclically. Since 
obtain that 
the ray 
, i.e. the ray is the bisector of .An easy extension. Let
be a circle what is tangent to 
in the point 
. Consider 
in this order so that 
.Denote the point
for which 
and 
. Define the intersection 
. Prove that the ray 
is the bisector of the angle 
.Proof. Denote the diameter
![$[AL]$](http://latex.artofproblemsolving.com/7/3/7/737a9b49658cbe239832952bb575c8bbdf7dbb35.png)
of and the midpoint , of the segments ![$[AE]$](http://latex.artofproblemsolving.com/0/e/e/0ee81ab9421e97824f7785dc2c44c0f5b04a03b1.png)
, respectively. Thus,
is an cyclical quadrilateral. Since 
is the bisector of the segment obtain that 
is the bisector of the angle .Lemma (Freeman). Let acute
with circumcircle 
. Denote the midpoints , of ![$[OA]$](http://latex.artofproblemsolving.com/c/8/8/c88c6759afc11d97267e95d4023b4419386ea20f.png)
and respectively. Then 
.Proof. Let
PP6. Let an acute
with circumcircle . Let midpoints , of and and 
. Suppose that 
and 
. Find 
.Proof 1. Since
from upper lemma obtain and 
.Proof 2. Let
, 
, 
. So 
, where 
. Therefore, 
, i.e. 
.PP7. Let
with orthocentre and circumcircle 
and the parallelogram 
, where 
and 
. Prove that 
.Proof.
.An easy extension. Let
with the circumcenter 
and let its interior point so that 
. Define 
. Prove that .Proof.
.PP8. Let an equilateral
and which belongs to the small 
. Denote 
. Prove that 
.Proof 1.
. Ptolemy's theorem 
. Thus, 
.Proof 2. From the second theorem of the
-bisector in 
obtain tat 
. Using the Ptolemy's theoremin
obtain that . Thus, .Proof 3. Apply in
relation for length of the -bisector: 
.PP9. Let
be three circles so that 
is interior tangent to the circles in the points 
,where
is equilateral. Ascertain the radius 
of the circle 
which is inside of and is exterior tangent to the circles 
.Proof. Apply the generalized Pythagoras' theorem:
Denote
. Apply theorem of Cosine in 
![$\cos \phi_3=\frac {\left(r+r_1\right)^2+\left(r+r_2\right)^2-\left[3R^2-3R\left(r_1+r_2\right) +r_1^2+r_2^2+r_1r_2\right]}{2\left(r+r_1\right)\left(r+r_2\right)}=$](http://latex.artofproblemsolving.com/c/a/9/ca92da3e1695ab252382777324bcf03752663698.png)
. Prove easily that
.PP10. Let
with 
and 
. Denote 
. Prove that 
.Proof 1 (metric). Apply the Stewart's relation to the cevian
. Thus, 
, what is the relation
. Observe that 
.Proof 2. Denote
and apply Sinus' theorem in 
, where 
.Therefore,
, what is truly.Remark. I"ll apply to the degenerate quadrilateral
one well-known property: if 
is a quadrilateral with 
, then there is the relation
, where 
. Obtain that 
. Otherwise. 
.PP11. In
denote the points of tangenty 
, 
, 
of the incircle 
.The bisector of
meet 
at 
and denote 
. Prove that 
bisects 
.Proof.
bisects .An easy extension. Let
be a triangle with incenter 
. For an interior point 
of the angle 
denote its projections , , to the sides of . The bisector of 
meet at and denote . Prove that bisects 
.Proof.
. Therefore, bisects 
. In conclusion, bisects . Remark that for 
obtain another interesting particular cases (Stergiu's problems) .PP12. Let a
-rightangled with 
and 
. For a mobile point ![$M\in [AC]$](http://latex.artofproblemsolving.com/7/b/1/7b1a07ab551be77c55343ddd04e6923c5592b7ab.png)
define the point ![$P\in [BC]$](http://latex.artofproblemsolving.com/0/8/7/087dcfc6e817021079bdd2436e980d096613bc7a.png)
so that 
. Find 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)
. Therefore, 
with equality iff 
. In conclusion, 
.Remark. I used only inequality
with equality iff 
. A shorter way to obtain relation 
is application of known relation for 
.Extension. Let a
-right with 
and 
. For a mobile let so that 
, so that 
. Find the range of Proof. Let
and 
. Apply an well-known relation for 
![$\frac {2ab\sin\phi}{b\left[\cos\phi -\cos (2x+\phi )\right]+a\left[\sin (2x+\phi )+\sin\phi\right]}\implies$](http://latex.artofproblemsolving.com/2/4/2/24221b5e48c94948bcacc3759fcbe1fe8bdc7b32.png)
![$\boxed{BP=\frac {2ab\sin\phi}{(b\cos\phi +a\sin\phi )+\left[a\sin (2x+\phi )-b\cos (2x+\phi )\right]}}$](http://latex.artofproblemsolving.com/f/5/6/f56fd19b47ccc590a046773483f8c1a9e029f81a.png)
. Thus, 
because 
.An interesting case. If
, then 
. If and 
, then 
.Remark.
. PP13. Let a square
and so that 
and which doesn't separate , . Define 
and 
. Prove that 
.Proof 1 (metric).
.
.Thus, the relations
mean that is the orthocenter of 
.Proof 2 (analytic). Let
, 
, 
and 
and 
, 
so that 
, i.e. 
.Thus,
. Thus, 
. Thus, 
.PP14. Let an
-isosceles with the incenter such that 
and 
. Find the length 
of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
.Proof 1. Let
, i.e. 
. Apply the theorem of Sines in 
. Since 
obtain that 
. Let 
, where 
, i.e.
.Prove easily
and 
.Proof 2 (metric).
. Thus,
.PP15. In
the ray 
is the bisector of the angle 
, where 
. For 
denote
and 
. Prove that 
.Proof. Let
. Apply Ceva's theorem to and 
.Thus,
.PP16. Let
be a rectangle with 
. For a point 
denote 
,the incircle
of 
and the -excircle 
of 
. Prove that 
.Proof.
.On other hands
. PP17 (IMO shortlist 1998). Let
be the incenter of . Let 
and be the points of tangency of the incircle of with 
and respectively. The line 
passes through and is parallel to 
. The lines 
and 
intersect at and respectively. Prove that 
is acute.Lemma.
Proof.
Therefore, 
.Proof of the proposed problem.
If 
, then 
is harmonic and belongs to the exterior of the circle with the diameter 
PP18. Prove that in any triangle
there are the identities 
.Proof.
because 
.In conclusion,
. Remark that 
. I"ll use the identity 
.Thus,
![$2R^3\cdot\left[3\sum_{\mathrm{cyc}}\sin A\cos (B-C)-\sum_{3\mathrm{cyc}}\sin 3A\cos (B-C)\right]\ \stackrel{1\wedge 2}{=}$](http://latex.artofproblemsolving.com/a/9/3/a93c4e237c35ea68d199022e57d86b1bb4084fa4.png)
.PP19. Let
be an -right triangle with the incenter . Prove that 
.Proof 1. Let
so that 
. Thus, 
and prove easily that 
. In conclusion, 
.Proof 2. I"ll use a known relation
a.s.o 
.Proof 3. Is well-known
. Thus, 
, what is truly.Proof 4.
and ![$ar=2\cdot [BIC]=IB\cdot IC\cdot\sin\widehat{BIC}\implies$](http://latex.artofproblemsolving.com/9/f/e/9fe81ebe652a5334657b5a6b14b9a56c84d5f2d4.png)
.Proof 5. Apply theorem of Sines in
.Remark. Denote the tangent points
of the incircle with 
respectively. Thus, 
is a square and 
. Apply an well-known property:
.
.Remark.
there is the chain 
.PP20. Let an isosceles
with 
and the midpoint of ![$[BC]\ ,\ AM=11\ .$](http://latex.artofproblemsolving.com/f/7/6/f765b2eb7030f765e494a744c8097d717635dba0.png)
Suppose thatthere is
with 
and 
Find the perimeter of 
Proof (without trigonometry). Denote
and 
. Thus, 
, 
and
. Thus, 
. So the perimeter of is 
, i.e. 
.
Remark. Prove easily that 
is true the equiuvalence : 
. See PP2 from hereThis post has been edited 132 times. Last edited by Virgil Nicula, Dec 1, 2015, 6:34 PM
August 2018
January 2018