431. Trigonometry in geometry II.
by Virgil Nicula, Nov 14, 2015, 3:55 PM
P1. Prove that in any 
with the incircle 
there is the identity 
.
Proof 1. I"ll use two well-known relations
. Therefore, 
.
Proof 2. I"ll use the remarkable relations
. Therefore, obtain that 
.
Proof 3. Let the tangent points
. Thus, ![$2S=2\sum [AFIE]=\sum AI\cdot EF=$](//latex.artofproblemsolving.com/1/c/1/1c1d251ccb100cccf7b6ddfbb978d6f04d692910.png)
.
Remark. I"ll use identity
. Indeed, 
P2. Let with the 
-symmedian 
, where 
. For a point 
denote 
.
Proof. Is well-known that
. Thus, 
![$\frac {\frac {[ABL]}{c}}{\frac {[ACL]}{b}}\cdot\frac bc=$](//latex.artofproblemsolving.com/e/8/9/e896e6c462b4c3338c109a788169997cabea1ef4.png)
![$\frac {[ABL]}{[ACL]}\cdot \frac {b^2}{c^2}=$](//latex.artofproblemsolving.com/7/f/7/7f738b66bf97c7ce01760f27e2cf6649a5ea6671.png)
. I used the notation 
- the projection on the line 
of the segment ![$[MN]$](//latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
.
P3. Let for which 
and there is 
so that 
and 
. Prove that 
.
Proof 1. Let
. Thus 
and 
. Otherwise. Apply the theorem of Sines in 
.
Proof 2. Let midpoint
of ![$[DC]$](//latex.artofproblemsolving.com/d/6/6/d6654344b54d0f19b705aee27adc9aa802739f16.png)
and the second intersection 
of the circumcircle of 
with
bisector of
. Prove easily that 
and 
.
Proof 3. Denote the midpoint of ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
and 
so that 
. Thus, and 
. Apply the theorem of Sines to
. Since 
obtain that 
.
P4 (Miguel Ochoa Sanchez). Let a trapezoid
with 
, 
and
midpoints , 
of ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, respectively. Prove that 
Proof.
. Thus, 
cyclic 
.
P5 (Ruben Dario Auqui). Let with -bisector 
where 
. Prove that
, where 
, 
Proof 1. Denote
and 
. Observe that 
. Apply the theorem of Sines in the triangles:
. I"ll use an well-known relation 
.
In conclusion,
![$d^2\left[(m+n)^2-4mn\cos^2\phi\right]=4m^2n^2\cos^2\phi\iff$](//latex.artofproblemsolving.com/9/c/a/9caae2b1d8be17712a9d4a56aa9c28ade4ccc90e.png)
![$d^2\left[(m+n)^2\sin ^2\phi +(m-n)^2\cos^2\phi\right]=4m^2n^2\cos^2\phi\iff$](//latex.artofproblemsolving.com/7/f/5/7f58db9bbaca18a01d8b457eb50237e36b6ace3a.png)
![$d^2\left[(m+n)^2\tan^2\phi +(m-n)^2\right]=4m^2n^2\iff$](//latex.artofproblemsolving.com/b/b/6/bb655e4bd0d3e135705215bcd0a9316cf6e6522f.png)
.
Proof 2.
.
Particular cases.
. Another cases: 
.
Remark. The relation
can write as 
.
P6. Let with incircle 
. Let 
and midpoints , , 
of ![$[EF]$](//latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
, ![$[FD]$](//latex.artofproblemsolving.com/0/4/3/043cba2a3f325e257f54812d8162fcfcce05424c.png)
, ![$[DE]$](//latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Prove that 
.
Proof. Let length
of inradius of 
, 
, 
and 
, 
and , 
, 
. Thus
, i.e. 
. Thus, 
.
Show analogously that
and 
. Therefore, 
, i.e. .
P7 (Sunken Rock). Let isosceles trapezoid with 
and 
. Find 
P7 bis (equivalent enunciation).Take on the side ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
of the rhombus with 
so that 
. Find 
.
Proof 1.
and 
, what is true. So 
.
Proof 2. Thus . Let 
for which 
. Prove easily that 
is an equilateral 
is isosceles and 
is incenter of 
.
Remark. Let a rhombus and ![$M\in [AD]$](//latex.artofproblemsolving.com/f/5/4/f54ed5809569e842aab68830010b6b84214017f0.png)
so that . Prove similarly that the ray 
is 
-bisector of 
.
P8. Let with 
and the incircle . Denote 
and 
, Consider a point 
so that 
separates and 
. Observe that
. Let 
, 
, 
be the distancies from to the sidelines 
, 
, 
respectively. Prove that 
.
Proof.![$\left\{\begin{array}{ccc}
x=r+r\sin \left[\pi-\phi-\left(\frac {\pi}{2}-C\right)\right] & \implies & x=2r\cos^2\frac {\phi -C}{2}\\\\
y=r-r\cos\phi & \implies & y=2r\sin^2\frac {\phi}{2}\\\\
z=r-r\cos (\pi-A-\phi ) & \implies & z=2r\cos^2\frac {A+\phi}{2}\end{array}\right\|\implies$](//latex.artofproblemsolving.com/7/3/5/73523009280c3097e60e2a78ddea7fd1105838b8.png)
what is truly because
.
P9 (Carrasco's lemma). Let and 
so that 
and 
. Prove that 
and 
.
Proof 1. Let so that 
and 
. I"ll apply the relation 
. The triangles 
and 
have same circumradius 
and common
-altitude. Therefore, 
. The triangles 
and 
have same circumradius 
and a common -altitude.
Therefore,
. In conclusion, from the relations 
and 
obtain that 
.
Proof 2. Denote
and 
. Apply an well-known relations 
.
Proof 3. I"ll use same notations. The circumcircles of
and 
are equivalently. Thus, 
. Apply theorem of Sines in 
.
Thus,
.
P10. Let a convex with 
and 
. Prove that 
Proof.
. Apply an well-known relation in 
Thus,
.
Therefore, 
.
Remark. Denote
. Thus, 
.
Therefore, 
a.s.o.
P11 (TST Peru, IMO 2007). Let so that 
with circumcircle 
and incircle . Let 
and 
so that 
. Prove that 
.
Proof 1. Suppose w.l.o.g.
. Denote the diameters ![$[CC']$](//latex.artofproblemsolving.com/a/8/1/a81ee27e589c3262d8fb37e72a84a363bca0eb34.png)
, ![$[PP']$](//latex.artofproblemsolving.com/c/8/5/c851691a4111d3eed5ac07f9d7ca3c25d061d0bf.png)
of 
and 
.
At first I"ll find the position of , i.e. the ratio 
. Thus, 
the division 
is harmonically, i.e.
.
At second I"ll find the positions of 
, 
and I"ll value the ratio 
. Observe that 
and 
. Apply the Menelaus' theorem to the transversal 
in:
. In conclusion, 
.
Proof 2.
is inscribed in the circle with the diameter ![$[IQ]\implies$](//latex.artofproblemsolving.com/a/e/3/ae3a7cedd308ae77f83303ab3f1407069b194d93.png)
.
In conclusion,
. Very nice proof!
P12. Fie-right si punctele ![$\{X,Y\}\subset [BC]$](//latex.artofproblemsolving.com/a/4/9/a494f805e4795a44cafa0d1f47d56ec804148dca.png)
astfel incat 
Sa se arate ca 
Generalizare 1 Fie si punctele 
astfel incat Sa se arate ca 
Se poate arata ca
Observatie. 
si 
Generalizare 2.
![$\frac{2bc\cdot\sin(A-x)\sin(A-z)}{1+\cos y} \le BY\cdot CX \le a^2 \cdot \frac{\sin(A-x)\cdot\sin(A-z)}{\left[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right]^2}\ .$](//latex.artofproblemsolving.com/f/7/3/f73f4b516fc79511d66f15e2f14d9fcf4ac18865.png)
![$ AX\cdot AY \le bc \cdot \frac{\sin^2 A}{\left[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right]^2}\ .$](//latex.artofproblemsolving.com/f/9/e/f9e5b7fb431e72216dfa772b72407433db1714a5.png)
![$2bc\cdot\frac{\sin x\cdot\sin z}{1+\cos y}\le BX\cdot CY \le a^2\cdot \frac{\sin x\cdot\sin z}{\left[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right]^2}\ \blacktriangleleft\ \bullet\ \blacktriangleright\ \left\{\begin{array}{c}
a\cdot XY=BX\cdot CY\cdot\frac{\sin y\cdot \sin A}{\sin x\cdot \sin z}\\\\
\frac{BX\cdot CY}{BY\cdot CX}=\frac{\sin x\cdot\sin z}{\sin(A-x)\cdot\sin(A-z)}\end{array}\right\|\ .$](//latex.artofproblemsolving.com/1/1/5/11503669ee144bf2d52a37bd08d471f79c3f0465.png)
P13 Prove that in any triangle
there is the equivalence 
Proof 1 (own). Denote
so that 
and denote 
Thus, 
i.e. 
and 
Apply an well-known property in 
where 
Proof 2 (own). Let
so that 
and 
where denote 
and 
Thus, 
Now apply an well-known property in 
where 
From the relations and get that 
Proof 3 (own). Construct the isosceles trapezoid
where 
and denote 
Thus, 
Apply an well-known property in
Denote the midpoint of and the projection 
of on 
Therefore, in there is the remarkable relation 
i.e. 
From the relations and obtain
that
Proof 4. Exists
so that 
i.e. 
I"ll use the theorem of Sines 
and the well known trigonometrical identity
Therefore, 
O.K.
Proof 5.
![$\sin (A-B)\cancel{\sin C}[\sin (A-B)\cancel{\sin C}+\sin A\cancel{\sin C}]=$](//latex.artofproblemsolving.com/0/3/1/0310007e940f10b823b0a01c37b046cbb7e40504.png)
![$\sin (A-B)[\sin (A-B)+\sin A]=\sin^2B\iff$](//latex.artofproblemsolving.com/1/e/5/1e55d2845e79852a0134db6d9d5530aa7b1b785c.png)
Remark. 
Proof 6 (Ruben Dario). Let so that 
i.e. 
Apply the well-known property to 
Thus, 
From the relations and obtain that
Proof 7 (M.O. Sanchez). Construct the-isosceles 
outside of 
where denote 
and 
Apply Stewart's relation in particular case for 
From and eliminate 
P14. Let a square with 
the circle where 
the distance 
where ![$r\in\left(0,\frac 12\right].$](//latex.artofproblemsolving.com/f/0/1/f017e95416efaa6f0b55a5233e8319c6c722aed9.png)
Let
so that 
is tangent to 
and 
Prove that 
Proof. Denote
Prove easily that 
Thus, 
Therefore, 
In conclusion, 
Particular case. If 
then 
i.e. 
P15. Let be a rectangle and be the its interior point for which 
Find and the ratio 
Proof 1 (trigonometric). Theorem of SINES in
and 
In conclusion, 
Hence
![$\frac {\sin \left(90^{\circ}-6x\right)}{\sin\left[x+\left(90^{\circ}-6x\right)\right]}=$](//latex.artofproblemsolving.com/5/a/1/5a14eeb49876743297c9991977aac3a2dbb72e81.png)
P16. Let
and the circumcircles 
of the triangles 
respectively. Prove that 
Proof 1. Observe that
In conclusion, 
Proof 2.
is a rhombus 
P17 (Ruben Dario). Let be a trapezoid with so that the semicircle with the diameter is tangent to
the side at 
Denote 
and 
where 
Prove that 
Proof (Cristian Tello). Denote
and 
i.e. 
Thus, prove easily that
and 
From 
obtain that 
In conclusion, 
P18 (Ruben Dario). Let a convex quadrilateral so that and 
Prove that 
Proof. Suppose w.l.o.g.
Let the midpoint of the side and 
Therefore, 
and apply the generalized
Pythagoras' theorem to![$[BC]/\triangle BMC\ :\ BC^2=MB^2+MC^2-2\cdot MB\cdot MC\cdot\cos\widehat{BMC}\iff$](//latex.artofproblemsolving.com/7/d/4/7d4679d3873de35cd0fd8777d6cbd853978044a5.png)
![$1+4\cos (x+y)[\cos (x-y)-\cos (x+y)]=2(1-\cos 2x)+2(1-\cos 2y)\iff$](//latex.artofproblemsolving.com/e/0/c/e0c82bb4b382a498cbdf96c5c48e3b208d6fd779.png)
![$3+2[1+\cos (2x+2y)]=2(\cos 2x+\cos 2y)+2(\cos 2x+\cos 2y)\iff$](//latex.artofproblemsolving.com/3/0/c/30c949425342847f4a07301ed6fa30a871bde7fa.png)
P19 (Mehmet Sahin).Let an acute with 
the midpoints 
of the ![$[BC]\ ,$](//latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
![$[CA]\ ,$](//latex.artofproblemsolving.com/7/9/0/79053f4e90a202762296b3a7020733fab1b27fe1.png)
respectively 
the projections 
of
on the opposite sides 
Denote 
and 
Prove that 
Proof 1. Apply generalized Pythagoras' theorem to
and
![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
a.s.o. Thus, 
Proof 2.
a.s.o.
with the incircle 
there is the identity 
.Proof 1. I"ll use two well-known relations
. Therefore, 
.Proof 2. I"ll use the remarkable relations
. Therefore, obtain that 
.Proof 3. Let the tangent points
. Thus, ![$2S=2\sum [AFIE]=\sum AI\cdot EF=$](http://latex.artofproblemsolving.com/1/c/1/1c1d251ccb100cccf7b6ddfbb978d6f04d692910.png)
.Remark. I"ll use identity
. Indeed, 
P2. Let
with the 
-symmedian 
, where 
. For a point 
denote 
.Proof. Is well-known that
. Thus, 
![$\frac {\frac {[ABL]}{c}}{\frac {[ACL]}{b}}\cdot\frac bc=$](http://latex.artofproblemsolving.com/e/8/9/e896e6c462b4c3338c109a788169997cabea1ef4.png)
![$\frac {[ABL]}{[ACL]}\cdot \frac {b^2}{c^2}=$](http://latex.artofproblemsolving.com/7/f/7/7f738b66bf97c7ce01760f27e2cf6649a5ea6671.png)
. I used the notation 
- the projection on the line 
of the segment ![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
.P3. Let
for which 
and there is 
so that 
and 
. Prove that 
.Proof 1. Let
. Thus 
and 
. Otherwise. Apply the theorem of Sines in 
.Proof 2. Let midpoint
of ![$[DC]$](http://latex.artofproblemsolving.com/d/6/6/d6654344b54d0f19b705aee27adc9aa802739f16.png)
and the second intersection 
of the circumcircle of 
withbisector of
. Prove easily that 
and 
.Proof 3. Denote the midpoint
of ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
and 
so that 
. Thus, and 
. Apply the theorem of Sines to
. Since 
obtain that 
. P4 (Miguel Ochoa Sanchez). Let a trapezoid
with 
, 
andmidpoints
, 
of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, respectively. Prove that 
Proof.
. Thus, 
cyclic 
.P5 (Ruben Dario Auqui). Let
with -bisector 
where 
. Prove that
, where 
, 
Proof 1. Denote
and 
. Observe that 
. Apply the theorem of Sines in the triangles:
. I"ll use an well-known relation 
.In conclusion,
![$d^2\left[(m+n)^2-4mn\cos^2\phi\right]=4m^2n^2\cos^2\phi\iff$](http://latex.artofproblemsolving.com/9/c/a/9caae2b1d8be17712a9d4a56aa9c28ade4ccc90e.png)
![$d^2\left[(m+n)^2\sin ^2\phi +(m-n)^2\cos^2\phi\right]=4m^2n^2\cos^2\phi\iff$](http://latex.artofproblemsolving.com/7/f/5/7f58db9bbaca18a01d8b457eb50237e36b6ace3a.png)
![$d^2\left[(m+n)^2\tan^2\phi +(m-n)^2\right]=4m^2n^2\iff$](http://latex.artofproblemsolving.com/b/b/6/bb655e4bd0d3e135705215bcd0a9316cf6e6522f.png)
.Proof 2.
.Particular cases.
. Another cases: 
.Remark. The relation
can write as 
.P6. Let
with incircle 
. Let 
and midpoints
, , 
of ![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
, ![$[FD]$](http://latex.artofproblemsolving.com/0/4/3/043cba2a3f325e257f54812d8162fcfcce05424c.png)
, ![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Prove that 
.Proof. Let length
of inradius of 
, 
, 
and 
, 
and , 
, 
. Thus
, i.e. 
. Thus, 
.Show analogously that
and 
. Therefore, 
, i.e. .P7 (Sunken Rock). Let isosceles trapezoid
with 
and 
. Find 
P7 bis (equivalent enunciation).Take
on the side ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
of the rhombus with 
so that 
. Find 
.Proof 1.
and 
, what is true. So 
.Proof 2. Thus
. Let 
for which 
. Prove easily that 
is an equilateral 
is isosceles and 
is incenter of 
.Remark. Let a rhombus
and ![$M\in [AD]$](http://latex.artofproblemsolving.com/f/5/4/f54ed5809569e842aab68830010b6b84214017f0.png)
so that . Prove similarly that the ray 
is 
-bisector of 
.P8. Let
with 
and the incircle . Denote 
and 
, Consider a point 
so that 
separates and 
. Observe that
. Let 
, 
, 
be the distancies from to the sidelines 
, 
, 
respectively. Prove that 
.Proof.
![$\left\{\begin{array}{ccc}
x=r+r\sin \left[\pi-\phi-\left(\frac {\pi}{2}-C\right)\right] & \implies & x=2r\cos^2\frac {\phi -C}{2}\\\\
y=r-r\cos\phi & \implies & y=2r\sin^2\frac {\phi}{2}\\\\
z=r-r\cos (\pi-A-\phi ) & \implies & z=2r\cos^2\frac {A+\phi}{2}\end{array}\right\|\implies$](http://latex.artofproblemsolving.com/7/3/5/73523009280c3097e60e2a78ddea7fd1105838b8.png)
what is truly because
.P9 (Carrasco's lemma). Let
and 
so that 
and 
. Prove that 
and 
.Proof 1. Let
so that 
and 
. I"ll apply the relation 
. The triangles 
and 
have same circumradius 
and common-altitude. Therefore, 
. The triangles 
and 
have same circumradius 
and a common -altitude.Therefore,
. In conclusion, from the relations 
and 
obtain that 
.Proof 2. Denote
and 
. Apply an well-known relations 
.Proof 3. I"ll use same notations. The circumcircles of
and 
are equivalently. Thus, 
. Apply theorem of Sines in 
.Thus,
. P10. Let a convex
with 
and 
. Prove that 
Proof.
. Apply an well-known relation in 
Thus,
.Therefore,
.Remark. Denote
. Thus, 
.Therefore,
a.s.o.P11 (TST Peru, IMO 2007). Let
so that 
with circumcircle 
and incircle . Let 
and 
so that 
. Prove that 
.Proof 1. Suppose w.l.o.g.
. Denote the diameters ![$[CC']$](http://latex.artofproblemsolving.com/a/8/1/a81ee27e589c3262d8fb37e72a84a363bca0eb34.png)
, ![$[PP']$](http://latex.artofproblemsolving.com/c/8/5/c851691a4111d3eed5ac07f9d7ca3c25d061d0bf.png)
of 
and 
.
At first I"ll find the position of , i.e. the ratio 
. Thus, 
the division 
is harmonically, i.e.
.
At second I"ll find the positions of 
, 
and I"ll value the ratio 
. Observe that 
and 
. Apply the Menelaus' theorem to the transversal 
in:
. In conclusion, 
.Proof 2.
is inscribed in the circle with the diameter ![$[IQ]\implies$](http://latex.artofproblemsolving.com/a/e/3/ae3a7cedd308ae77f83303ab3f1407069b194d93.png)
.In conclusion,
. Very nice proof!P12. Fie
-right si punctele ![$\{X,Y\}\subset [BC]$](http://latex.artofproblemsolving.com/a/4/9/a494f805e4795a44cafa0d1f47d56ec804148dca.png)
astfel incat 
Sa se arate ca 
Generalizare 1 Fie
si punctele 
astfel incat Sa se arate ca 
Se poate arata ca
Observatie. 
si 
Generalizare 2.
![$\frac{2bc\cdot\sin(A-x)\sin(A-z)}{1+\cos y} \le BY\cdot CX \le a^2 \cdot \frac{\sin(A-x)\cdot\sin(A-z)}{\left[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right]^2}\ .$](http://latex.artofproblemsolving.com/f/7/3/f73f4b516fc79511d66f15e2f14d9fcf4ac18865.png)
![$ AX\cdot AY \le bc \cdot \frac{\sin^2 A}{\left[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right]^2}\ .$](http://latex.artofproblemsolving.com/f/9/e/f9e5b7fb431e72216dfa772b72407433db1714a5.png)
![$2bc\cdot\frac{\sin x\cdot\sin z}{1+\cos y}\le BX\cdot CY \le a^2\cdot \frac{\sin x\cdot\sin z}{\left[\sqrt{\sin x\cdot \sin z}+\sqrt{\sin(A-x)\cdot\sin(A-z)}\right]^2}\ \blacktriangleleft\ \bullet\ \blacktriangleright\ \left\{\begin{array}{c}
a\cdot XY=BX\cdot CY\cdot\frac{\sin y\cdot \sin A}{\sin x\cdot \sin z}\\\\
\frac{BX\cdot CY}{BY\cdot CX}=\frac{\sin x\cdot\sin z}{\sin(A-x)\cdot\sin(A-z)}\end{array}\right\|\ .$](http://latex.artofproblemsolving.com/1/1/5/11503669ee144bf2d52a37bd08d471f79c3f0465.png)
P13 Prove that in any triangle
there is the equivalence 
Proof 1 (own). Denote
so that 
and denote 
Thus, 
i.e. 
and 
Apply an well-known property in 
where 
Proof 2 (own). Let
so that 
and 
where denote 
and 
Thus, 
Now apply an well-known property in 
where 
From the relations and get that 
Proof 3 (own). Construct the isosceles trapezoid
where 
and denote 
Thus, 
Apply an well-known property in
Denote the midpoint of and the projection 
of on 
Therefore, in
there is the remarkable relation 
i.e. 
From the relations and obtainthat
Proof 4. Exists
so that 
i.e. 
I"ll use the theorem of Sines 
and the well known trigonometrical identity
Therefore, 
O.K.Proof 5.
![$\sin (A-B)\cancel{\sin C}[\sin (A-B)\cancel{\sin C}+\sin A\cancel{\sin C}]=$](http://latex.artofproblemsolving.com/0/3/1/0310007e940f10b823b0a01c37b046cbb7e40504.png)
![$\sin (A-B)[\sin (A-B)+\sin A]=\sin^2B\iff$](http://latex.artofproblemsolving.com/1/e/5/1e55d2845e79852a0134db6d9d5530aa7b1b785c.png)
Remark. 
Proof 6 (Ruben Dario). Let
so that 
i.e. 
Apply the well-known property to 
Thus, 
From the relations and obtain that
Proof 7 (M.O. Sanchez). Construct the
-isosceles 
outside of 
where denote 
and 
Apply Stewart's relation in particular case for 
From and eliminate 
P14. Let a square
with 
the circle where 
the distance 
where ![$r\in\left(0,\frac 12\right].$](http://latex.artofproblemsolving.com/f/0/1/f017e95416efaa6f0b55a5233e8319c6c722aed9.png)
Let
so that 
is tangent to 
and 
Prove that 
Proof. Denote
Prove easily that 
Thus, 
Therefore, 
In conclusion, 
Particular case. If 
then 
i.e. 
P15. Let
be a rectangle and be the its interior point for which 
Find and the ratio 
Proof 1 (trigonometric). Theorem of SINES in
and 
In conclusion, 
Hence
![$\frac {\sin \left(90^{\circ}-6x\right)}{\sin\left[x+\left(90^{\circ}-6x\right)\right]}=$](http://latex.artofproblemsolving.com/5/a/1/5a14eeb49876743297c9991977aac3a2dbb72e81.png)
P16. Let
and the circumcircles 
of the triangles 
respectively. Prove that 
Proof 1. Observe that
In conclusion, 
Proof 2.
is a rhombus 
P17 (Ruben Dario). Let
be a trapezoid with so that the semicircle with the diameter is tangent tothe side
at 
Denote 
and 
where 
Prove that 
Proof (Cristian Tello). Denote
and 
i.e. 
Thus, prove easily that
and 
From 
obtain that 
In conclusion, 
P18 (Ruben Dario). Let a convex quadrilateral
so that and 
Prove that 
Proof. Suppose w.l.o.g.
Let the midpoint of the side and 
Therefore, 
and apply the generalizedPythagoras' theorem to
![$[BC]/\triangle BMC\ :\ BC^2=MB^2+MC^2-2\cdot MB\cdot MC\cdot\cos\widehat{BMC}\iff$](http://latex.artofproblemsolving.com/7/d/4/7d4679d3873de35cd0fd8777d6cbd853978044a5.png)
![$1+4\cos (x+y)[\cos (x-y)-\cos (x+y)]=2(1-\cos 2x)+2(1-\cos 2y)\iff$](http://latex.artofproblemsolving.com/e/0/c/e0c82bb4b382a498cbdf96c5c48e3b208d6fd779.png)
![$3+2[1+\cos (2x+2y)]=2(\cos 2x+\cos 2y)+2(\cos 2x+\cos 2y)\iff$](http://latex.artofproblemsolving.com/3/0/c/30c949425342847f4a07301ed6fa30a871bde7fa.png)
P19 (Mehmet Sahin).Let
an acute with 
the midpoints 
of the ![$[BC]\ ,$](http://latex.artofproblemsolving.com/6/5/6/6563ca77d31eb676ef646fa9c37ea181b812d864.png)
![$[CA]\ ,$](http://latex.artofproblemsolving.com/7/9/0/79053f4e90a202762296b3a7020733fab1b27fe1.png)
respectively 
the projections 
of
on the opposite sides 
Denote 
and 
Prove that 
Proof 1. Apply generalized Pythagoras' theorem to
and
![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
a.s.o. Thus, 
Proof 2.
a.s.o.This post has been edited 245 times. Last edited by Virgil Nicula, Nov 30, 2017, 3:39 PM
August 2018
January 2018