406. Trigonometry in geometry (middle or high school) II.
by Virgil Nicula, Nov 19, 2014, 10:20 AM
P1. Let 
-right 
, 
, 
so that 
and 
. Prove 
.
Proof.
.
P2 (Ruben Dario). Let an equilateral and an interior point 
so that 
, where 
. Prove that 
.
Proof 1. Observe that
. Apply the trigonometrical form of the Ceva's theorem 
, where 
is an increasing
function for which 
and 
. In conclusion 
.
Proof 2.
. Denote 
. Thus 
and our equation becomes
![$\left(t-\sqrt 3\right)\left(\frac {2t}{1-t^2}-\sqrt 3\right)\left[\frac {t\left(3-t^2\right)}{1-3t^2}-\sqrt 3\right]=0\iff$](//latex.artofproblemsolving.com/5/7/8/578771f5d398c7133052170c10b2c71c1d4cf3f0.png)
,
where
. In conclusion, 
. Indeed, 
PP3 (Ruben Dario). Let a square
with 
and the incenter 
. Prove that 
.
Proof. Let the midpoints
of the sides ![$[AB]\ ,\ [BC]\ ,\ [CD]\ ,\ [DA]$](//latex.artofproblemsolving.com/2/b/9/2b97ba9aaf3d39d5030a94f025d328c7de70d01f.png)
respectively and suppose w.l.o.g. that 
and 
belongs to the small ![$\overarc[]{KL}$](//latex.artofproblemsolving.com/5/e/a/5eac210b0792bf28989983c1c0496b4fa8d06ed1.png)
. Let
the projections
of the point on the sides respectively and denote 
.
Thus,
where 
. Observe that 
.
Hence
![$36+\left[9-4(u-v)^2\right]+\left[9-4(u+v)^2\right]=$](//latex.artofproblemsolving.com/5/8/9/589266bbc07783061da61d8331fab3776bc636e1.png)
. In conclusion, ![$\sum PA^4=\left(\sum PA^2\right)^2-2\cdot \left[\left(PA^2+PC^2\right)\left(PB^2+PD^2\right)+PA^2\cdot PC^2+PB^2\cdot PD^2\right]=$](//latex.artofproblemsolving.com/8/b/0/8b0cbd7008f234bf3e0a9a311f2a120e28efb632.png)
, i.e. 
. Thus, 
![$6\left[52-9+4(u-v)^2-9+4(u+v)^2\right]=$](//latex.artofproblemsolving.com/7/f/4/7f480491d4af250ae2be2b4b60ac5ffdf9f5c9bd.png)
, i.e. 
. Observe that 
![$\left[9-4(u-v)^2\right]\cdot\left[9-4(u+v)^2\right]=$](//latex.artofproblemsolving.com/3/a/8/3a8911c84feae1883f6657234cccde67ddcda5e0.png)
![$81-36\left[(u-v)^2+(u+v)^2\right]+16\left(u^2-v^2\right)^2=$](//latex.artofproblemsolving.com/2/2/3/223927c8a6965669899f6ca15944333c05ce31b8.png)
with equality
. In conclusion, 
and in this case the minimum of the product is 
.
P4. Let , 
and the sentencies 
. Prove that 
.
Proof. I"ll use the metrical form of the Ceva's theorem
and the trigonometrical form of same theorem 
. I"ll use now the well-known identities 
![$\left[\frac {AN}{AP}\cdot \frac {BP}{BM}\cdot \frac {CM}{CN}\right]\cdot$](//latex.artofproblemsolving.com/e/f/f/effbb474d4bb3bd32509c71129918a9ee1f8c74b.png)
![$\left[\frac {\sin\widehat{XAN}}{\sin\widehat{XAP}}\cdot \frac {\sin\widehat{YBP}}{\sin\widehat{YBM}}\cdot \frac {\sin\widehat{ZCM}}{\sin\widehat{ZCN}}\right]\implies$](//latex.artofproblemsolving.com/2/a/a/2aa0fddedf09f8590492c6e63f7b451dcc20b4d4.png)
, what means that the required implications are truly.
P5. Let with the incircle which touches given triangle at 
, 
and 
respectively. Prove that
.
Proof. I"ll use the well-known identity:
, where the given point 
has the barycentric coordinates
w.r.t. the circumcircle 
of and 
is the power of w.r.t. the same circle. Indeed, for the 
, where 
we have
, i.e. 
. Stewart's relation to the cevians in the mentioned triangles
.
P6. Prove that for any
there are the identities 
.
Proof. Can prove easily the identities
. In the case 
and 
obtain that
. In conclusion, 
. Obtain analogously that
.
Application. Let an equilateral with 
, its interior 
and lengths 
of the projections of ![$[PQ]$](//latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
on the sides of . Then 
.
P7 (Cristian Tello). Let an-right with the circumcircle 
and 
. The circle 
is tangent
to
, 
and is interior tangent to 
. The circle 
is tangent to , 
and is interior tangent to . Prove that 
.
Proof. Suppose w.l.o.g.
. Denote 
and 
. Thus, 
.
Hence
. Therefore,
.
.
In conclusion,
. .
Otherwise. From the relations
and 
obtain that 
. Prove easily that 
.
Remark. In the particular case
obtain that 
.
P8 (Claudiu Mandrila). Let an acute triangle
with 
where 
are the midpoints of the sides ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
. Prove that 
with equality iff 
.
Proof. I"ll use an well-known property (or prove easily)
. From other well-known relation 
where![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
- the area of obtain with the relation 
that 
.
P9. Let an equilateral , 
so that 
and 
. Ascertain 
.
Proof 1. Denote
, where 
and 
. Thus,
Therefore,
![$3\left[(a-x)(2a+b-x)+(b-x)(2b+a-x)\right]=$](//latex.artofproblemsolving.com/5/6/9/5696d8385d33e6b812466b29c89ad658df6f0b17.png)
with 
and 
.
Proof 2 (Baris Altay). Let so that 
separates 
and 
. Thus, 
. Apply the generalized Pythagoras' theorem
a.s.o. Very nice proof !
An easy extension. Let an-isosceles and so that and
. Prove that ![$\boxed{\left(\sin A+\cos A\tan\phi\right)x^2-(a+b)\left[\sin A+\left(1+\cos A\right)\tan\phi\right]x+\left(a^2+b^2+2ab\cos A\right)\tan\phi}$](//latex.artofproblemsolving.com/5/a/c/5ac0b02b2e515e8b326bf3d62a50b7c072d88b58.png)
.
P10 Prove that
there is the equivalence 
Proof 1. Apply the identity
Therefore, 
Remark. 
Proof 2.
Proof 3. Apply the well known identity
a.s.o. Thus, 
Proof 4. Denote the second intersection of the line 
with the circumcircle of 
I"ll use the well known identities or prove easily that 
and 
Therefore, 
Remark. 
-right 
, 
, 
so that 
and 
. Prove 
.Proof.
. P2 (Ruben Dario). Let an equilateral
and an interior point 
so that 
, where 
. Prove that 
.Proof 1. Observe that
. Apply the trigonometrical form of the Ceva's theorem 
, where 
is an increasing
function for which 
and 
. In conclusion 
.Proof 2.
. Denote 
. Thus 
and our equation becomes
![$\left(t-\sqrt 3\right)\left(\frac {2t}{1-t^2}-\sqrt 3\right)\left[\frac {t\left(3-t^2\right)}{1-3t^2}-\sqrt 3\right]=0\iff$](http://latex.artofproblemsolving.com/5/7/8/578771f5d398c7133052170c10b2c71c1d4cf3f0.png)
,where
. In conclusion, 
. Indeed, 
PP3 (Ruben Dario). Let a square
with 
and the incenter 
. Prove that 
.Proof. Let the midpoints
of the sides ![$[AB]\ ,\ [BC]\ ,\ [CD]\ ,\ [DA]$](http://latex.artofproblemsolving.com/2/b/9/2b97ba9aaf3d39d5030a94f025d328c7de70d01f.png)
respectively and suppose w.l.o.g. that 
and 
belongs to the small ![$\overarc[]{KL}$](http://latex.artofproblemsolving.com/5/e/a/5eac210b0792bf28989983c1c0496b4fa8d06ed1.png)
. Letthe projections
of the point on the sides respectively and denote 
.Thus,
where 
. Observe that 
.Hence
![$36+\left[9-4(u-v)^2\right]+\left[9-4(u+v)^2\right]=$](http://latex.artofproblemsolving.com/5/8/9/589266bbc07783061da61d8331fab3776bc636e1.png)
. In conclusion, ![$\sum PA^4=\left(\sum PA^2\right)^2-2\cdot \left[\left(PA^2+PC^2\right)\left(PB^2+PD^2\right)+PA^2\cdot PC^2+PB^2\cdot PD^2\right]=$](http://latex.artofproblemsolving.com/8/b/0/8b0cbd7008f234bf3e0a9a311f2a120e28efb632.png)
, i.e. 
. Thus, 
![$6\left[52-9+4(u-v)^2-9+4(u+v)^2\right]=$](http://latex.artofproblemsolving.com/7/f/4/7f480491d4af250ae2be2b4b60ac5ffdf9f5c9bd.png)
, i.e. 
. Observe that 
![$\left[9-4(u-v)^2\right]\cdot\left[9-4(u+v)^2\right]=$](http://latex.artofproblemsolving.com/3/a/8/3a8911c84feae1883f6657234cccde67ddcda5e0.png)
![$81-36\left[(u-v)^2+(u+v)^2\right]+16\left(u^2-v^2\right)^2=$](http://latex.artofproblemsolving.com/2/2/3/223927c8a6965669899f6ca15944333c05ce31b8.png)
with equality
. In conclusion, 
and in this case the minimum of the product is 
. P4. Let
, 
and the sentencies 
. Prove that 
.Proof. I"ll use the metrical form of the Ceva's theorem
and the trigonometrical form of same theorem 
. I"ll use now the well-known identities 
![$\left[\frac {AN}{AP}\cdot \frac {BP}{BM}\cdot \frac {CM}{CN}\right]\cdot$](http://latex.artofproblemsolving.com/e/f/f/effbb474d4bb3bd32509c71129918a9ee1f8c74b.png)
![$\left[\frac {\sin\widehat{XAN}}{\sin\widehat{XAP}}\cdot \frac {\sin\widehat{YBP}}{\sin\widehat{YBM}}\cdot \frac {\sin\widehat{ZCM}}{\sin\widehat{ZCN}}\right]\implies$](http://latex.artofproblemsolving.com/2/a/a/2aa0fddedf09f8590492c6e63f7b451dcc20b4d4.png)
, what means that the required implications are truly.P5. Let
with the incircle which touches given triangle at 
, 
and 
respectively. Prove that
.Proof. I"ll use the well-known identity:
, where the given point 
has the barycentric coordinates
w.r.t. the circumcircle 
of and 
is the power of w.r.t. the same circle. Indeed, for the 
, where 
we have 
, i.e. 
. Stewart's relation to the cevians in the mentioned triangles 
.P6. Prove that for any
there are the identities 
.Proof. Can prove easily the identities
. In the case 
and 
obtain that
. In conclusion, 
. Obtain analogously that
. Application. Let an equilateral
with 
, its interior 
and lengths 
of the projections of ![$[PQ]$](http://latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
on the sides of . Then 
.P7 (Cristian Tello). Let an
-right with the circumcircle 
and 
. The circle 
is tangentto
, 
and is interior tangent to 
. The circle 
is tangent to , 
and is interior tangent to . Prove that 
.Proof. Suppose w.l.o.g.
. Denote 
and 
. Thus, 
.Hence
. Therefore,
.
.In conclusion,
. .Otherwise. From the relations
and 
obtain that 
. Prove easily that 
.Remark. In the particular case
obtain that 
.P8 (Claudiu Mandrila). Let an acute triangle
with 
where 
are the midpoints of the sides ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
. Prove that 
with equality iff 
.Proof. I"ll use an well-known property (or prove easily)
. From other well-known relation 
where
![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
- the area of obtain with the relation 
that 
. P9. Let an equilateral
, 
so that 
and 
. Ascertain 
.Proof 1. Denote
, where 
and 
. Thus,
Therefore,
![$3\left[(a-x)(2a+b-x)+(b-x)(2b+a-x)\right]=$](http://latex.artofproblemsolving.com/5/6/9/5696d8385d33e6b812466b29c89ad658df6f0b17.png)
with 
and 
.Proof 2 (Baris Altay). Let
so that 
separates 
and 
. Thus, 
. Apply the generalized Pythagoras' theorem
a.s.o. Very nice proof ! An easy extension. Let an
-isosceles and so that and
. Prove that ![$\boxed{\left(\sin A+\cos A\tan\phi\right)x^2-(a+b)\left[\sin A+\left(1+\cos A\right)\tan\phi\right]x+\left(a^2+b^2+2ab\cos A\right)\tan\phi}$](http://latex.artofproblemsolving.com/5/a/c/5ac0b02b2e515e8b326bf3d62a50b7c072d88b58.png)
.P10 Prove that
there is the equivalence 
Proof 1. Apply the identity
Therefore, 
Remark. 
Proof 2.
Proof 3. Apply the well known identity
a.s.o. Thus, 
Proof 4. Denote the second intersection
of the line 
with the circumcircle of 
I"ll use the well known identities or prove easily that 
and 
Therefore, 
Remark. 
This post has been edited 183 times. Last edited by Virgil Nicula, Sep 3, 2016, 3:37 PM
August 2018
January 2018