306. Some easy and nice "slicing" problems.
by Virgil Nicula, Jul 30, 2011, 2:16 PM
PP1. Let 
be a triangle with 
. Let 
be an interior point for which 
. Find 
.
Proof 1 (trigonometric). Denote
. Construct the equilateral triangle 
so that the sideline 
separates 
, 
. Observe that 
,
and 
, 
and 
. Therefore, 
Proof 2 (synth. and bit trig.). Let
be midpoint of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. Observe that 
, what is truly. Therefore, 
. Since 
is
the bisector of
obtain that 
the ray 
is the bisector of 
.
Proof 3 (synthetic). Let be midpoint of the side and . Denote 
and 
for which 
.
Observe that
and 
. Thus, ![$SB\cdot SA=[ABC]=\frac 12\cdot AC\cdot BY=AX\cdot BY\implies$](//latex.artofproblemsolving.com/a/a/f/aafc39d631827c2a71143657419510edd1ddba16.png)
.
Since
and using the relations from 
obtain that 
, i.e. 
. Since
is the bisector of obtain that 
the ray is the bisector of .
PP2. Let be a triangle . Prove that 
.
Proof 1 (metric). Suppose w.l.o.g.
. Thus, 
, what is falsely.
An easy extension. Let be a triangle . Prove that 
we have 
.
PP3. In
with 
exists 
so that 
. Find .
Lemma. In a quadrilateral
given 
and 
, 
. Find 
in terms of 
.
Proof 1 (synthetic). Denote
. Let the isosceles triangle 
so that 
doesn't separate 
, 
and 
. Thus,
.
Proof 2 (trigonometric). Using and the Sinus' theorem in and 
obtain that 
.
Therefore,
and 
. Nice problem !
PP4. the point
is taken inside of so that 
. Find 
.
Proof 1 (synthetic). Denote
, 
so that 
and 
. Since 
(exterior), 
and
(exterior) obtain that 
, i.e. the quadrilateral 
is cyclic. Since 
is isosceles obtain that the quadrilateral
is cyclic, so 
. Finally 
and 
.
Proof 2 (trigonometric). Denote
. Apply an well-known property 
.
PP5. Let be a quadrilateral so that is -right isosceles, 
and 
. Find 
.
Proof 1 (synthetic).
is equilateral 
.
Proof 2 (trigonometric). Denote
and apply an well-known property 
.
PP6. In a
point 
is taken on 
so that 
and 
find 
Proof 1 (synthetic). Denote
so that 
and 
and 
. Since 
is also isosceles, yielding 
. Hence 
.From the relations 
and 
yield
. Thus 
is isosceles. Hence 
.
Proof 2 (trigonometric).
.
Otherwise.
.
PP7. Let be a triangle, where 
, 
and 
. Suppose that exists such that 
and 
. Find .
Proof 1 (synthetic). Draw
and let 
be the midpoint of 
. Then is circumcenter of 
and is circumcenter
of
. On the other hand, quadrilateral 
is cyclic, so 
is right angled isosceles, and 
.
Proof 2 (trigonometric). Apply theorem of Sinus :
.
PP8. Let be a triangle with the incenter 
. Denote 
and 
. Prove that if 
.
Proof 1 (trigonometric). I"ll apply well-known property in quadrilateral
![\[\boxed{\ \begin{array}{ccc}
\sin\left(A+\frac C2\right)\sin\frac B2\sin\frac C2\sin\left(60^{\circ}-\frac A2\right) & = & \sin\frac B2\sin\frac C2\sin\left(A+\frac B2\right)\sin 30^{\circ}\\\\
\sin\left(A+\frac C2\right)\sin\left(60^{\circ}-\frac A2\right) & = & \sin\left(A+\frac B2\right)\sin 30^{\circ}\\\\
2\sin\left(A+\frac C2\right)\sin\left(60^{\circ}-\frac A2\right) & = & \sin\left(A+\frac B2\right)\\\\
2\cos\frac {A-B}{2}\cos\left(30^{\circ}+\frac A2\right) & = & \cos\frac {A-C}{2}\\\\
\cos\frac {2A-B+60^{\circ}}{2}+\cos\frac {B+60^{\circ}}{2} & = & \cos\frac {A-C}{2}\\\\
\cos\frac {B+60^{\circ}}{2} & = & \cos\frac {A-C}{2}-\cos\frac {2A-B+60^{\circ}}{2}\\\\
\cos\frac {B+60^{\circ}}{2} & = & 2\sin \frac {3A-B-C+60^{\circ}}{4}\sin\frac {A+C-B+60^{\circ}}{4}\\\\
\cos\frac {B+60^{\circ}}{2} & = & 2\sin \frac {4A-120^{\circ}}{4}\sin\frac {240^{\circ}-2B}{4}\\\\
\sin\left(60^{\circ}-\frac B2\right) & = & 2\sin (A-30^{\circ} )\sin\left(60^{\circ}-\frac B2\right)\end{array}\ }\]](//latex.artofproblemsolving.com/e/5/1/e51daf579f96b1682fc88aab4ee729f5fa99916c.png)
In conclusion,
.
PP9. Let be an interior point of for which 
. Prove that is isosceles.
Proof 1 (trigonometric). Denote
. Thus, 
. I shall aplly the Ceva's theorem (in the trigonometrical form):
.
Proof 2 (synthetic). Reflect across 
and name it . Then 
is equilateral and 
. Denote 
. Prove easily that 
is cyclically. Therefore
. Note that 
. Since 
obtain that 
.
PP10. Let
be a triangle and 
, 
be two points so that 
,
, 
and 
. Find 
.
Proof 1 (trigonometric). Denote
. Thus, 
. Apply in 
the trigonometrical form of the Ceva's theorem:
.
Proof 2 (metric). Prove easily that
. Hence the 
-bisector in 
and the 
-bisector in 
are concurrently in
. Denote 
so that 
and 
. Observe that the points , belong to the circle with diameter ![$ [KL]$](//latex.artofproblemsolving.com/a/e/f/aef79871691703ff93e83466aa0c5e9bcc1fe14f.png)
.
Denote
. Since 
and 
, 
obtain 
, i.e. 
.
Extension. Let
, 
so that 
Prove that 
PP11. Let and be points on the sides 
and of a triangle such that the lines 
and 
are the interior angle bisectors of the
angles and 
respectively. Given that 
and 
. Determine the angles 
, 
, 
of triangle .
Proof. Prove easily that
and 
. Hence and 
are both acute. By the sine theorem for the 
,
obtain that 
. Thus, 
. Using 
obtain that 
. In conclusion, 
and 
.
=======================================================================================================
, what is truly.
Very nice ! Thank you, Yetti.
PP12. Let be a triangle with 
. Denote the midpoint of ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and suppose that 
. Find .
Proof 1(trigonometric). Observe that
. Therefore,
![\[\boxed{\begin{array}{c}
1=\frac {DA}{DC}=\frac {BA}{BC}\cdot\frac {\sin \widehat{DBA}}{\sin \widehat{DBC}}=\frac {\sin C}{\sin 3C}\cdot\frac {\sin (45^{\circ}+3C)}{\sin (45^{\circ}-C)}\\\\
\sin C\sin (45^{\circ}+3C)=\sin 3C\sin (45^{\circ}-C)\\\\
\cos (45^{\circ}+2C)-\cos (45^{\circ}+4C)=\cos (45^{\circ}-4C)-\cos (45^{\circ}+2C)\\\\
2\cos (45^{\circ}+2C)=\cos (45^{\circ}+4C)+\cos (45^{\circ}-4C)\\\\
\sqrt 2\cdot\cos (45^{\circ}+2C)=\cos 4C\\\\
\cos 2C-\sin 2C=\cos 4C=\cos^22C-\sin^22C\\\\
\cos 2C=\sin 2C\ \ \vee\ \ \cos 2C+\sin 2C=1\\\\
C=\frac {\pi}{8}\ \ \vee\ \ C=\frac {\pi}{4}\\\\
C=\frac {\pi}{8}\end{array}}\]](//latex.artofproblemsolving.com/b/7/5/b758cfca8755fd06361a9befaf306e655e395bbe.png)
.
Proof 2 (trigonometric). Denote
and 
. Thus 
and 
, 
. Therefore,
.
PP13. Let and an interior so that 
. Find 
.
Proof 1 (synthetic). Construct the symmetrical point of w.r.t. the line 
. Observe that is
a cyclical quadrilateral,
is an equilateral triangle and 
. Finally, 
.
Proof 2 (trigonometric). Denote
and apply the trigonometric form of the Ceva's theorem for the point 
![\[\boxed{\begin{array}{c}
\sin (40^{\circ}+x)\sin 10^{\circ}\sin 30^{\circ}=\sin (x-60^{\circ})\sin 40^{\circ}\sin 20^{\circ}\\\\
\sin (40^{\circ}+x)=4\sin (x-60)\sin 40^{\circ}\cos 10^{\circ}\\\\
\sin (40^{\circ}+x)=2\sin (x-60^{\circ})\left(\sin 50^{\circ}+\frac 12\right)\\\\
\sin (40^{\circ}+x)=2\cos 40^{\circ}\sin (x-60^{\circ})+\sin (x-60^{\circ})\\\\
\sin (40^{\circ}+x)=\sin (x-20^{\circ})-\sin (100^{\circ}-x)+\sin (x-60^{\circ})\\\\
\sin (40^{\circ}+x)+\sin (80^{\circ}+x)=\sin (x-20^{\circ})+\sin (x-60^{\circ})\\\\
\sin (60^{\circ}+x)\cos 20^{\circ}=\sin (x-40^{\circ})\cos 20^{\circ}\\\\
\sin (60^{\circ}+x)=\sin (x-40^{\circ})\\\\
(60^{\circ}+x)+(x-40^{\circ})=180^{\circ}\\\\
x=80^{\circ}\end{array}}\]](//latex.artofproblemsolving.com/4/f/9/4f9266c9abc80a9ce2450502f2680d289cb66471.png)
PP14 (Langley's problem). Let an-isosceles with 
and the points 
Find the value of the angle 
See and here.
Proof 1. Let
so that 
Hence 
because 
Since 
obtain 
From
and 
obtain 
From obtain 
In conclusion 
and 
An easy extension.. Let with 
and the points 
Find the value of the angle
Proof. Let so that 
From and 
obtain that 
and 
Hence
From 
obtain that 
Thus, 
i.e. 
is equilateral. From 
obtain that
and 
In conclusion, 
and 
be a triangle with 
. Let 
be an interior point for which 
. Find 
.Proof 1 (trigonometric). Denote
. Construct the equilateral triangle 
so that the sideline 
separates 
, 
. Observe that 
,
and 
, 
and 
. Therefore, 
Proof 2 (synth. and bit trig.). Let
be midpoint of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
. Observe that 
, what is truly. Therefore, 
. Since 
isthe bisector of
obtain that 
the ray 
is the bisector of 
.Proof 3 (synthetic). Let
be midpoint of the side and . Denote 
and 
for which 
.Observe that
and 
. Thus, ![$SB\cdot SA=[ABC]=\frac 12\cdot AC\cdot BY=AX\cdot BY\implies$](http://latex.artofproblemsolving.com/a/a/f/aafc39d631827c2a71143657419510edd1ddba16.png)
.Since
and using the relations from 
obtain that 
, i.e. 
. Since is the bisector of
obtain that 
the ray is the bisector of .PP2. Let
be a triangle . Prove that 
.Proof 1 (metric). Suppose w.l.o.g.
. Thus, 
, what is falsely.An easy extension. Let
be a triangle . Prove that 
we have 
.PP3. In
with 
exists 
so that 
. Find .Lemma. In a quadrilateral
given 
and 
, 
. Find 
in terms of 
.
, 
.Proof 1 (synthetic). Denote
. Let the isosceles triangle 
so that 
doesn't separate 
, 
and 
. Thus,
.Proof 2 (trigonometric). Using
and the Sinus' theorem in and 
obtain that 
.Therefore,
and 
. Nice problem !PP4. the point
is taken inside of so that 
. Find 
.Proof 1 (synthetic). Denote
, 
so that 
and 
. Since 
(exterior), 
and
(exterior) obtain that 
, i.e. the quadrilateral 
is cyclic. Since 
is isosceles obtain that the quadrilateral
is cyclic, so 
. Finally 
and 
.Proof 2 (trigonometric). Denote
. Apply an well-known property 
.PP5. Let
be a quadrilateral so that is -right isosceles, 
and 
. Find 
.Proof 1 (synthetic).
is equilateral 
.Proof 2 (trigonometric). Denote
and apply an well-known property 
.PP6. In a
point 
is taken on 
so that 
and 
find 
Proof 1 (synthetic). Denote
so that 
and 
and 
. Since 
is also isosceles, yielding 
. Hence 
.From the relations 
and 
yield
. Thus 
is isosceles. Hence 
.Proof 2 (trigonometric).
.Otherwise.
. PP7. Let
be a triangle, where 
, 
and 
. Suppose that exists such that 
and 
. Find .Proof 1 (synthetic). Draw
and let 
be the midpoint of 
. Then is circumcenter of 
and is circumcenterof
. On the other hand, quadrilateral 
is cyclic, so 
is right angled isosceles, and 
.Proof 2 (trigonometric). Apply theorem of Sinus :
.
PP8. Let be a triangle with the incenter 
. Denote 
and 
. Prove that if 
.Proof 1 (trigonometric). I"ll apply well-known property in quadrilateral
![\[\boxed{\ \begin{array}{ccc}
\sin\left(A+\frac C2\right)\sin\frac B2\sin\frac C2\sin\left(60^{\circ}-\frac A2\right) & = & \sin\frac B2\sin\frac C2\sin\left(A+\frac B2\right)\sin 30^{\circ}\\\\
\sin\left(A+\frac C2\right)\sin\left(60^{\circ}-\frac A2\right) & = & \sin\left(A+\frac B2\right)\sin 30^{\circ}\\\\
2\sin\left(A+\frac C2\right)\sin\left(60^{\circ}-\frac A2\right) & = & \sin\left(A+\frac B2\right)\\\\
2\cos\frac {A-B}{2}\cos\left(30^{\circ}+\frac A2\right) & = & \cos\frac {A-C}{2}\\\\
\cos\frac {2A-B+60^{\circ}}{2}+\cos\frac {B+60^{\circ}}{2} & = & \cos\frac {A-C}{2}\\\\
\cos\frac {B+60^{\circ}}{2} & = & \cos\frac {A-C}{2}-\cos\frac {2A-B+60^{\circ}}{2}\\\\
\cos\frac {B+60^{\circ}}{2} & = & 2\sin \frac {3A-B-C+60^{\circ}}{4}\sin\frac {A+C-B+60^{\circ}}{4}\\\\
\cos\frac {B+60^{\circ}}{2} & = & 2\sin \frac {4A-120^{\circ}}{4}\sin\frac {240^{\circ}-2B}{4}\\\\
\sin\left(60^{\circ}-\frac B2\right) & = & 2\sin (A-30^{\circ} )\sin\left(60^{\circ}-\frac B2\right)\end{array}\ }\]](http://latex.artofproblemsolving.com/e/5/1/e51daf579f96b1682fc88aab4ee729f5fa99916c.png)
In conclusion,
. PP9. Let be an interior point of for which 
. Prove that is isosceles.Proof 1 (trigonometric). Denote
. Thus, 
. I shall aplly the Ceva's theorem (in the trigonometrical form):
.Proof 2 (synthetic). Reflect
across 
and name it . Then 
is equilateral and 
. Denote 
. Prove easily that 
is cyclically. Therefore
. Note that 
. Since 
obtain that 
.PP10. Let
be a triangle and 
, 
be two points so that 
,
, 
and 
. Find 
.Proof 1 (trigonometric). Denote
. Thus, 
. Apply in 
the trigonometrical form of the Ceva's theorem:
.Proof 2 (metric). Prove easily that
. Hence the 
-bisector in 
and the 
-bisector in 
are concurrently in
. Denote 
so that 
and 
. Observe that the points , belong to the circle with diameter ![$ [KL]$](http://latex.artofproblemsolving.com/a/e/f/aef79871691703ff93e83466aa0c5e9bcc1fe14f.png)
.Denote
. Since 
and 
, 
obtain 
, i.e. 
.Extension. Let
, 
so that 
Prove that 
PP11. Let
and be points on the sides 
and of a triangle such that the lines 
and 
are the interior angle bisectors of theangles
and 
respectively. Given that 
and 
. Determine the angles 
, 
, 
of triangle .Proof. Prove easily that
and 
. Hence and 
are both acute. By the sine theorem for the 
,
obtain that 
. Thus, 
. Using 
obtain that 
. In conclusion, 
and 
.=======================================================================================================
, what is truly.Very nice ! Thank you, Yetti.
PP12. Let
be a triangle with 
. Denote the midpoint of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and suppose that 
. Find .Proof 1(trigonometric). Observe that
. Therefore,![\[\boxed{\begin{array}{c}
1=\frac {DA}{DC}=\frac {BA}{BC}\cdot\frac {\sin \widehat{DBA}}{\sin \widehat{DBC}}=\frac {\sin C}{\sin 3C}\cdot\frac {\sin (45^{\circ}+3C)}{\sin (45^{\circ}-C)}\\\\
\sin C\sin (45^{\circ}+3C)=\sin 3C\sin (45^{\circ}-C)\\\\
\cos (45^{\circ}+2C)-\cos (45^{\circ}+4C)=\cos (45^{\circ}-4C)-\cos (45^{\circ}+2C)\\\\
2\cos (45^{\circ}+2C)=\cos (45^{\circ}+4C)+\cos (45^{\circ}-4C)\\\\
\sqrt 2\cdot\cos (45^{\circ}+2C)=\cos 4C\\\\
\cos 2C-\sin 2C=\cos 4C=\cos^22C-\sin^22C\\\\
\cos 2C=\sin 2C\ \ \vee\ \ \cos 2C+\sin 2C=1\\\\
C=\frac {\pi}{8}\ \ \vee\ \ C=\frac {\pi}{4}\\\\
C=\frac {\pi}{8}\end{array}}\]](http://latex.artofproblemsolving.com/b/7/5/b758cfca8755fd06361a9befaf306e655e395bbe.png)
.Proof 2 (trigonometric). Denote
and 
. Thus 
and 
, 
. Therefore,
. PP13. Let
and an interior so that 
. Find 
. Proof 1 (synthetic). Construct the symmetrical point
of w.r.t. the line 
. Observe that isa cyclical quadrilateral,
is an equilateral triangle and 
. Finally, 
.Proof 2 (trigonometric). Denote
and apply the trigonometric form of the Ceva's theorem for the point 
![\[\boxed{\begin{array}{c}
\sin (40^{\circ}+x)\sin 10^{\circ}\sin 30^{\circ}=\sin (x-60^{\circ})\sin 40^{\circ}\sin 20^{\circ}\\\\
\sin (40^{\circ}+x)=4\sin (x-60)\sin 40^{\circ}\cos 10^{\circ}\\\\
\sin (40^{\circ}+x)=2\sin (x-60^{\circ})\left(\sin 50^{\circ}+\frac 12\right)\\\\
\sin (40^{\circ}+x)=2\cos 40^{\circ}\sin (x-60^{\circ})+\sin (x-60^{\circ})\\\\
\sin (40^{\circ}+x)=\sin (x-20^{\circ})-\sin (100^{\circ}-x)+\sin (x-60^{\circ})\\\\
\sin (40^{\circ}+x)+\sin (80^{\circ}+x)=\sin (x-20^{\circ})+\sin (x-60^{\circ})\\\\
\sin (60^{\circ}+x)\cos 20^{\circ}=\sin (x-40^{\circ})\cos 20^{\circ}\\\\
\sin (60^{\circ}+x)=\sin (x-40^{\circ})\\\\
(60^{\circ}+x)+(x-40^{\circ})=180^{\circ}\\\\
x=80^{\circ}\end{array}}\]](http://latex.artofproblemsolving.com/4/f/9/4f9266c9abc80a9ce2450502f2680d289cb66471.png)
PP14 (Langley's problem). Let an
-isosceles with 
and the points 
Find the value of the angle 
See and here.Proof 1. Let
so that 
Hence 
because 
Since 
obtain 
From
and 
obtain 
From obtain 
In conclusion 
and 
An easy extension.. Let
with 
and the points 
Find the value of the angle Proof. Let
so that 
From and 
obtain that 
and 
Hence
From 
obtain that 
Thus, 
i.e. 
is equilateral. From 
obtain that
and 
In conclusion, 
and 
This post has been edited 162 times. Last edited by Virgil Nicula, Apr 13, 2017, 11:03 AM
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