find the value of an angles

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

find the value of an angles

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 1st National Women´s Contest of Mexican Mathematics Olympiad 2022, problem 2 teams

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

241 posts

AlanLG

#1 h Jul 23, 2023, 11:23 PM • 2 Y

Y by Rounak_iitr, FrancoGiosefAG

Consider $\triangle ABC$ an isosceles triangle such that $AB = BC$ . Let $P$ be a point satisfying

$\angle ABP = 80^\circ, \angle CBP = 20^\circ, \textrm{and} \hspace{0.17cm} AC = BP$
Find all possible values of $\angle BCP$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1135 posts

#2 h Jul 23, 2023, 11:43 PM

Y by

If $P$ is not within the angle formed by $AB$ and $BC$ that is part of triangle $ABC$ , $\angle ABC = 60^{\circ}$ so $\triangle ABC$ is equilateral. Then, $\triangle BPC$ is isosceles so we get $80^{\circ}$ .

second case wrong

This post has been edited 2 times. Last edited by sixoneeight, Jul 25, 2023, 4:37 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

49 posts

#3 h Jul 25, 2023, 12:32 AM

Y by

As in #2, if $P$ and $A$ are in different half-planes w.r.t the line $BC$ , $\angle{ABC}=\angle{ABP}-\angle{CBP}=60^\circ$ , which shows that $\triangle {ABC}$ is equilateral. Then $BP=AC=BC$ , so $\triangle {BPC}$ is isosceles, with $\angle{CBP} = 20^\circ$ . We obtain $\angle{BCP}=80^\circ$ .

Instead, if $P$ and $A$ are in the same half-plane w.r.t the line $BC$ , it can be easily seen that ray $[BP$ must lie in the interior of the angle $\angle{ABC}$ . Therefore, $\angle{ABC}=\angle{ABP}+\angle{CBP}=100^\circ$ , whence the angles of $\triangle {ABC}$ are $100^\circ-40^\circ-40^\circ$ . Let $BP\cap AC=\{E\}$ and let $F$ be on the segment $[AC]$ such that lines $BE$ and $BF$ are isogonal (and thus isotomic since $BA=BC$ ). We get $CE=FA$ . By symmetry, we have $BE=BF$ . Moreover, $\angle{EBF}=100^\circ-2\cdot 20^\circ=60^\circ$ . Consequently, $\triangle{BEF}$ is equilateral and $BE=EF$ . It follows that $BP-BE=AC-EF\Longleftrightarrow PE=CE+FA$ , which rewrites as $PE=2\cdot CE$ . Now look at triangle $\triangle{PCE}$ : $\angle{CEP}=60^\circ$ and $PE=2\cdot CE$ . This is a well-known configuration in which $\angle{PCE}$ turns to be $90^\circ$ ; if one is not familiar with said configuration, one may take the reflection $E'$ of $E$ w.r.t $C$ and effortlessly prove that $\triangle{EPE'}$ is equilateral in order to reach the same conclusion in regards to $\angle{PCE}$ .

Finally, since $\angle{PCA}+\angle{BCA}=90^\circ+40^\circ$ , we conclude that $\angle{BCP}=130^\circ$ .

Attachments:

This post has been edited 2 times. Last edited by speedymind22, Jul 25, 2023, 5:16 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9 posts

#4 h Mar 26, 2025, 9:58 AM

Y by

We have two cases: when $P$ is inside the angle $\angle ABC$ and when it is not. If $P$ is outside angle $\angle ABC$ , then $\angle ABC=60^\circ$ , so $\triangle ABC$ is equilateral and $BP=AC=BC$ , so $\angle BCP=80^\circ$ .Now let's work with $P$ inside angle $ABC$ .

Synthetic Solution: Let $D$ the point such that $\triangle ACD$ is equilateral and $B,D$ an opposite sides od $AC$ . It's clare that $BD$ is perpendicular bisector of $AC$ , so $\angle ADB=30^\circ$ and $\angle DBP=50^\circ-20^\circ=30^\circ$ , hence $AD \| BP$ . Since $AD=AC=BP$ , $ABPD$ is a parallelogram. Then $\triangle BPD \cong \triangle DAB \cong \triangle DCB$ , hence $BCPD$ is an isosceles trapezoid. Finally, $\angle BCP=180^\circ-\angle DBC=180^\circ-50^\circ=130^\circ$ .

Trigo Solution: Now triangle $ABC$ has angle $40^\circ-100^\circ-40^\circ$ . Let $x = \angle BPC$ , hence $\angle BCP=160^\circ-x$ . By Sin Law's
$\frac{BP}{BC}=\frac{AC}{BC} \Longrightarrow \frac{\sin(160^\circ-x)}{\sin(x)}=\frac{\sin(100^\circ)}{\sin(40^\circ)}=\frac{\sin(80^\circ)}{\sin(40^\circ)}=2\sin(50^\circ)=\frac{\sin(50^\circ)}{\sin(30^\circ)}=\frac{\sin(130^\circ)}{\sin(30^\circ)}.$ By "La Hack", $x=30^\circ$ . Finally, $\angle BCP=160^\circ-\angle BPC=160^\circ-30^\circ=130^\circ$ .

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4378 posts

#5 h Mar 27, 2025, 7:42 PM

Y by

My synthetic solution at my blog https://stanfulger.blogspot.com/2025/03/problem-aops-httpsartofproblemsolvingco_27.html.

Best regards,
sunken rock

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a