Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Stay ahead of learning milestones! Enroll in a class over the summer!
geometry
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
ISI UGB 2025 P2
SomeonecoolLovesMaths   2
N 12 minutes ago by SomeonecoolLovesMaths
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2 C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
2 replies
SomeonecoolLovesMaths
5 hours ago
SomeonecoolLovesMaths
12 minutes ago
J
H
angle chasing in RMO, cyclic ABCD, 2 circumcircles, incenter, right wanted
parmenides51   5
N 14 minutes ago by Krishijivi
Source: CRMO 2015 region 1 p1
In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$ . If $X$ is the incentre of triangle $ABY$ , show that $\angle CAD = 90^o$.
5 replies
parmenides51
Sep 30, 2018
Krishijivi
14 minutes ago
J
No more topics!
H
J
H
The incircle problem
danil_e   1
N Dec 1, 2023 by ancamagelqueme
Given triangle \(ABC\) inscribed in circle \((O)\) and circumscribed about circle \((I)\). A circle passing through \(B\) and \(C\) is tangent to \((I)\) at \(N_a\) and intersects \(AB\) and \(AC\) at \(A_c\) and \(A_b\) respectively. Similarly, define \(B_c, B_a, N_b\) and \(C_b, C_a, N_c\) correspondingly. Let \(XYZ\) be the triangle formed by the radical axis of circles \((N_aBC)\), \((N_bCA)\), \((N_cBA)\) (as shown in the figure); \(MNP\) is the triangle formed by the intersection of lines \(A_cC_a\), \(B_cC_b\), \(A_bB_a\) (as shown in the figure).

a) Prove that: Triangle \(XYZ\) and triangle \(MNP\) are in perspective axially.
b) Let \(S\) be the point of concurrence of \(XM\), \(YN\), \(ZP\). Prove that: \(S\), \(I\), \(O\) are collinear.
1 reply
danil_e
Dec 1, 2023
ancamagelqueme
Dec 1, 2023
J
The incircle problem
G H J
Reveal topic tags
geometry
incircle
L
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
danil_e
25 posts
danil_e
#1 h  Dec 1, 2023, 4:23 PM
Y by
Given triangle \(ABC\) inscribed in circle \((O)\) and circumscribed about circle \((I)\). A circle passing through \(B\) and \(C\) is tangent to \((I)\) at \(N_a\) and intersects \(AB\) and \(AC\) at \(A_c\) and \(A_b\) respectively. Similarly, define \(B_c, B_a, N_b\) and \(C_b, C_a, N_c\) correspondingly. Let \(XYZ\) be the triangle formed by the radical axis of circles \((N_aBC)\), \((N_bCA)\), \((N_cBA)\) (as shown in the figure); \(MNP\) is the triangle formed by the intersection of lines \(A_cC_a\), \(B_cC_b\), \(A_bB_a\) (as shown in the figure).

a) Prove that: Triangle \(XYZ\) and triangle \(MNP\) are in perspective axially.
b) Let \(S\) be the point of concurrence of \(XM\), \(YN\), \(ZP\). Prove that: \(S\), \(I\), \(O\) are collinear.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ancamagelqueme
104 posts
ancamagelqueme
#2 h  Dec 1, 2023, 11:44 PM
Y by
The point $S$ is NOT aligned with $I=X_1$ and $O=X_3$. If it is on the line $X_{57}X_{2346}$.

The barycentric coordinates of $S$, with respect to $ABC$, are given by the triangle center function

f(a,b,c)= a (a+b-c) (a-b+c)(a(b+c)-(b-c)^2)((b-c)^6 (b+c)^2 (b^2-3 b c+c^2)-2 (b-c)^4 (4 b^5-b^4 c-11 b^3 c^2-11 b^2 c^3-b c^4+4 c^5) a+(b-c)^2 (27 b^6+4 b^5 c-39 b^4 c^2-112 b^3 c^3-39 b^2 c^4+4 b c^5+27 c^6) a^2-2 (b-c)^2 (24 b^5+55 b^4 c+97 b^3 c^2+97 b^2 c^3+55 b c^4+24 c^5) a^3+2 (21 b^6+56 b^5 c+141 b^4 c^2+140 b^3 c^3+141 b^2 c^4+56 b c^5+21 c^6) a^4-2 b c (49 b^3+167 b^2 c+167 b c^2+49 c^3) a^5-2 (21 b^4+7 b^3 c-34 b^2 c^2+7 b c^3+21 c^4) a^6+2 (24 b^3+35 b^2 c+35 b c^2+24 c^3) a^7+(-27 b^2-41 b c-27 c^2) a^8+8 (b+c) a^9-a^10)
Z K Y
N Quick Reply
G
H
=
a