Taking antipode on isosceles triangle's circumcenter

by Nuran2010, May 11, 2025, 11:46 AM

In isosceles triangle, the condition $AB=AC>BC$ is satisfied. Point $D$ is taken on the circumcircle of $ABC$ such that $\angle CAD=90^{\circ}$.A line parallel to $AC$ which passes from $D$ intersects $AB$ and $BC$ respectively at $E$ and $F$.Show that circumcircle of $ADE$ passes from circumcenter of $DFC$.

ISI UGB 2025 P7

by SomeonecoolLovesMaths, May 11, 2025, 11:28 AM

Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -8.44, xmax = 9.4, ymin = -5.34, ymax = 5.46;  /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0); pen ttqqqq = rgb(0.2,0,0); 

draw((-1.14,4.36)--(-4.46,-1.28)--(3.32,-2.78)--cycle, linewidth(2) + zzttqq); 
 /* draw figures */
draw((-1.14,4.36)--(-4.46,-1.28), linewidth(2) + ttqqqq); 
draw((-4.46,-1.28)--(3.32,-2.78), linewidth(2) + ttqqqq); 
draw((3.32,-2.78)--(-1.14,4.36), linewidth(2) + ttqqqq); 
draw((-1.4789869126960866,-1.8547454538503692)--(-3.0139955173701907,1.1764654463952184), linewidth(2) + linetype("4 4"),EndArrow(6)); 
draw((-3.0139955173701907,1.1764654463952184)--(0.7266612107598007,1.3716679271692873), linewidth(2) + linetype("4 4"),EndArrow(6)); 
draw((0.726661210759801,1.3716679271692873)--(-1.4789869126960864,-1.8547454538503694), linewidth(2) + linetype("4 4"),EndArrow(6)); 
 /* dots and labels */
dot((-1.14,4.36),dotstyle); 
label("$A$", (-1.06,4.56), NE * labelscalefactor); 
dot((-4.46,-1.28),dotstyle); 
label("$B$", (-4.74,-1.14), NE * labelscalefactor); 
dot((3.32,-2.78),dotstyle); 
label("$C$", (3.4,-2.58), NE * labelscalefactor); 
dot((-1.4789869126960866,-1.8547454538503692),dotstyle); 
label("$D$", (-1.6,-2.3), NE * labelscalefactor); 
dot((0.726661210759801,1.3716679271692873),dotstyle); 
label("$E$", (0.8,1.58), NE * labelscalefactor); 
dot((-3.0139955173701907,1.1764654463952184),dotstyle); 
label("$F$", (-3.44,1.14), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]
This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, 4 hours ago

ISI UGB 2025 P2

by SomeonecoolLovesMaths, May 11, 2025, 11:16 AM

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.
This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, 4 hours ago

Drawing equilateral triangle

by xeroxia, May 11, 2025, 7:14 AM

Equilateral triangle $ABC$ is given. Let $M_a$ and $M_c$ be the midpoints of $BC$ and $AB$, respectively.
A point $D$ on segment $BM_c$ is given. Draw equilateral $\triangle DEF$ such that $E$ is on $BC$ and $F$ is on $AM_a$.
This post has been edited 1 time. Last edited by xeroxia, Today at 7:15 AM

The familiar right angle from the orthocenter

by buratinogigle, May 11, 2025, 5:33 AM

Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
Attachments:

Calculus

by youochange, May 10, 2025, 2:38 PM

Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$

find angle

by TBazar, May 8, 2025, 6:57 AM

Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$

JBMO Shortlist 2021 G5

by Lukaluce, Jul 2, 2022, 9:13 PM

Let $ABC$ be an acute scalene triangle with circumcircle $\omega$. Let $P$ and $Q$ be interior points of the sides $AB$ and $AC$, respectively, such that $PQ$ is parallel to $BC$. Let $L$ be a point on $\omega$ such that $AL$ is parallel to $BC$. The segments $BQ$ and $CP$ intersect at $S$. The line $LS$ intersects $\omega$ at $K$. Prove that $\angle BKP = \angle CKQ$.

Proposed by Ervin Macić, Bosnia and Herzegovina
This post has been edited 1 time. Last edited by Lukaluce, Jul 2, 2022, 10:08 PM

D'B, E'C and l are congruence.

by cronus119, May 22, 2022, 7:03 PM

Let $E$ and $F$ on $AC$ and $AB$ respectively in $\triangle ABC$ such that $DE || BC$ then draw line $l$ through $A$ such that $l || BC$ let $D'$ and $E'$ reflection of $D$ and $E$ to $l$ respectively prove that $D'B, E'C$ and $l$ are congruence.

a set of $9$ distinct integers

by N.T.TUAN, Mar 31, 2007, 5:13 AM

Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube.

"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein

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