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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by Philippine 2025
sqing   1
N 3 minutes ago by sqing
Source: Own
Let $ a,b,c $ be real numbers . Prove that
$$\frac{(a-1)(b-1)(c-1)}{(a^2+1)(b^2+1)(c^2+1)} \ge -\frac{7+5\sqrt 2}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+2)(b^2+2)(c^2+2)} \ge -\frac{5+3\sqrt 3}{32}$$
1 reply
1 viewing
sqing
11 minutes ago
sqing
3 minutes ago
My Unsolved Problem
ZeltaQN2008   1
N 13 minutes ago by Funcshun840
Source: IDK
Let triangle \(ABC\) be inscribed in circle \((O)\). Let \((I_a)\) be the \(A\)-excircle of triangle \(ABC\), which is tangent to \(BC\), the extension of \(AB\), and the extension of \(AC\). Let \(BE\) and \(CF\) be the angle bisectors of triangle \(ABC\). Let \(EF\) intersect \((O)\) at two points \(S\) and \(T\).

a) Prove that circle \((O)\) bisects the segments \(I_aT\) and \(I_aS\).
b) Prove that \(S\) and \(T\) are the points of tangency of the common external tangents of circles \((O)\) and \((I_a)\) .

1 reply
ZeltaQN2008
Yesterday at 3:07 PM
Funcshun840
13 minutes ago
Random walk
EthanWYX2009   2
N 28 minutes ago by mathematical-forest
As shown in the graph, an ant starts from $4$ and walks randomly. The probability of any point reaching all adjacent points is equal. Find the probability of the ant reaching $1$ without passing through $6.$
2 replies
EthanWYX2009
3 hours ago
mathematical-forest
28 minutes ago
Inspired by old results
sqing   2
N 35 minutes ago by pooh123
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
2 replies
sqing
6 hours ago
pooh123
35 minutes ago
No more topics!
Circles tangent to BC at B and C
MarkBcc168   9
N Mar 30, 2025 by channing421
Source: ELMO Shortlist 2024 G3
Let $ABC$ be a triangle, and let $\omega_1,\omega_2$ be centered at $O_1$, $O_2$ and tangent to line $BC$ at $B$, $C$ respectively. Let line $AB$ intersect $\omega_1$ again at $X$ and let line $AC$ intersect $\omega_2$ again at $Y$. If $Q$ is the other intersection of the circumcircles of triangles $ABC$ and $AXY$, then prove that lines $AQ$, $BC$, and $O_1O_2$ either concur or are all parallel.

Advaith Avadhanam
9 replies
MarkBcc168
Jun 22, 2024
channing421
Mar 30, 2025
Circles tangent to BC at B and C
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO Shortlist 2024 G3
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MarkBcc168
1595 posts
#1 • 3 Y
Y by Rounak_iitr, GeoKing, farhad.fritl
Let $ABC$ be a triangle, and let $\omega_1,\omega_2$ be centered at $O_1$, $O_2$ and tangent to line $BC$ at $B$, $C$ respectively. Let line $AB$ intersect $\omega_1$ again at $X$ and let line $AC$ intersect $\omega_2$ again at $Y$. If $Q$ is the other intersection of the circumcircles of triangles $ABC$ and $AXY$, then prove that lines $AQ$, $BC$, and $O_1O_2$ either concur or are all parallel.

Advaith Avadhanam
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MarkBcc168
1595 posts
#2 • 2 Y
Y by GeoKing, farhad.fritl
Solution
This post has been edited 1 time. Last edited by MarkBcc168, Jun 22, 2024, 3:47 PM
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DottedCaculator
7357 posts
#3 • 1 Y
Y by GeoKing
By Monge on $\omega_1$, $\omega_2$, and the circle passing through $A$ tangent to the line through $A$ parallel to $BC$ with radius $0$, we need to show $\frac{BQ}{QC}=\frac{BX}{CY}$, which follows from spiral similarity as $\triangle QBX\sim\triangle QCY$.
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BlizzardWizard
108 posts
#4 • 1 Y
Y by GeoKing
Let $b=-1$, $c=1$, $x=-1+s$, and $y=1+t$.
We have $\frac{a+1}s,\frac{a-1}t\in\mathbb R$, so $\frac{\overline sa+\overline s-s}s=\overline a=\frac{\overline ta-\overline t+t}t$. Solving, $a=\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}$.
We have $q=\frac{cx-by}{c+x-b-y}=\frac{s+t}{s-t}$.
Also, letting $q_1=o_1+1$ and $q_2=o_2-1$, we have $(q_1-s)(\overline{q_1}-\overline s)=q_1\overline{q_1}$; substituting $\overline{q_1}=-q_1$ and solving gives $q_1=\frac{s\overline s}{\overline s-s}$.
The exsimilicenter, which is the intersection of $O_1O_2$ and $BC$, is given by
$z=\frac{o_1\overline{o_2}-\overline{o_1}o_2}{o_1-o_2-\overline{o_1}+\overline{o_2}}=\frac{(q_1-1)(-q_2+1)-(-q_1-1)(q_2+1)}{(q_1-1)-(q_2+1)-(-q_1-1)+(-q_2+1)}=\frac{2(q_1+q_2)}{2(q_1-q_2)}=\frac{s\overline s(\overline t-t)+t\overline t(\overline s-s)}{s\overline s(\overline t-t)-t\overline t(\overline s-s)}$.
We are done because this equals the intersection of $AQ$ and $BC$,
$\frac{a\overline q-\overline aq}{a-q-\overline a+\overline q}=\frac{\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}\cdot\frac{\overline s+\overline t}{\overline s-\overline t}-\frac{-2\overline{st}+s\overline t+\overline st}{\overline st-s\overline t}\cdot\frac{s+t}{s-t}}{\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}-\frac{s+t}{s-t}-\frac{-2\overline{st}+s\overline t+\overline st}{\overline st-s\overline t}+\frac{\overline s+\overline t}{\overline s-\overline t}}=\frac{(s-t)(2st-s\overline t-\overline st)(\overline s+\overline t)-(\overline s-\overline t)(-2\overline{st}+s\overline t+\overline st)(s+t)}{(s-t)(\overline s-\overline t)((2st-2s\overline t-2\overline st+2\overline{st}))+(\overline st-s\overline t)((s-t)(\overline s+\overline t)(\overline s-\overline t)(s+t))}=\frac{2(s-t-\overline s+\overline t)(s\overline s(\overline t-t)+t\overline t(\overline s-s))}{2(s-t-\overline s+\overline t)(s\overline s(\overline t-t)-t\overline t(\overline s-s))}=z$
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CyclicISLscelesTrapezoid
372 posts
#5 • 1 Y
Y by GeoKing
Lemma: Let $A$, $B$, $C$, and $D$ be points on a circle, and let $\overline{AB}$ and $\overline{CD}$ intersect at $X$. Then, $\tfrac{CX}{XD}=\tfrac{CA}{AD} \cdot \tfrac{CB}{BD}$.

Proof: We have
\[\frac{CX}{XD}=\frac{[ABC]}{[ABD]}=\frac{\frac{1}{2} CA \cdot CB \cdot \sin \angle ACB}{\frac{1}{2} DA \cdot DB \cdot \sin \angle ADB}=\frac{CA}{AD} \cdot \frac{CB}{BD},\]as desired. $\square$

Let $\overline{AQ}$ and $\overline{BC}$ intersect at $Z$. Notice that $QBX \sim QCY$, so $\tfrac{QB}{QC}=\tfrac{BX}{CY}$. By the lemma, we have
\[\frac{ZB}{ZC}=\frac{BA}{AC} \cdot \frac{BQ}{QC}=\frac{AB}{AC} \cdot \frac{BX}{CY}=\frac{AB}{AC} \cdot \frac{2BO_1 \cdot \cos \angle O_1BA}{2CO_2 \cdot \cos \angle O_2CA}=\frac{AB}{AC} \cdot \frac{BO_1}{CO_2} \cdot \frac{\sin \angle ABC}{\sin \angle ACB}=\frac{BO_1}{CO_2}.\]Thus, we have $ZBO_1 \sim ZCO_2$, so $Z$, $O_1$, and $O_2$ are collinear, as desired. $\square$

Remark: This problem has 3 main components: the circles and intersections, the definition of $Q$, and the definition of $Z$. These can be dealt with mostly independently of each other: using the lemma, length relations involving $Z$ decompose to those involving $Q$, which decompose by spiral similarity to those involving $X$ and $Y$, which are easily calculated.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Jul 9, 2024, 6:23 PM
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Pyramix
419 posts
#6 • 1 Y
Y by GeoKing
Here's a solution using degree 1 moving points / spiral similarity.

It is well-known that spiral similarities are linear maps. In fact, the following lemma is true.

Lemma. (well-known) Let $l_1,l_2$ denote two lines intersecting in point $Z$. Let $O$ be a fixed point. If $X_t$ is a linearly moving point on $l_1$ and $Y_t=l_2\cap (AOX_t)$, then $\frac{X_0X_t}{Y_0Y_t}$ is constant.
Proof. $O$ is the center of the spiral similarity sending $X_t$ to $Y_t$. Hence, $\triangle QX_0X_t\sim QY_0Y_t$, which means $\frac{X_0X_t}{Y_0Y_t}=\frac{QX_0}{QY_0}$, which is indeed a constant. $\blacksquare$

Fix point $Q$ on $(ABC)$. Choose a point $X$ line $AB$ and define $Y=(AQX)\cap AC$. Then, $X\rightarrow Y$ is a linear map (spiral similarity centered at $Q$). From our lemma, we may write $\frac{BX}{CY}=\frac{QB}{QC}$.
Since $\omega_1,\omega_2$ are tangent to $BC$, we have $\measuredangle BO_1X=2\measuredangle CBX$ and $\measuredangle CO_2Y=2\measuredangle BCY$. As a result, we have
\[\frac{O_1B}{O_2C}=\frac{BX}{CY}\cdot\frac{\sin(\measuredangle CBA)}{\sin(\measuredangle ABC)}=\frac{QB}{QC}\cdot\frac{AC}{AB}\]Let $T=AQ\cap BC$. For $T,O_1,O_2$ to be collinear, it suffices to show that \[\frac{TB}{TC}=\frac{O_1B}{O_2C}=\frac{QB}{QC}\cdot\frac{AC}{AB},\]which is a well-known identity (sketch: use sine rule in $\triangle TAB,\triangle TAC$). $\blacksquare$
This post has been edited 2 times. Last edited by Pyramix, Jul 9, 2024, 6:21 PM
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Deadline
13 posts
#7
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another way to obtain ratio
This post has been edited 1 time. Last edited by Deadline, Jul 10, 2024, 8:45 AM
Reason: $
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mcmp
53 posts
#8
Y by
Oh bruh why am I so washed at geometry again :censored: my solution is wayyy overcomplicated oops.

So we start by defining $T=\overline{AQ}\cap\overline{BC}$. Note $|\operatorname{Pow}_\omega(T)|=TA\cdot TQ=TB\cdot TC=k^2$, so we can consider an inversion $I$ centred at $T$ with radius $\sqrt{k}$. Let $Y’=I(X)$ and $X’=I(Y)$ under this inversion. Note that $T-X-Y’$ and $T-X’-Y$. However I would like to show that $Y’\in\omega_2$ and $X’\in\omega_1$ for reasons to be disclosed later on. This is not hard; notice that $\omega_1\iff\omega_2$ under $I$ which is clear since they are both tangent to $\overline{BC}$ and $B\iff C$.

Now I can disclose why I wanted $X’$ and $Y’$ to satisfy these properties; it’s quite clear now that under $I$ we must have $\omega_1\iff\omega_2$, so $\overline{O_1O_2}$ also passes through $T$.

Edit: I just realised I proved wayyyy too much again (I proved that five lines concurred at $T$ instead of your normal three). To prove your normal three just take the inversion $I$ centred at $T$ with power $\operatorname{Pow}_\omega(T)$ and then note that $\omega_1\iff\omega_2$, done.
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OMD_MHB
4 posts
#9 • 2 Y
Y by Taha.kh, Radin.AmirAslani
Beautiful problem!
Lemma : In a circle $w$ , we have two lines such that one of them intersect $w$ at points $C$ and $B$ and the other intersect $w$ at points $X$ and $Y$ . That two lines intersect at $Z$ ( $Z$ is out of the circle) . Suppose that $B$ lies between $Z$ and $C$ and $X$ lies between $Z$ and $Y$ .
I claim that $$\frac{XB}{XC} . \frac{YB}{YC} = \frac{ZB}{ZC} $$( It follows from similarity)

Now call $T$ the intersection point of AQ and BC . Now applying the lemma :
$$\frac{AB}{AC} . \frac{QB}{QC} = \frac{TB}{TC} $$Denote the radius of circle tangent to $BC$ at $B$ R1 and the other one R2
So if we show $$\frac{TB}{TC} = \frac{R1}{R2} $$the problem will solve by homothety and that three lines are concur.
By similarity , $$\frac{QB}{QC} = \frac{BX}{CY}$$We know that $$\frac{a}{sin\angle A} = 2R $$So $$\frac{2R1}{2R2} = \frac{BX}{CY} . \frac{sin\angle C}{sin\angle B}$$
And the RHS is equivalent to $$\frac{QB}{QC} . \frac{AB}{AC} $$. So the problem solved

Note that if R1 = R2 that lines will be parallel and we can check it easily .
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channing421
1353 posts
#10
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wow advaith orz
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