Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

G
Topic
First Poster
Last Poster
IMO 2017 Problem 4
Amir Hossein   116
N 3 hours ago by cj13609517288
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
116 replies
Amir Hossein
Jul 19, 2017
cj13609517288
3 hours ago
A sharp one with 3 var
mihaig   10
N 3 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
10 replies
mihaig
May 13, 2025
mihaig
3 hours ago
Another right angled triangle
ariopro1387   1
N 3 hours ago by lolsamo
Source: Iran Team selection test 2025 - P7
Let $ABC$ be a right angled triangle with $\angle A=90$. Point $M$ is the midpoint of side $BC$ And $P$ be an arbitrary point on $AM$. The reflection of $BP$ over $AB$ intersects lines $AC$ and $AM$ at $T$ and $Q$, respectively. The circumcircles of $BPQ$ and $ABC$ intersect again at $F$. Prove that the center of the circumcircle of $CFT$ lies on $BQ$.
1 reply
ariopro1387
Yesterday at 4:13 PM
lolsamo
3 hours ago
four points lie on a circle
pohoatza   78
N 4 hours ago by ezpotd
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
78 replies
pohoatza
Jun 28, 2007
ezpotd
4 hours ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   2
N 4 hours ago by Stear14
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
2 replies
FishkoBiH
Yesterday at 1:38 PM
Stear14
4 hours ago
Does there exist 2011 numbers?
cyshine   8
N 4 hours ago by TheBaiano
Source: Brazil MO, Problem 4
Do there exist $2011$ positive integers $a_1 < a_2 < \ldots < a_{2011}$ such that $\gcd(a_i,a_j) = a_j - a_i$ for any $i$, $j$ such that $1 \le i < j \le 2011$?
8 replies
cyshine
Oct 20, 2011
TheBaiano
4 hours ago
D1036 : Composition of polynomials
Dattier   1
N 4 hours ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Saturday at 1:52 PM
Dattier
4 hours ago
number sequence contains every large number
mathematics2003   3
N 4 hours ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
4 hours ago
IMO ShortList 2002, geometry problem 2
orl   28
N 4 hours ago by ezpotd
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
28 replies
orl
Sep 28, 2004
ezpotd
4 hours ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   56
N 4 hours ago by ezpotd
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
56 replies
ike.chen
Jul 9, 2023
ezpotd
4 hours ago
Non-linear Recursive Sequence
amogususususus   4
N 5 hours ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
amogususususus
Jan 24, 2025
GreekIdiot
5 hours ago
Find all possible values of BT/BM
va2010   53
N Apr 25, 2025 by ja.
Source: 2015 ISL G4
Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.
53 replies
va2010
Jul 7, 2016
ja.
Apr 25, 2025
Find all possible values of BT/BM
G H J
G H BBookmark kLocked kLocked NReply
Source: 2015 ISL G4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bjump
1035 posts
#48
Y by
DottedCaculator wrote:
[asy]
unitsize(1cm);
pair A, B, C, M, P, Q, T;
A=(2,2.5sqrt(5));
B=(0,0);
C=(8,0);
M=(B+C)/2;
pair X;
X=(1,0);
P=intersectionpoints(A--B,circumcircle(A,M,X))[1];
Q=intersectionpoints(A--C,circumcircle(A,M,X))[1];
T=P+Q-A;
draw(A--B--C--A--P--T--Q--A);
draw(circumcircle(A,M,X));
draw(circumcircle(A,B,C));
label("$B$", A, N);
label("$A$", B, SW);
label("$C$", C, SE);
label("$M$", M, S);
label("$P$", P, W);
label("$Q$", Q, E);
label("$T$", T, S);
label("$X$", X, NE);
[/asy]

bashy bary
less bashy
best solution

How do you learn to make diagrams like this also this got liked by tapir orz orz orz
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
fuzimiao2013
3311 posts
#50 • 2 Y
Y by Mango247, Mango247
oh, great, here we go again. i am now thoroughly convinced that in this white wasteland of geometry, bary is just problem suicide - your last resort, unless it's not.

We use barycentric coordinates.

Preliminaries:
Let $ABC$ be the reference triangle, and $P = (p, 1-p, 0), Q = (0, 1-q, q)$. The circle through $B, M$ is characterized by \begin{align*}-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+wz) &= 0,\\ u+w &= \frac{b^2}{2}.\end{align*}So after plugging in $P$ and $Q$ and heavily simplifying, \begin{align*}c^2p &= c^2-u \\ a^2q &= a^2-w.\end{align*}
Actual Problem:
By distance formula, we can now proceed:
\begin{align*}\frac{BT^2}{BM^2} &= \frac{a^2p-wp+a^2q-wq+c^2p-up+c^2q-uq-b^2pq}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{(a^2-w+c^2-u)(p+q)+a^2q-a^2pq-a^2q^2+c^2p-c^2p^2-c^2pq}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)(p+q)+(a^2q+c^2p)(1-p-q)}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)(p+q)+\left(a^2+c^2-\frac{b^2}{2}\right)(1-p-q)}{\frac 12\left(a^2+c^2-\frac{b^2}{2}\right)} \\
&= \frac{\left(a^2+c^2-\frac{b^2}{2}\right)}{\frac 12 \left(a^2+c^2-\frac{b^2}{2}\right)} = 2.
\end{align*}Since length is positive, the only possible value is $\sqrt 2$, done. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Knty2006
50 posts
#51
Y by
G #1/100

Let the midpoint of $BT$ be $N$, Let $PQ$ intersect $AC$ at $D$ and the Miquel point of $APQC$ be $R$

Claim: $DRNM$ cyclic

Proof: Due to the definition of a Miquel point,
$APRD$ cyclic, $QCDR$ cyclic
So, $R$ is the center of the spiral similarity sending $QC$ to $PA$, which means that it is the center of the spiral similarity sending $NM$ to $PA$

Hence, our claim must be true.

Claim: $MB$ tangent to $MNRT$

Proof: $$\angle BMR=\angle BPR$$$$=180-\angle APR$$$$=\angle MDR$$
Hence, our claim must be true.

Using the tangency, we have
$MB^2=(NB)(BT)=\frac{1}{2} BT^2$

So, $$\frac{BT}{MB}=\sqrt{2}$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Taco12
1757 posts
#52 • 1 Y
Y by Mango247
Solved with a hint from Evan Chen's solution.

Let $X$ be the second intersect of $\omega$. Set $T=(p,q,r)$. The parallelogram condition yields $P=(p:q+r:0)$ and $Q=(0:p+q:r)$. Then, we have $$b^2=qc^2+rc^2+a^2p+a^2q.$$By distance formula, we get $BT^2=a^2r+c^2p$, which gives us $BT^2=2BM^2$, so $\sqrt2$ is the only value.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
john0512
4190 posts
#53
Y by
We will use barycentric coordinates. Let $$T=(r,s,t)$$for which $r+s+t=1.$ Then, from the parallelogram, $$P=(r,s+t,0),Q=(0,r+s,t).$$Consider the equation of $(BQMP)$. By putting in $B$, $v=0$. By putting in $P$, we have $u=c^2(1-r)$. By plugging in $Q$, we have $w=a^2(1-t).$ Therefore, the equation of the circle is $$(x+y+z)(c^2(1-r)x+a^2(1-t)z)=a^2yz+b^2xz+c^2xy.$$Plugging in $(1/2,0,1/2)$ for $M$ reveals that $$b^2=2(c^2(1-r)+a^2(1-t)) (*).$$We will return to the starred equation later.

For now, note that $$\overrightarrow{BT}=(r,s-1,t),$$so $$BT^2=-(a^2(s-1)t+b^2rt+c^2r(s-1)).$$Let's expand this: $$-(\mathbf{a^2st+b^2rt+c^2rs}-a^2t-c^2r).$$Since $(r,s,t)$ lies on the circumcircle, the bolded things sum to 0, so this is just $$BT^2=a^2t+c^2r.$$Let's now go back to the starred equation: $$b^2=2(c^2(1-r)+a^2(1-t)).$$Expanding, this becomes $$b^2=2(a^2+c^2-a^2t-c^2r)$$$$b^2=2(a^2+c^2-BT^2)$$$$BT^2=a^2+c^2-\frac{1}{2}b^2.$$It is well known that $$BM^2=\frac{1}{2}a^2+\frac{1}{2}c^2-\frac{1}{4}b^2,$$so our answer is $\sqrt{2}$ and we are done.
This post has been edited 1 time. Last edited by john0512, Jan 26, 2023, 3:42 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ibrahim_K
62 posts
#54
Y by
Similar bary bash:
$A=(1,0,0),B=(0,1,0),C=(0,0,1),M=(\frac{1}{2},0,\frac{1}{2}),T=(m,n,k)$. Then equation of $PT$ and $QT$ is:
$$PT:(n+k)x-m(y+z)=0 \implies P=(m,1-m,0) \qquad QT:(m+n)z-k(x+y)=0 \implies Q=(0,1-k,k)$$Let $(BPMQ):-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+wz)=0.$ By plugging coordinates of $M,P$and $Q$ we get:
$$u+w=\frac{b^2}{2} \qquad c^2m=c^2-u \qquad a^2k=a^2-w$$Also since $T \in (ABC)$ we have $a^2nk+b^2mk+c^2mn=0$. Thus by distance formula:
$$BT^2=-a^2(n-1)k-b^2mk-c^2m(n-1)=-a^2nk-b^2mk-c^2mn+a^2k+c^2m=a^2k+c^2m=a^2+c^2-(u+w)=a^2+c^2-\frac{b^2}{2}$$On the other hand $BM^2=\frac{1}{2}(a^2+c^2-\frac{b^2}{2}).$ Hence $\frac{BT}{BM}=\sqrt 2 \qquad \blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
awesomeming327.
1735 posts
#55 • 1 Y
Y by GeoKing
Let $X$ be $BM$ intersect $(ABC)$. Let $Z$ be center of $BPTQ$ and $Y$ be $BZ$ intersect $(BPQ)$. Let $(ABC)$ and $(BPQ)$ intersect again at $E$ then $E$ is center of spiral similarity taking $ATXCM$ to $PYMQZ$. Clearly, $\angle ZMB$ is the angle of the spin about $E$, since $ZM$ is taken to $MX$. Since $ZY$ is taken to $MT$, $\angle ZTM$ is equal to that. Thus, $BM$ is tangent to $ZMT$ and so the answer is just $\sqrt{2}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
vsamc
3789 posts
#57
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
popop614
272 posts
#58 • 1 Y
Y by GeoKing
Great. The answer is $\sqrt{2}$ only.
Let $E$ and $F$ lie on $BA$ and $BC$ such that $M$ is the $B$-Dumpty point of $BEF$, and let $K$ be the reflection of $B$ over $M$, hence on $(BEF)$.

We now claim that $T$ lies on $EF$. Let $P'$ and $Q'$ denote the images of $P$ and $Q$ under a scale 2 homothety at $B$, so it suffices to prive $EF$ bisects $P'Q'$. But applying phantom points we observe that the intersection of $P'Q'$ and $EF$ is on $(KEP')$ and $(KFQ')$. The result is true after a very brief law of sines computation.

Take an inversion at $B$ with radius $BM\sqrt{2}$. $BEF$ becomes through a line antiparallel to $EF$ through $M$, whence $(E, A)$ amd $(F, C)$ swap. Hence line $EF$ maps to $(ABC)$. But this implies if $T$ lies on said circumcircle, $BT = BM\sqrt{2}$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
642 posts
#59
Y by
Actually trivial by bary. The answer is $\sqrt{2}$ only. We set $A=(1,0,0)$ and etc. Then, $M=(1:0:1)$. Now, let $T=(r,s,t)$ for real numbers $r,s,t$ such that $r+s+t=1$. Now, since $\overline{PT} \parallel \overline{BC}$, $P=(r,t_1,0)$ for some real number $t_1$ which then implies $t_1=s+t$ so $P=(r,s+t,0)$. Similarly, $Q=(0,r+s,t)$. Now, by Stewart's theorem, we can compute $BM^2 = \frac{2a^2-b^2+2c^2}{4}$. We consider the displacement vector $\overrightarrow{BT} = (r,s-1,t)=(r,-(r+t),t)$. Then, by the barycentric distance formula, we have that
\[BT^2 = a^2(r+t)t-b^2rt+c^2(r+t)t = a^2t+c^2r\]where the second equality follows from the fact that $a^2st+b^2rt+c^2rs=0$ since $T$ lies on the circumcircle. Now, since $BPMQ$ is cyclic, we consider the equation of this circle
\[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]where $v=0$ immediately since $B$ lies on the circle. Further, since $P$ lies on the circle we have
\begin{align*}
    -c^2r(s+t) + ur &=0\\
    u &= c^2(s+t)
\end{align*}similarly, we also have that since $Q$ lies on the circle, $w=a^2(r+s)$. Thus, the equation of the circle is simply,
\[-a^2yz-b^2xz-c^2xy+(c^2(s+t)x+a^2(r+s)z)(x+y+z)=0\]which since $M$ lies on this circle implies,
\begin{align*}
    -b^2 + 2(c^2(s+t)+a^2(r+s)) &=0\\
    -b^2 + 2(c^2 (1-r)+ a^2 (1-t)) &= 0\\
    c^2 + a^2 - (a^2t+c^2r) &= \frac{b^2}{2}\\
    a^2t + c^2 r &= \frac{2a^2 - b^2 + 2c^2}{2}
\end{align*}But this is precisely the length of $BT^2$! So, we have that
\[BT^2 = a^2t + c^2 r =  \frac{2a^2 - b^2 + 2c^2}{2} = 2\left(\frac{2a^2-b^2+2c^2}{4}\right) = 2BM^2\]Thus, $\frac{BT}{BM} = \sqrt{2}$ which finishes the problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Aiden-1089
297 posts
#60
Y by
I changed it to being $A$-centered for convenience, $P$ lies on $AB$ and $Q$ lies on $AC$.

Let $(A)$ be the circle with centre $A$ and radius $0$. Denote by $Pow_{\omega}(P)$ as the power of a point $P$ wrt circle $\omega$.
Define $f(X)=Pow_{(A)}(X)-Pow_{(ABC)}(X)$.
By linearity of pop, we have that $f(A)+f(T)=f(P)+f(Q)$. So $$AT^2=PA^2-(-PA \cdot PB)+QA^2-(-QA \cdot QB)=PA \cdot AB + QA \cdot AC.$$Now let $R \neq M$ be the second intersection of $\omega$ with $BC$, then $$PB \cdot AB + QC \cdot AC = BM \cdot BR + CM \cdot CR = \frac{BC^2}{2}.$$It follows that $AT^2+\frac{BC^2}{2}=AB^2+AC^2 \implies AT^2=AB^2+AC^2-\frac{BC^2}{2}$. By Stewart's theorem, $\frac{BC}{2}(AB^2 +AC^2)=BC \cdot (AM^2+ \frac{BC^2}{4}) \implies 2AM^2=AB^2+AC^2-\frac{BC^2}{2}=AT^2$.
Hence $\frac{AT}{AM}=\sqrt{2}$, so we are done. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cj13609517288
1923 posts
#61
Y by
WHAT

We employ barycentric coordinates wrt $ABC$. The equation of the circle is
\[-a^2yz-b^2xz-c^2xy+(ux+wz)(x+y+z)=0\]where $u+w=\frac{b^2}{2}$.

Let $P=(p,1-p,0)$ and $Q=(0,1-q,q)$. Then
\[-c^2p(1-p)+up=0\Longrightarrow c^2(1-p)=u\]\[-a^2q(1-q)+wq=0\Longrightarrow a^2(1-p)=w\]Therefore,
\[c^2(1-p)+a^2(1-q)=\frac{b^2}{2}\Longrightarrow c^2p+a^2q=c^2+a^2-\frac{b^2}{2}.\]Also, note that $T=(p,1-p-q,q)$, so
\[a^2(1-p-q)q+b^2pq+c^2(1-p-q)p=0\]\[-a^2(-p-q)q-b^2pq-c^2(-p-q)p=a^2q+c^2p=c^2+a^2-\frac{b^2}{2}.\]That looks suspiciously like the distance formula! Indeed, we get that
\[BT=\sqrt{c^2+a^2-\frac{b^2}{2}}.\]Also, it is well known that
\[BM=\frac{1}{\sqrt2}\cdot\sqrt{c^2+a^2-\frac{b^2}{2}}.\]Therefore, $\frac{BT}{BM}=\boxed{\sqrt2}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SomeonesPenguin
129 posts
#62 • 1 Y
Y by zzSpartan
bary
This post has been edited 1 time. Last edited by SomeonesPenguin, Nov 7, 2024, 3:41 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eg4334
636 posts
#63
Y by
Use bary wrt to $ABC$. The equation of $\omega$ is $-a^2yz-b^2xz-c^2xy+(ux+wz)(x+y+z) = 0$ for $u+w=\frac{b^2}{2}$. Then its trivial to find that $Q = (0, \frac{w}{a^2}, 1 - \frac{w}{a^2}), P = (1 - \frac{u}{c^2}, \frac{u}{c^2}, 0)$. But $P+Q=B+T$, so $T = (1 - \frac{u}{c^2}, \frac{w}{a^2} + \frac{u}{c^2} - 1, 1 - \frac{w}{a^2})$. For simplicity, let this be $T = (x, y, z)$. Then $a^2yz+b^2xz+c^2xy=0$. But by bary distance formula \begin{align*}
\frac{BT}{BM} &= \sqrt{\frac{-a^2(y-1)z - b^2xz - c^2x(y-1)}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}}} \\
&= \sqrt{\frac{c^2x+a^2z}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}}} \\
&= \sqrt{\frac{c^2+a^2-(u+w)}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}}} \\
&= \frac{a^2+c^2-\frac{b^2}{2}}{\frac{a^2}{2} - \frac{b^2}{4} + \frac{c^2}{2}} \\
&= \boxed{\sqrt{2}}
\end{align*}
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ja.
23 posts
#64 • 1 Y
Y by navi_09220114
Let $N$ be the midpoint of $PQ$. Then, as $P$ moves, the shape of $MPNQ$ is constant, and the locus of $P$ is a line so the locus of $N$ Is a line. Then, the locus of $T$ must be a line too. Let $\omega_1$ be the circle through $M$ tangent to $BC$ at $B$, and $\omega_2$ is tangent to $BA$ at $B$. Take $P=B$, $Q=B$ gives that $T$ lies on line $DE$ where $D$ is the intersection of $\omega_1$ with $BA$ and $E$ is the intersection of $\omega_2$ with $BC$.

Let $B'$ be the reflection of $B$ in $M$, then $\angle{AB'M}=\angle{MBC}=\angle{ADM}$ hence $AMB'D$ cyclic, then $BA\cdot BD=2BM^2$. Similarly, we will have $BC\cdot BE=2BM^2$.

Finally, consider the inversion at $B$ with radius $BM\sqrt2$, then $(BAC)$ is swapped with $DE$, so $T$ is fixed. Therefore, $BT=BM\sqrt2$.
This post has been edited 1 time. Last edited by ja., Apr 25, 2025, 8:09 AM
Z K Y
N Quick Reply
G
H
=
a