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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by qrxz17
sqing   6
N 10 minutes ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
6 replies
+1 w
sqing
4 hours ago
sqing
10 minutes ago
Geometry problem
Whatisthepurposeoflife   2
N 14 minutes ago by Whatisthepurposeoflife
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
2 replies
Whatisthepurposeoflife
Yesterday at 1:45 PM
Whatisthepurposeoflife
14 minutes ago
A Sequence of +1's and -1's
ike.chen   36
N 18 minutes ago by maromex
Source: ISL 2022/C1
A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and $$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$
36 replies
1 viewing
ike.chen
Jul 9, 2023
maromex
18 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   1
N 31 minutes ago by grupyorum
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
1 reply
Maths_VC
Tuesday at 7:54 PM
grupyorum
31 minutes ago
No more topics!
Internally tangent with the circumscribed circle
orl   23
N Dec 27, 2023 by shendrew7
Source: IMO LongList, USA 1, IMO 1978, Day 2, Problem 4
In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$
23 replies
orl
Nov 12, 2005
shendrew7
Dec 27, 2023
Internally tangent with the circumscribed circle
G H J
Source: IMO LongList, USA 1, IMO 1978, Day 2, Problem 4
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orl
3647 posts
#1 • 3 Y
Y by Adventure10, Mango247, Rounak_iitr
In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$
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yetti
2643 posts
#2 • 2 Y
Y by Adventure10, Mango247
Denote a = BC, b = AB = AC the triangle sides, r, R the triangle inradius and circumradius and s, $\triangle$ the triangle semiperimeter and area. Let S, T be the tangency points of the incircle (I) with the sides AB, AC. Let K, L be the midpoints of of BC, PQ and M the midpoint of the arc BC of the circumcircle (O) opposite to the vertex A. The isosceles triangles $\triangle AST \sim \triangle APQ \sim \triangle ABC$ are all centrally similar with the homothety center A. The homothety coefficient for the triangles $\triangle AST \sim \triangle ABC$ is $h_{13} = \frac{AS}{AB} = \frac{s - a}{b}$. The homothety coefficient for the triangles $\triangle AST \sim \triangle APQ$ is $h_{12} = \frac{AS}{AP}$. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles $\triangle ASK \sim \triangle APM$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles $\triangle AST \sim \triangle APQ$: $h_{12} = \frac{AS}{AP} = \frac{AK}{AM} = \frac{h}{2R}$, where h = AK is the A-altitude of the triangle $\triangle ABC$. The homothety coefficient of the triangles $\triangle APQ \sim \triangle ABC$ is then $h_{23} = \frac{h_{12}}{h_{13}} = \frac{s - a}{b} \cdot \frac{2R}{h}$ Denote h' = AL the A-altitude of the triangle $\triangle APQ$. Then

$\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}$

$KL = h - h' = h - 2R\ \frac{s - a}{b}$

Substituting $h = \frac{2 \triangle}{a}$, $2R = \frac{ab^2}{2 \triangle}$ and $s - a = \frac{\triangle^2}{s(s - b)^2}$, we get

$KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =$

(substituting 2s = 2b + a and $s - b = \frac a 2$ for an isosceles triangle)

$= \frac{\triangle}{s}\left(\frac{2s}{a} - \frac{ab}{2(s - b)^2}\right) = \frac{\triangle}{s} \left(\frac{2b}{a} + 1 - \frac{2b}{a}\right) = \frac{\triangle}{s} = r$

which means that the point $L \equiv I$ is identical with the incenter of the triangle $\triangle ABC$.
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Virgil Nicula
7054 posts
#3 • 2 Y
Y by Adventure10, Adventure10
I note: the circumcircle $C(O,R)$, the incircle $C(I,r)$ and the circle $w=C(I_1,r_1)$ interior tangent to the circumcircle in the point $T$ and tangent to the sides $AB,AC$ in the points $P,Q$ respectively. From the a wellknown property, the points $U\in CI\cap TP$, $V\in BI\cap TQ$ belong to the circumcircle. Thus, the quadrilaterals $AUPI$, $AVQI$ are cyclic and
$UA\perp UT$, $VA\perp VT\Longrightarrow PI\perp AI,\ QI\perp AI\Longrightarrow I\in PQ$.

Remark. The property $I\in PQ$ is true in any triangle $ABC$. See
http://www.mathlinks.ro/Forum/viewtopic.php?t=46418
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Virgil Nicula
7054 posts
#4 • 2 Y
Y by Adventure10, Mango247
A generalization of this problem: "The circle $w=C(I_1,r_1)$ is internally tangent (in the point $T$) with the circumcircle of the triangle $ABC$ and is also tangent to the sides $[AB],[AC]$ in the points $P,Q$. Prove that the incenter $I$ of the triangle $ABC$ belongs to the segment $[PQ]$ and $AP=\frac{bc}{s},\ \ \frac{r_1}{r}=\frac{bc}{s(s-a)}.$

My solution. From a wellknown property, the points $M\in BI\cap TQ$, $N\in CI\cap TP$ belong to the circumcircle of the triangle $ABC$. From the Pascal's theorem applied to the inscribed hexagon $BACNTM$ results that the points $P\in BA\cap NT$, $Q\in AC\cap TM$, $I\in CN\cap MB$ are collineary, i.e. $I\in PQ$ (The other mentioned relations are easily solved).
See http://www.mathlinks.ro/Forum/viewtopic.php?t=46418
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Erken
1363 posts
#5 • 2 Y
Y by Adventure10, Mango247
A well-known consequence of the Pascal theorem.
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feliz
290 posts
#6 • 2 Y
Y by Adventure10, Mango247
Hm... Can you give more details, Erken?
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Erken
1363 posts
#7 • 2 Y
Y by Adventure10, Mango247
See Virgil Nicula's post above.
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ElChapin
486 posts
#8 • 2 Y
Y by Adventure10, Mango247
there is yet another proof of the generalization were you construc the circumference and then prove it is unique and that the theorem holds, it is not hard but I am to lazy to write it now I'll just write the beggining





Click to reveal hidden text


:)
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xeroxia
1140 posts
#9 • 2 Y
Y by Adventure10, Mango247
Let $ R$ be the tangency of circles and $ S$ be the center of inner circle. $ \angle PSA = 2k$ and $ \angle PRS = k$. $ \angle ABR = 90$. The circumcenter and $ S$ lie on $ AR$. $ AR \perp PQ$ at $ M$ and $ AR \perp BC$ at $ N$ . $ PMRB$ is cyclic. $ \angle PBM = \angle PRS = k$. $ PBNS$ is cyclic. $ \angle PBN = \angle PSA = 2k$. So $ \angle MBN = k$. $ BM$ is angle bisector.
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jayme
9801 posts
#10 • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
for an extension of the initial problem, see

http://perso.orange.fr/jl.ayme/ vol. 4 A new mixtilinear incircle adventure I.

Sincerely
jean-Louis
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Mithril
28 posts
#11 • 1 Y
Y by Adventure10
Virgil Nicula wrote:
A generalization of this problem: "The circle $ w = C(I_1,r_1)$ is internally tangent (in the point $ T$) with the circumcircle of the triangle $ ABC$ and is also tangent to the sides $ [AB],[AC]$ in the points $ P,Q$. Prove that the incenter $ I$ of the triangle $ ABC$ belongs to the segment $ [PQ]$

Let $ B'$ and $ C'$ be points in $ AC$ and $ AB$, respectively, such that $ AB = AB'$ and $ AC = AC'$. Let $ I_a$ be the center of the excircle of $ ABC$ tangent to $ AB$ and $ AC$ at $ D$ and $ E$ respectively ($ I_a$ lies in the bisector of $ \angle A$). We will show that the midpoint $ M$ of $ PQ$ is $ I$.

Consider the inversion with center $ A$ and radius $ AB \times AC$. It maps $ B$ and $ C$ into $ C'$ and $ B'$ respectively. Then, the circumcircle of $ ABC$ becomes line $ B'C'$. Therefore, $ w$ is mapped into a circle tangent to a circle tangent to $ AB$, $ AC$ and $ B'C'$.

Notice $ B'C'$ is tangent to both the incircle and $ (I_a)$ (because it is the reflection of $ BC$ over the bisector of $ \angle A$). Then $ w$ becomes $ (I_a)$ (it is easy to see that it cannot be the inverse of the incircle).


Now, if we call $ M'$ the inverse of $ M$, we have $ \angle ADM' = \angle AMP = 90^o$. As $ M'$ is on the bisector of $ \angle A$, we get that $ M' = I_a$. This implies that $ AM \times AI_a = AB \times BC$.

We will show that $ AI \times AI_a = AB \times BC$. Consider the circle $ K$ of diameter $ II_a$. It is known that $ \angle ICI_a = \angle IBI_a = 90^o$. Then both $ B$ and $ C$ belong to $ K$. By simmetry, we also get that $ B'$ and $ C'$ are in $ K$.

Then, we know that $ AI \times AI_a$ is the power of $ A$ with respect to $ K$. Then $ AI \times AI_a = AB \times AC' = AB \times AC$. As we knew that $ AM \times AI_a = AB \times AC$, this implies that $ I = M$ (because both points are on the bisector of $ \angle A$, inside $ ABC$), then $ I \in PQ$, as we wanted to prove.

Virgil Nicula wrote:
$ AP = \frac {bc}{s},\ \ \frac {r_1}{r} = \frac {bc}{s(s - a)}.$

The first equality follows, because $ D$ is the inverse of $ P$ and $ AD = s$ is known.

For the second one, an homotethy with center in $ A$ and ratio $ \frac {AP}{s-a}$ maps the incircle into $ w$. The ratio must be equal to $ \frac {r_1}r$ and the result follows using the first equality.
Attachments:
P61021.pdf (27kb)
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SnowEverywhere
801 posts
#12 • 2 Y
Y by Adventure10, Mango247
Let $\Gamma$ denote the circumcircle of $ABC$ and $\Omega$ denote the internally tangent circle through $P$ and $Q$. Let $M$ denote the point of tangency between $\Gamma$ and $\Omega$, $I$ denote the midpoint of $PQ$ and let $O$ denote the center of $\Omega$. Let $\angle{ABC}=a$.

First note that $A$, $M$, $I$ and $O$ are collinear and the entire figure is symmetrical around line $AM$. Therefore $I$ lies on the angle bisector of $\angle{BAC}$. We now have that $\angle{POQ}=180-\angle{BAC}=2a$. Hence we have that $\angle{PMQ}=a$ and by symmetry, $\angle{PMI}=a/2$.

Now note that $\angle{PIM}=\angle{PBM}=90$. Hence $PIMB$ is cyclic. Therefore $\angle{PBI}=\angle{PMI}=a/2$. Hence $I$ is also on the angle bisector of $\angle{ABC}$ and $I$ is the incenter of the triangle.
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sunken rock
4401 posts
#13 • 2 Y
Y by Adventure10, Mango247
The 'standard' proof uses Casey's for the 'circles' $A, \;B, \;C$ and $\Omega$, all internally tangent to $\Gamma$, followed by the reciprocal of the Transversal Theorem (Cristea's).

Best regards,
sunken rock
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Altheman
6194 posts
#14 • 1 Y
Y by Adventure10
Could you explain what you mean in detail? I do not know what Cristea's theorem is... and how casey's theorem is used here (i.e. what selection of +/- do we use).
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sunken rock
4401 posts
#15 • 2 Y
Y by Adventure10, Mango247
To Altheman:

For Cristea's theorem see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=323170&p=1741338&hilit=cristea#p1741338

Regarding the use of Casey's:
We have the 'circles' $A,B, \Omega, C$, all internally tangent to $\Gamma$, hence the following equality holds: $AP\cdot BC=AB\cdot CQ+AC\cdot BP$ $(\;1\; )$. If $BC=a$, etc, $a+b+c=2s$ and $AP=AQ=m$, from $(1)$ we get $a\cdot m=c(b-m)+b(c-m)$, or $m=\frac{bc}{s}$ $(\; 2 \;)$

If $\{ D  \} \equiv AI \cap BC$, then: $\frac{DI}{AI}=\frac{a}{b+c} \; ( \; 3 \; )$, $CD=\frac{ab}{b+c} \; ( \; 4 \; )$ and $BD=\frac{ac}{b+c} \; ( \; 5 \; )$.

Next, as per Cristea's, we have to prove:
$CD\cdot \frac{BP}{PA}+BD\cdot \frac{CQ}{AQ}=BC\cdot \frac{DI}{AI}$ and, with the relations 2-5:
$\frac{ab}{b+c} \cdot \frac{c-\frac{bc}{s}}{\frac{bc}{s}}+\frac{ac}{b+c}\cdot \frac{b-\frac{bc}{s}}{\frac{bc}{s}}=a \cdot \frac{a}{b+c}$, or $2s-b-c=a$, which is true.

Best regards,
sunken rock
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spikerboy
83 posts
#16 • 1 Y
Y by Adventure10
This solution is motivated from snowEverywhere.
Let do as usual notations midpoint of $PQ$ be $I.$ and let the angle bisector of $\angle A$ hits the $\odot ABC$
at $M$.note that $MA$ contains $I$.If you rotate the figure about $A$ with angle $90^{\circ}$ then $C \mapsto B$.so by the symmetry we can say that $AM$ is the diameter of the $\odot ABC$.
our motive is to show that, $P,I,M,B$ is cyclic.$M$ is the midpoint of the arc $PMQ \implies$ $PM$ bisects $\angle BPQ$.From above we can infer $\angle ABM =90^{\circ}$ and $\angle PIM =90^{\circ}$.so concyclic proved. it implies that, $\angle IBM =\angle IPM$ and $\angle MPB=\angle BIM \implies \angle IBM =\angle BIM \implies MI=MB$ so it follows that $I$ is the incenter.
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aayush-srivastava
137 posts
#18 • 1 Y
Y by Adventure10
Let $O$ be the center of the circle. Let the circle be tangent to the circumcircle of $ABC$ at $D$. Let $I $be the midpoint of $PQ$. Then $A, I, O, D$ are
collinear by symmetry. Consider the $homothety$ with center $A$ that sends $ABC$ to $AB’C’$ such that$ D$ is on $B’C’$. Thus, $k=AB’/AB$. As right triangles $AIP, ADB’, ABD, APO$ are similar, we have $AI /AO = (AI / AP)(AP / AO) = (AD /AB’)(AB /AD) = AB/AB’=1/k$.Hence the homothety sends $I$ to $O$.
Then $O$ being the incenter of $AB’C’$
implies$ I$ is the incenter of $ABC$.
QED :D
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anantmudgal09
1980 posts
#19 • 2 Y
Y by Adventure10, Mango247
This is true in general.

Use $\sqrt[2]{bc} $ inversion. The mixitilinear Incircle becomes the excircle and the incenter the excenter.

Now, it's trivial thereafter.

Sorry, if this is resembling some previously posted solution.
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viperstrike
1198 posts
#20 • 2 Y
Y by Adventure10, Mango247
By incenter/excenter lemma, it suffices to show $JM=JB$. Let $\angle MOP=2x$. Then $\angle MJP=x$ and $\angle JAB=90-2x \to \angle BAJ=2x$. Thus $\angle BJP=\angle MJP$. Since $\angle JBP=\angle JMP=90$, $BPMJ$ is a kite, done.
This post has been edited 1 time. Last edited by viperstrike, Mar 22, 2016, 2:24 AM
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wasikgcrushedbi
52 posts
#21
Y by
$\text{Midpoint of}$ $PQ=M$; $\text{line parallel to BC through}$ $I \cap AB=S$; $\text{line parallel to BC through}$ $I \cap AC=R$
Its apparent that $\triangle ASI \cong \triangle ARI \implies \angle AIS=90$ but $\angle AMP=90$ too and $PQ||SR$ so a contradiction thus $M=I$ (hopefully not a wrong solution)
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howcanicreateacccount
13 posts
#22
Y by
Is not this true for all triangles? how AB=AC is used?
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EpicBird08
1755 posts
#23 • 1 Y
Y by howcanicreateacccount
howcanicreateacccount wrote:
Is not this true for all triangles? how AB=AC is used?

yeah it is true for all triangles. It is something called Mixtillinear Incircles. See Evan Chen's handout on it (or EGMO chapter 4).
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HamstPan38825
8868 posts
#24 • 1 Y
Y by centslordm
This fact is actually true for all triangles.

First I claim $I$ actually lies on $\overline{PQ}$. Consider a $\sqrt{bc}$-inversion which swaps this circle $\omega$ with the $A$-excircle. The conclusion is equivalent to showing $E, F, A, I_A$ cyclic, where $E, F$ are the excentral touchpoints, which is obvious.

Now $AK=AL$ which implies the result.
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shendrew7
799 posts
#25
Y by
Label the touch points to $AB$, $AC$, and $(ABC)$ as $K$, $L$, and $T$, and the midpoints of arcs $AB$ and $AC$ as $X$ and $Y$.

Homothety tells us $TKX$ and $TLY$ collinear. Pascal on $BACXTY$ then tells us that $KIL$ collinear, and noting $\triangle AKI \cong \triangle ALI$ implies $IK = IL$. $\blacksquare$
Z K Y
N Quick Reply
G
H
=
a