Internally tangent with the circumscribed circle

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Internally tangent with the circumscribed circle

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: IMO LongList, USA 1, IMO 1978, Day 2, Problem 4

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3647 posts

orl

#1 h Nov 12, 2005, 7:46 PM • 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 $ABC.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2643 posts

yetti

#2 h Nov 13, 2005, 7:47 AM • 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 $\triangle ABC$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

7054 posts

#3 h Nov 13, 2005, 1:20 PM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

7054 posts

#4 h Nov 14, 2005, 10:35 PM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1363 posts

Erken

#5 h Jan 9, 2009, 2:41 PM • 2 Y

Y by Adventure10, Mango247

A well-known consequence of the Pascal theorem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

290 posts

feliz

#6 h Jan 10, 2009, 10:06 PM • 2 Y

Y by Adventure10, Mango247

Hm... Can you give more details, Erken?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1363 posts

Erken

#7 h Jan 11, 2009, 5:33 AM • 2 Y

Y by Adventure10, Mango247

See Virgil Nicula's post above.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

486 posts

#8 h Jan 11, 2009, 6:31 AM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1140 posts

#9 h Jul 31, 2009, 9:42 AM • 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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9801 posts

jayme

#10 h Jul 31, 2009, 2:51 PM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

28 posts

#11 h Jul 31, 2009, 5:17 PM • 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)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

801 posts

#12 h Jul 22, 2010, 4:03 PM • 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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4401 posts

#13 h Jul 22, 2010, 4:54 PM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6194 posts

#14 h Jul 22, 2010, 9:11 PM • 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).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4401 posts

#15 h Jul 23, 2010, 8:33 AM • 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

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

83 posts

#16 h Apr 14, 2014, 12:57 PM • 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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

aayush-srivastava

137 posts

aayush-srivastava

#18 h Dec 4, 2015, 8:02 AM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1980 posts

#19 h Dec 4, 2015, 4:06 PM • 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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1198 posts

#20 h Mar 22, 2016, 2:23 AM • 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

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

wasikgcrushedbi

52 posts

wasikgcrushedbi

#21 h Mar 7, 2022, 2:27 PM

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)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

howcanicreateacccount

13 posts

howcanicreateacccount

#22 h Apr 25, 2023, 11:31 AM

Y by

Is not this true for all triangles? how AB=AC is used?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1755 posts

#23 h Apr 25, 2023, 2:23 PM • 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).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

8868 posts

#24 h Aug 16, 2023, 8:01 PM • 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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

799 posts

#25 h Dec 27, 2023, 4:22 AM

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

=

© 2025 AoPS Incorporated

a