Easy Geometry

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TheOverlord

97 posts

TheOverlord

#1 h May 10, 2015, 1:18 PM • 2 Y

Y by Dadgarnia, Adventure10

$I_b$ is the $B$ -excenter of the triangle $ABC$ and $\omega$ is the circumcircle of this triangle. $M$ is the middle of arc $BC$ of $\omega$ which doesn't contain $A$ . $MI_b$ meets $\omega$ at $T\not =M$ . Prove that
$TB\cdot TC=TI_b^2.$

This post has been edited 2 times. Last edited by djmathman, Sep 10, 2015, 3:59 AM
Reason: latex/formatting

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Luis González

4149 posts

Luis González

#2 h May 10, 2015, 5:10 PM • 6 Y

Y by phantranhuongth, AlastorMoody, lahmacun, Eliot, Adventure10, Mango247

Let $I$ and $I_a,I_c$ be the incenter and the excenters of $\triangle ABC$ againts $A,C.$ Since $\omega$ is 9-point circle of $\triangle I_aI_bI_c,$ then $\omega$ cuts $II_a,I_aI_b$ at their midpoints $M,N$ $\Longrightarrow$ $MN \parallel BI_b$ $\Longrightarrow$ $\angle BI_bT=\angle TMN=\angle TCI_b.$ Since $\angle BTI_b=\angle CTI_b$ ( $TM$ bisects $\angle BTC$ ), then $\triangle TBI_b \sim \triangle TI_bC$ $\Longrightarrow$ $\tfrac{TB}{TI_b}=\tfrac{TI_b}{TC}$ $\Longrightarrow$ $TB \cdot TC={TI_b}^2.$

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aditya21

717 posts

aditya21

#3 h May 18, 2015, 2:04 PM • 2 Y

Y by Adventure10, Mango247

nice and easy!
my solution = as $ABMT,ATCM$ are concylic hence
$\angle MTB=\angle MTC=\frac{\angle A}{2}$ which is due to $AM$ bisecting $\angle BAC$
and thus $\angle CTI_b=\angle I_bTC=180-\frac{\angle A}{2}$
also now let $\angle TCI_b=x$ than $\angle TCB=90+\frac{\angle C}{2}-x$
and thus $\angle CBT=180-\angle TCB=90-\frac{\angle C}{2}+x$
and so $\angle TBI_b=\frac{\angle B}{2}-\angle CBT=\frac{\angle A}{2}-x=\angle CI_bT$
and thus $\triangle TBI_b\sim \triangle TI_bC \Longrightarrow TB.TC={TI_b}^2$

so we are done

This post has been edited 2 times. Last edited by aditya21, May 19, 2015, 6:56 AM
Reason: e

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tranquanghuy7198

253 posts

tranquanghuy7198

#4 h May 18, 2015, 3:41 PM • 2 Y

Y by AlastorMoody, Adventure10

My solution:
Let $D$ be the midpoint of arc $BC$ containing $A$
$\Rightarrow$ $D$ is the circumcenter of $\triangle{I_bBC}$ and $MB, MC$ are tangent to $(I_bBC)$
$\Rightarrow$ $I_bM$ is the symmedian of $\triangle{I_bBC}$
Moreover, $T$ lies on the symmedian such that $TI_b$ bisects $\angle{BTC}$ $\Rightarrow$ $\triangle{TBI_b}$ $\sim$ $\triangle{TI_bC}$ and the conclusion follows

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colinhy

751 posts

colinhy

#5 h May 23, 2015, 3:10 AM • 1 Y

Y by Adventure10

Here's a solution:
Observe that $\angle BTC = \angle A$ and since $MI_B$ is an angle bisector of $\angle BTC$ , $\angle BTI_B = \angle CTI_B = 180 - \frac{1}{2}\angle A$ . Now, let $D$ be the second intersection of the circumcircle $\tau$ of $CTI_B$ and $BC$ , so we have $\angle CDI_B = \frac{1}{2} \angle A$ , which means that $\triangle BCI_B \sim \triangle BI_BD$ and $BI_B$ is tangent to $\tau$ , so $\angle TCI_B = \angle TI_BB$ , so by AA symmetry, $\triangle BTI_B \sim \triangle I_BTC$ , and the result follows.

Here's a slightly more difficult problem based on the above:
Let $I$ be the incenter of $\triangle ABC$ . Show that it is possible to construct a right triangle with side lengths $IM, MT, TI_B$ .

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infiniteturtle

1131 posts

infiniteturtle

#6 h May 23, 2015, 3:41 AM • 2 Y

Y by Adventure10, Mango247

Extravert (mostly for convenience) to get

Quote:

$ABC$ is a triangle with circumcircle $w$ . Let $I$ be the incenter, $M$ be the midpoint of arc $BC$ not including $A$ , and let $N$ be the midpoint of arc $BAC$ . If $NI\cap w=T$ , show that $TI^2=TB\cdot TC.$

The problem now falls easily after drawing $(BIC)$ : Let $IT\cap (BIC)=X, CT\cap (BIC)=Y$ . It's trivial by angle chasing that $BT=TY$ and that $IT=TX$ , now PoP kills it.

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Dukejukem

695 posts

Dukejukem

#7 h Sep 10, 2015, 3:56 AM • 1 Y

Y by Adventure10

Let $N$ be the midpoint of arc $\widehat{BAC}$ of $\omega$ and let $I_c$ be the $C$ -excenter. It is well-known that $B, C, I_b, I_c$ are inscribed in the circle $\Gamma$ of diameter $\overline{I_bI_c}$ with center $N.$ Moreover, note that $M$ is the midpoint of arc $\widehat{BC}$ , implying that $TM$ and $TN$ are the internal and external bisectors of $\angle BTC$ , respectively. Therefore, the inversion with power $TB \cdot TC$ combined with a reflection in $TM$ swaps $B$ and $C$ fixes lines $TM$ and $TN.$ It follows that the center of $\Gamma'$ (the image of $\Gamma$ ) is a point on line $TN.$ However, the center of $\Gamma'$ must also lie on the perpendicular bisector of $\overline{BC}$ , implying that the center is $N$ itself. Therefore, $\Gamma$ is fixed under this inversion, and hence $I_b$ is fixed as well. Thus, $TI_b^2 = TB \cdot TC$ as desired. $\square$

This post has been edited 1 time. Last edited by Dukejukem, Sep 10, 2015, 9:54 PM

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hayoola

123 posts

hayoola

#8 h Sep 10, 2015, 5:40 AM • 2 Y

Y by Adventure10, Mango247

the interestion between IbC and w is thw midpoint of arc ABC

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soroush.MG

158 posts

soroush.MG

#9 h Feb 19, 2017, 8:30 PM • 2 Y

Y by Adventure10, Mango247

Let $D$ be the midpoint of arc $BAC$ . And $c$ is a circle with centre $D$ and radius $DB$ . $c \cap I_bT=E$ .
As $ATM=90$ and $AI_b=AE$ so $TE=TI_b$ and it's enough to show that $TE^2=TB.TC$ We know that $MC$ and $MB$ are tangent to $c$ . So we can simply show that $\triangle TCE \sim \triangle TEB$ and $TE^2=TB.TC$ .

This post has been edited 1 time. Last edited by soroush.MG, Feb 19, 2017, 8:30 PM

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tenplusten

1000 posts

tenplusten

#10 h Feb 20, 2017, 11:57 AM • 1 Y

Y by Adventure10

An easy barycentric problem.
We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$ .It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$ .So we finish problem using Distance formula.

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artsolver

139 posts

artsolver

#11 h Feb 20, 2017, 12:13 PM • 1 Y

Y by Adventure10

Murad.Aghazade wrote:

An easy barycentric problem.
We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$ .It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$ .So we finish problem using Distance formula.

Well, if you are about to bash, then bash to the end, since it is known that you can always finish problem that way, no matter how you start, but if you are doing it synthetic then it's nice to write just sketch of the proof...

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Anar24

475 posts

Anar24

#12 h Apr 10, 2018, 5:58 PM • 3 Y

Y by tenplusten, Adventure10, Mango247

artsolver wrote:

Murad.Aghazade wrote:

An easy barycentric problem.
We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$ .It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$ .So we finish problem using Distance formula.

Well, if you are about to bash, then bash to the end, since it is known that you can always finish problem that way, no matter how you start, but if you are doing it synthetic then it's nice to write just sketch of the proof...

Tenplusten gave the sketch of bary solution,and it is good idea to allow others to finish by themselves.For that reason it is not correct to deny others for skipping the long calculations.

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WolfusA

1900 posts

WolfusA

#13 h Apr 10, 2018, 6:31 PM • 2 Y

Y by Adventure10, Mango247

In math publications leaving long calculations is acceptable, but you have to give final result or conclusion following from it.

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Taha1381

816 posts

Taha1381

#14 h Apr 28, 2019, 12:29 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

It is just sum of some lemmas:).First note that $MC,MB$ are both tangent to circumcircle of $I_bBC$ by angle chasing.Also $BMCT$ is cyclic so $T$ is the midpoint of $I_b$ symmedian in triangle $BCI_B$ and so two triangles $TI_bC$ is similar to $TBI_b$ from which the relation follows.

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FISHMJ25

293 posts

FISHMJ25

#16 h Jun 9, 2019, 2:23 PM • 2 Y

Y by Adventure10, Mango247

Here is complex bash solution. $\omega$ is unit circle, $a=x^2$ $b=y^2$ $c=z^2$ . So $m=-yz$ and $Ib=xy+yz-xz$ . We now want $t$ . But its easy just solve $\frac{m-t}{\bar m -\bar t }=\frac{m-Ib}{\bar m -\bar Ib }$ .We get $t=\frac{x(xy+2yz-xz)}{z+2x-y}$ And now we need to prove $|(t-y^2)(t-z^2)|=|Ib-t|^2$ . After substituting $t$ and $Ib$ we see that both sides are equal to $\frac{(z+x)^2(x-y)^2(x-z)^2}{(z+2x-y)^2}$ and we are done.
Calculations are not bad can be done under 20 minutes.

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ComiCabE

25 posts

ComiCabE

#17 h Jun 20, 2019, 4:02 PM • 1 Y

Y by Adventure10

Here is a different approach.

Denote by $N$ the midpoint of $\widehat{BCA}$ . With elementary angle chasing we may show that $MN \parallel BI_B$ . This implies $\angle TCI_B =\angle TMN =\angle TI_BB$ . This combined with $\angle MTC =\angle MTB$ gives $\triangle I_BTC \sim \triangle I_BBT$ , and the desired equality follows $\square$

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Krits

147 posts

Krits

#18 h Apr 11, 2020, 9:43 AM

Y by

observe that $TM$ bisects $\angle CTB$ so $\angle CTI_B=\angle BTI_B$ let $(BCI_B)$ intersect $MT$ at $U$ so $\angle CUB=180-\frac{\angle A}{2}$ combining with $\angle CI_B U=\angle CBU$ we are done because we have $\triangle I_B TC \sim \triangle TBI_B$

This post has been edited 1 time. Last edited by Krits, Apr 11, 2020, 9:43 AM

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mathaddiction

308 posts

mathaddiction

#19 h Jul 12, 2020, 5:50 AM

Y by

We will show that $T$ is the $I_b-$ dumpty point of $\triangle BI_bC$ which implies the result since the dumpty point is the center of spiral similarity sending $\overline{CI_b}$ to $\overline{I_bC}$ .
Notice that the dumpty point is the intersection of the $I_b-$ symmedian and $(BOC)$ , where $O$ is the circumcenter of $\triangle BI_bC$ . Now by Ceva's theorem
$\frac{\sin\angle BI_bM}{\sin\angle CI_bT}\cdot\frac{\sin\angle I_bCM}{\sin\angle MCB}\cdot\frac{\sin\angle MBC}{\sin\angle MBI_b}=1$ Since $\angle MBC=\angle MCB$ ,
$\frac{\sin\angle BI_bM}{\sin\angle CI_bT}=\frac{\sin\angle MBI_b}{\sin\angle I_bCM}=\frac{\cos\frac{C}{2}}{\sin\frac{B}{2}}=\frac{\sin\angle I_bCB}{\sin\angle I_bBC}$ which implies that $MT$ is a symmedian. Moreover,
$\angle BTC=\angle A=2\angle IAC=2\angle II_bC=\angle BOC$ Hence $BI_BOC$ is cyclic which completes the proof.

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Gaussian_cyber

162 posts

Gaussian_cyber

#20 h Jul 26, 2020, 6:47 PM • 2 Y

Y by amar_04, Eliot

Storage

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jayme

9801 posts

jayme

#21 h Aug 20, 2020, 8:41 AM

Y by

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20produit%20AB.AC.pdf p. 35-38.

Sincerely
Jean-Louis

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TheUltimate123

1740 posts

TheUltimate123

#22 h Oct 11, 2020, 10:59 PM • 1 Y

Y by snakeaid

Solved with nukelauncher.

First observe $\measuredangle BTI_B=\measuredangle I_BTC$ since $\overline{TM}$ bisects $\angle BTC$ .

$[asy] size(5cm); defaultpen(fontsize(10pt)); pair B,A,C,I,M,NN,IB,T,X; B=dir(110); A=dir(210); C=dir(330); I=incenter(A,B,C); M=extension( (B+C)/2,origin,A,I); NN=extension( (C+A)/2,origin,B,I); IB=2NN-I; T=2*foot(origin,M,IB)-M; X=2*foot(origin,IB,C)-C; draw(circumcircle(A,C,IB),gray); draw(M--A,gray); draw(B--IB,dashed); draw(C--IB,dashed); draw(unitcircle); draw(A--B--C--cycle); draw(M--IB); dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$M$",M,M); dot("$N$",NN,SW); dot("$I_B$",IB,S); dot("$T$",T,SE); dot("$I$",I,NW); [/asy]$

In addition, $\measuredangle I_BTC=\measuredangle MTC=\measuredangle IAC=\measuredangle BI_BC=\measuredangle BI_BT+\measuredangle TI_BC,$ thus $\measuredangle BI_BT=\measuredangle I_BTC+\measuredangle CI_BT=\measuredangle I_BCT$ , so $\triangle TBI_B\sim\triangle TI_BC.$ It follows that $TB\cdot TC=TI_B^2$ , as needed

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snakeaid

125 posts

snakeaid

#23 h Oct 19, 2020, 6:37 PM

Y by

I really need to learn directed angles

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rafaello

1079 posts

rafaello

#25 h Dec 2, 2020, 2:29 PM

Y by

We claim that $\triangle BTI_B\sim\triangle I_bTC$ .
Obviously, we have that $\angle BTM=\angle BAM=\angle MAC=\angle MTC\implies \angle I_bTB=\angle CTI_b.$ We obtain that $\angle I_bCT=\angle I_bCA-\angle TCA$ and
$\angle BI_bT=\angle AII_b-\angle AMI_b=180^\circ-\angle AIB -\angle TCA=90^\circ-0.5\angle ACB-\angle TCA=\angle I_bCT-\angle TCA,$ hence we conclude that $\angle I_bCT=\angle BI_bT$ .
$TB\cdot TC=TI_b^2$ follows directly from $\triangle BTI_B\sim\triangle I_bTC$ .

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AMN300

563 posts

AMN300

#26 h Mar 27, 2021, 6:08 AM

Y by

We claim triangles $CTI_B, I_BTB$ are similar, which clearly implies the result. Relabel $M \rightarrow M_A$ and let $M_B$ be the arc midpoint of minor arc $BC$ . First remark $\angle CTM_A = \angle BTM_A = A/2$ so $\angle CTI_B = \angle BTI_B$ . Next let $\angle M_B BT = x$ . Straightforward angle chasing yields $\angle ACI_B = 90$ , so $\angle TCI_B = 90-\frac{C}{2}-\angle ACT = 90-\frac{C}{2}-\frac{A}{2}-x = \frac{B}{2}-x$ . But $\angle TI_B B = \frac{2(B/2)-2x}{2}=\frac{B}{2}-x$ by standard angle facts. So $\angle TCI_B = \angle TI_B B$ , so we're done.

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hakN

429 posts

hakN

#27 h Mar 27, 2021, 2:18 PM

Y by

It is easy to see that $\angle BTI_B = \angle CTI_B$ . Let $N$ be the midpoint of the arc $\widehat{AC}$ not containing $B$ . Becuase $NIMC$ is a kite, we get that $NM\perp IC$ and we know that $I_BC\perp IC \implies NM\parallel I_BC$ . So we get $\angle NBT = \angle NMT = \angle TI_BC \implies \triangle BTI_B \sim \triangle I_BTC \implies TB\cdot TC = TI_B^2$ .

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nixon0630

31 posts

nixon0630

#28 h Oct 19, 2021, 5:39 AM • 1 Y

Y by javohirsultanov20

It's enough to show, that: $\triangle BTI_B\sim\triangle CTI_B$ Claim. $MT$ bisects $\angle CTB$ .
Proof.
Claim. $\angle TCI_B = \angle TI_BB$ .
Proof.
Therefore $\triangle BTI_B\sim\triangle CTI_B$ by $ASA$ . $\blacksquare$

Attachments:

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trying_to_solve_br

191 posts

trying_to_solve_br

#29 h Oct 20, 2021, 1:07 AM

Y by

Nice and cute, notice first that, as $M$ is the midpoint of the arc, then $\angle BTI_b=\angle CTI_b$ but notice we want $TI_b/TC=TB/TI_b$ and thus we just need one more angle to prove $BTI_b \sim I_bTC$ . We'll show $\angle I_bBT=\angle TI_bC$ , but $\angle I_bBT=B/2-\angle TBC$ and as $\angle BTI_b=90+B/2+C/2=\angle TBC+\angle TI_bC+\angle BCI_b=\angle TBC+\angle TI_bC+90+C/2$ and thus $\angle TBC \angle TI_bC=B/2=\angle TBI_b+\angle TBC$ , proving our claim.

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JAnatolGT_00

559 posts

JAnatolGT_00

#30 h Oct 22, 2021, 10:21 AM

Y by

Note that $\angle CTI_b=$ $\angle I_bTB=$ $\pi$ $-\frac{\angle BAC}{2}=$ $\pi$ $-\angle BI_bC$ $\implies$ $\angle TBI_b=$ $\angle TI_bC.$
Hence $\triangle BTI_b\sim \triangle I_bTC\implies |TB|\cdot |TC|=|TI_b|^2$ $\blacksquare$

This post has been edited 2 times. Last edited by JAnatolGT_00, Oct 22, 2021, 10:28 AM

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Mahdi_Mashayekhi

697 posts

Mahdi_Mashayekhi

#31 h Jan 18, 2022, 4:21 PM

Y by

We will prove BTIb and IbTC are similar and it will follow TB . TC = TIb^2.
∠BTIb = 180 - ∠MTB = 180 - ∠CTM = ∠IbTC. Let ∠TBIb = x. ∠ACT = ∠B/2 + x and ∠ACIb = 90 - ∠C/2 so ∠TCIb = 90 - ∠C/2 - ∠B/2 - x.
∠CIbT = ∠A/2 - ∠TCIb = ∠A/2 - 90 + ∠C/2 + ∠B/2 + x = x = ∠TBIb so BTIb and IbTC are similar.
we're Done.

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827681

163 posts

827681

#34 h May 27, 2022, 10:15 AM • 1 Y

Y by Mango247

Overcomplicated headsolve

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David_Kim_0202

384 posts

David_Kim_0202

#35 h Jun 5, 2023, 6:04 PM

Y by

Too easy for Iran TST

1. think about using simillar triangle theory
2. just use angle moving for the evidemce of proof.

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math_comb01

662 posts

math_comb01

#36 h Dec 9, 2023, 5:51 PM

Y by

Easy problem
Claim 1: $MB$ and $MC$ are tangent to $(BI_BC)$
Proof
Now this combined with $\measuredangle BTI_B = \measuredangle CTI_B$ gives that $T$ is $I_B-$ dumpty point so this implies the conclusion as
$\triangle BTI_B \sim \triangle I_BTC$

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