Prove that AW=WT

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Prove that AW=WT

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Source: MEMO 2018 T6

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sqing

#1 h Sep 3, 2018, 12:02 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle . The internal bisector of $ABC$ intersects the side $AC$ at $L$ and the circumcircle of $ABC$ again at $W \neq B.$ Let $K$ be the perpendicular projection of $L$ onto $AW.$ the circumcircle of $BLC$ intersects line $CK$ again at $P \neq C.$ Lines $BP$ and $AW$ meet at point $T.$ Prove that $AW=WT.$

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#3 h Sep 5, 2018, 9:22 AM • 4 Y

Y by HappyMathEducation, PRMOisTheHardestExam, Adventure10, Mango247

Solution

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#4 h Sep 5, 2018, 9:44 AM • 4 Y

Y by HappyMathEducation, PRMOisTheHardestExam, Adventure10, Mango247

Problem : Let $ABC$ be a triangle. The angle bisector of $\angle BAC$ meets $BC$ and $\odot(ABC)$ at $D, M$ respectively. Let $X$ be the projection from $D$ onto $BM$ and $\odot(ADC)$ meets $CX$ at $Y$ . Line $AY$ and $BM$ meet at point $T$ . Prove that $BM = MT$ .

Solution : Let $K$ be the midpoint of $BC$ . Notice that $DXMK$ is concyclic thus $\angle XKC = 180^{\circ} - \angle DMX = \angle AMT$ . Moreover, $\angle MAT= \angle XCK$ , we get $\triangle AMT\sim\triangle CKX$ . Thus combining with $\triangle BKX\sim\triangle BMD$ yields
$\frac{MT}{MA} = \frac{KX}{KC} = \frac{KX}{MD}\cdot\frac{MD}{KC} = \frac{BK}{BM}\cdot\frac{MD}{KC} = \frac{MD}{MB} = \frac{MB}{MA}$ hence we are done.

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#5 h Jan 12, 2022, 1:37 PM

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Sinus law:
$\frac{sin \angle ABW}{ sin \angle WBT} = \frac{sin \angle BAW}{sin \angle BTW}$

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#6 h Jan 13, 2022, 1:45 PM • 2 Y

Y by HappyMathEducation, PRMOisTheHardestExam

Let $TC$ intersect $(BLC)$ at $X$ then Pascal on $CCXLBP$ gives that $W-T- XL \cap PC = K'$ , so $PC$ intersects $WT$ at $K'$ which forces $K' \equiv K$ and in particular $K-L-X$ . $\measuredangle CTA = 90^\circ - \measuredangle KXC = 90^\circ - \measuredangle LBC =90^\circ - \measuredangle TAC$ So $\measuredangle ACT =90^\circ$ which means $W$ is the circumcenter of $\triangle ACT$ and hence the result.

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#7 h Jan 14, 2022, 2:45 AM • 1 Y

Y by PRMOisTheHardestExam

Let the circumcircle of $\triangle BLC$ meet $AB$ at $D$ . Then $\angle DBL = \angle CBL \implies DL=CL$ .
Also $\angle DCL = \angle DBL = \angle ABL = \angle CBW = \angle CAW$ , so $DC \parallel AW$ .
Let the parallel through $B$ to $CD$ meet the circumcircle of $\triangle BCD$ at $E$ , then $BECD$ is an isosceles trapezoid.
Let $CE$ intersect $AT$ at $F$ , then $ADCF$ is an isosceles trapezoid so since $DL=CL$ , by symmetry $AK=KF$ .
Now projecting through $B$ onto the circle $(BCD)$ ,
$(A,T;W,AT_{\infty})=(D,P;L,E)$
Projecting through $C$ on $AT$ ,
$(D,P;L,E)=(AT_{\infty},K;A,F)=(A,F;K,AT_{\infty})=1$ (since $AK=KF$ ).
Thus $(A,T;W,AT_{\infty})=1$ , so $AW=WT$ .

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#8 h Jan 14, 2022, 5:15 AM

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We start with a claim -
Claim - $WC$ is tangent to $(BLC)$ -
Proof - Note that $\angle LCW = \angle ACW = \angle ABW = \angle CBW = \angle CBL$ $\blacksquare$

Now suppose $TC$ intersect $(LBC)$ again at $Q$ , then by Pascal Theorem of the hexagon $CQLBPC$ , we have $\overline{L-K-Q}$ , so this implies that from angle chase that $QL$ is diameter of $(BLC)$ , which implies $\angle ACT = 90$ ,which implies that $(LCKT)$ is cyclic.

Claim - $WT$ is tangent to $(BLT)$ .
Proof - Note that $\angle ATL = \angle KTL = \angle KCL = \angle PCL = \angle PBL = \angle TBL$ $\blacksquare$ .

Now from Power of Point we have $WT^2 = WL.WB = WC^2 = WA^2$ , where at last we uses Fact-5 , so $TW = AW$ and we are done. $\blacksquare$

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#9 h Jan 14, 2022, 6:57 AM • 2 Y

Y by HappyMathEducation, PRMOisTheHardestExam

Let $T$ be the reflection of $A$ over $W,$ and let $P'=AT\cap KC$ , we´ll show $BLP'C$ is cyclic.
To do this note that from equal sides we know $\angle{ACT}$ is right so $LKTC$ is cyclic, then $\angle{LCK}=\angle{LTK}=\angle{LTW}=\angle{LBT}$ (last quality follows from noticing that $WT$ is tangent to $(BLT)$ , because $WL\cdot WB=WA^2=WT^2$ ), hence $BLP'C$ is cyclic, as desired.

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#10 h Aug 25, 2023, 11:37 AM

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Let $T'$ be the point such that $\angle ACT'=90$ and let $BT'\cap KC=P'$ . Then we have $T'W^2=AW^2=WL*WB$ , which means that $\angle WBT'=\angle LT'W$ , and since $(L K T' C)$ is cyclic, we have $\angle LT'W=\angle LCK$ , which means that $\angle LCP'=\angle LBP'$ , then $(B L P' C)$ is cyclic, then $P=P'$ which completes the solution.

This post has been edited 4 times. Last edited by X.Allaberdiyev, Aug 25, 2023, 3:09 PM
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