Prove line tangent to circle given segment is bisected

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Prove line tangent to circle given segment is bisected

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Source: CSEMO 2018 Grade 10 Q3

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#1 h Jul 30, 2018, 5:42 AM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the circumcenter of acute $\triangle ABC$ ( $AB<AC$ ), the angle bisector of $\angle BAC$ meets $BC$ at $T$ and $M$ is the midpoint of $AT$ . Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$ . $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$ . If $AO$ bisects segment $DE$ , prove that $AO$ is tangent to the circumcircle of $\triangle AMP$ .

This post has been edited 1 time. Last edited by mofumofu, Jul 30, 2018, 5:42 AM

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#2 h Jul 30, 2018, 8:00 AM • 2 Y

Y by ThE-dArK-IOrD, Adventure10

Let $\ell$ be the line perpendicular to $AP$ at $P$ . Let $\ell \cap AO=Q$ and we get that $Q$ is the midpoint of $DE$ .
Easy angle chasing gives us $DB\perp EC$ , let their intersection be $R$ .
Suppose the tangent line at $A$ of $(ABC)$ intersects $BC$ at $K$ . Note that $\angle{OAK}=90^{\circ}$ .
Easy to see that $KA=KT$ and so $\angle{KMA}=90^{\circ}$ .

Then, we'll proof that $K,D,E$ collinear.
This is true by considering radical axises of circles $(APQ),(ABC),$ and $(PBC)$ , we get

Radical axis of $(APQ)$ and $(ABC)$ is the tangent line at $A$ of $(ABC)$ (by angle chasing.)
Radical axis of $(ABC)$ and $(PBC)$ is $BC$ .
Radical axis of $(APQ)$ and $(PBC)$ is $PQ=\ell$ if one can show that $Q$ lie on $(BPC)$ .

For the last line, we have to show that $Q$ lie on $(BPC)$ , this is true if one can prove the following lemma:

Lemma wrote:

Given a right triangle $ERD$ with $\angle{ERD}=90^{\circ}$ . Let $K$ be the midpoint of $ED$ and $P$ be an arbitrary point on the side $DE$ . Draw perpendicular bisector of $DP$ and $EP$ intersects the sides $DR$ and $ER$ at $B$ and $C$ , respectively. Then $P,Q,C,B$ concyclic.

Which I'm too lazy to find synthetic proof.

Now, from $K,D,E$ collinear, we get that $\angle{KPA}=90^{\circ}$ and then $K,M,P,A$ concyclic.
This finish the problem by angle chasing that $\angle{OAP}=\angle{PKA}=\angle{PMA}$ .

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Jul 30, 2018, 8:08 AM

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#3 h Aug 3, 2018, 9:17 AM • 2 Y

Y by Adventure10, Mango247

Let $K = DE\cap BC$ and $N$ is the midpoint of $DE$ .

The key step is to prove that $B, C, P, N$ are concyclic. To do that, let $X, Y$ be the projections from $B, C$ to $DE$ respectively, which are midpoints of $PD, PE$ respectively. Since $\angle BPC = 90^{\circ}$ , triangles $PBX$ and $CPY$ are similar. Moreover, simple side-chasing gives $XN=YP$ thus
$\frac{XN}{XB} = \frac{PY}{BX} = \frac{PX}{CY} = \frac{NY}{YC}$ or triangles $NBX$ and $CNY$ are similar, which implies $\angle BNC=90^{\circ}$ as claimed.

Since $K$ is radical center of $\odot(APN), \odot(BCPN), \odot(ABC)$ , it lies on radical axis of $\odot(APN), \odot(ABC)$ . However, these circles are obviously tangent so $AK$ is tangent to $\odot(ABC)$ .

Finally, recall that $KA=KT$ so $\angle AMK = 90^{\circ} = \angle APK$ . Hence $\odot(AMP)$ is circle with diameter $AK$ , which is obviously orthogonal to $\odot(ABC)$ , implying the conclusion.

This post has been edited 1 time. Last edited by MarkBcc168, Aug 3, 2018, 9:18 AM

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#4 h Mar 26, 2021, 8:05 AM

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CSEMO 2018 Grade 10 Q3 wrote:

Let $O$ be the circumcenter of acute $\triangle ABC$ ( $AB<AC$ ), the angle bisector of $\angle BAC$ meets $BC$ at $T$ and $M$ is the midpoint of $AT$ . Point $P$ lies inside $\triangle ABC$ such that $PB\perp PC$ . $D,E$ distinct from $P$ lies on the perpendicular to $AP$ through $P$ such that $BD=BP, CE=CP$ . If $AO$ bisects segment $DE$ , prove that $AO$ is tangent to the circumcircle of $\triangle AMP$ .

Great problem.
Let $F$ be the midpoint of $DE$ , $I$ be the intersection of $BD$ and $CE$ , and $J$ be the intersection of $BC$ and $DE$ .
Claim 01. $BPFC$ is cyclic.
Proof. Notice that $DB \perp CE$ . Since $F$ is the midpoint of $DE$ , we have
$\measuredangle FIB \equiv \measuredangle FID = \measuredangle IDF \equiv \measuredangle BDP = \measuredangle DPB = \measuredangle FPB$ Therefore $BPFI$ is cyclic. Similarly, we get $BFCI$ being cyclic. Since $\measuredangle BIC = \measuredangle BPC = 90^{\circ}$ as well, we have $BPCI$ is cyclic. Therefore, we conclude that $BPFC$ is cyclic.

Claim 02. $AJ$ tangent to $(ABC)$ .
Proof. By radical axis theorem on $(ABC), (APF), (BPFC)$ . We have $PF \cap BC = J$ lie on the radical axis of both $(ABC)$ and $(APF)$ . However, $(APF)$ has diameter $AF \equiv AO$ , and therefore these circles are tangent to each other at $A$ . Hence, $AJ$ is the radical axis of these circle, proving that $AJ$ is tangent to $(ABC)$ .

Claim 03. $APMJ$ is cyclic.
Proof. Since $JA$ is tangent to circle and $AT$ is the angle bisector of $\angle BAC$ , we have $JA = JT$ . Since $M$ is the midpoint of $AT$ , we have $JM \perp AT$ . This forces
$\measuredangle JMA = 90^{\circ} = \measuredangle JPA$ proving $APMJ$ is cyclic.

Claim 04. $AO$ is tangent to $(AMP)$ .
Proof. Notice that $AJ$ is tangent to $(ABC)$ , and therefore $AJ \perp AO$ . Since $(APMJ)$ has diameter $AJ$ and $(ABC)$ has radius $AO$ . We conclude that $(APMJ)$ and $(ABC)$ are orthogonal, forcing $AO$ is tangent to $(AMP)$ .

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