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Source: Bosnia and Herzegovina JBMO TST 2018

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#1 h Jul 14, 2018, 10:03 PM • 2 Y

Y by Adventure10, Mango247

Let $\Gamma$ be circumscribed circle of triangle $ABC$ $(AB \neq AC)$ . Let $O$ be circumcenter of the triangle $ABC$ . Let $M$ be a point where angle bisector of angle $BAC$ intersects $\Gamma$ . Let $D$ $(D \neq M)$ be a point where circumscribed circle of the triangle $BOM$ intersects line segment $AM$ and let $E$ $(E \neq M)$ be a point where circumscribed circle of triangle $COM$ intersects line segment $AM$ . Prove that $BD+CE=AM$ .

This post has been edited 2 times. Last edited by MathsLion, Feb 25, 2023, 5:26 PM
Reason: Fixing latex

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#2 h Jul 14, 2018, 10:38 PM • 2 Y

Y by Adventure10, Mango247

Note that $\angle BDM = \angle BAC, \angle DMB = \angle BCA$ , so $BDM \sim BAC$ . So $\frac{BD}{AB} = \frac{BM}{BC}= \frac{CM}{BC}$ , so $BD + CE = \frac{AB \cdot CM + AC \cdot BM}{BC} = \frac{AM \cdot BC}{BC} = AM$ .

Alternatively, note that $\angle DBA = \angle MBC = \angle MCB = \angle DAB$ , so $DAB, EAC$ are $D, E$ isosceles. So $OD \perp AB, OE \perp AC$ . Thus $OD, OE$ make equal angles with $AM$ , so $D, E$ are isotomic wrt $AM$ . Then we have $BD + CE = AD + AE = AD + DM = AM$ .

This post has been edited 1 time. Last edited by WizardMath, Jul 14, 2018, 10:49 PM

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jayme

#3 h Jul 17, 2018, 10:50 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

this is an application of

http://www.artofproblemsolving.com/community/c6t48f6h1080717_easy_geometry

noting also that the (BOM) and (COM) are equal...ODE is O-isoceles.

Sincerely
Jean-Louis

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#4 h Dec 9, 2022, 11:13 PM

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Let $\Gamma$ be circumscribed circle of triangle $ABC$ ( $AB \ne AC$ ). Let O be circumcenter of triangle $ABC$ . Let $M$ be a point where angle bisector of angle $\angle BAC$ intersects $\Gamma$ . Let $D$ ( $D\ne M$ ) be a point where circumscribed circle of triangle $BOM$ intersects line segment $AM$ and let $E$ ( $E\ne M$ ) be a point where circumscribed circle of triangle $COM$ intersects line segment $AM$ . Prove that $BD+CE=AM$ .

This post has been edited 2 times. Last edited by parmenides51, Dec 10, 2022, 7:03 PM

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#5 h May 25, 2023, 4:33 PM

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Let $\angle{BCE}=\beta$ and $\angle{ECM}=\alpha$ . $\implies \angle{BOM}=\angle{COM}=2\alpha+ 2\beta$ .
Since $DOMB$ and $COEM$ cyclic $\implies \angle{BDM}=\angle{CEM}= {2\alpha + 2\beta}$ , and $\angle{BAM}=\angle{CAM}=\alpha+\beta$ this means $AD=DB$ and $AE=EC$ .So $BD+CE=AD+AE$ $\implies$ we need to find $MD=AE$ .From easy angle chasing we get $\triangle DBM \sim \triangle EMC$ and since $MB=MC$ we get $MD=CE=AE$ and we're done!

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