Geometry Again!

Following are some geometry questions I couldnot solve. They may be from other olympiads, and if their solutions are available online please point me to them.

For a pentagon $ABCDE,\ AE=ED,\ AB+CD=BC,$ and $\angle BAE + \angle CDE=180 ^{\circ} $. P.T. $\angle AED=2\angle BEC$.
Let P be a point inside a triangle with $\angle C=90 ^{\circ}$ such that $AP=AC$ , and let M be the midpt of AB and CH be the altitude. Prove that PM bisects $\angle BPH$ iff $\angle A=60 ^{\circ}$ .

#Geometry #EuclideanGeometry #OlympiadMath #Puregeometry

Easy Math Editor

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Comments

For Problem 1. Consider the point $F$ in the ray $BA$ such that $AF=CD$ ( $A$ between $B$ and $F$ ), then triangles $FAE$ and $CDE$ are congruent (using $AE=ED$ and $\angle BAE+\angle CDE=180^\circ$ . Note that the triangles $CEB$ and $FEB$ are congruent because $EC=EF$ and $BC=AB+CD=AB+AF=BF$ . Finally, $\angle BEC=\angle BEF=\angle BEA+\angle AEF=\angle BEA+\angle DEC$ therefore $\angle AED=2\angle BEC$ .

Jorge Tipe - 7 years, 3 months ago

Thanks.

Megh Parikh - 7 years, 3 months ago

This problem wouldn't be so hard once you notice a property that $P$ carries from $AC=AP$ .

WLOG say that $H$ is between $A,M$ . Since $AC=AH\times AB$ , therefore $AP=AH*AB$ which is enough to show that $\triangle APH\sim \triangle ABP\Rightarrow \frac {PH}{PB}=\frac {AP}{AB}=\frac {AC}{AB}$ . Since $PM$ bisects $\angle BPH\iff \frac {PH}{PB}=\frac {HM}{BM}$ combining the results above we continue the equivalence: $\iff \frac {AC}{AB}=\frac {HM}{BM}=\frac {2HM}{AB}$ $\iff AC=2HM$ .

Instead to computing for $HM$ , we can geometrically show that $AC=2HM\iff \angle A$ . To relate $AC$ and $HM$ , we introduce the midpoint of $AC$ (I actually originally connected $GH$ , but then I found a simpler way). Since $\angle AGM=\angle CHM=90$ ( $CM=AM$ ) and $AG=\frac {AC}{2}=HM$ , by $HL$ congruence we have $\angle GAM=\angle CMA=\angle ACM\Rightarrow \angle CAB=60$ . Q.E.D

For the first one I have the same proof as Jorge :)

Xuming Liang - 7 years, 3 months ago

As a crazy thought for this problem, we could also argue by considering when Apollonius' circle of $H,B;M$ and the circle centered at $A$ with radius $AC$ intersect.

Xuming Liang - 7 years, 3 months ago

Thanks.

Megh Parikh - 7 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$