Bosnia and Herzegovina JBMO TST 2015 Problem 3

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Bosnia and Herzegovina JBMO TST 2015 Problem 3

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Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2015

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#1 h Sep 16, 2018, 1:55 PM • 1 Y

Y by Adventure10

Let $AD$ be an altitude of triangle $ABC$ , and let $M$ , $N$ and $P$ be midpoints of $AB$ , $AD$ and $BC$ , respectively. Furthermore let $K$ be a foot of perpendicular from point $D$ to line $AC$ , and let $T$ be point on extension of line $KD$ (over point $D$ ) such that $\mid DT \mid = \mid MN \mid + \mid DK \mid$ . If $\mid MP \mid = 2 \cdot \mid KN \mid$ , prove that $\mid AT \mid = \mid MC \mid$ .

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#2 h May 26, 2023, 8:06 AM • 1 Y

Y by preparingfor_icho

Im surprised to see that no one has posted a solution

First notice that $MP=2KN \implies MP=AD$ because $\angle{AKD}=90, N \to$ midpoint.And $MP$ is the midsegment so $AC=2MP \implies 2AD=AC$ , and this means $\triangle ADC$ is a $30-60-90$ triangle, and this also implies $2DK=DC$ since $\triangle ADC \sim \triangle DKC$ . We were given that $MN+DK=DT$ , and since $2DK=DC$ and $2MN=BD$ , it follows that $2DT=BC \implies DT=PC$ .Let $\angle{ACM}=\alpha \implies \angle{MCP}= 30-\alpha$ , and since $MP\parallel AC \to \angle{PMC}=\alpha$ , so $\angle{MPC}=150$ .Similarly, $\angle{KDC}=60=\angle{BDT}$ and $\angle{ADB}=90$ , so $\triangle MPC \cong \triangle ADT$ and we're done!

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