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Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 2"

 
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==Problem==
 
==Problem==
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Let <math>\displaystyle ABC</math> and <math>\displaystyle DBC</math> be isosceles triangle with the base <math>\displaystyle BC</math>. We know that <math>\displaystyle \measuredangle ABD = \frac{\pi}{2}</math>. Let <math>\displaystyle M</math> be the midpoint of <math>\displaystyle BC</math>. The points <math>\displaystyle E,F,P</math> are chosen such that <math>\displaystyle E \in (AB)</math>, <math>\displaystyle P \in (MC)</math>, <math>\displaystyle C \in (AF)</math>, and <math>\displaystyle \measuredangle BDE = \measuredangle ADP = \measuredangle CDF</math>. Prove that <math>\displaystyle P</math> is the midpoint of <math>\displaystyle EF</math> and <math>\displaystyle DP \perp EF</math>.
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Let <math>ABC</math> and <math>DBC</math> be [[isosceles triangle]]s with the base <math>BC</math>. We know that <math>\angle ABD = \frac{\pi}{2}</math>. Let <math>M</math> be the [[midpoint]] of <math>BC</math>. The points <math>E,F,P</math> are chosen such that <math>E \in (AB)</math>, <math>P \in (MC)</math>, <math>C \in (AF)</math>, and <math>\angle BDE = \angle ADP = \angle CDF</math>. Prove that <math>P</math> is the midpoint of <math>EF</math> and <math>DP \perp EF</math>.
 
==Solution==
 
==Solution==
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Since <math>\triangle BDC</math> is isosceles, <math>BD=DC</math>. Since <math>\angle ABD = \dfrac{\pi}{2}</math>, <math>\dfrac{\pi}{2} = \angle ABD = \angle ABC + \angle DBC = \angle ACB + \angle DCB = \angle ACD</math>, which means that <math>\angle DCF = \dfrac{\pi}{2}</math>, too. Thus <math>\angle EBD = \angle DCF</math>, so by ASA, <math>\triangle EBD \cong \triangle FCD</math>. This means that <math>ED = FD</math>. Since <math>AB = AC</math>, <math>BD = DC</math>, and <math>\angle ABD = \angle ACD</math>, by SAS, <math>\triangle ABD \cong \triangle ACD</math>, so <math>\angle BDA = \angle ADC</math>. Since <math>\angle BDE = \angle MDP</math>, <math>\angle EDA = \angle BDA - \angle BDE = \angle ADC - \angle MDP = \angle CDP</math>. Thus <math>\angle EDP = \angle EDA + \angle MDP = \angle PDC + \angle CDF = \angle PDF</math>. <math>DP</math> is the angle bisector of <math>\angle EDF</math>, and <math>ED = DF</math>. This means that <math>EF \perp DP</math>. <math>AM = AM</math>, <math>BM = MC</math>, and <math>AB = AC</math>, so by SSS, <math>\triangle ABM \cong \triangle ACM</math>. Thus <math>\angle AMC = \dfrac{\pi}{2}</math> and <math>\angle DMP = \dfrac{\pi}{2}</math>. <math>\angle MDP = \angle BDE</math>, so by AA, <math>\triangle BDE \sim \triangle MDP</math>. Thus <math>\dfrac{ED}{BD} = \dfrac{DP}{MD}</math>. Also, <math>\angle EDF = \angle EDC + \angle CDF = \angle EDC + \angle EDB = \angle BDC</math>. <math>BD = DC</math> and <math>ED = DF</math>, so <math>\dfrac{BD}{ED} = \dfrac{DC}{DF}</math>. By SAS similarity, <math>\triangle BDC \sim \triangle EDF</math>. <math>MD</math> is a median and an angle bisector of <math>\triangle BDC</math>. Now assume that P' is the point such that DP' is a median of <math>\triangle EDF</math> (it is on <math>EF</math>). It is on DP, the angle bisector, and since <math>\triangle BDC \sim \triangle EDF</math>, <math>\dfrac{ED}{BD} = \dfrac{DP'}{MD}</math>, but we also showed that <math>\dfrac{ED}{BD} = \dfrac{DP}{MD}</math>. Thus <math>DP' = DP</math>. Since P and P' are on the same ray (<math>DP</math>), P = P' and P is the midpoint of <math>EF</math>.
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==See also==
 
==See also==
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*[[2006 Romanian NMO Problems/Grade 9/Problem 1 | Previous problem]]
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*[[2006 Romanian NMO Problems/Grade 9/Problem 3 | Next problem]]
 
*[[2006 Romanian NMO Problems]]
 
*[[2006 Romanian NMO Problems]]
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 13:05, 15 December 2022

Problem

Let $ABC$ and $DBC$ be isosceles triangles with the base $BC$. We know that $\angle ABD = \frac{\pi}{2}$. Let $M$ be the midpoint of $BC$. The points $E,F,P$ are chosen such that $E \in (AB)$, $P \in (MC)$, $C \in (AF)$, and $\angle BDE = \angle ADP = \angle CDF$. Prove that $P$ is the midpoint of $EF$ and $DP \perp EF$.

Solution

Since $\triangle BDC$ is isosceles, $BD=DC$. Since $\angle ABD = \dfrac{\pi}{2}$, $\dfrac{\pi}{2} = \angle ABD = \angle ABC + \angle DBC = \angle ACB + \angle DCB = \angle ACD$, which means that $\angle DCF = \dfrac{\pi}{2}$, too. Thus $\angle EBD = \angle DCF$, so by ASA, $\triangle EBD \cong \triangle FCD$. This means that $ED = FD$. Since $AB = AC$, $BD = DC$, and $\angle ABD = \angle ACD$, by SAS, $\triangle ABD \cong \triangle ACD$, so $\angle BDA = \angle ADC$. Since $\angle BDE = \angle MDP$, $\angle EDA = \angle BDA - \angle BDE = \angle ADC - \angle MDP = \angle CDP$. Thus $\angle EDP = \angle EDA + \angle MDP = \angle PDC + \angle CDF = \angle PDF$. $DP$ is the angle bisector of $\angle EDF$, and $ED = DF$. This means that $EF \perp DP$. $AM = AM$, $BM = MC$, and $AB = AC$, so by SSS, $\triangle ABM \cong \triangle ACM$. Thus $\angle AMC = \dfrac{\pi}{2}$ and $\angle DMP = \dfrac{\pi}{2}$. $\angle MDP = \angle BDE$, so by AA, $\triangle BDE \sim \triangle MDP$. Thus $\dfrac{ED}{BD} = \dfrac{DP}{MD}$. Also, $\angle EDF = \angle EDC + \angle CDF = \angle EDC + \angle EDB = \angle BDC$. $BD = DC$ and $ED = DF$, so $\dfrac{BD}{ED} = \dfrac{DC}{DF}$. By SAS similarity, $\triangle BDC \sim \triangle EDF$. $MD$ is a median and an angle bisector of $\triangle BDC$. Now assume that P' is the point such that DP' is a median of $\triangle EDF$ (it is on $EF$). It is on DP, the angle bisector, and since $\triangle BDC \sim \triangle EDF$, $\dfrac{ED}{BD} = \dfrac{DP'}{MD}$, but we also showed that $\dfrac{ED}{BD} = \dfrac{DP}{MD}$. Thus $DP' = DP$. Since P and P' are on the same ray ($DP$), P = P' and P is the midpoint of $EF$.

See also

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