Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 2"
m |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Let <math> | + | 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== | ||
− | {{ | + | |
+ | 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>\anlge 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>. | ||
+ | |||
==See also== | ==See also== | ||
*[[2006 Romanian NMO Problems/Grade 9/Problem 1 | Previous problem]] | *[[2006 Romanian NMO Problems/Grade 9/Problem 1 | Previous problem]] |
Revision as of 22:48, 1 June 2011
Problem
Let and be isosceles triangles with the base . We know that . Let be the midpoint of . The points are chosen such that , , , and . Prove that is the midpoint of and .
Solution
Since is isosceles, . Since , , which means that , too. Thus , so by ASA, . This means that . Since , , and , by SAS, , so . Since , $\anlge EDA = \angle BDA - \angle BDE = \angle ADC - \angle MDP = \angle CDP$ (Error compiling LaTeX. Unknown error_msg). Thus . is the angle bisector of , and . This means that . , , and , so by SSS, . Thus and . , so by AA, . Thus . Also, . and , so . By SAS similarity, . is a median and an angle bisector of . Now assume that P' is the point such that DP' is a median of (it is on ). It is on DP, the angle bisector, and since , , but we also showed that . Thus . Since P and P' are on the same ray (), P = P' and P is the midpoint of .