Difference between revisions of "2006 Romanian NMO Problems/Grade 7/Problem 3"
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b) <math>CH=DE</math>. | b) <math>CH=DE</math>. | ||
==Solution== | ==Solution== | ||
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a) Note that quadrilateral <math>ABA_1B_1</math> is cyclic, because <math>\angle{AB_1B}=\angle{AA_1B}=90^{\circ}</math>. Thus <math>\angle{B_1A_1B}=180-\angle{B_1AB}</math> and <math>\angle{B_1A_1C}=\angle{B_1AB}</math>. Similarly <math>\angle{A_1B_1C}=\angle{ABC}</math>. Therefore <math>\triangle{A_1B_1C} \sim \triangle{ABC}</math> and <math>\dfrac{A_1B_1}{AB}=\dfrac{B_1C}{BC}</math>. However <math>\triangle{BB_1C}</math> is a <math>45-45-90</math> triangle, so <math>\dfrac{B_1C}{BC}=\dfrac{1}{\sqrt{2}}</math> and <math>AB=A_1B_1\sqrt{2}</math>. By Pythagorean theorem, <math>AB=\sqrt{A_1B^2+AA_1^2}</math>. However <math>AA_1=A_1C</math>, so <math>AB=\sqrt{A_1B^2+A_1C^2}</math>, and thus <math>A_1B_1\sqrt{2}=\sqrt{A_1B^2+A_1C^2}</math>, or <math>A_1B_1=\sqrt{\dfrac{A_1B^2+A_1C^2}{2}}</math>. | a) Note that quadrilateral <math>ABA_1B_1</math> is cyclic, because <math>\angle{AB_1B}=\angle{AA_1B}=90^{\circ}</math>. Thus <math>\angle{B_1A_1B}=180-\angle{B_1AB}</math> and <math>\angle{B_1A_1C}=\angle{B_1AB}</math>. Similarly <math>\angle{A_1B_1C}=\angle{ABC}</math>. Therefore <math>\triangle{A_1B_1C} \sim \triangle{ABC}</math> and <math>\dfrac{A_1B_1}{AB}=\dfrac{B_1C}{BC}</math>. However <math>\triangle{BB_1C}</math> is a <math>45-45-90</math> triangle, so <math>\dfrac{B_1C}{BC}=\dfrac{1}{\sqrt{2}}</math> and <math>AB=A_1B_1\sqrt{2}</math>. By Pythagorean theorem, <math>AB=\sqrt{A_1B^2+AA_1^2}</math>. However <math>AA_1=A_1C</math>, so <math>AB=\sqrt{A_1B^2+A_1C^2}</math>, and thus <math>A_1B_1\sqrt{2}=\sqrt{A_1B^2+A_1C^2}</math>, or <math>A_1B_1=\sqrt{\dfrac{A_1B^2+A_1C^2}{2}}</math>. | ||
− | b) <math>DE=\sqrt{A_1E^2+A_1D^2}</math>, <math>A_1E=A_1D=A_1B_1</math>, <math>DE=\sqrt{2A_1B_1^2}=A_1B_1\sqrt{2}=\sqrt{A_1B^2+A_1C^2}=\sqrt{A_1H^2+A_1C^2}=CH</math>. | + | b) <math>DE=\sqrt{A_1E^2+A_1D^2}</math>, <math>A_1E=A_1D=A_1B_1</math>, <math>DE=\sqrt{2A_1B_1^2}=A_1B_1\sqrt{2}=\sqrt{A_1B^2+A_1C^2}=\sqrt{A_1H^2+A_1C^2}=CH</math>. |
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+ | {{alternate solutions}} | ||
==See also== | ==See also== | ||
*[[2006 Romanian NMO Problems]] | *[[2006 Romanian NMO Problems]] | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 22:24, 26 November 2023
Problem
In the acute-angle triangle we have
. The points
and
are the feet of the altitudes from
and
, and
is the orthocenter of the triangle. We consider the points
and
on the segments
and
such that
. Prove that
a) ;
b) .
Solution
a) Note that quadrilateral is cyclic, because
. Thus
and
. Similarly
. Therefore
and
. However
is a
triangle, so
and
. By Pythagorean theorem,
. However
, so
, and thus
, or
.
b) ,
,
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.