Difference between revisions of "2008 IMO Problems/Problem 1"
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== Solution == | == Solution == | ||
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Let <math>M_A</math>, <math>M_B</math>, and <math>M_C</math> be the midpoints of sides <math>BC</math>, <math>CA</math>, and <math>AB</math>, respectively. It's not hard to see that <math>M_BM_C\parallel BC</math>. We also have that <math>AH\perp BC</math>, so <math>AH \perp M_BM_C</math>. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that <math>H</math> is on the radical axis of the circles centered at <math>M_B</math> and <math>M_C</math>, so <math>A</math> is too. We then have <math>AC_1\cdot AC_2=AB_2\cdot AB_1\Rightarrow \frac{AB_2}{AC_1}=\frac{AC_2}{AB_1}</math>. This implies that <math>\triangle AB_2C_1\sim \triangle AC_2B_1</math>, so <math>\angle AB_2C_1=\angle AC_2B_1</math>. Therefore <math>\angle C_1B_2B_1=180^{\circ}-\angle AB_2C_1=180^{\circ}-\angle AC_2B_1</math>. This shows that quadrilateral <math>C_1C_2B_1B_2</math> is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments <math>C_1C_2</math> and <math>B_1B_2</math>. However, these are just the perpendicular bisectors of <math>AB</math> and <math>CA</math>, which meet at the circumcenter of <math>ABC</math>, so the circumcenter of <math>C_1C_2B_1B_2</math> is the circumcenter of triangle <math>ABC</math>. Similarly, the circumcenters of <math>A_1A_2B_1B_2</math> and <math>C_1C_2A_1A_2</math> are coincident with the circumcenter of <math>ABC</math>. The desired result follows. | Let <math>M_A</math>, <math>M_B</math>, and <math>M_C</math> be the midpoints of sides <math>BC</math>, <math>CA</math>, and <math>AB</math>, respectively. It's not hard to see that <math>M_BM_C\parallel BC</math>. We also have that <math>AH\perp BC</math>, so <math>AH \perp M_BM_C</math>. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that <math>H</math> is on the radical axis of the circles centered at <math>M_B</math> and <math>M_C</math>, so <math>A</math> is too. We then have <math>AC_1\cdot AC_2=AB_2\cdot AB_1\Rightarrow \frac{AB_2}{AC_1}=\frac{AC_2}{AB_1}</math>. This implies that <math>\triangle AB_2C_1\sim \triangle AC_2B_1</math>, so <math>\angle AB_2C_1=\angle AC_2B_1</math>. Therefore <math>\angle C_1B_2B_1=180^{\circ}-\angle AB_2C_1=180^{\circ}-\angle AC_2B_1</math>. This shows that quadrilateral <math>C_1C_2B_1B_2</math> is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments <math>C_1C_2</math> and <math>B_1B_2</math>. However, these are just the perpendicular bisectors of <math>AB</math> and <math>CA</math>, which meet at the circumcenter of <math>ABC</math>, so the circumcenter of <math>C_1C_2B_1B_2</math> is the circumcenter of triangle <math>ABC</math>. Similarly, the circumcenters of <math>A_1A_2B_1B_2</math> and <math>C_1C_2A_1A_2</math> are coincident with the circumcenter of <math>ABC</math>. The desired result follows. | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2008|before=First Problem|num-a=2}} |
Latest revision as of 01:08, 19 November 2023
Problem
An acute-angled triangle has orthocentre
. The circle passing through
with centre the midpoint of
intersects the line
at
and
. Similarly, the circle passing through
with centre the midpoint of
intersects the line
at
and
, and the circle passing through
with centre the midpoint of
intersects the line
at
and
. Show that
,
,
,
,
,
lie on a circle.
Solution
Let ,
, and
be the midpoints of sides
,
, and
, respectively. It's not hard to see that
. We also have that
, so
. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that
is on the radical axis of the circles centered at
and
, so
is too. We then have
. This implies that
, so
. Therefore
. This shows that quadrilateral
is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments
and
. However, these are just the perpendicular bisectors of
and
, which meet at the circumcenter of
, so the circumcenter of
is the circumcenter of triangle
. Similarly, the circumcenters of
and
are coincident with the circumcenter of
. The desired result follows.
See Also
2008 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |