Difference between revisions of "2014 IMO Problems/Problem 4"
Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
+ | [asy] | ||
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
import graph; size(10.60000000000002cm); | import graph; size(10.60000000000002cm); | ||
Line 55: | Line 56: | ||
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
/* end of picture */ | /* end of picture */ | ||
+ | [/asy] | ||
We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. | We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. |
Revision as of 01:04, 14 December 2015
Problem
Points and
lie on side
of acute-angled
so that
and
. Points
and
lie on lines
and
, respectively, such that
is the midpoint of
, and
is the midpoint of
. Prove that lines
and
intersect on the circumcircle of
.
Solution
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10.60000000000002cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.740000000000007, xmax = 16.46000000000002, ymin = -7.520000000000004, ymax = 4.140000000000004; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000);
draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001)--cycle, zzttqq); draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006)--(7.660000000000009,-1.140000000000001)--cycle, red); draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073)--(-0.2200000000000002,-1.200000000000001)--cycle, qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813)--(1.800000000000002,3.640000000000004)--cycle, qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944)--(1.800000000000002,3.640000000000004)--cycle, red);
/* draw figures */
draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001), zzttqq); draw((-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001), zzttqq); draw((7.660000000000009,-1.140000000000001)--(1.800000000000002,3.640000000000004), zzttqq); draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006), red); draw(arc((7.660000000000009,-1.140000000000001),0.5000000000000008,140.7958863822920,180.4362538499006), red); draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073), qqwuqq); draw(arc((-0.2200000000000002,-1.200000000000001),0.5000000000000008,0.4362538499004549,67.34652063545073), qqwuqq); draw(arc((-0.2200000000000002,-1.200000000000001),0.4000000000000006,0.4362538499004549,67.34652063545073), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-106.1141136177082,-39.20411361770813), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.4000000000000006,-106.1141136177082,-39.20411361770813), qqwuqq); draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944), red); draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-112.6534793645495,-73.01347936454944), red); draw((-1.022670636276736,-6.130338243877306)--(7.660000000000009,-1.140000000000001)); draw((4.740746205921980,-5.986847110107199)--(-0.2200000000000002,-1.200000000000001)); draw((1.800000000000002,3.640000000000004)--(4.740746205921980,-5.986847110107199)); draw((1.800000000000002,3.640000000000004)--(-1.022670636276736,-6.130338243877306)); draw(circle((3.711084749329270,0.0008695880898521494), 4.110415438128883));
/* dots and labels */
dot((1.800000000000002,3.640000000000004),dotstyle);
label("", (1.880000000000002,3.760000000000004), NE * labelscalefactor);
dot((-0.2200000000000002,-1.200000000000001),dotstyle);
label("
", (-0.1400000000000008,-1.080000000000000), NE * labelscalefactor);
dot((7.660000000000009,-1.140000000000001),dotstyle);
label("
", (7.740000000000012,-1.020000000000000), NE * labelscalefactor);
dot((0.3886646818616330,-1.245169121938651),dotstyle);
label("
", (0.4600000000000001,-1.120000000000000), NE * labelscalefactor);
dot((3.270373102960991,-1.173423555053598),dotstyle);
label("
", (3.360000000000004,-1.060000000000000), NE * labelscalefactor);
dot((4.740746205921980,-5.986847110107199),dotstyle);
label("
", (4.820000000000006,-5.860000000000003), NE * labelscalefactor);
dot((-1.022670636276736,-6.130338243877306),dotstyle);
label("
", (-0.9400000000000020,-6.020000000000004), NE * labelscalefactor);
dot((2.709057008802497,-3.985539257126989),dotstyle);
label("
", (2.780000000000003,-3.860000000000002), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
We are trying to prove that the intersection of BM and CN, call it point D, is on the circumcircle of triangle ABC. In other words, we are trying to prove angle BAC plus angle BDC is 180 degrees. Let the intersection of BM and AN be point E, and the intersection of AM and CN be point F. Let us assume (angle BDC) + (angle BAC) = 180. Note: This is circular reasoning. If angle BDC plus angle BAC is 180, then angle BAC should be equal to angles BDN and CDM. We can quickly prove that the triangles ABC, APB, and AQC are similar, so angles BAC = AQC = APB. We also see that angles AQC = BQN = APB = CPF. Also because angles BEQ and NED, MFD and CFP are equal, the triangles BEQ and NED, MDF and FCP must be two pairs of similar triangles. Therefore we must prove angles CBM and ANC, AMB and BCN are equal. We have angles BQA = APC = NQC = BPM. We also have AQ = QN, AP = PM. Because the triangles ABP and ACQ are similar, we have EC/EN = BF/FM, so triangles BFM and NEC are similar. So the angles CBM and ANC, BCN and AMB are equal and we are done.
Solution 2
Let be the midpoint of
. Easy angle chasing gives
. Because
is the midpoint of
, the cotangent rule applied on triangle
gives us
Hence, by the cotangent rule on
, we have
Because the period of cotangent is
, but angles are less than
, we have
Similarly, we have Hence, if
and
intersect at
, then
by the Angle Sum in a Triangle Theorem. Hence,
is cyclic, which is equivalent to the desired result.
--Suli 23:27, 7 February 2015 (EST)
Solution 3
Let be the midpoint of
. By AA Similarity, triangles
and
are similar, so
and
. Similarly,
, and so triangle
is isosceles. Thus,
, and so
. Dividing both sides by 2, we have
, or
But we also have
, so triangles
and
are similar by
similarity. In particular,
. Similarly,
, so
. In addition, angle sum in triangle
gives
. Therefore, if we let lines
and
intersect at
, by Angle Sum in quadrilateral
concave
, and so convex
, which is enough to prove that
is cyclic. This completes the proof.
--Suli 10:38, 8 February 2015 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2014 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |