Difference between revisions of "2007 IMO Problems/Problem 4"
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Solution by Ilikeapos | Solution by Ilikeapos | ||
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+ | ==Solution 4== | ||
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+ | <math>[\triangle{RPK}]=[\triangle{RQL}], LQ*RQ*\sin\angle{LQR}=KP*PR*\sin\angle{RPK}</math> | ||
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+ | Since <math>CR</math> bisects <math>\angle{ACB},\angle{QCL}=\angle{PCK}</math>, <math>OL,OK</math> are perpendicular to sides <math>AC,BC</math> separately, <math>\angle{QLC}=\angle{PKC}=90^{\circ}, \angle{CQL}=\angle{CPK}, \angle{LQR}=\angle{RPK}</math> | ||
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+ | So now, we only have to prove <math>RP*PK=LQ*QR</math>, which is <math>RP*(CQ+QP)*\cos\angle{CPK}=CQ*(QP+PR)*\cos\angle{CQL}</math>, as mentioned above, the two angles are the same, we have to prove that <math>RP(CQ+QP)=CQ(QP+PR)</math>, which is equivalent to <math>RP*QP=QP*CQ</math>, we have to prove <math>RP=CQ</math> | ||
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+ | Now notice that <math>\triangle{OQP}</math> is isosceles. <math>\angle{CQL}=\angle{OQP}=\angle{OPQ}</math>, construct <math>OJ \bot CR</math>, <math>CJ=JR,JQ=JP,CQ=QR</math> as desired | ||
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+ | ~bluesoul | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 02:08, 25 February 2022
Contents
Problem
In the bisector of
intersects the circumcircle again at
, the perpendicular bisector of
at
, and the perpendicular bisector of
at
. The midpoint of
is
and the midpoint of
is
. Prove that the triangles
and
have the same area.
Diagram
~KingRavi
Solution 1 (Efficient)
, and similarly
. Therefore,
. Using the triangle area formula
yields
after cancelling the sines and constant. Draw line
perpendicular to
that intersects
at
, then
because the perpendicular bisectors are congruent, (or alternatively
). This presents us
by similar triangles; now, we have only to prove
, or
.
Since , we have
is isosceles. Draw the perpendicular from
to
, intersecting at
. Then
for a real
, now because the perpendicular from the center of a circle to a chord bisects that chord,
. Let
, and then
, proving our claim.
Alternative END Solution (Power of a Point)
Since , we have
. Let the radius of the circumcircle be
, then the diameter through
is divided by point
into lengths of
and
. By power of point,
. Similarly,
. Therefore
.
Solution by ~KingRavi
Alternate Solution by ~mathdummy
Edifying edits made by ~TheGrandioseGeometrician
Solution 2
The area of is given by
and the area of
is
. Let
,
, and
. Now
and
, thus
.
, so
, or
. The ratio of the areas is
. The two areas are only equal when the ratio is 1, therefore it suffices to show
. Let
be the center of the circle. Then
, and
. Using law of sines on
we have:
so
.
by law of sines, and
, thus 1)
. Similarly, law of sines on
results in
or
. Cross multiplying we have
or 2)
. Dividing 1) by 2) we have
Solution 3
WLOG, let the diameter of be
We see that and
from right triangles
and
We now look at By the Extended Law of Sines on
we get that
Similarly,
We now look at By Ptolemy's Theorem, we have
which gives us
This means that
We now seek to relate the lengths computed with the areas.
To do this, we consider the altitude from to
This is to find the area of
Finding the area of
is similar.
We claim that In order to prove this, we will prove that
In other words, we wish to prove that
This is equivalent to proving that
Note that and
Therefore, we get that
Thus,
In this way, we get that the altidude from
to
has length
Therefore, we see that
and
so the two areas are equal.
Solution by Ilikeapos
Solution 4
Since bisects
,
are perpendicular to sides
separately,
So now, we only have to prove , which is
, as mentioned above, the two angles are the same, we have to prove that
, which is equivalent to
, we have to prove
Now notice that is isosceles.
, construct
,
as desired
~bluesoul
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
2007 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |