Difference between revisions of "2007 IMO Problems/Problem 4"
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In <math>\triangle ABC</math> the bisector of <math>\angle{BCA}</math> intersects the circumcircle again at <math>R</math>, the perpendicular bisector of <math>BC</math> at <math>P</math>, and the perpendicular bisector of <math>AC</math> at <math>Q</math>. The midpoint of <math>BC</math> is <math>K</math> and the midpoint of <math>AC</math> is <math>L</math>. Prove that the triangles <math>RPK</math> and <math>RQL</math> have the same area. | In <math>\triangle ABC</math> the bisector of <math>\angle{BCA}</math> intersects the circumcircle again at <math>R</math>, the perpendicular bisector of <math>BC</math> at <math>P</math>, and the perpendicular bisector of <math>AC</math> at <math>Q</math>. The midpoint of <math>BC</math> is <math>K</math> and the midpoint of <math>AC</math> is <math>L</math>. Prove that the triangles <math>RPK</math> and <math>RQL</math> have the same area. | ||
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− | ==Solution | + | ==Solution 1 (Efficient)== |
<math>\angle{RQL}=90+\angle{QCL}=90+\dfrac{C}{2}</math>, and similarly <math>\angle{RPK}=90+\angle{PCK}=90+\dfrac{C}{2}</math>, we have <math>\angle{RQL}=\angle{RPK}</math>. Using triangle area formula <math>A=\dfrac{1}{2}bc\sin{\angle{A}}</math>, the problem is equivalent to proving <math>RQ*QL=RP*PK</math>, or <math>\dfrac{PK}{QL}=\dfrac{RQ}{RP}</math>. Draw line <math>QM</math> perpendicular to BC and intersects BC at <math>M</math>, then <math>QM=QL</math>, and <math>\dfrac{PC}{QC}=\dfrac{PK}{QM}=\dfrac{PK}{QL}</math>. Now the problem is equivalent to proving <math>\dfrac{PC}{QC}=\dfrac{RQ}{RP}</math>, or <math>RQ*QC=RP*PC</math>. Since <math>\angle{OPQ}=180-\angle{RPK}=180-\angle{RQL}=\angle{OQP}</math>, we have <math>OQ=OP=x</math>. Let the radius of the circumcircle be <math>r</math>, then the diameter through <math>P</math> is divided by point <math>P</math> into lengths of <math>r+x</math> and <math>r-x</math>. By power of point, <math>RP*PC=(r+x)(r-x)</math>. Similarly, <math>RQ*QC=(r+x)(r-x)</math>. Therefore <math>RP*PC=RQ*QC</math>. <math>\square</math> | <math>\angle{RQL}=90+\angle{QCL}=90+\dfrac{C}{2}</math>, and similarly <math>\angle{RPK}=90+\angle{PCK}=90+\dfrac{C}{2}</math>, we have <math>\angle{RQL}=\angle{RPK}</math>. Using triangle area formula <math>A=\dfrac{1}{2}bc\sin{\angle{A}}</math>, the problem is equivalent to proving <math>RQ*QL=RP*PK</math>, or <math>\dfrac{PK}{QL}=\dfrac{RQ}{RP}</math>. Draw line <math>QM</math> perpendicular to BC and intersects BC at <math>M</math>, then <math>QM=QL</math>, and <math>\dfrac{PC}{QC}=\dfrac{PK}{QM}=\dfrac{PK}{QL}</math>. Now the problem is equivalent to proving <math>\dfrac{PC}{QC}=\dfrac{RQ}{RP}</math>, or <math>RQ*QC=RP*PC</math>. Since <math>\angle{OPQ}=180-\angle{RPK}=180-\angle{RQL}=\angle{OQP}</math>, we have <math>OQ=OP=x</math>. Let the radius of the circumcircle be <math>r</math>, then the diameter through <math>P</math> is divided by point <math>P</math> into lengths of <math>r+x</math> and <math>r-x</math>. By power of point, <math>RP*PC=(r+x)(r-x)</math>. Similarly, <math>RQ*QC=(r+x)(r-x)</math>. Therefore <math>RP*PC=RQ*QC</math>. <math>\square</math> | ||
<math>(mathdummy)</math> | <math>(mathdummy)</math> | ||
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+ | ==Solution 2== | ||
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+ | The area of <math>\triangle{RQL}</math> is given by <math>\dfrac{1}{2}QL*RQ\sin{\angle{RQL}}</math> and the area of <math>\triangle{RPK}</math> is <math>\dfrac{1}{2}RP*PK\sin{\angle{RPK}}</math>. Let <math>\angle{BCA}=C</math>, <math>\angle{BAC}=A</math>, and <math>\angle{ABC}=B</math>. Now <math>\angle{KCP}=\angle{QCL}=\dfrac{C}{2}</math> and <math>\angle{PKC}=\angle{QLC}=90</math>, thus <math>\angle{RPK}=\angle{RQL}=90+\dfrac{C}{2}</math>. <math>\triangle{PKC} \sim \triangle{QLC}</math>, so <math>\dfrac{PK}{QL}=\dfrac{KC}{LC}</math>, or <math>\dfrac{PK}{QL}=\dfrac{BC}{AB}</math>. The ratio of the areas is <math>\dfrac{[RPK]}{[RQL]}=\dfrac{BC*RP}{AC*RQ}</math>. The two areas are only equal when the ratio is 1, therefore it suffices to show <math>\dfrac{RP}{RQ}=\dfrac{AC}{BC}</math>. Let <math>O</math> be the center of the circle. Then <math>\angle{ROK}=A+C</math>, and <math>\angle{ROP}=180-(A+C)=B</math>. Using law of sines on <math>\triangle{RPO}</math> we have: <math>\dfrac{RP}{\sin{B}}=\dfrac{OR}{\sin{(90+\dfrac{C}{2})}}</math> so <math>RP*\sin{(90+\dfrac{C}{2})}=OR*\sin{B}</math>. <math>OR*\sin{B}=\dfrac{1}{2}AC</math> by law of sines, and <math>\sin{(90+\dfrac{C}{2})}=\cos{\dfrac{C}{2}}</math>, thus 1) <math>2RP\cos{\dfrac{C}{2}}=AC</math>. Similarly, law of sines on <math>\triangle{ROQ}</math> results in <math>\dfrac{RQ}{\sin{(180-A)}}=\dfrac{OR}{\sin{(90-\dfrac{C}{2})}}</math> or <math>\dfrac{RQ}{\sin{A}}=\dfrac{OR}{\cos{\dfrac{C}{2}}}</math>. Cross multiplying we have <math>RQ\cos{\dfrac{C}{2}}=OR*\sin{A}</math> or 2) <math>2RQ\cos{\dfrac{C}{2}}=BC</math>. Dividing 1) by 2) we have <math>\dfrac{RP}{RQ}=\dfrac{AC}{BC}</math> <math>\square</math> | ||
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+ | <math>(tkhalid)</math> | ||
==Solution 3== | ==Solution 3== |
Revision as of 11:59, 24 February 2022
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.
Solution 1 (Efficient)
, and similarly
, we have
. Using triangle area formula
, the problem is equivalent to proving
, or
. Draw line
perpendicular to BC and intersects BC at
, then
, and
. Now the problem is equivalent to proving
, or
. 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 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
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 | ||
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