Mixing
Q3 from British Mathematical Olympiad Paper 2 1996
$begingroup$
I have been trying to practice my maths by going through old BMO papers and have come across this question for which I have no idea. They do not display answers for questions this old anymore.
Distinct points A, P, Q, R, and S lie on the circumference of a circle and AP, AQ, AR, and AS are
chords with the property that:
∠PAQ = ∠QAR = ∠RAS.
Prove that
AR(AP + AR) = AQ (AQ + AS).
SOURCE:
https://bmos.ukmt.org.uk/home/bmolot.pdf Page 4
NOTE: If needed could people edit the tags because I am not fully sure what tags are most appropriate.
geometry vectors contest-math euclidean-geometry
edited Feb 3 at 0:32
tomasz
24.1k23482
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
$endgroup$
add a comment |
$begingroup$
I have been trying to practice my maths by going through old BMO papers and have come across this question for which I have no idea. They do not display answers for questions this old anymore.
Distinct points A, P, Q, R, and S lie on the circumference of a circle and AP, AQ, AR, and AS are
chords with the property that:
∠PAQ = ∠QAR = ∠RAS.
Prove that
AR(AP + AR) = AQ (AQ + AS).
SOURCE:
https://bmos.ukmt.org.uk/home/bmolot.pdf Page 4
NOTE: If needed could people edit the tags because I am not fully sure what tags are most appropriate.
geometry vectors contest-math euclidean-geometry
edited Feb 3 at 0:32
tomasz
24.1k23482
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
$endgroup$
1
$begingroup$
This amounts to showing that the parallelopipeds spanned by AR and AP and AQ and AS (respectively) have the same area. The area of each is proportional to the product times the sine of the angles (which is equal by assumption). Also, you have two chords trisecting the angle PAS (in the nondegenerate case). I'm not sure if it helps.
$endgroup$
– tomasz
Feb 3 at 0:39
$begingroup$
My guess is that you should be able to do it by pure tirgonometry. Maybe using complex numbers to ease the calculations a little.
$endgroup$
– tomasz
Feb 3 at 0:43
add a comment |
$begingroup$
I have been trying to practice my maths by going through old BMO papers and have come across this question for which I have no idea. They do not display answers for questions this old anymore.
Distinct points A, P, Q, R, and S lie on the circumference of a circle and AP, AQ, AR, and AS are
chords with the property that:
∠PAQ = ∠QAR = ∠RAS.
Prove that
AR(AP + AR) = AQ (AQ + AS).
SOURCE:
https://bmos.ukmt.org.uk/home/bmolot.pdf Page 4
NOTE: If needed could people edit the tags because I am not fully sure what tags are most appropriate.
geometry vectors contest-math euclidean-geometry
edited Feb 3 at 0:32
tomasz
24.1k23482
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
$endgroup$
I have been trying to practice my maths by going through old BMO papers and have come across this question for which I have no idea. They do not display answers for questions this old anymore.
Distinct points A, P, Q, R, and S lie on the circumference of a circle and AP, AQ, AR, and AS are
chords with the property that:
∠PAQ = ∠QAR = ∠RAS.
Prove that
AR(AP + AR) = AQ (AQ + AS).
SOURCE:
https://bmos.ukmt.org.uk/home/bmolot.pdf Page 4
NOTE: If needed could people edit the tags because I am not fully sure what tags are most appropriate.
geometry vectors contest-math euclidean-geometry
geometry vectors contest-math euclidean-geometry
edited Feb 3 at 0:32
tomasz
24.1k23482
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
edited Feb 3 at 0:32
tomasz
24.1k23482
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
edited Feb 3 at 0:32
tomasz
24.1k23482
edited Feb 3 at 0:32
tomasz
24.1k23482
edited Feb 3 at 0:32
tomasz
24.1k23482
24.1k23482
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
asked Feb 3 at 0:03
wesdrxvrtgfwesdrxvrtgf
908
908
1
$begingroup$
This amounts to showing that the parallelopipeds spanned by AR and AP and AQ and AS (respectively) have the same area. The area of each is proportional to the product times the sine of the angles (which is equal by assumption). Also, you have two chords trisecting the angle PAS (in the nondegenerate case). I'm not sure if it helps.
$endgroup$
– tomasz
Feb 3 at 0:39
$begingroup$
My guess is that you should be able to do it by pure tirgonometry. Maybe using complex numbers to ease the calculations a little.
$endgroup$
– tomasz
Feb 3 at 0:43
add a comment |
1
$begingroup$
This amounts to showing that the parallelopipeds spanned by AR and AP and AQ and AS (respectively) have the same area. The area of each is proportional to the product times the sine of the angles (which is equal by assumption). Also, you have two chords trisecting the angle PAS (in the nondegenerate case). I'm not sure if it helps.
$endgroup$
– tomasz
Feb 3 at 0:39
$begingroup$
My guess is that you should be able to do it by pure tirgonometry. Maybe using complex numbers to ease the calculations a little.
$endgroup$
– tomasz
Feb 3 at 0:43
1
1
$begingroup$
This amounts to showing that the parallelopipeds spanned by AR and AP and AQ and AS (respectively) have the same area. The area of each is proportional to the product times the sine of the angles (which is equal by assumption). Also, you have two chords trisecting the angle PAS (in the nondegenerate case). I'm not sure if it helps.
$endgroup$
– tomasz
Feb 3 at 0:39
$begingroup$
This amounts to showing that the parallelopipeds spanned by AR and AP and AQ and AS (respectively) have the same area. The area of each is proportional to the product times the sine of the angles (which is equal by assumption). Also, you have two chords trisecting the angle PAS (in the nondegenerate case). I'm not sure if it helps.
$endgroup$
– tomasz
Feb 3 at 0:39
$begingroup$
My guess is that you should be able to do it by pure tirgonometry. Maybe using complex numbers to ease the calculations a little.
$endgroup$
– tomasz
Feb 3 at 0:43
$begingroup$
My guess is that you should be able to do it by pure tirgonometry. Maybe using complex numbers to ease the calculations a little.
$endgroup$
– tomasz
Feb 3 at 0:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$begin{align}
angle PAQ=angle QAR=angle RAS &quadimpliesquad |PQ|=|QR|=|RS| &=: a \
angle PAR=angle QAS &quadimpliesquad |PR|=|QS| &=: b
end{align}$$
Now, we simply invoke Ptolemy's Theorem on two inscribed quadrilaterals:
$$left.begin{align}
square APQR: &quad p a + r a = q b \
square AQRS: &quad q a + s a = r b
end{align}quadright}quadimpliesquad frac{p+r}{q} = frac{b}{a} = frac{q+s}{r}quadimpliesquad r(p+r) = q(q+s)$$
answered Feb 3 at 9:24
BlueBlue
49.7k870158
$endgroup$
add a comment |
$begingroup$
A synthetic solution:
The basic principle here? Keep adding things to the picture until we have enough to make it work.
First, the basic diagram with our five points $A,P,Q,R,S$ on a circle $Gamma$ and some equal angles marked. Also, for future use, I've labeled the segments $AQ$ and $AR$.
Second, that expression $AR(AP+AR)$ kind of reminds me of Power of a Point. Let's make it actually into a case of that - extend $PA$ to a point $B$ such that $AB=AR$, and extend $SA$ to $C$ such that $AC=AQ$. Also, draw in $BR$ and $CQ$ to close those isosceles triangles:
Well, now we have two different points that we're computing powers from. It's the same circle $Gamma$, so we win if they're the same distance from the center. That's not much use directly, but the new lines we've drawn in are interesting. The isosceles triangles make $angle AQC =angle QCA=frac12angle QAS = angle QAR$, so $CQ$ and $AR$ are parallel. Similarly, $BR$ and $AQ$ are parallel. Extend $BR$ and $CQ$ to meet at $D$ and complete the parallelogram:
We have $angle CDB = angle ACD = angle DBA$, so $ACDR$ and $AQDB$ are isosceles trapezoids. The perpendicular bisector of $AQ$, which passes through the center of $Gamma$, also bisects $BD$, so $D$ is the reflection of $B$ across this bisector. In particular, if $R'$ is the second intersection of $BD$ with $Gamma$, then $BR'=DR$ and $BR=DR'$, so $DRcdot DR'=BR'cdot BR=BAcdot BP=ARcdot (AR+AP)$. Similarly, let $Q'$ be the second intersection of $CD$ with $Gamma$. $D$ is the reflection of $C$ across the perpendicular bisector of $AR$, which also passes through the center of $Gamma$ and bisects $CD$, so $CQ'=DQ$ and $CQ=DQ'$. Thus $DQcdot DQ' = CQ'cdot CQ=CAcdot CS = AQcdot (AQ+AS)$. But of course, $DQcdot DQ'=DRcdot DR'$, so $ARcdot (AR+AP)=AQcdot (AQ+AS)$. Done.
As it happens, $QQ'=AP$ and $RR'=AS$. We didn't use these directly, but in the end we can mark off an awful lot of equal lengths in the picture. My method is certainly not the only one that works well.
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$$begin{align}
angle PAQ=angle QAR=angle RAS &quadimpliesquad |PQ|=|QR|=|RS| &=: a \
angle PAR=angle QAS &quadimpliesquad |PR|=|QS| &=: b
end{align}$$
Now, we simply invoke Ptolemy's Theorem on two inscribed quadrilaterals:
$$left.begin{align}
square APQR: &quad p a + r a = q b \
square AQRS: &quad q a + s a = r b
end{align}quadright}quadimpliesquad frac{p+r}{q} = frac{b}{a} = frac{q+s}{r}quadimpliesquad r(p+r) = q(q+s)$$
answered Feb 3 at 9:24
BlueBlue
49.7k870158
$endgroup$
add a comment |
$begingroup$
$$begin{align}
angle PAQ=angle QAR=angle RAS &quadimpliesquad |PQ|=|QR|=|RS| &=: a \
angle PAR=angle QAS &quadimpliesquad |PR|=|QS| &=: b
end{align}$$
Now, we simply invoke Ptolemy's Theorem on two inscribed quadrilaterals:
$$left.begin{align}
square APQR: &quad p a + r a = q b \
square AQRS: &quad q a + s a = r b
end{align}quadright}quadimpliesquad frac{p+r}{q} = frac{b}{a} = frac{q+s}{r}quadimpliesquad r(p+r) = q(q+s)$$
answered Feb 3 at 9:24
BlueBlue
49.7k870158
$endgroup$
add a comment |
$begingroup$
$$begin{align}
angle PAQ=angle QAR=angle RAS &quadimpliesquad |PQ|=|QR|=|RS| &=: a \
angle PAR=angle QAS &quadimpliesquad |PR|=|QS| &=: b
end{align}$$
Now, we simply invoke Ptolemy's Theorem on two inscribed quadrilaterals:
$$left.begin{align}
square APQR: &quad p a + r a = q b \
square AQRS: &quad q a + s a = r b
end{align}quadright}quadimpliesquad frac{p+r}{q} = frac{b}{a} = frac{q+s}{r}quadimpliesquad r(p+r) = q(q+s)$$
answered Feb 3 at 9:24
BlueBlue
49.7k870158
$endgroup$
$$begin{align}
angle PAQ=angle QAR=angle RAS &quadimpliesquad |PQ|=|QR|=|RS| &=: a \
angle PAR=angle QAS &quadimpliesquad |PR|=|QS| &=: b
end{align}$$
Now, we simply invoke Ptolemy's Theorem on two inscribed quadrilaterals:
$$left.begin{align}
square APQR: &quad p a + r a = q b \
square AQRS: &quad q a + s a = r b
end{align}quadright}quadimpliesquad frac{p+r}{q} = frac{b}{a} = frac{q+s}{r}quadimpliesquad r(p+r) = q(q+s)$$
answered Feb 3 at 9:24
BlueBlue
49.7k870158
answered Feb 3 at 9:24
BlueBlue
49.7k870158
answered Feb 3 at 9:24
BlueBlue
49.7k870158
answered Feb 3 at 9:24
BlueBlue
49.7k870158
49.7k870158
add a comment |
add a comment |
$begingroup$
A synthetic solution:
The basic principle here? Keep adding things to the picture until we have enough to make it work.
First, the basic diagram with our five points $A,P,Q,R,S$ on a circle $Gamma$ and some equal angles marked. Also, for future use, I've labeled the segments $AQ$ and $AR$.
Second, that expression $AR(AP+AR)$ kind of reminds me of Power of a Point. Let's make it actually into a case of that - extend $PA$ to a point $B$ such that $AB=AR$, and extend $SA$ to $C$ such that $AC=AQ$. Also, draw in $BR$ and $CQ$ to close those isosceles triangles:
Well, now we have two different points that we're computing powers from. It's the same circle $Gamma$, so we win if they're the same distance from the center. That's not much use directly, but the new lines we've drawn in are interesting. The isosceles triangles make $angle AQC =angle QCA=frac12angle QAS = angle QAR$, so $CQ$ and $AR$ are parallel. Similarly, $BR$ and $AQ$ are parallel. Extend $BR$ and $CQ$ to meet at $D$ and complete the parallelogram:
We have $angle CDB = angle ACD = angle DBA$, so $ACDR$ and $AQDB$ are isosceles trapezoids. The perpendicular bisector of $AQ$, which passes through the center of $Gamma$, also bisects $BD$, so $D$ is the reflection of $B$ across this bisector. In particular, if $R'$ is the second intersection of $BD$ with $Gamma$, then $BR'=DR$ and $BR=DR'$, so $DRcdot DR'=BR'cdot BR=BAcdot BP=ARcdot (AR+AP)$. Similarly, let $Q'$ be the second intersection of $CD$ with $Gamma$. $D$ is the reflection of $C$ across the perpendicular bisector of $AR$, which also passes through the center of $Gamma$ and bisects $CD$, so $CQ'=DQ$ and $CQ=DQ'$. Thus $DQcdot DQ' = CQ'cdot CQ=CAcdot CS = AQcdot (AQ+AS)$. But of course, $DQcdot DQ'=DRcdot DR'$, so $ARcdot (AR+AP)=AQcdot (AQ+AS)$. Done.
As it happens, $QQ'=AP$ and $RR'=AS$. We didn't use these directly, but in the end we can mark off an awful lot of equal lengths in the picture. My method is certainly not the only one that works well.
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
$endgroup$
add a comment |
$begingroup$
A synthetic solution:
The basic principle here? Keep adding things to the picture until we have enough to make it work.
First, the basic diagram with our five points $A,P,Q,R,S$ on a circle $Gamma$ and some equal angles marked. Also, for future use, I've labeled the segments $AQ$ and $AR$.
Second, that expression $AR(AP+AR)$ kind of reminds me of Power of a Point. Let's make it actually into a case of that - extend $PA$ to a point $B$ such that $AB=AR$, and extend $SA$ to $C$ such that $AC=AQ$. Also, draw in $BR$ and $CQ$ to close those isosceles triangles:
Well, now we have two different points that we're computing powers from. It's the same circle $Gamma$, so we win if they're the same distance from the center. That's not much use directly, but the new lines we've drawn in are interesting. The isosceles triangles make $angle AQC =angle QCA=frac12angle QAS = angle QAR$, so $CQ$ and $AR$ are parallel. Similarly, $BR$ and $AQ$ are parallel. Extend $BR$ and $CQ$ to meet at $D$ and complete the parallelogram:
We have $angle CDB = angle ACD = angle DBA$, so $ACDR$ and $AQDB$ are isosceles trapezoids. The perpendicular bisector of $AQ$, which passes through the center of $Gamma$, also bisects $BD$, so $D$ is the reflection of $B$ across this bisector. In particular, if $R'$ is the second intersection of $BD$ with $Gamma$, then $BR'=DR$ and $BR=DR'$, so $DRcdot DR'=BR'cdot BR=BAcdot BP=ARcdot (AR+AP)$. Similarly, let $Q'$ be the second intersection of $CD$ with $Gamma$. $D$ is the reflection of $C$ across the perpendicular bisector of $AR$, which also passes through the center of $Gamma$ and bisects $CD$, so $CQ'=DQ$ and $CQ=DQ'$. Thus $DQcdot DQ' = CQ'cdot CQ=CAcdot CS = AQcdot (AQ+AS)$. But of course, $DQcdot DQ'=DRcdot DR'$, so $ARcdot (AR+AP)=AQcdot (AQ+AS)$. Done.
As it happens, $QQ'=AP$ and $RR'=AS$. We didn't use these directly, but in the end we can mark off an awful lot of equal lengths in the picture. My method is certainly not the only one that works well.
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
$endgroup$
add a comment |
$begingroup$
A synthetic solution:
The basic principle here? Keep adding things to the picture until we have enough to make it work.
First, the basic diagram with our five points $A,P,Q,R,S$ on a circle $Gamma$ and some equal angles marked. Also, for future use, I've labeled the segments $AQ$ and $AR$.
Second, that expression $AR(AP+AR)$ kind of reminds me of Power of a Point. Let's make it actually into a case of that - extend $PA$ to a point $B$ such that $AB=AR$, and extend $SA$ to $C$ such that $AC=AQ$. Also, draw in $BR$ and $CQ$ to close those isosceles triangles:
Well, now we have two different points that we're computing powers from. It's the same circle $Gamma$, so we win if they're the same distance from the center. That's not much use directly, but the new lines we've drawn in are interesting. The isosceles triangles make $angle AQC =angle QCA=frac12angle QAS = angle QAR$, so $CQ$ and $AR$ are parallel. Similarly, $BR$ and $AQ$ are parallel. Extend $BR$ and $CQ$ to meet at $D$ and complete the parallelogram:
We have $angle CDB = angle ACD = angle DBA$, so $ACDR$ and $AQDB$ are isosceles trapezoids. The perpendicular bisector of $AQ$, which passes through the center of $Gamma$, also bisects $BD$, so $D$ is the reflection of $B$ across this bisector. In particular, if $R'$ is the second intersection of $BD$ with $Gamma$, then $BR'=DR$ and $BR=DR'$, so $DRcdot DR'=BR'cdot BR=BAcdot BP=ARcdot (AR+AP)$. Similarly, let $Q'$ be the second intersection of $CD$ with $Gamma$. $D$ is the reflection of $C$ across the perpendicular bisector of $AR$, which also passes through the center of $Gamma$ and bisects $CD$, so $CQ'=DQ$ and $CQ=DQ'$. Thus $DQcdot DQ' = CQ'cdot CQ=CAcdot CS = AQcdot (AQ+AS)$. But of course, $DQcdot DQ'=DRcdot DR'$, so $ARcdot (AR+AP)=AQcdot (AQ+AS)$. Done.
As it happens, $QQ'=AP$ and $RR'=AS$. We didn't use these directly, but in the end we can mark off an awful lot of equal lengths in the picture. My method is certainly not the only one that works well.
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
$endgroup$
A synthetic solution:
The basic principle here? Keep adding things to the picture until we have enough to make it work.
First, the basic diagram with our five points $A,P,Q,R,S$ on a circle $Gamma$ and some equal angles marked. Also, for future use, I've labeled the segments $AQ$ and $AR$.
Second, that expression $AR(AP+AR)$ kind of reminds me of Power of a Point. Let's make it actually into a case of that - extend $PA$ to a point $B$ such that $AB=AR$, and extend $SA$ to $C$ such that $AC=AQ$. Also, draw in $BR$ and $CQ$ to close those isosceles triangles:
Well, now we have two different points that we're computing powers from. It's the same circle $Gamma$, so we win if they're the same distance from the center. That's not much use directly, but the new lines we've drawn in are interesting. The isosceles triangles make $angle AQC =angle QCA=frac12angle QAS = angle QAR$, so $CQ$ and $AR$ are parallel. Similarly, $BR$ and $AQ$ are parallel. Extend $BR$ and $CQ$ to meet at $D$ and complete the parallelogram:
We have $angle CDB = angle ACD = angle DBA$, so $ACDR$ and $AQDB$ are isosceles trapezoids. The perpendicular bisector of $AQ$, which passes through the center of $Gamma$, also bisects $BD$, so $D$ is the reflection of $B$ across this bisector. In particular, if $R'$ is the second intersection of $BD$ with $Gamma$, then $BR'=DR$ and $BR=DR'$, so $DRcdot DR'=BR'cdot BR=BAcdot BP=ARcdot (AR+AP)$. Similarly, let $Q'$ be the second intersection of $CD$ with $Gamma$. $D$ is the reflection of $C$ across the perpendicular bisector of $AR$, which also passes through the center of $Gamma$ and bisects $CD$, so $CQ'=DQ$ and $CQ=DQ'$. Thus $DQcdot DQ' = CQ'cdot CQ=CAcdot CS = AQcdot (AQ+AS)$. But of course, $DQcdot DQ'=DRcdot DR'$, so $ARcdot (AR+AP)=AQcdot (AQ+AS)$. Done.
As it happens, $QQ'=AP$ and $RR'=AS$. We didn't use these directly, but in the end we can mark off an awful lot of equal lengths in the picture. My method is certainly not the only one that works well.
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
answered Feb 3 at 2:37
jmerryjmerry
17.1k11633
17.1k11633
add a comment |
add a comment |
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$begingroup$
This amounts to showing that the parallelopipeds spanned by AR and AP and AQ and AS (respectively) have the same area. The area of each is proportional to the product times the sine of the angles (which is equal by assumption). Also, you have two chords trisecting the angle PAS (in the nondegenerate case). I'm not sure if it helps.
$endgroup$
– tomasz
Feb 3 at 0:39
$begingroup$
My guess is that you should be able to do it by pure tirgonometry. Maybe using complex numbers to ease the calculations a little.
$endgroup$
– tomasz
Feb 3 at 0:43