JEMC 2016/2: Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2...
$begingroup$
Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2 respectively,
such that |AP| = |AQ|. The segment P Q intersects circles C1 and C2 in points M, N respectively. Let C be
the center of the arc BP of C1 which does not contain point A and let D be the center of arc BQ of C2 which
does not contain point A. Let E be the intersection of CM and DN. Prove that AE is perpendicular to CD.
--- Europeon Mathematical Cup 2016: Junior cathegory question 2:
http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Juniors_ENG.pdf
When I tried to solve this problem, I managed to prove that $$ E in AB, $$ which reduces the question to proving that $$ CD perp AE = AB perp O_1O_2 Rightarrow CDparallel O_1O_2 \ text{ (with } O_1 text{ and } O_2 text{ the centers of } C_1 text{ and } C_2 text{ respectively).} $$
How can I do this?
geometry contest-math
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
$endgroup$
add a comment |
$begingroup$
Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2 respectively,
such that |AP| = |AQ|. The segment P Q intersects circles C1 and C2 in points M, N respectively. Let C be
the center of the arc BP of C1 which does not contain point A and let D be the center of arc BQ of C2 which
does not contain point A. Let E be the intersection of CM and DN. Prove that AE is perpendicular to CD.
--- Europeon Mathematical Cup 2016: Junior cathegory question 2:
http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Juniors_ENG.pdf
When I tried to solve this problem, I managed to prove that $$ E in AB, $$ which reduces the question to proving that $$ CD perp AE = AB perp O_1O_2 Rightarrow CDparallel O_1O_2 \ text{ (with } O_1 text{ and } O_2 text{ the centers of } C_1 text{ and } C_2 text{ respectively).} $$
How can I do this?
geometry contest-math
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
$endgroup$
add a comment |
$begingroup$
Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2 respectively,
such that |AP| = |AQ|. The segment P Q intersects circles C1 and C2 in points M, N respectively. Let C be
the center of the arc BP of C1 which does not contain point A and let D be the center of arc BQ of C2 which
does not contain point A. Let E be the intersection of CM and DN. Prove that AE is perpendicular to CD.
--- Europeon Mathematical Cup 2016: Junior cathegory question 2:
http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Juniors_ENG.pdf
When I tried to solve this problem, I managed to prove that $$ E in AB, $$ which reduces the question to proving that $$ CD perp AE = AB perp O_1O_2 Rightarrow CDparallel O_1O_2 \ text{ (with } O_1 text{ and } O_2 text{ the centers of } C_1 text{ and } C_2 text{ respectively).} $$
How can I do this?
geometry contest-math
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
$endgroup$
Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2 respectively,
such that |AP| = |AQ|. The segment P Q intersects circles C1 and C2 in points M, N respectively. Let C be
the center of the arc BP of C1 which does not contain point A and let D be the center of arc BQ of C2 which
does not contain point A. Let E be the intersection of CM and DN. Prove that AE is perpendicular to CD.
--- Europeon Mathematical Cup 2016: Junior cathegory question 2:
http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Juniors_ENG.pdf
When I tried to solve this problem, I managed to prove that $$ E in AB, $$ which reduces the question to proving that $$ CD perp AE = AB perp O_1O_2 Rightarrow CDparallel O_1O_2 \ text{ (with } O_1 text{ and } O_2 text{ the centers of } C_1 text{ and } C_2 text{ respectively).} $$
How can I do this?
geometry contest-math
geometry contest-math
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
edited Dec 15 '18 at 17:05
Quang Hoang
13.1k1233
13.1k1233
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
asked Dec 15 '18 at 13:49
Jonas De SchouwerJonas De Schouwer
3458
3458
add a comment |
add a comment |
1 Answer
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$begingroup$
We can solve that $E$ is the orthocenter of $triangle ACD$. In the picture above,
$$angle BAD =frac{angle BAQ}2, angle BAC = frac{angle BAP}2.$$
So $angle CAD = frac12 angle PAQ$. Moreover,
$$angle MCA = angle QPA = angle PQA = frac{180^circ - angle PAQ}2.$$
So $angle MCA + angle CAD = 90^circ$, or $CMperp AD$. Similarly $DNperp AC$.
Note: it's not immediate for me to see that $A,E,B$ are colinear. Would be nice if you can include the proof in your question.s
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
$endgroup$
$begingroup$
I assumed that $N in AC$ and $M in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N in AC$ and $M in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven.
$endgroup$
– Jonas De Schouwer
Dec 16 '18 at 12:11
add a comment |
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1 Answer
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votes
$begingroup$
We can solve that $E$ is the orthocenter of $triangle ACD$. In the picture above,
$$angle BAD =frac{angle BAQ}2, angle BAC = frac{angle BAP}2.$$
So $angle CAD = frac12 angle PAQ$. Moreover,
$$angle MCA = angle QPA = angle PQA = frac{180^circ - angle PAQ}2.$$
So $angle MCA + angle CAD = 90^circ$, or $CMperp AD$. Similarly $DNperp AC$.
Note: it's not immediate for me to see that $A,E,B$ are colinear. Would be nice if you can include the proof in your question.s
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
$endgroup$
$begingroup$
I assumed that $N in AC$ and $M in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N in AC$ and $M in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven.
$endgroup$
– Jonas De Schouwer
Dec 16 '18 at 12:11
add a comment |
$begingroup$
We can solve that $E$ is the orthocenter of $triangle ACD$. In the picture above,
$$angle BAD =frac{angle BAQ}2, angle BAC = frac{angle BAP}2.$$
So $angle CAD = frac12 angle PAQ$. Moreover,
$$angle MCA = angle QPA = angle PQA = frac{180^circ - angle PAQ}2.$$
So $angle MCA + angle CAD = 90^circ$, or $CMperp AD$. Similarly $DNperp AC$.
Note: it's not immediate for me to see that $A,E,B$ are colinear. Would be nice if you can include the proof in your question.s
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
$endgroup$
$begingroup$
I assumed that $N in AC$ and $M in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N in AC$ and $M in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven.
$endgroup$
– Jonas De Schouwer
Dec 16 '18 at 12:11
add a comment |
$begingroup$
We can solve that $E$ is the orthocenter of $triangle ACD$. In the picture above,
$$angle BAD =frac{angle BAQ}2, angle BAC = frac{angle BAP}2.$$
So $angle CAD = frac12 angle PAQ$. Moreover,
$$angle MCA = angle QPA = angle PQA = frac{180^circ - angle PAQ}2.$$
So $angle MCA + angle CAD = 90^circ$, or $CMperp AD$. Similarly $DNperp AC$.
Note: it's not immediate for me to see that $A,E,B$ are colinear. Would be nice if you can include the proof in your question.s
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
$endgroup$
We can solve that $E$ is the orthocenter of $triangle ACD$. In the picture above,
$$angle BAD =frac{angle BAQ}2, angle BAC = frac{angle BAP}2.$$
So $angle CAD = frac12 angle PAQ$. Moreover,
$$angle MCA = angle QPA = angle PQA = frac{180^circ - angle PAQ}2.$$
So $angle MCA + angle CAD = 90^circ$, or $CMperp AD$. Similarly $DNperp AC$.
Note: it's not immediate for me to see that $A,E,B$ are colinear. Would be nice if you can include the proof in your question.s
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
answered Dec 15 '18 at 19:04
Quang HoangQuang Hoang
13.1k1233
13.1k1233
$begingroup$
I assumed that $N in AC$ and $M in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N in AC$ and $M in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven.
$endgroup$
– Jonas De Schouwer
Dec 16 '18 at 12:11
add a comment |
$begingroup$
I assumed that $N in AC$ and $M in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N in AC$ and $M in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven.
$endgroup$
– Jonas De Schouwer
Dec 16 '18 at 12:11
$begingroup$
I assumed that $N in AC$ and $M in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N in AC$ and $M in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven.
$endgroup$
– Jonas De Schouwer
Dec 16 '18 at 12:11
$begingroup$
I assumed that $N in AC$ and $M in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N in AC$ and $M in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven.
$endgroup$
– Jonas De Schouwer
Dec 16 '18 at 12:11
add a comment |
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