Mixing
Easy Compass Construction Problem
$begingroup$
Here is a tricky compass and straightedge construction problem.
Given triangle $triangle ABC$ and point $D$ on segment $overline{AB}$, construct point $P$ on line $overleftrightarrow{CD}$ such that $angle APB = angle BAC$.
This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.
euclidean-geometry geometric-construction
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
$endgroup$
add a comment |
$begingroup$
Here is a tricky compass and straightedge construction problem.
Given triangle $triangle ABC$ and point $D$ on segment $overline{AB}$, construct point $P$ on line $overleftrightarrow{CD}$ such that $angle APB = angle BAC$.
This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.
euclidean-geometry geometric-construction
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
$endgroup$
$begingroup$
Never mind this is easy. Simply construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
$endgroup$
– Joshua Benabou
Jul 21 '15 at 0:30
add a comment |
$begingroup$
Here is a tricky compass and straightedge construction problem.
Given triangle $triangle ABC$ and point $D$ on segment $overline{AB}$, construct point $P$ on line $overleftrightarrow{CD}$ such that $angle APB = angle BAC$.
This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.
euclidean-geometry geometric-construction
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
$endgroup$
Here is a tricky compass and straightedge construction problem.
Given triangle $triangle ABC$ and point $D$ on segment $overline{AB}$, construct point $P$ on line $overleftrightarrow{CD}$ such that $angle APB = angle BAC$.
This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.
euclidean-geometry geometric-construction
euclidean-geometry geometric-construction
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
edited Jul 21 '15 at 0:30
Joshua Benabou
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
asked Jul 20 '15 at 23:52
Joshua BenabouJoshua Benabou
2,557625
2,557625
$begingroup$
Never mind this is easy. Simply construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
$endgroup$
– Joshua Benabou
Jul 21 '15 at 0:30
add a comment |
$begingroup$
Never mind this is easy. Simply construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
$endgroup$
– Joshua Benabou
Jul 21 '15 at 0:30
$begingroup$
Never mind this is easy. Simply construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
$endgroup$
– Joshua Benabou
Jul 21 '15 at 0:30
$begingroup$
Never mind this is easy. Simply construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
$endgroup$
– Joshua Benabou
Jul 21 '15 at 0:30
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I misread the question the first time I answered this, so let me try again.
The condition that you gave $angle APB = angle BAC$ is the degenerate case of the "two angles that inscribe the same arc are equal" theorem. More specifically, if you were to draw the circumcircle of $triangle APB$, you would see that the tangent to $A$ is $AC$. Also, we know that
$$angle APB = 180^circ - angle AID = angle AIC$$ where $I$ is the intersection of the circumcircle of $APB$ with $BC$. This implies that
$$triangle ABC sim triangle IAC.$$ The construction of $I$ isn't that difficult (just copy $angle B$ onto side $AC$), and everything can be reversed to find $P$.
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
$endgroup$
add a comment |
$begingroup$
Construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
$endgroup$
add a comment |
$begingroup$
There are 2 possible cases – $P$ can be outside or inside of $triangle ABC$. Maybe that is why the OP claims that this is trick problem.
Case-1 ($P$ is outside, easier to start with for my $ABC$)
1) Draw line $g$, the perpendicular bisector of $AB$.
2) Draw line $h$, through $A$ and perpendicular to $AC$, cutting $g$ at $O$.
3) Draw circle $k$ using $O$ as center and $OA$ as radius.
4) Let circle $k$ cuts $CD$ (extended) at $P$.
Proof: By angles in alternate segment, $alpha = beta$.
Case-2 ($P$ is inside triangle ABC.)
[Continuing from the above]
1) Locate $P’$, the mirror refection of $P$ about $AB$. It should be clear that $theta = alpha$.
2) Draw circle $m$, passing through $A, B, P’$, cutting $CD$ at $P”$.
Proof: By angles in the same segment, $phi = theta$.
Result follows.
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
$endgroup$
add a comment |
$begingroup$
Construct line $AK$ with $K$ on $BC$ (produced, if necessary) and $angle BAK = angle ACB$. The circumcircle of $triangle ABK$ intersects $CD$ (again, produced if needed) at two points $P$ and $P1$. Without loss of generality, let $P$ be on the same side of $AB$ as $K$. Then, $angle APB = angle AKB$ and by construction $angle AKB = angle BAC$. (since$angle ABC + angle BCA + angle BAC = angle ABK + angle BAK + angle AKB) $.
Thus we are done
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1368180%2feasy-compass-construction-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I misread the question the first time I answered this, so let me try again.
The condition that you gave $angle APB = angle BAC$ is the degenerate case of the "two angles that inscribe the same arc are equal" theorem. More specifically, if you were to draw the circumcircle of $triangle APB$, you would see that the tangent to $A$ is $AC$. Also, we know that
$$angle APB = 180^circ - angle AID = angle AIC$$ where $I$ is the intersection of the circumcircle of $APB$ with $BC$. This implies that
$$triangle ABC sim triangle IAC.$$ The construction of $I$ isn't that difficult (just copy $angle B$ onto side $AC$), and everything can be reversed to find $P$.
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
$endgroup$
add a comment |
$begingroup$
I misread the question the first time I answered this, so let me try again.
The condition that you gave $angle APB = angle BAC$ is the degenerate case of the "two angles that inscribe the same arc are equal" theorem. More specifically, if you were to draw the circumcircle of $triangle APB$, you would see that the tangent to $A$ is $AC$. Also, we know that
$$angle APB = 180^circ - angle AID = angle AIC$$ where $I$ is the intersection of the circumcircle of $APB$ with $BC$. This implies that
$$triangle ABC sim triangle IAC.$$ The construction of $I$ isn't that difficult (just copy $angle B$ onto side $AC$), and everything can be reversed to find $P$.
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
$endgroup$
add a comment |
$begingroup$
I misread the question the first time I answered this, so let me try again.
The condition that you gave $angle APB = angle BAC$ is the degenerate case of the "two angles that inscribe the same arc are equal" theorem. More specifically, if you were to draw the circumcircle of $triangle APB$, you would see that the tangent to $A$ is $AC$. Also, we know that
$$angle APB = 180^circ - angle AID = angle AIC$$ where $I$ is the intersection of the circumcircle of $APB$ with $BC$. This implies that
$$triangle ABC sim triangle IAC.$$ The construction of $I$ isn't that difficult (just copy $angle B$ onto side $AC$), and everything can be reversed to find $P$.
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
$endgroup$
I misread the question the first time I answered this, so let me try again.
The condition that you gave $angle APB = angle BAC$ is the degenerate case of the "two angles that inscribe the same arc are equal" theorem. More specifically, if you were to draw the circumcircle of $triangle APB$, you would see that the tangent to $A$ is $AC$. Also, we know that
$$angle APB = 180^circ - angle AID = angle AIC$$ where $I$ is the intersection of the circumcircle of $APB$ with $BC$. This implies that
$$triangle ABC sim triangle IAC.$$ The construction of $I$ isn't that difficult (just copy $angle B$ onto side $AC$), and everything can be reversed to find $P$.
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
edited Jul 21 '15 at 0:37
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
answered Jul 21 '15 at 0:14
thkim1011thkim1011
817517
817517
add a comment |
add a comment |
$begingroup$
Construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
$endgroup$
add a comment |
$begingroup$
Construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
$endgroup$
add a comment |
$begingroup$
Construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
$endgroup$
Construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
answered Jul 21 '15 at 0:33
Joshua BenabouJoshua Benabou
2,557625
2,557625
add a comment |
add a comment |
$begingroup$
There are 2 possible cases – $P$ can be outside or inside of $triangle ABC$. Maybe that is why the OP claims that this is trick problem.
Case-1 ($P$ is outside, easier to start with for my $ABC$)
1) Draw line $g$, the perpendicular bisector of $AB$.
2) Draw line $h$, through $A$ and perpendicular to $AC$, cutting $g$ at $O$.
3) Draw circle $k$ using $O$ as center and $OA$ as radius.
4) Let circle $k$ cuts $CD$ (extended) at $P$.
Proof: By angles in alternate segment, $alpha = beta$.
Case-2 ($P$ is inside triangle ABC.)
[Continuing from the above]
1) Locate $P’$, the mirror refection of $P$ about $AB$. It should be clear that $theta = alpha$.
2) Draw circle $m$, passing through $A, B, P’$, cutting $CD$ at $P”$.
Proof: By angles in the same segment, $phi = theta$.
Result follows.
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
$endgroup$
add a comment |
$begingroup$
There are 2 possible cases – $P$ can be outside or inside of $triangle ABC$. Maybe that is why the OP claims that this is trick problem.
Case-1 ($P$ is outside, easier to start with for my $ABC$)
1) Draw line $g$, the perpendicular bisector of $AB$.
2) Draw line $h$, through $A$ and perpendicular to $AC$, cutting $g$ at $O$.
3) Draw circle $k$ using $O$ as center and $OA$ as radius.
4) Let circle $k$ cuts $CD$ (extended) at $P$.
Proof: By angles in alternate segment, $alpha = beta$.
Case-2 ($P$ is inside triangle ABC.)
[Continuing from the above]
1) Locate $P’$, the mirror refection of $P$ about $AB$. It should be clear that $theta = alpha$.
2) Draw circle $m$, passing through $A, B, P’$, cutting $CD$ at $P”$.
Proof: By angles in the same segment, $phi = theta$.
Result follows.
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
$endgroup$
add a comment |
$begingroup$
There are 2 possible cases – $P$ can be outside or inside of $triangle ABC$. Maybe that is why the OP claims that this is trick problem.
Case-1 ($P$ is outside, easier to start with for my $ABC$)
1) Draw line $g$, the perpendicular bisector of $AB$.
2) Draw line $h$, through $A$ and perpendicular to $AC$, cutting $g$ at $O$.
3) Draw circle $k$ using $O$ as center and $OA$ as radius.
4) Let circle $k$ cuts $CD$ (extended) at $P$.
Proof: By angles in alternate segment, $alpha = beta$.
Case-2 ($P$ is inside triangle ABC.)
[Continuing from the above]
1) Locate $P’$, the mirror refection of $P$ about $AB$. It should be clear that $theta = alpha$.
2) Draw circle $m$, passing through $A, B, P’$, cutting $CD$ at $P”$.
Proof: By angles in the same segment, $phi = theta$.
Result follows.
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
$endgroup$
There are 2 possible cases – $P$ can be outside or inside of $triangle ABC$. Maybe that is why the OP claims that this is trick problem.
Case-1 ($P$ is outside, easier to start with for my $ABC$)
1) Draw line $g$, the perpendicular bisector of $AB$.
2) Draw line $h$, through $A$ and perpendicular to $AC$, cutting $g$ at $O$.
3) Draw circle $k$ using $O$ as center and $OA$ as radius.
4) Let circle $k$ cuts $CD$ (extended) at $P$.
Proof: By angles in alternate segment, $alpha = beta$.
Case-2 ($P$ is inside triangle ABC.)
[Continuing from the above]
1) Locate $P’$, the mirror refection of $P$ about $AB$. It should be clear that $theta = alpha$.
2) Draw circle $m$, passing through $A, B, P’$, cutting $CD$ at $P”$.
Proof: By angles in the same segment, $phi = theta$.
Result follows.
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
answered Jul 21 '15 at 15:17
MickMick
11.9k21641
11.9k21641
add a comment |
add a comment |
$begingroup$
Construct line $AK$ with $K$ on $BC$ (produced, if necessary) and $angle BAK = angle ACB$. The circumcircle of $triangle ABK$ intersects $CD$ (again, produced if needed) at two points $P$ and $P1$. Without loss of generality, let $P$ be on the same side of $AB$ as $K$. Then, $angle APB = angle AKB$ and by construction $angle AKB = angle BAC$. (since$angle ABC + angle BCA + angle BAC = angle ABK + angle BAK + angle AKB) $.
Thus we are done
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
$endgroup$
add a comment |
$begingroup$
Construct line $AK$ with $K$ on $BC$ (produced, if necessary) and $angle BAK = angle ACB$. The circumcircle of $triangle ABK$ intersects $CD$ (again, produced if needed) at two points $P$ and $P1$. Without loss of generality, let $P$ be on the same side of $AB$ as $K$. Then, $angle APB = angle AKB$ and by construction $angle AKB = angle BAC$. (since$angle ABC + angle BCA + angle BAC = angle ABK + angle BAK + angle AKB) $.
Thus we are done
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
$endgroup$
add a comment |
$begingroup$
Construct line $AK$ with $K$ on $BC$ (produced, if necessary) and $angle BAK = angle ACB$. The circumcircle of $triangle ABK$ intersects $CD$ (again, produced if needed) at two points $P$ and $P1$. Without loss of generality, let $P$ be on the same side of $AB$ as $K$. Then, $angle APB = angle AKB$ and by construction $angle AKB = angle BAC$. (since$angle ABC + angle BCA + angle BAC = angle ABK + angle BAK + angle AKB) $.
Thus we are done
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
$endgroup$
Construct line $AK$ with $K$ on $BC$ (produced, if necessary) and $angle BAK = angle ACB$. The circumcircle of $triangle ABK$ intersects $CD$ (again, produced if needed) at two points $P$ and $P1$. Without loss of generality, let $P$ be on the same side of $AB$ as $K$. Then, $angle APB = angle AKB$ and by construction $angle AKB = angle BAC$. (since$angle ABC + angle BCA + angle BAC = angle ABK + angle BAK + angle AKB) $.
Thus we are done
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
answered Jan 15 at 15:37
PRITHU PURKAITPRITHU PURKAIT
11
11
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1368180%2feasy-compass-construction-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Never mind this is easy. Simply construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
$endgroup$
– Joshua Benabou
Jul 21 '15 at 0:30