Ignoring the tail of the letter â€œQâ€ in the node of a TikZ diagram

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In the following TikZ diagram, $triangleABC$ is an acute triangle, and $trianglePQR$ is inscribed in it. L is the point on the line through P and Q that is closest to A.

I would like to typeset the label for Q at the intersection of the green lines along AC and LP as if I were typesetting the letter "O." The tail of the letter "Q" is pushing it too far from the point it represents.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$Q$;

endtikzpicture

enddocument

asked 44 mins ago

Adelyn

1,381923

add a commentÂ |Â

up vote
2
down vote

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In the following TikZ diagram, $triangleABC$ is an acute triangle, and $trianglePQR$ is inscribed in it. L is the point on the line through P and Q that is closest to A.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$Q$;

endtikzpicture

enddocument

asked 44 mins ago

Adelyn

1,381923

add a commentÂ |Â

up vote
2
down vote

favorite

In the following TikZ diagram, $triangleABC$ is an acute triangle, and $trianglePQR$ is inscribed in it. L is the point on the line through P and Q that is closest to A.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$Q$;

endtikzpicture

enddocument

asked 44 mins ago

Adelyn

1,381923

In the following TikZ diagram, $triangleABC$ is an acute triangle, and $trianglePQR$ is inscribed in it. L is the point on the line through P and Q that is closest to A.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$Q$;

endtikzpicture

enddocument

tikz-pgf

asked 44 mins ago

Adelyn

1,381923

asked 44 mins ago

Adelyn

1,381923

asked 44 mins ago

Adelyn

1,381923

asked 44 mins ago

Adelyn

1,381923

asked 44 mins ago

Adelyn

1,381923

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
2
down vote

Here is a proposal.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize,text=blue] 
(aux) at (label_Q) $O$;
path node[anchor=north west, inner sep=0, font=footnotesize] at (aux.north west)
$Q$;
endtikzpicture
enddocument

enter image description here

Of course, you may replace text=blue by opacity=0 in the aux node, I just keep it to prove that the Q is precisely where the O is in the sense that the little wiggle that distinguishes these glyphs is not respected.

answered 36 mins ago

marmot

67.2k473143

What is (aux.north west)? I see that (aux) is the label for the coordinate for the label 'O' if the letter 'O' were to be typeset at the intersection of the two green lines.
â€“Â Adelyn
23 mins ago

1

@Adelyn it is the north east corner of the aux node. All I do is to make sure that the north east corner of the Q node is at the same position as the north east corner of the O node. That way I make sure that the Q sits at the same position as the O, and the wiggle gets ignored. And O is set (in the present version), even in blue, but you can't see it because the Q sits precisely on top. BTW, if you add Bernards proposal instead of mine, you will see that his Q does not sit precisely on top of the O (but his proposal looks nice, too, IMHO).
â€“Â marmot
21 mins ago

Thanks for the careful explanation.
â€“Â Adelyn
19 mins ago

add a commentÂ |Â

up vote
1
down vote

You can simply replace the last line of code for the tikzpicture with

path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$smashQ$;

enter image description here

answered 33 mins ago

Bernard

159k764192

The tail of the "Q" is pushing the letter away from the intersection of the two green lines.
â€“Â Adelyn
28 mins ago

I don't think so since Q is Ã¢Â€Â˜smashedÃ¢Â€Â™, which amounts to make it have 0 height and depth. Change it for a letter without descenders like N, and you'll see it's placed the same way: the base line is about 1 pt from the green line.
â€“Â Bernard
19 mins ago

Dear Bernard If you replace the last node in my code by your proposal, you will see that @Adelyn is right.
â€“Â marmot
17 mins ago

I agree - it "should" be a command that gives me the letter Q typeset where I want it. It does not, though. Compare your diagram with marmot's diagram.
â€“Â Adelyn
16 mins ago

My code was just a work-around as simple as possible. Marmot knowsTiKZ much better than I do Ã¢Â€Â“ indeed I know pstricks better than TiKZ, and could place the letter Q anywhere you want, with orientation you want.
â€“Â Bernard
4 mins ago

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

Here is a proposal.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize,text=blue] 
(aux) at (label_Q) $O$;
path node[anchor=north west, inner sep=0, font=footnotesize] at (aux.north west)
$Q$;
endtikzpicture
enddocument

enter image description here

answered 36 mins ago

marmot

67.2k473143

What is (aux.north west)? I see that (aux) is the label for the coordinate for the label 'O' if the letter 'O' were to be typeset at the intersection of the two green lines.
â€“Â Adelyn
23 mins ago

1

@Adelyn it is the north east corner of the aux node. All I do is to make sure that the north east corner of the Q node is at the same position as the north east corner of the O node. That way I make sure that the Q sits at the same position as the O, and the wiggle gets ignored. And O is set (in the present version), even in blue, but you can't see it because the Q sits precisely on top. BTW, if you add Bernards proposal instead of mine, you will see that his Q does not sit precisely on top of the O (but his proposal looks nice, too, IMHO).
â€“Â marmot
21 mins ago

Thanks for the careful explanation.
â€“Â Adelyn
19 mins ago

add a commentÂ |Â

up vote
2
down vote

Here is a proposal.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize,text=blue] 
(aux) at (label_Q) $O$;
path node[anchor=north west, inner sep=0, font=footnotesize] at (aux.north west)
$Q$;
endtikzpicture
enddocument

enter image description here

answered 36 mins ago

marmot

67.2k473143

What is (aux.north west)? I see that (aux) is the label for the coordinate for the label 'O' if the letter 'O' were to be typeset at the intersection of the two green lines.
â€“Â Adelyn
23 mins ago

1

@Adelyn it is the north east corner of the aux node. All I do is to make sure that the north east corner of the Q node is at the same position as the north east corner of the O node. That way I make sure that the Q sits at the same position as the O, and the wiggle gets ignored. And O is set (in the present version), even in blue, but you can't see it because the Q sits precisely on top. BTW, if you add Bernards proposal instead of mine, you will see that his Q does not sit precisely on top of the O (but his proposal looks nice, too, IMHO).
â€“Â marmot
21 mins ago

Thanks for the careful explanation.
â€“Â Adelyn
19 mins ago

add a commentÂ |Â

up vote
2
down vote

Here is a proposal.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize,text=blue] 
(aux) at (label_Q) $O$;
path node[anchor=north west, inner sep=0, font=footnotesize] at (aux.north west)
$Q$;
endtikzpicture
enddocument

enter image description here

answered 36 mins ago

marmot

67.2k473143

Here is a proposal.

documentclassamsart
usepackageamsmath
usepackageamsfonts

usepackagetikz
usetikzlibrarycalc,intersections

begindocument

begintikzpicture

%A triangle and its orthic triangle are drawn.
path (-1.75,0) coordinate (A) (2.75,0) coordinate (B) (0,3.25) coordinate (C);
draw (A) -- (B) -- (C) -- cycle;
%
path node[anchor=north, inner sep=0, font=footnotesize] at ($(A) +(0,-0.15)$)$A$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(B) +(0,-0.15)$)$B$;
path node[anchor=south, inner sep=0, font=footnotesize] at ($(C) +(0,0.15)$)$C$;
%
path ($(B)!(A)!(C)$) coordinate (P) ($(A)!(B)!(C)$) coordinate (Q) (0,0) coordinate (R);
draw (P) -- (Q) -- (R) -- cycle;
%
%
%The foot of the altitude from A onto the line through P and Q is labeled L.
coordinate (L) at ($(P)!(A)!(Q)$);
draw[dashed] (L) -- (Q);
draw let p1=($(L)-(P)$), n1=atan(y1/x1) in node[anchor=n1, inner sep=0, font=footnotesize] at ($(L) +(n1+180:0.15)$)$L$;

draw let p1=($(A)-(P)$), n1=atan(y1/x1) in node[anchor=n1+180, inner sep=0, font=footnotesize] at ($(P) +(n1:0.15)$)$P$;
path node[anchor=north, inner sep=0, font=footnotesize] at ($(R) +(0,-0.15)$)textitR;


draw[name path=path_1, green] ($(A)!0.15cm!90:(C)$) -- ($(C)!0.15cm!-90:(A)$);
draw[name path=path_2, green] ($(P)!0.15cm!-90:(L)$) -- ($(L)!0.15cm!90:(P)$);
coordinate[name intersections=of=path_1 and path_2, by=label_Q];
path node[anchor=south east, inner sep=0, font=footnotesize,text=blue] 
(aux) at (label_Q) $O$;
path node[anchor=north west, inner sep=0, font=footnotesize] at (aux.north west)
$Q$;
endtikzpicture
enddocument

enter image description here

answered 36 mins ago

marmot

67.2k473143

answered 36 mins ago

marmot

67.2k473143

answered 36 mins ago

marmot

67.2k473143

answered 36 mins ago

marmot

67.2k473143

What is (aux.north west)? I see that (aux) is the label for the coordinate for the label 'O' if the letter 'O' were to be typeset at the intersection of the two green lines.
â€“Â Adelyn
23 mins ago

1

@Adelyn it is the north east corner of the aux node. All I do is to make sure that the north east corner of the Q node is at the same position as the north east corner of the O node. That way I make sure that the Q sits at the same position as the O, and the wiggle gets ignored. And O is set (in the present version), even in blue, but you can't see it because the Q sits precisely on top. BTW, if you add Bernards proposal instead of mine, you will see that his Q does not sit precisely on top of the O (but his proposal looks nice, too, IMHO).
â€“Â marmot
21 mins ago

Thanks for the careful explanation.
â€“Â Adelyn
19 mins ago

add a commentÂ |Â

What is (aux.north west)? I see that (aux) is the label for the coordinate for the label 'O' if the letter 'O' were to be typeset at the intersection of the two green lines.
â€“Â Adelyn
23 mins ago

1

@Adelyn it is the north east corner of the aux node. All I do is to make sure that the north east corner of the Q node is at the same position as the north east corner of the O node. That way I make sure that the Q sits at the same position as the O, and the wiggle gets ignored. And O is set (in the present version), even in blue, but you can't see it because the Q sits precisely on top. BTW, if you add Bernards proposal instead of mine, you will see that his Q does not sit precisely on top of the O (but his proposal looks nice, too, IMHO).
â€“Â marmot
21 mins ago

Thanks for the careful explanation.
â€“Â Adelyn
19 mins ago

What is (aux.north west)? I see that (aux) is the label for the coordinate for the label 'O' if the letter 'O' were to be typeset at the intersection of the two green lines.
â€“Â Adelyn
23 mins ago

@Adelyn it is the north east corner of the aux node. All I do is to make sure that the north east corner of the Q node is at the same position as the north east corner of the O node. That way I make sure that the Q sits at the same position as the O, and the wiggle gets ignored. And O is set (in the present version), even in blue, but you can't see it because the Q sits precisely on top. BTW, if you add Bernards proposal instead of mine, you will see that his Q does not sit precisely on top of the O (but his proposal looks nice, too, IMHO).
â€“Â marmot
21 mins ago

Thanks for the careful explanation.
â€“Â Adelyn
19 mins ago

add a commentÂ |Â

up vote
1
down vote

You can simply replace the last line of code for the tikzpicture with

path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$smashQ$;

enter image description here

answered 33 mins ago

Bernard

159k764192

The tail of the "Q" is pushing the letter away from the intersection of the two green lines.
â€“Â Adelyn
28 mins ago

I don't think so since Q is Ã¢Â€Â˜smashedÃ¢Â€Â™, which amounts to make it have 0 height and depth. Change it for a letter without descenders like N, and you'll see it's placed the same way: the base line is about 1 pt from the green line.
â€“Â Bernard
19 mins ago

Dear Bernard If you replace the last node in my code by your proposal, you will see that @Adelyn is right.
â€“Â marmot
17 mins ago

I agree - it "should" be a command that gives me the letter Q typeset where I want it. It does not, though. Compare your diagram with marmot's diagram.
â€“Â Adelyn
16 mins ago

My code was just a work-around as simple as possible. Marmot knowsTiKZ much better than I do Ã¢Â€Â“ indeed I know pstricks better than TiKZ, and could place the letter Q anywhere you want, with orientation you want.
â€“Â Bernard
4 mins ago

add a commentÂ |Â

up vote
1
down vote

You can simply replace the last line of code for the tikzpicture with

path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$smashQ$;

enter image description here

answered 33 mins ago

Bernard

159k764192

The tail of the "Q" is pushing the letter away from the intersection of the two green lines.
â€“Â Adelyn
28 mins ago

I don't think so since Q is Ã¢Â€Â˜smashedÃ¢Â€Â™, which amounts to make it have 0 height and depth. Change it for a letter without descenders like N, and you'll see it's placed the same way: the base line is about 1 pt from the green line.
â€“Â Bernard
19 mins ago

Dear Bernard If you replace the last node in my code by your proposal, you will see that @Adelyn is right.
â€“Â marmot
17 mins ago

I agree - it "should" be a command that gives me the letter Q typeset where I want it. It does not, though. Compare your diagram with marmot's diagram.
â€“Â Adelyn
16 mins ago

My code was just a work-around as simple as possible. Marmot knowsTiKZ much better than I do Ã¢Â€Â“ indeed I know pstricks better than TiKZ, and could place the letter Q anywhere you want, with orientation you want.
â€“Â Bernard
4 mins ago

add a commentÂ |Â

up vote
1
down vote

You can simply replace the last line of code for the tikzpicture with

path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$smashQ$;

enter image description here

answered 33 mins ago

Bernard

159k764192

You can simply replace the last line of code for the tikzpicture with

path node[anchor=south east, inner sep=0, font=footnotesize] at (label_Q)$smashQ$;

enter image description here

answered 33 mins ago

Bernard

159k764192

answered 33 mins ago

Bernard

159k764192

answered 33 mins ago

Bernard

159k764192

answered 33 mins ago

Bernard

159k764192

The tail of the "Q" is pushing the letter away from the intersection of the two green lines.
â€“Â Adelyn
28 mins ago

I don't think so since Q is Ã¢Â€Â˜smashedÃ¢Â€Â™, which amounts to make it have 0 height and depth. Change it for a letter without descenders like N, and you'll see it's placed the same way: the base line is about 1 pt from the green line.
â€“Â Bernard
19 mins ago

Dear Bernard If you replace the last node in my code by your proposal, you will see that @Adelyn is right.
â€“Â marmot
17 mins ago

I agree - it "should" be a command that gives me the letter Q typeset where I want it. It does not, though. Compare your diagram with marmot's diagram.
â€“Â Adelyn
16 mins ago

My code was just a work-around as simple as possible. Marmot knowsTiKZ much better than I do Ã¢Â€Â“ indeed I know pstricks better than TiKZ, and could place the letter Q anywhere you want, with orientation you want.
â€“Â Bernard
4 mins ago

add a commentÂ |Â

The tail of the "Q" is pushing the letter away from the intersection of the two green lines.
â€“Â Adelyn
28 mins ago

I don't think so since Q is Ã¢Â€Â˜smashedÃ¢Â€Â™, which amounts to make it have 0 height and depth. Change it for a letter without descenders like N, and you'll see it's placed the same way: the base line is about 1 pt from the green line.
â€“Â Bernard
19 mins ago

Dear Bernard If you replace the last node in my code by your proposal, you will see that @Adelyn is right.
â€“Â marmot
17 mins ago

I agree - it "should" be a command that gives me the letter Q typeset where I want it. It does not, though. Compare your diagram with marmot's diagram.
â€“Â Adelyn
16 mins ago

My code was just a work-around as simple as possible. Marmot knowsTiKZ much better than I do Ã¢Â€Â“ indeed I know pstricks better than TiKZ, and could place the letter Q anywhere you want, with orientation you want.
â€“Â Bernard
4 mins ago

The tail of the "Q" is pushing the letter away from the intersection of the two green lines.
â€“Â Adelyn
28 mins ago

I don't think so since Q is Ã¢Â€Â˜smashedÃ¢Â€Â™, which amounts to make it have 0 height and depth. Change it for a letter without descenders like N, and you'll see it's placed the same way: the base line is about 1 pt from the green line.
â€“Â Bernard
19 mins ago

Dear Bernard If you replace the last node in my code by your proposal, you will see that @Adelyn is right.
â€“Â marmot
17 mins ago

I agree - it "should" be a command that gives me the letter Q typeset where I want it. It does not, though. Compare your diagram with marmot's diagram.
â€“Â Adelyn
16 mins ago

My code was just a work-around as simple as possible. Marmot knowsTiKZ much better than I do Ã¢Â€Â“ indeed I know pstricks better than TiKZ, and could place the letter Q anywhere you want, with orientation you want.
â€“Â Bernard
4 mins ago

add a commentÂ |Â

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