mathz

[glad burrow]

Look at the picture

[unreal jacinth]

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+close

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[vestal peak]

,rccw

[humble junco]

well, what is the distance from B to EF

[glad burrow]

answer b

[humble junco]

yeah b is right

[glad burrow]

yuh but how i get to that

[humble junco]

well you could coordinaet bash

set D to be the origin

find the centre of the circle based on R, and the midpoint of EF
those should be R apart

[humble junco]

yes that

[vestal peak]

an even better picture

,tikz[scale=.5]
\pgfmathsetmacro\r{4};
\coordinate[label=below right:$D$] (D) at (0,0);
\coordinate[label=above left:$B$] (B) at (-16,16);
\coordinate[label=above right:$C$] (C) at (0,16);
\coordinate[label=below left:$A$] (A) at (-16,0);
\coordinate[label=above:$F$] (F) at (-4,16);
\coordinate[label=left:$E$] (E) at (-16,4);
\coordinate[label=left:$O$] (O) at (-\r,\r);
\draw (B) rectangle (D);
\draw (E) -- (F) coordinate[pos=.5, label=above:$T$] (T);
\draw (B) -- (T) node[pos=.5, sloped, font=\small, above] {height calculated};
\draw (T) -- (O) node[pos=.5, above] {$r$} -- (D) node[pos=.5, yellow, draw, anchor=west] {an expression in terms of $r$};
\draw[dashed,yellow, opacity=.8] (-\r,0) -- (O) node[pos=.5, left] {$r$} -- ++(\r,0) node[pos=.5, above] {$r$};

[icy grotto]

vin100

[humble junco]

actualy BT is useless that was when i was trying to figure out how to solve it

[glad burrow]

ahhhhh im so confuised

[vestal peak]

OT : OD

= 1 : √2

[humble junco]

oh yeah, that works too

[vestal peak]

skill: if you have equal ratio conditions like

,,\frac{a}{b} = \frac{c}{d}

[icy grotto]

vin100

[vestal peak]

u might wanna introduce a parameter (say t)

,,\frac{a}{b} = \frac{c}{d} = t

[icy grotto]

vin100

[vestal peak]

and re-write a and c in terms of t

,align
a&=bt \
c&= dt

[icy grotto]

vin100

[vestal peak]

,,\frac{OT}{OD} = \frac{1}{\sqrt2}

[icy grotto]

vin100

[vestal peak]

u dun needa intro a new var

[humble junco]

yeah you can just add 1 to both sides to get (OT + OD)/OD = 1 + 1/sqrt(2)

which is nice

[vestal peak]

lemme expand cute's comments

,align
\frac{OT}{OD} &= \frac{1}{\sqrt2} \
\frac{OT + OD}{OD} &= \frac{1}{\sqrt2} + 1 \tag{add 1 on both sides} \
\frac{DT}{OD} &= \frac{1}{\sqrt2} + 1 \tag{$O, D, T$ are collinear} \
\frac{?}{OD} &= \frac{1}{\sqrt2} + 1 \tag{$DT = ?$ calculated} \

[icy grotto]

vin100

[vestal peak]

,,\frac{a}{a} = 1 \text{ and } \frac{b}{a} + \frac{c}{a} = \frac{b+c}{a}

[icy grotto]

vin100

[topaz jungle]

@vestal peak thank you, excellent solution 🙂

[sweet scarab]

@topaz jungle has given 1 rep to @vestal peak

[brisk sandal]

the exact image was in another help post what the heck

nvm its you #1159163926890684447 message

@glad burrow are you still conmfused by the solution?

im guessing you reopened the thread because you werent satisfied with the previous solution given in the earlier thread

[vestal peak]

oh i didn't know this im new

,,\frac{ka}{kb+kc} = \frac{\cancel{k}a}{\cancel{k}b+\cancel{k}c}

[icy grotto]

vin100

[brisk sandal]

i doub tanyone notcied lol

[vestal peak]

it turns out that OP has problems with basic fraction arithmetic

,,=\frac{a}{b+c}

[icy grotto]

vin100

[vestal peak]

,,\ne\frac{a}{b+{\color{red!40}k}c}

[icy grotto]

vin100

[vestal peak]

remarks: a math Olympiad geometer once said, the best illustration is the one that is between right and wrong. (Take the proof Ceva's Theorem for example.) In this spirit, I set a wrong r at will.

[glad burrow]

+close