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$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.

How can I approach this problem, preferably without trigonometry?

enter image description here

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  • $\begingroup$ Actually, I think it is quite easy with trigonometry : you can write the law of cosines in triangles $ECG$, $EFC$ and $EFG$. From this you get a relation between $x$ and the length $L$ of the side of your square. Then you have also $(7+x)^2 = 2L^2$, and the two relations together give you $x$. But without trigonometry, I have no idea... $\endgroup$ – TheSilverDoe Feb 25 at 16:21
  • $\begingroup$ One reflection proves that G is on AD. A second reflection proves that AF=AC/4. $\endgroup$ – amI Feb 25 at 19:52
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    $\begingroup$ Seeing the lengths 3cm, 4cm and half of a right angle, I somehow think of x=5 $\endgroup$ – QBrute Feb 25 at 21:53
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    $\begingroup$ This is quite a common pattern in elementary geometry: half a right angle, stuck in a frame of right angles. Generally speaking, this can be solved by constructing 'rotationally similar' triangles. It even has a name in Chinese geometry education: the half-angle model. $\endgroup$ – Trebor Mar 4 at 2:16
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    $\begingroup$ Take any Chinese geometry book for middle school students, and there will be mountains of similar problems. For problems on this level, I do not have much information for English materials. $\endgroup$ – Trebor Mar 4 at 2:42

12 Answers 12

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Following solution uses no concept of concyclic or similar triangles, only rotation and Pythagoras theorem.

Let $O$ be a midpoint of a segment $AC$.

Rotate $G$ for $-90^{\circ}$ around $E$ to a new point $K$. This rotation takes $C$ to $O$ and $O$ to $B$, so $K$ is on a segment $BO$ (which is a part of diagonal $BD$)

Then $EK=EG$ so $E$ is on perpendicular bisector of $KG$. Since $\angle KEF = \angle FEG = 45^{\circ}$ line $EG$ is perpendicular bisector for $KG$ so $FK = FG = x$.

Since $ FO = {x+1\over 2}$ we can use Pythagoras theorem in triangle $\triangle FOK$:

$$ x^2 = 4^2+\Big({x+1\over 2}\Big)^2 \implies x=5$$

enter image description here

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    $\begingroup$ +1. Nicer still! Interestingly, I had just returned to post an addendum to my answer, recognizing a Pythagorean relation I had originally missed. I hadn't had time to give much thought to its geometric interpretation, but then I saw that your answer provided exactly that interpretation. Thanks for saving me the trouble! :) $\endgroup$ – Blue Mar 2 at 14:46
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enter image description here

Let $H$ be the midpoint of $AC$ and $\angle EIC= 90^{\circ}$. We can observe that $$FH+3=HG+4,\quad FH+HG=x.$$ So we obtain $HG=\frac{x-1}2$. Since two corresponding angles are congruent; $\angle FEG =\angle EHG=45^{\circ}$ and $\angle EGF=\angle HGE$, we have that $\triangle FEG$ and $\triangle EHG$ are similar to each other. This gives $$FG:EG=EG:HG\implies EG^2 = FG\cdot HG=\frac{x(x-1)}2.$$ Now, note that $EI=\frac14 AC=\frac{x+7}4$ and $IG=IH-GH=\frac{x+7}{4}-\frac{x-1}2=\frac{9-x}{4}$. Since $\triangle EIG$ is a right triangle, by Pythagorean theorem, we find that $$ EG^2=\frac{x(x-1)}{2}=EI^2+IG^2=\frac{(x+7)^2}{16}+\frac{(9-x)^2}{16}, $$ which implies $x=5 $ or $x=-\frac{13}3$. Since $x>0$, we get $x=5$.

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  • $\begingroup$ I would suggest to remove the angle also from the drawing (instead both angles of $45^{\circ}$ can be shown). By the way what a nice program do you use for the drawings? $\endgroup$ – user Feb 25 at 21:39
  • $\begingroup$ @user I just updated my figure, thank you. By the way, the program I used is the Geogebra. It provides an online workspace and even cloud services if you sign up! $\endgroup$ – Song Feb 25 at 22:12
  • $\begingroup$ I'm surprised we didn't draw the circle through EFG whose center, O, is the right angle of the triangle FOG. I can see that the perpendicular bisector of OC is EG, transporting length 4 cm to one leg. Transporting 3 cm to the other leg thwarts me. $\endgroup$ – Eric Towers Feb 26 at 1:25
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Let $O$ be a midpoint of a segment $AC$.

Rotate $F$ for $90^{\circ}$ around $E$ to a new point $H$. This rotation takes $O$ to $C$.

Then $EF=EH$ and $\angle FEH = 90^{\circ}$ so $EG$ is angle bisector for $\angle FEH$ so $GH = GF = x$.

It is easy to see that $$CH = FO = {x+1\over 2}$$

Now $FHCE$ is cyclic quadrilateral since $\angle CFE = \angle CHE$ so $$\angle FCH = \angle FEH =90^{\circ}$$

Finaly we use Pythagoras theorem in triangle $\triangle CGH$:

$$ x^2 = 4^2+\Big({x+1\over 2}\Big)^2 \implies x=5$$

enter image description here

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  • $\begingroup$ Your solution encourages me to think that there may be even more geometric solutions. $\endgroup$ – blackened Feb 27 at 18:30
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With the figure as labeled below, define $$a := |AA^\prime| \qquad b := |BB^\prime| \qquad c := |A^\prime B^\prime| \qquad d := |A^\prime M|$$

Note that the perpendicular from midpoint $M$ to the diagonal necessarily quadrisects that diagonal; moreover, $\overline{MM^\prime}\cong\overline{BM^\prime}$.

enter image description here

$$\begin{align} \triangle A^\prime M^\prime M \text{ is a right triangle} &\quad\to\quad d^2 = \left(c+b-\frac14(a+b+c)\right)^2+\left(\frac14(a+b+c)\right)^2 \tag{1}\\[4pt] \triangle A^\prime B^\prime M \sim \triangle A^\prime M B \text{ (Angle-Angle)} &\quad\to\quad \frac{|A^\prime M|}{|A^\prime B^\prime|} = \frac{|A^\prime B|}{|A^\prime M|} \quad\to\quad d^2 = c(b+c) \tag{2} \end{align}$$

Equating $d^2$ with $d^2$ yields $$3c^2 + 2c(a-b) - a^2 + 2 a b - 5 b^2 = 0 \tag{3}$$ which suggests that there's nothing inherently Pythagorean about the general values $a$, $b$, $c$ (but see the Addendum). However, for the specific values $a=3$ and $b=4$, we have $$0 = 3 c^2 - 2c - 65 = (c-5)(3c+13) \quad\to\quad c = 5 \tag{4}$$ where we have ignored the "obviously"-extraneous solution $c=-13/3$. $\square$


Note. The extraneous solution from $(4)$ becomes valid if we interpret the problem as saying that "lines $\overleftrightarrow{MA^\prime}$ and $\overleftrightarrow{MB^\prime}$ make a $45^\circ$ angle".

enter image description here


Addendum. I notice that there is actually something inherently Pythagorean about general $a$, $b$, $c$. We can write $(3)$ as $$\left(\frac{- a + b + c}{2} \right)^2 + b^2 = c^2 \tag{3'}$$ implying that $|-a+b+c|/2$, $|b|$, $|c|$ make a right triangle. (With $a=3$ and $b=4$, the two solutions to $(4)$ correspond to a $3$-$4$-$5$ triangle, and a $5$-$12$-$13$ triangle scaled by $1/3$.) I hadn't seen a way to make $(3')$ obvious in the figure (I was just playing with the algebra); however, by interesting coincidence, @greedoid's new answer seems to show exactly this! Here's a version of @greedoid's figure, with a more-arbitrary relationship between lengths $a$ and $b$.

enter image description here

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    $\begingroup$ Very nice solution +1. You could get a formula (2) also by the power of the point $A'$ with respect to circumcircle of the triangle $BB'M$. $\endgroup$ – Aqua Mar 1 at 15:30
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    $\begingroup$ @greedoid: I thought of that. But, with the power-of-a-point approach, I'd have to explain how I knew that $\overline{A^\prime M}$ is actually tangent to that circumcircle. Of course, that tangency follows from some angle considerations; but then, so does the similarity argument, and the latter seems a bit more immediate (especially since the power-of-a-point result is effectively a consequence of that similarity, anyway). $\endgroup$ – Blue Mar 1 at 15:38
  • $\begingroup$ @Blue Please kindly state your choice of best answer. $\endgroup$ – blackened Mar 7 at 11:06
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    $\begingroup$ @blackened: I believe greedoid revealed the true, beating heart of the problem by constructing the right triangle corresponding to the Pythagorean relation I've called $(3^\prime)$. $\endgroup$ – Blue Mar 7 at 12:59
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Here's an attempt using the method of "have a hunch, then confirm it works", which probably follows the method by which the problem was first created. Intuiting where some missing lines ought to be can be helpful sometimes!


Begin with the following figure, with dimensions as given; we will show that filling in $z=4$ will force $y=3$, and also $x=5$, which is the answer we're looking for.

Figure with relevant lines added

First observe that the angle at $E$ is equal to $\pi/4$ or $45^{\circ}$. [see below]

Now, because we have a bunch of similar triangles (the two meeting vertically at $G$, the two meeting vertically at $F$, and the right-angled ones along the right-hand side), we observe:

$2d/d = (x+y)/z$

$d/ (\frac{1}{3}d) = (x+z)/y$

$d/2d = (z/\sqrt{2})/(\frac{4}{3}d)$

Now if we set $z=4$, a little bit of linear algebra (or simple substitution if you will) gives us $d=3\sqrt{2}$, $y=3$, and $x=5$.

Notice how removing the continuations of the lines $EF$ and $EG$ makes everything look really hard!


Proof of the claim about the angle: This is because of a "fun fact" you may have discovered by playing around on squared graphing paper: the vectors $(2,1)$ and $(3,-1)$ (or suitable rotations of these) meet at that angle. You can prove this by reflecting one in the other using the standard formula from vector geometry: let $v=(2,1)$ and $n=(1,3)$ (which is orthogonal to $(3,-1)$). The standard reflection-in-a-line formula $v-2\frac{v\cdot n}{n\cdot n} n$ gives us

$(2,1) - 2\frac{5}{10}(1,3) = (2,1) - (1,3) = (1,-2)$

which is orthogonal to $(2,1)$; this implies that the reflection line $(3,-1)$ bisects that right angle, i.e. is at $\pi/4$ (or $45^{\circ}$) to $(2,1)$. $\square$

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Calling

$$ AF = a\\ FG=x\\ GC = b\\ EC = c\\ E = (X_0, Y_0)\\ O = \left(a+\frac x2, \frac x2\right) $$

Considering in the plane $X\times Y$ the square diagonal as the $X$ axis with $A$ at the origin, we have

$$ \frac{\sqrt 2}{2}(a+x+b)= 2c\\ x = \sqrt2 r\\ \left(X_0-\left(a+\frac x2\right)\right)^2+\left(Y_0-\frac x2\right)^2= r^2\\ X_0 = a+x+b -\frac{\sqrt 2}{2}c\\ Y_0 = \frac{\sqrt 2}{2}c $$

Here $r$ represents the radius for the circle which intersects the $X$ axis at $F, G$ such that $\angle FEG = \frac{\pi}{4}$ is the angle subtended by arc $AB$

Solving for $x,r,X_0,Y_0$ we get at

$$ \left\{ \begin{array}{rcl} x& =& \frac{1}{3} \left(b-a+2 \sqrt{a^2-2 b a+4 b^2}\right) \\ c& =& \frac{a+2 b+\sqrt{a^2-2 b a+4 b^2}}{3 \sqrt{2}} \\ r& =& \frac{b-a+2 \sqrt{a^2-2 b a+4 b^2}}{3 \sqrt{2}} \\ X_0&=&\frac{1}{2} \left(a+2 b+\sqrt{a^2-2 b a+4 b^2}\right) \\ Y_0&=&\frac{1}{6} \left(a+2 b+\sqrt{a^2-2 b a+4 b^2}\right) \\ \end{array} \right. $$

giving after substitution $x = 5$

enter image description here

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Let $DE$ intersect $AC$ at $G'$. Then $G'C:G'A=CE:AD=1:2$, hence $3G'C=AC$.

Let the perpendicular bisector of $BE$ intersect $AC$ at $F'$. Easy to see that $4AF'=AC$.

Note that $F'E=F'B=F'D$. Hence $F'$ is the circumcenter of $\triangle BED$, so $\angle DF'E = 2\angle DBE = 2\cdot 45^\circ=90^\circ$. It follows that $\angle F'EG'=45^\circ$.

Observe that $$\frac{AF'}{G'C} = \frac{AC/4}{AC/3}=\frac 34 = \frac{AF}{GC}.$$

If $AF'>AF$ then by the above equality we also have $CG'>CG$, hence $45^\circ = \angle FEG > \angle F'EG' = 45^\circ$, a contradiction. Similarly we rule out the possibility $AF'<AF$.

Hence $AF'=AF$, which means that $CG'=CG$, so $F'=F$ and $G'=G$. Therefore $AC = 4AF = 12 \mathrm{cm}$, and

$$FG = AC - AF - GC = 12\mathrm{cm} - 3\mathrm{cm} - 4\mathrm{cm} = 5\mathrm{cm}.$$

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Drop perpendiculars $FH, EI$ and $FJ$ and label $BH=y$, $EH=z$ and $\angle GEI=\alpha$:

$\hspace{2cm}$enter image description here

Use the similarity of $\triangle FEH$ and $\triangle EIG$: $$\frac{GI}{HE}=\frac{EI}{FH} \Rightarrow \\ \frac{4-CI}{z}=\frac{CI}{y+2z}\Rightarrow \\ \frac{4-\frac{y+z}{\sqrt{2}}}{z}= \frac{\frac{y+z}{\sqrt{2}}}{y+2z}\Rightarrow \\ \frac{4\sqrt{2}-y-z}{z}=\frac{y+z}{y+2z}\Rightarrow\\ 3z^2+(4y-8\sqrt{2})z+y^2-4\sqrt{2}y=0 \stackrel{y=\frac3{\sqrt{2}}}{\Rightarrow} \\ 6z^2-4\sqrt{2}z-15=0 \Rightarrow \\ z=\frac3{\sqrt{2}}.$$ Hence: $$x+7=2(y+z)\sqrt{2}=12 \Rightarrow x=5.$$

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I do not like it. This is not a geometry problem; it is indeed a puzzle where you have to find this coincidence, which occurs only if $AF:GC=3:4$.

enter image description here

If we can find/prove that $GDF\angle=45^\circ$ in some different way, then we can define the point $P$ that as the reflection of $C$ about $DE$ and the reflection of $A$ about $DF$. Due to $FPD\angle=DPG\angle=45^\circ$, the triangle $FGP$ is right-angled and its legs are $PF=AF=3$, $PG=CG=4$, so $FG=5$.

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  • $\begingroup$ I don't understand it. Is this solution or just an idea how to prove it? $\endgroup$ – Aqua Mar 1 at 18:28
  • $\begingroup$ No solution, no idea. Just a demonstration, that somebody took this configuration and converted it to a puzzle. $\endgroup$ – user141614 Mar 1 at 18:36
  • $\begingroup$ No, this configuration has no free parameter, it is a concrete picture. $\endgroup$ – user141614 Mar 1 at 18:40
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Without trigonometry or algebra, to find the length of $FG$.

With the figure and conditions as posted, construct circles with centers $F$, $G$ and radii $FA$, $GC$. Let the circle centered on $G$ cross $AC$, $BC$, $DC$ at $J$, $K$, $L$, and join $GK$ and $GL$.

Collinear $GK, GL$ are thus a diameter of the circle about $G$ and perpendicular to $AC$.

On $BA$ make $BM=BK$, and join $MK$, $MJ$, and $JK$.

Since $MBK$ and $ABC$ are isosceles right triangles sharing the angle at $B$ $$MK\parallel AJ$$ length of <span class=$FG$"> And since$$AM=CK=JK$$and$$AM\parallel JK$$then $AJKM$ is a parallelogram and$$AJ=MK=CG=4$$

Therefore$$FJ=AJ-AF=4-3=1$$ and$$FG=FJ+JG=1+4=5$$

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    $\begingroup$ How did you ascertain that $GJMK$ is a rectangle? You don't know $\angle KMJ$ or $\angle MJG$, do you? $\endgroup$ – Jens Mar 6 at 20:54
  • $\begingroup$ @Jens-point well taken. See if revised argument is better. $\endgroup$ – Edward Porcella Mar 6 at 22:19
  • $\begingroup$ @Blue--Right, I had tried making tangents at $J$ and $K$, but couldn't prove $M$ must be on $AB$. So I made $BM=BK$ and tried to go from there. But I agree the argument was still incomplete. See if you think the revised argument meets the objection. $\endgroup$ – Edward Porcella Mar 6 at 22:26
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    $\begingroup$ Your argument cannot work because you are not using the $45^\circ$ angle info. Essentially you are trying to prove that any square whose diagonal has a $3$ cm bit in one corner and a $4$ cm bit in the other corner, must have a $5$ cm bit in the middle. This is not true. $\endgroup$ – Jens Mar 7 at 9:36
  • $\begingroup$ You're right. $AJ=MK$ for other positions of $G$, but $MK$ although parallel is not generally equal to $GC$. So I can't conclude $AJ=GC=4$. Thanks for the light. $\endgroup$ – Edward Porcella Mar 7 at 16:57
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Ok after some calculus, I figured out to solve this problem. Let's draw the vertical V that goes through E, and call $\alpha$ and $\frac{\pi}{4}-\alpha$ the angles we get on the left and right of the 45° angle cut by V and call c the length of the side of the square. We can then do some trigonometry to get :

  • $c \cdot tan(\alpha) - (c-\frac{3}{\sqrt{2}}) = c \cdot tan(\alpha)$ (Continue the line EF to cut AD)
  • $ tan(\frac{\pi}{4}-\alpha) = c \cdot \frac{\sqrt{2}}{8}-1$ (G is $\frac{4}{\sqrt{2}}$ away vertically from the right side and $\frac{c}{2}-\frac{4}{\sqrt{2}}$ away horizontally from the middle)

With a little trigo, we have 2 equations in $c$ and $tan(\alpha)$ that we can solve, which allow us to solve the whole problem. This makes me think that there is no "simple" answer without trigo :(

Edit : Well, seems there was a simple answer afterall. Really nice proof, I'm glad I was wrong !

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Draw the red lines:

$\hspace{3cm}$enter image description here

The steps:

1) Draw $EO=OK$ and $KP$ parallel to $EG$ so that $\triangle CGE$ and $\triangle OPK$ are congruent, in particular, $OP=4$.

2) Draw $KM$ so that $MO=OG$, hence $\triangle KMO$ and $\triangle OGE$ are congruent (two sides and angle between them are equal, i.e. $MO=OG, KO=OE, \angle KOM=\angle EOG=45^\circ$).

3) $\triangle AKM$ and $\triangle CEG$ are congruent (because $\triangle KMO$ and $OGE$ are congruent)

4) $\triangle KPM$ is isosceles ($KP=KM=EG$).

5) $KT$ is perpendicular to $PM$ (because $\triangle AKM$ and $\triangle KTO$ are symmetric (reflective) around it.

6) If we label $PT=z$, then $AP=3-z,TM=z,MO=4-2z,OG=3-z$

7) Since $MO=OG \Rightarrow 4-2z=3-z \Rightarrow z=1$ and $x=z+4-2z+3-z=7-2z=5.$

Note: No formula (or theorem) is used.

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  • $\begingroup$ @greedoid, I am curious if you find this solution worthwhile (I'm not pretending to the bounty). Obviously, the first step should be $EC=EO=OK$. Any flaws in reasoning? $\endgroup$ – farruhota Mar 9 at 11:46
  • $\begingroup$ First let me apologize for my late reply, I just don't now why i wasn't notified about this comment. Anyway I'm not sure about 5). Why is $T $ a midpoint of $PM$? In whole analysis before you never mentioned the $T$. How is it related to $PM$? $\endgroup$ – Aqua Mar 9 at 16:01
  • $\begingroup$ because $\triangle AKM$ and $\triangle KPO$ (sorry, $\triangle KTO$ was a typo) are congruent. $\endgroup$ – farruhota Mar 9 at 16:06
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    $\begingroup$ Yes, I understand this. But anyway I still do not understand why is T midpoint of PM. If $T$ is a midpoint then this triangles are symmetric, but vice versa? I'm not sure. $\endgroup$ – Aqua Mar 9 at 16:10

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