Trapezoid and squares

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a_507_bc

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a_507_bc

#1 h Apr 14, 2023, 5:02 PM

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Outside of the trapezoid $ABCD$ with the smaller base $AB$ are constructed the squares $ADEF$ and $BCGH$ . Prove that the perpendicular bisector of $AB$ passes through the midpoint of $FH$ .

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Assassino9931

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Assassino9931

#2 h Apr 14, 2023, 5:14 PM

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Trivial by coordinate bash.

Choose coordinates so that $A(-1,0)$ , $B(1,0)$ , $C(c,a)$ , $D(d,a)$ - then the $y$ -axis is the perpendicular bisector of $AB$ . It suffices to compute the $x$ -coordinates of $F$ and $H$ and to check that their sum (and hence their arithmetic mean) is $0$ .

This can be done in tons of ways, here is the fastest. Let $K$ and $L$ be the feet of the perpendiculars from $D$ and $F$ to $AB$ . Then $\triangle ADK \cong \triangle FAL$ by hypotenuse and angles, so $AL = DK = a$ , thus $L(-a-1,0)$ and the $x$ -coordinate of $F$ is $-a-1$ . Analogously the $x$ -coordinate of $H$ is $a+1$ and we are done.

@2below Remark on synthetic solution

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yofro

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yofro

#3 h Apr 14, 2023, 7:39 PM

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Complex numbers is very clean. Note $h-b=(b-c)i$ , $f-a=(a-d)i$ so the midpoint of $FH$ is $\frac{a+b}{2}+\frac{(a+b)-(c+d)}{2}i.$ Orienting so that $\Im(a)=\Im(b)$ now solves the problem. This is because the perpendicular bisector of $AB$ is $\Re(z)=\frac{\Re(a)+\Re(b)}{2}$ and $(a+b)-(c+d)$ is purely real.

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newinolympiadmath

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newinolympiadmath

#4 h Apr 14, 2023, 8:28 PM

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is there synthetic solution?

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HaO-R-Zhe

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HaO-R-Zhe

#5 h Apr 15, 2023, 12:11 AM

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I swear this was on some Chinese middle school final exam...

Construct point $X = AD \cap BC$ , and $M$ the midpoint of $FG$ . Reflect $A$ and $B$ over $M$ to get $A'$ and $B'$ . We see that $FB'=BH=BC$ and $FA=AD$ . Moreover, $\angle B'FA = \angle HFA + \angle BHF = 360^{\circ}-\angle FAB - \angle ABH = \angle BAD + \angle CBA - 180^{\circ} = \angle CXD$ . Because $AB \parallel CD$ , then $\frac{FB'}{FA} = \frac{BC}{AD} = \frac{XC}{XD}$ . So $\triangle B'FA \sim \triangle CXD$ . Finally, $\angle BAB' = 270^{\circ} - \angle FAB' - \angle DAB = 90^{\circ}$ , so $BB' \perp AB$ . Similarly, $AA' \perp AB$ , meaning $ABA'B'$ is a rectangle with $M$ as its center. It is evident now that $X$ lies on the perpendicular bisector of $AB$ .

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sunken rock

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sunken rock

#6 h Apr 15, 2023, 10:04 AM

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There is no need of $E, G$ . Translate $\triangle ADF$ of vector $\stackrel{\longrightarrow}{AB}$ , $A$ goes to $B$ , getting $D',F'$ so that $FF'\stackrel{\parallel}{=}DD'\stackrel{\parallel}{=}AB$ . Well known, perpendicular from $B$ to $CD'$ goes through $P$ , midpoint of $HF'$ .
Call $N$ midpoint of $AB$ , construct rectangle $PBNM$ , see that $H-M-F$ are collinear; with $MP\parallel FF'$ we are done.

Best regards,
sunken rock

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Helixglich

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Helixglich

#7 h Apr 16, 2023, 9:20 PM

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yofro wrote:

Complex numbers is very clean. Note $h-b=(b-c)i$ , $f-a=(a-d)i$

Uhh correct me if I am wrong here. The idea is correct. The approach is definitely clean. But you have ( Probably because the solution is elementary and you rushed it ) a mistake in computing one of $f$ and $h$ . (depends on orientation ). You can further check that

yofro wrote:

This is because the perpendicular bisector of $AB$ is $\Re(z)=\frac{\Re(a)+\Re(b)}{2}$ and $(a+b)-(c+d)$ is purely real.

$(a+b)-(c+d)$ is never real since $AB \parallel CD$

Corrected it should be $(a+c)-(b+d)$

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yofro

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yofro

#8 h Apr 17, 2023, 12:54 AM

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Helixglich wrote:

yofro wrote:

Complex numbers is very clean. Note $h-b=(b-c)i$ , $f-a=(a-d)i$

Uhh correct me if I am wrong here. The idea is correct. The approach is definitely clean. But you have ( Probably because the solution is elementary and you rushed it ) a mistake in computing one of $f$ and $h$ . (depends on orientation ). You can further check that

yofro wrote:

This is because the perpendicular bisector of $AB$ is $\Re(z)=\frac{\Re(a)+\Re(b)}{2}$ and $(a+b)-(c+d)$ is purely real.

$(a+b)-(c+d)$ is never real since $AB \parallel CD$

Corrected it should be $(a+c)-(b+d)$

Sorry, one of $i$ should be replaced with $-i$ , and I meant $(a+c)-(b+d)$ . Everything else should check out

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Ferum_2710

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Ferum_2710

#10 h Apr 29, 2023, 7:27 PM

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I the middle of the base AB, we choose the points M and Q on the base CD such that DM = QC = AI. Obviously, AIMD and BIQC are parallelograms, so IM = AD, IM || AD, and IQ = BC, IQ || BC. Outside the triangle IMQ, we construct the squares IMNP and IQRS. We consider the points U and V such that IPUS is a parallelogram, and V is the projection of I on CD. We have ∠PIS + ∠PIM + ∠MIQ + ∠QIS = 360°, and since IPUS is a parallelogram, we deduce ∠PIS + ∠MIQ = 180° = ∠IPU + ∠PIS, so ∠IPU = ∠MIQ. Since IP = IM and PU = IS = IQ, the triangles IPUS and MIQ are congruent (L.U.L.), so ∠PIU = ∠IMQ. The triangle IMV is right-angled at V, so ∠MIV = 90° - ∠IMV = 90° - ∠PIU. We obtain ∠MIV + ∠PIU + ∠MIP = 180°, so the points U, I, and V are collinear. Thus, UI is the median of the segment AB. Since IPUS is a parallelogram, the line UI passes through the midpoint J of the diagonal PS. From ∠DAF = ∠MIP = 90° and AD || IM, it follows that AF || IP and AF = IP, so the quadrilateral AIPF is a parallelogram. Thus, FP || AI and FP = AI. Similarly, we can show that SH || IB and SH = IB. Since I is the midpoint of the segment AB, we deduce that FP || HS and FP = SH, so FPHS is a parallelogram. Consequently, the diagonal FH passes through the midpoint J of the segment PS, so the lines UI and FH are concurrent

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Ferum_2710

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Ferum_2710

#11 h Apr 29, 2023, 7:27 PM

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Ferum_2710 wrote:

Let ∠AIP = ∠MIP = ∠MNP = ∠AFI = 90°. Let I be the midpoint of the base AB. We choose points M and Q on the base CD such that DM = QC = AI.

Clearly, AIMN and BIQC are parallelograms, so IM = AD, IM ∥ AD, and IQ = BC, IQ ∥ BC. Outside the triangle IMQ, we construct the squares IMNP and IQRS. Let U and V be the points such that IPUS is a parallelogram, and V is the projection of I onto CD.

We have ∠PIS + ∠PIM + ∠MIQ + ∠QIS = 360°. Since IPUS is a parallelogram, we deduce that ∠PIS + ∠MIQ = 180° = ∠IPU + ∠PIS, so ∠IPU = ∠MIQ. Since IP = IM and PU = IS = IQ, it follows that triangles IP U and MIQ are congruent (by L.U.L.), so ∠PIU = ∠IMQ.

Triangle IMV is right-angled at V, so ∠MIV = 90° − ∠IMV = 90° − ∠PIU. We obtain ∠MIV + ∠PIU + ∠MIP = 180°, so the points U, I, and V are collinear. Thus, UI is the perpendicular bisector of segment AB.

Since IPUS is a parallelogram, the line UI passes through the midpoint J of the diagonal PS. Since ∠DAF = ∠MIP = 90° and AD ∥ IM, it follows that AF ∥ IP, and since AF = IP, the quadrilateral AIPF is a parallelogram. Thus, FP ∥ AI and FP = AI. Analogously, it can be shown that SH ∥ IB and SH = IB.

Since I is the midpoint of segment AB, we deduce that FP ∥ SH and FP = SH, so FPHS is a parallelogram. Consequently, the diagonal FH passes through the midpoint J of segment PS, so the lines UI and FH are concurrent.

Please turn it into LaTeX

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EHoTuK

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EHoTuK

#12 h Apr 12, 2025, 11:20 PM

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We fix points $A$ and $B$ and the line $\ell$ containing $CD$ .

If we start moving $D$ along $\ell$ , $F$ will move along a line obtained by rotating $\ell$ around $A$ by $90^\circ$ , thus, a line perpendicular to AB. Similarly, if we start moving $C$ along $\ell$ , $H$ will move along a line perpendicular to $AB$ . Therefore, middle of $FH$ is projecting on the same point on $AB$ .

It suffices to note that the condition holds when $AD$ and $BC$ are perpendicular to $AB$ .

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