given right & isosceles triange, another isosceles, 2 midpoints, square wanted

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given right & isosceles triange, another isosceles, 2 midpoints, square wanted

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Source: Lusophon 2018 CPLP P2 , May Olympiad (Olimpiada de Mayo) 2002 L2

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#1 h Sep 13, 2018, 7:54 PM • 3 Y

Y by donotoven, Adventure10, Mango247

In a triangle $ABC$ , right in $A$ and isosceles, let $D$ be a point on the side $AC$ ( $A \ne D \ne C$ ) and $E$ be the point on the extension of $BA$ such that the triangle $ADE$ is isosceles. Let $P$ be the midpoint of segment $BD$ , $R$ be the midpoint of the segment $CE$ and $Q$ the intersection point of $ED$ and $BC$ . Prove that the quadrilateral $ARQP$ is a square.

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#2 h Sep 14, 2018, 8:01 AM • 3 Y

Y by donotoven, Adventure10, Mango247

Clearly the coordinates of all points are linear when $D$ moves (linearly) along the line $AC$ . So it suffices to prove the claim for two distinct cases. But in the cases $D=A$ and $D=C$ the claim is trivially true. Done.

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#3 h Sep 14, 2018, 8:06 AM • 3 Y

Y by donotoven, Adventure10, Mango247

Another way: If we rotate the whole figure by $90^\circ$ around $A$ , $B \mapsto C, D \mapsto E$ and hence $P \mapsto R$ .
So $AP=AR$ and $\angle RAP=90^\circ$ .
Moreover, since $R$ is the midpoint of $CE$ and $\angle EAC=\angle EQC=90^\circ$ , $EQAC$ is cyclic with circumcenter $R$ . So $QR=AR=AP$ .
But similarly $ABQD$ is cyclic with circumcenter $P$ and so $PQ=AP=AR=QR$ so $ARQP$ is a square.

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#4 h Sep 14, 2018, 8:10 AM • 3 Y

Y by donotoven, Adventure10, Mango247

Finally: (Explicit) Coordinates. W.l.o.g. take $A=(0,0), B=(0,1), C=(1,0)$ . Then $D=(x,0)$ for some $x \in [0,1], E=(0,-x), R=\left(\frac{1}{2}, -\frac{x}{2}\right), P=\left(\frac{x}{2}, \frac{1}{2}\right), Q=\left(\frac{x+1}{2}, \frac{1-x}{2}\right)$ . Then one can easily check that $ARQP$ is indeed a square.

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#5 h Sep 14, 2018, 9:31 PM • 3 Y

Y by donotoven, Adventure10, Mango247

the problem also appeared 16 years ago, in May Olympiad (Olimpiada de Mayo) 2002 L2 as P3

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#7 h Jul 16, 2021, 10:22 PM • 1 Y

Y by donotoven

$Lemma:$ If $ABC$ is a right triangle, right in $B$ , and $D$ is the intersection of the median of $B$ to $AC$ , so $AD \cong DC \cong BD$ .
$Proof: AD \cong DC$ by definition. Tracing a perpendicular from $D$ to $BC$ , intersecting at $E$ . It's not hard to notice that $DE$ is a midsegment of $ABC$ . So the triangle $BED \cong CDE$ by side-angle-side congruence, which give us $BD \cong DC$ as we desired.

The triangle $ABC$ of the question is necessarily a 90-45-45 triangle.
Applying the $Lemma$ in the triangle $ABD$ , give us that $BP \cong PD \cong PA.$ Note that $AD \cong AE \Leftrightarrow \angle ADE = 45 \Leftrightarrow \angle QDC = 45 \Leftrightarrow \angle DQC = 90 \Leftrightarrow \angle BQD = 90$ .
Applying the $Lemma$ in the triangle $AQD$ , we get that $PA \cong BP \cong PD \cong PQ$ , so $PA \cong PQ$ .
Applying the $Lemma$ in the triangles $ACE$ and $EQC$ , we get $AR \cong ER \cong RC \cong RQ$ .
Since $AD \cong AE, BA \cong AC$ , and $\angle BAC = \angle EAC$ , we have that $AEC \cong ADB$ . So $PA \cong PQ \cong PD \cong \frac{BD}{2} \cong \frac{EC}{2} \cong RE \cong RA \cong RQ$
We concluded that $AP \cong PQ \cong QR \cong RA$
Let $\angle ABD = \beta$ and $\angle BDA = \alpha$
$\angle AER = \angle EAR = \alpha \Rightarrow \angle DAR = 90- \alpha = \beta.$
$\angle PAD = \angle PDA = \alpha.$ So the $\angle PAR = \angle RAD + \angle PAD = \alpha + \beta = 90$
Look at the $\angle PBQ = PQB = 45 - \angle PAB = 45-\beta,$ once that $PB \cong PQ$ . And so $\angle DPQ = 2(45-\beta) = 90 -2\beta$ .
The $\angle APD = \angle BAP + \angle PBA = 2 \beta$ . So $\angle APQ = \angle APD + \angle DPQ = 2\beta +90 - 2\beta = 90$ .
$AP \cong PQ \cong QR \cong RA$ and $\angle PAR = \angle APQ \Leftrightarrow APQR$ is a square.

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Reason: Completing the sol.

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#8 h Aug 11, 2021, 2:02 PM

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This problem is very similar to ISL G1 2020

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#9 h Jun 3, 2024, 4:15 PM

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rhombus + 1 right angle

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