Given points $P$ and $Q$, Jaqueline has a ruler that allows tracing the line $PQ$. Jaqueline also has a special object that allows the construction of a circle of diameter $PQ$. Also, always when two circles (or a circle and a line, or two lines) intersect, she can mark the points of the intersection with a pencil and trace more lines and circles using these dispositives by the points marked. Initially, she has an acute scalene triangle $ABC$. Show that Jaqueline can construct the incenter of $ABC$.
Problem
Source: 2023 Girls in Mathematics Tournament- Level B, Problem 4
Tags: geometry, incenter
CyclicISLscelesTrapezoid
29.10.2023 05:52
Solved with CT17. Let $(PQ)$ denote the circle with diameter $\overline{PQ}$. We claim that any straightedge and compass construction can be constructed using a straightedge and Jaqueline's tool. Then, there are many ways to construct the incenter (for example, by constructing the angle bisectors). Claim 1: For any point $A$ and line $\ell$ not passing through $A$, Jaqueline can construct the altitude from $A$ to $\ell$. Proof: Choose a random point $B$ on $\ell$, let $(AB)$ intersect $\ell$ again at $C$, and draw $\overline{AC}$. $\square$ Claim 2: For any point $A$ and line $\ell$ passing through $A$, Jaqueline can construct the altitude from $A$ to $\ell$. Proof: Choose a random point $B$, let $(AB)$ intersect $\ell$ again at $C$, and draw $\overline{BC}$. Choose a random point $D$ and use Claim 1 to draw the altitude $m$ from $D$ to $\overline{BC}$ and the altitude from $A$ to $m$. $\square$ Claim 3: For points $A$ and $B$, Jaqueline can construct the reflection of $A$ over $B$. Proof: By Claim 2, draw the line $m$ through $B$ perpendicular to $\overline{AB}$. Choose a random point $C$ on $m$ and use Claim 1 to construct the point $D$ on $m$ with $\angle CAD=90^\circ$. Intersect $\overline{AB}$ again with $(CD)$ to get the reflection of $A$ over $B$. $\square$ Thus, given points $A$ and $B$, Jaqueline can construct the reflection $B'$ of $B$ over $A$ and the circle $(BB')$ centered at $A$. Thus, she can construct any straightedge and compass construction using a straightedge and her tool, as desired. $\square$
ohhh
23.06.2024 20:11
sketch $1)$ intersect $(AB)$ and $(AC)$ to obtain the ortocenter $H$. $2)$ let $BH \cap AC = E$ and $CH \cap AB = F$. $3)$ From $EF \cap BC$, we gain the $A-ex$ point X. $4)$ Doing $AX \cap (AH)$ we gain the $A-queue$ point Q. $5)$ Using $(AE) \cap EF$, draw the perpendicular line from $A$ to $EF$ and intersect with $QH$, it's the $A-antipode$ $T$. $6)$ Get $(ABC)$ from $(AT)$. $7)$ Obtain the midpoints of the sides using the queue-points. $8)$ Use them to draw the perpendicular bissector of $BC$ $9)$ Get the $A$ angle bissector from $6)$ and $8)$.