# 2017 Indonesia MO Problems/Problem 1

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## Problem

$ABCD$ is a parallelogram. $g$ is a line passing $A$. Prove that the distance from $C$ to $g$ is either the sum or the difference of the distance from $B$ to $g$, and the distance from $D$ to $g$.

## Solution

In order to prove that the distance from $C$ to $g$ is either the sum or the difference of the distance from $B$ to $g$, and the distance from $D$ to $g$, we will use casework to show that the statement is true for all scenarios.

Case 1: $g$ passes through $D$ or $B$
If $g$ passes through $D$, then the distance from $g$ to $D$ is zero. The distance from $g$ to $B$ is the same as the distance from $g$ to $C$ because $BC$ is parallel to $g$. That means the distance from $C$ to $g$ is the sum of the distance from from $B$ to $g$ and from $D$ to $g$. By using similar steps, if $g$ passes through $B$, the distance from $C$ to $g$ is the sum of the distance from from $B$ to $g$ and from $D$ to $g$.

Case 2: $g$ passes through $C$
By SSS Congruency, $\triangle ACD \cong \triangle CAB$. Since the area of the two triangles is the same, the distance from $D$ to $g$ equals the distance from $B$ to $g$. Because the distance from $C$ to $g$ is zero, the distance from $C$ to $g$ is the difference of the distance from from $B$ to $g$ and from $D$ to $g$.

Case 3: $g$ passes through $CD$ or $BC$
$[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NE); dot(D); label("D",D,NW); pair e=(48,40),XA=(55.082,45.902),XB=(29.508,24.59),XC=(25.574,21.311); draw(A--XA--C); draw(D--XC); draw(B--XB); dot(e); label("E",e,N); dot(XA); label("X_1",XA,N); dot(XB); label("X_2",XB,E); dot(XC); label("X_3",XC,S); [/asy]$

Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $CX_1, BX_2, DX_3 \perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_2$, making $EC = a-x$.

By Alternate Interior Angles Theorem and Vertical Angle Theorem, $\angle EAB = \angle AED = \angle CEX_1$. Thus, by AA Similarity, $\triangle X_2BA \sim \triangle X_3DE \sim \triangle X_1CE$.

Using similar triangle ratios, we have $DX_3 = \tfrac{bx}{a}$ and $X_1 = \tfrac{ba-bx}{a}$. Thus, we have $BX_2 - DX_3 = \frac{ba-bx}{a} = CX_1$, so the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$. By symmetry, we can also show that the distance from $C$ to $g$ is the difference between the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $BC$.

Case 4: $g$ does not intersect the parallelogram at any other points
$[asy] pair A=(0,0),B=(50,0),C=(60,40),D=(10,40); draw(A--B--C--D--A); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NE); dot(D); label("D",D,NW); pair e=(-40,40),XA=(-15,15),XB=(10,-10),XC=(25,-25); draw(D--e--XC--B); draw(D--XA); draw(C--XB); draw(B--XC); dot(e); label("E",e,N); dot(XA); label("X_1",XA,S); dot(XB); label("X_2",XB,S); dot(XC); label("X_3",XC,S); [/asy]$

Let $E$ be the intersection of lines $DC$ and $g$, and let $X_1, X_2, X_3$ be points on $g$ such that $DX_1, CX_2, BX_3 \perp g$. Also, let $a = AB$, $x = DE$, and $b = BX_3$, making $EC = a+x$.

By Alternate Interior Angles Theorem, $\angle CEA = \angle BAX_3$. Thus, by AA Similarity, $\triangle EDX_1 \sim \triangle ECX_2 \sim \triangle ABX_3$.

Using similar triangle ratios, we have $DX_1 = \tfrac{bx}{a}$ and $CX_2 = \tfrac{ba+bx}{a}$. Thus, we have $BX_3 + DX_1 = \frac{ba+bx}{a} = CX_2$, so the distance from $C$ to $g$ is the sum of the distance from $D$ to $g$ and the distance from $B$ to $g$ if $g$ passes through $CD$.

In all of the cases, the sum or difference of the distance from $B$ to $g$ and the distance from $D$ to $g$ is equal to the distance from $C$ to $g$.