Source: 2017 Canadian Open Math Challenge, Problem C3 Let $XYZ$ be an acute-angled triangle. Let $s$ be the side-length of the square which has two adjacent vertices on side $YZ$, one vertex on side $XY$ and one vertex on side $XZ$. Let $h$ be the distance from $X$ to the side $YZ$ and let $b$ be the distance from $Y$ to $Z$. [asy][asy] pair S, D; D = 1.27; S = 2.55; draw((2, 4)--(0, 0)--(7, 0)--cycle); draw((1.27,0)--(1.27+2.55,0)--(1.27+2.55,2.55)--(1.27,2.55)--cycle); label("$X$",(2,4),N); label("$Y$",(0,0),W); label("$Z$",(7,0),E); [/asy][/asy] (a) If the vertices have coordinates $X = (2, 4)$, $Y = (0, 0)$ and $Z = (4, 0)$, find $b$, $h$ and $s$. (b) Given the height $h = 3$ and $s = 2$, find the base $b$. (c) If the area of the square is $2017$, determine the minimum area of triangle $XYZ$.
Problem
Source:
Tags: Comc, 2017 COMC
13.10.2018 00:20
a)Since both Y and Z are on the x-axis, the height of the triangle is 2-0=2. b=base of the triangle=4-0=4 Since 2 = (4-0)/2, the triangle is isosceles. Hence s =4/2=2. Therefore, b=4, h=2, s=2.
13.10.2018 00:21
BTW I'm also preparing for this.
13.10.2018 00:32
b) Suppose the triangle is isosceles and it's easy!(BTW if you have a better way, tell me.)
13.10.2018 01:02
B) the triangle above the square has base 2 an height 1. The triangle XYZ has a height of 3, and by similar triangles, we get that the base is 6. So the area is 9
13.10.2018 01:07
LucaTu1 wrote: a)Since both Y and Z are on the x-axis, the height of the triangle is 2-0=2. b=base of the triangle=4-0=4 Since 2 = (4-0)/2, the triangle is isosceles. Hence s =4/2=2. Therefore, b=4, h=2, s=2. I thought s<h so that's not possible, I got 4/3 instead
13.10.2018 01:11
30.09.2019 09:23
cooljoseph wrote:
Fixed that for you.