Waste less time on Facebook — follow Brilliant.
×

Seeking Help

In triangle find points X,Y,Z on AB,BC,CA such that AXYZ is a rhombus.Show that the area of rhombus AXYZ≤(1/2) triangle ABC. Please show me the proof

Note by Kalpok Guha
3 years ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Here's an algebraic proof:

Let \(K =\) area of \(ABC, [\:] =\) the area of a polygon, \(b = AC,\) and \(c = AB.\) Note that, since \(AXYZ\) is a rhombus, \(AY\) is an angle bisector of \(\angle A.\) From the angle bisector theorem, we get \(\dfrac{BY}{YC} = \dfrac{c}{b}.\) From this, we get \(\dfrac{BY}{BC} = \dfrac{c}{b+c}\) and \(\dfrac{YC}{BC} = \dfrac{b}{b+c}.\)

\[\frac{[ZYC]}{K} = \left(\frac{b}{b+c}\right)^2 \Rightarrow [ZYC] = K \left(\frac{b}{b+c}\right)^2\] \[\frac{[XYB]}{K} = \left(\frac{c}{b+c}\right)^2 \Rightarrow [XYB] = K \left(\frac{c}{b+c}\right)^2.\]

Now, \([AXYZ] = K - [ZYC] - [XYB] = K - K \left(\dfrac{b}{b+c}\right)^2 - K \left(\dfrac{c}{b+c}\right)^2 = K \left(\dfrac{2bc}{(b+c)^2}\right).\) Finally, we are left to prove that \(\dfrac{2bc}{(b+c)^2} \leq \dfrac{1}{2}.\) Rearranging this gives us \((b - c)^2 \geq 0,\) which is clearly always true. Thus, \([AXYZ] \leq \dfrac{1}{2}K,\) and our proof is complete.

Steven Yuan - 3 years ago

Log in to reply

Thanks .But I think AY is bisector because AXYZ is rhombus,a diagonal is not a bisector in parallelogram.

Kalpok Guha - 3 years ago

Log in to reply

Oops, sorry about that! I've fixed my solution.

Steven Yuan - 3 years ago

Log in to reply

Let ABCD be a rhombus such that A: (-a,0), B; (0,b), C: (a,0) & D: (0,-b) with centre of the rhombus as (0,0). Now draw a line: y=mx+c passing through D and intersecting extended BA & BC in P & Q respectively. We now have a triangle PQB with a rhombus whose vertices are on the triangle sides. Since y=mx+c passes through D, c = -b. We can now determine P as intersection of y=mx -b & x/a+y/b=1 and thus, P:[2ab/(am-b), b(am+b)/(am-b)]. Similarly, Q:[2ab/(am+b), b(am-b)/(am+b)]. Further, we determine the area of triangle PQB taking the vertices in anti-clockwise direction. Tr. PQB = (1/2){(2ab/(am-b))[b(am-b)/(am+b) -b)+(2ab/(am+b))[b -b(am-b)/(am-b)]} This simplifies to, Tr. PQB = 4ab³/(b²-a²m² ) whereas atra pf Rhombus ABCD = 2ab. Note that if m=0 then (1/2) Tr.PQB = 2ab = Area of Rhombus ABCD.
Otherwise (1/2)*4ab³/(b²-a²m² ) > 2ab if b²>b² - a²m² or if a²m²>0 which is always true.
Hence the proposition stands proven.

One Top - 3 years ago

Log in to reply

Really thanks for the proof.But is there any prove without using coordinate geometry?

Kalpok Guha - 3 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...