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

#Geometry #Rhombus #TriangleArea #StraightLines

Note by Kalpok Guha
6 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

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.

Ajit Athle - 6 years, 6 months ago

Log in to reply

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

Kalpok Guha - 6 years, 6 months ago

Here's an algebraic proof:

Let K=K = area of ABC,[ ]=ABC, [\:] = the area of a polygon, b=AC,b = AC, and c=AB.c = AB. Note that, since AXYZAXYZ is a rhombus, AYAY is an angle bisector of ∠A.\angle A. From the angle bisector theorem, we get BYYC=cb.\dfrac{BY}{YC} = \dfrac{c}{b}. From this, we get BYBC=cb+c\dfrac{BY}{BC} = \dfrac{c}{b+c} and YCBC=bb+c.\dfrac{YC}{BC} = \dfrac{b}{b+c}.

[ZYC]K=(bb+c)2⇒[ZYC]=K(bb+c)2\frac{[ZYC]}{K} = \left(\frac{b}{b+c}\right)^2 \Rightarrow [ZYC] = K \left(\frac{b}{b+c}\right)^2 [XYB]K=(cb+c)2⇒[XYB]=K(cb+c)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(bb+c)2−K(cb+c)2=K(2bc(b+c)2).[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 2bc(b+c)2≤12.\dfrac{2bc}{(b+c)^2} \leq \dfrac{1}{2}. Rearranging this gives us (b−c)2≥0,(b - c)^2 \geq 0, which is clearly always true. Thus, [AXYZ]≤12K,[AXYZ] \leq \dfrac{1}{2}K, and our proof is complete.

Steven Yuan - 6 years, 6 months 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 - 6 years, 6 months ago

Log in to reply

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

Steven Yuan - 6 years, 6 months ago
×

Problem Loading...

Note Loading...

Set Loading...