Geometry Challenge 1614: Share your proof or solution in the comments below.
Target Audience: K-12, Honors Geometry, and College Mathematics Education.
Explore the full theorem and interactive diagrams by clicking the illustration below.
Target Audience: K-12, Honors Geometry, and College Mathematics Education.
Explore the full theorem and interactive diagrams by clicking the illustration below.
Click for additional details.
Proposed Solution
We invite students, teachers, and enthusiasts to share their proofs. This classic challenge can be approached using synthetic geometry.
How to contribute:
Post your step-by-step proof in the comments below. Feel free to:
We invite students, teachers, and enthusiasts to share their proofs. This classic challenge can be approached using synthetic geometry.
How to contribute:
Post your step-by-step proof in the comments below. Feel free to:
- Describe your construction and properties applied (centroid/medians).
- Provide a link to a diagram (GeoGebra, Desmos, etc.) if you have one.
Ready to contribute?
Please use the box below to Enter your Comment or Solution. You can use plain text or provide links to your diagrams.
Please use the box below to Enter your Comment or Solution. You can use plain text or provide links to your diagrams.
Trigonometry Solution
ReplyDeleteLet EG = b and EM = MF = FG = a
In Triangle EMG, if < MGE = @,
Cos@ = (2a^2 + 5a^2 - a^2)/(2*V2a*V5a) = 3/(V10)
and so, sin@ = 1/V10
Hence in Triangle OME, a/2 = R*sin@ = R/V10 (since < EOM = 2@)
So S(EFGH) = 2a^2 = 8R^2/10 = 4R^2/5 = S(ABCD)
And so, the result follows
Sumith Peiris
Moratuwa
Sri Lanka
We need to prove that Area of Rectangle ABCD = Area of EFGH
ReplyDeleteArea of ABCD =AB*AD = AB^2/5, if radius of circle is R then Area of Rectangle=4R^2/5
We can see that Quad. MFKG is square hence Angle EMG=135 deg,
Hence EG=√2R, Let FG=x and EF=2x, Then x^2 + 4x^2=2R^2 or x^2=2R^2/5
Area of Rectangle EFGH=EF*FG=2x^2=4R^2/5 Q.E.D.
Hi Pradyumna
DeleteHow is Quadrilateral MKFG a square?
You are right, MKFG may not be square, but Triangle MFG is right isosceles triangle as MF=FG, hence Angle FMG = 45 deg and Angle EMG=135 deg.
DeleteThanks
Pradyumna