Geometry Challenge 1612: 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.
https://photos.app.goo.gl/QhRyp1W2EjrEdgdw9
ReplyDeleteConnect MN and define angle u per sketch
Apply cosine formula in triangle AMN
AN^2= MN^2+ AM^2- 2. MN. AM.cos(u)
Replace AN= c/2 ; AM=3/2.x ; MN= b/2 ; cos(u)=x/b
We get c^2= b^2+3.x^2
Geometry Solution
ReplyDeleteLet CG = y so that GN = y/2
Applying Pythagoras twice,
c^2/4 - x^2 = y^2/4 = (b^2 - x^2)/4
Hence c^2 - b^2 = 3x^2
It can further be proved that in Triangle ABC,
a^2 + c^2 = 5b^2
Sumith Peiris
Moratuwa
Sri Lanka